Quantum_book.pdf

Mini book: Chapter planning1: 1. Basic concepts in Quantum Chemistry 1.1. Wave-particle duality : In general, waves and particles are considered as two different things. A wave is considered as a disturbance that transfers energy through matter and space. A wave is always looked upon as an oscillation, that can be created by applying a force in a medium. For example, when you walk in your house, believe it or not, you create oscillations i.e. waves. Snakes can feel these vibrations. You can also feel the vibration of your movement if you walk on a not-so-hard surface for example how about walking on a rope bridge, as shown in Figure ??.

Figure 1: Rope bridge On the other hand, a particle is considered as a localized thing with which you can associate properties like mass and volume. Localized means having a definite boundary. In general, Waves lack these properties. We don’t really apply the concepts of mass and volume to waves. Thus we see that waves and particles are two different things.

Figure 2: Wave and particle natures The above clear distinction between wave and particle natures, however, exist only for macroscopic particles. When one starts decreasing the size of the particle the concepts of wave and particle behaviors start mixing with each other and for very-very-small particles like electrons, protons and neutrons the two concepts completely intermix and one cannot really tell whether these very-very-small particles behave as wave or as particle. In the following sections, we’ll get evidence of how, light which is otherwise considered as a wave, can also behave as particle and an electron, which we think as a particle can also behave as a wave.

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1.1.1. light as particle: Black-body radiation, Photo-electric effect and Compton effect What is light? For most of us, light is nothing but a kind of energy or wave. Whatever we think of light, it is sure that it does not look like a particle. So, to be on the safe side, we say light is not a particle. When you say this you are basically supporting the concept of great scholars/philosophers like Robert Hook and Christiaan Huygens, who proposed that light is a wave. But at the same time you are opposing great scholars/philosophers like Pierre Gassendi and Isaac Newton who considered light as a particle. This topic belongs to a different field – beyond the scope of this mini book – and we don’t want to go into these details. So, let’s talk about one of the most celebrated concept of 19th century – electromagnetic concept of light. At the end of 19th century, to be precisely in 1865, James Clerk Maxwell unified the ideas of electricity, magnetism and light and came up with a new idea. He showed, through his equations, that light is an electromagnetic wave consisting of two fields – an electric and a magnetic field – oscillating perpendicular to each other. Light is thus an electromagnetic wave generated by the oscillations of the two fields and propagates in a direction perpendicular to the direction of the two fields. This description clearly classifies light as a wave. Although this is one of the most celebrated concept in physical science it is not suitable for explaining some of the experimental observations including photoelectric effect, Raman effect and Compton effect. In fact, these experiments clearly demonstrate that light cannot be considered as a wave rather it has corpuscular nature. 1.1.1.1. Photo-electric effect: As the name suggests, photo-electric effect is a phenomenon related to “photo” that means light and “electric” means electricity. In fact, photo-electric effect is the ejection of electrons from a material when the latter is exposed to a light of frequency greater than a threshold frequency. The electrons ejected are thus called photo-electrons and the current produced by them is called photo-current. Let us quickly go through the characteristics of photo-electric effect Characteristics of photo-electric effect: • Electrons are ejected out of the metal surface only when the frequency (ν) of incident light is greater than a particular frequency. The minimum frequency for which electrons are ejected is called threshold frequency (ν0 ) – a characteristic of the metal. The electrons are never ejected if ν < ν0 . The minimum energy required to eject the photo-electrons is called work function (W0 ). • The kinetic energy of the photo-electrons is directly proportional to frequency, ν (> ν0 ) of the incident light. • Kinetic energy of the photo-electrons is independent of the intensity of incident light. • For a given frequency, ν (> ν0 ), the number of photo-electrons ejected from the metal surface is directly proportional to the intensity of incident light. Now, let us try to explain these characteristics by considering 2

the wave-nature of light. •• Intensity of light is simply the energy transfered per unit time per unit area perpendicular to the direction of propagation of the light. Also, the energy of a wave is directly proportional to the square of its amplitude. Thus intensity of light is directly proportional to the amplitude of light. Thus, in wave-picture, a highly intense light should have high energy and hence electrons can be made to eject from a metal surface by using any highly intense light irrespective of its frequency. This is contrary to the first characteristic mentioned above. •• Based on the same argument, the kinetic energy of the photoelectrons should be proportional to the intensity of light – contrary to the second and third characteristics above. •• The number of photoelectrons ejected from a metal surface and hence the photoelectric current cannot be related with the amplitude or frequency in wave-picture of light. Thus we see that the wave-picture of light is not able to describe any characteristic of the photo-electric effect. Photo-electric effect was first described successfully by Albert Einstein, using a corpuscular picture of light. Einstein proposed that X light behaves as consisting of small energy packets (called photon). X The energy of each photon of light is directly proportional to the frequency of light. Thus, E∝ν =⇒ E = hν,

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where h is the Planck’s constant. X When the surface is illuminated with light, each photon is absorbed by one electron. A part of the energy of the absorbed photon is used by the electron to overcome the binding energy (i.e. work function) in the atom and the remaining energy appears as the kinetic energy of the photo-electron. According to Principle of conservation of energy, we can write hν = KE + W0

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=⇒ W0 = hν0 =⇒ KE = h(ν − ν0 )

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when ν = ν0 , KE = 0

Thus, Einstein proposed that light is no longer a continuous thing like wave but it consist of several small particles each of which has energy proportional to frequency of light. Now, let us see how this particle nature of light explains the photo-electric effect. XX From Eqs. ?? and ??, it is obvious that when ν < ν0 kinetic energy becomes negative. That means no electron will be ejected if frequency of light is less than the threshold frequency. The equations has no intensity term i.e. the ejection 3

section no.

of electron does not depend upon the intensity of light. [Characteristic 1] XX Kinetic energy is proportional to the frequency and is independent of intensity of light, as evident from Eq. ?? [Characteristics 2 and 3] XX In this particle-picture of light, intensity is simply number of photons emitted per unit time per unit area perpendicular to the direction of propagation of light. That means, larger the intensity larger will be the number of photons and hence larger will be the number of electrons emitted. Thus, for a given frequency (> ν0 ) the number of electron emitted and hence the amount of photoelectric current is proportional to the intensity of light. [Characteristic 4] Thus, we see that photoelectric effect can be described only if we consider light as a particle and not as wave. This is a clear evidence that light is indeed a particle. Now, we’ll go for another evidence – Compton effect. 1.1.1.2. Compton effect [1923]: Compton effect is the increase in wavelength (equivalently, decrease in frequency) of an x-ray when it is scattered by an electron. The amount by which wavelength increases is called Compton shift. Arthur H. Compton, while studying the interaction of x-ray with matter noticed that after scattering the wavelength of the scattered x-ray beam is larger than that of the incident x-ray beam. If one consider light (here, x-ray) as wave then this increase in wavelength is unexpected. Why? Well, since the energy of wave is proportional to the intensity and independent of the frequency (or wavelength), after scattering (i.e. after exchanging energy with the electron) the intensity only should decrease and frequency (or wavelength) should remain intact. Thus, this observation also demonstrate a case where wave-picture of light is inadequate. Compton explained this observation by considering the particle-nature of light. If one considers the particle nature of light – as considered in case of photo-electric effect – one can easily realize that after interaction with the electron a part of energy (which is proportional to the frequency of light) is absorbed by the electron and hence the energy of the scattered photon becomes less than that of the incident photon and hence the scattered photon should have smaller frequency (larger wavelength) than the incident photon. Now, let us move to do some mathematics related to Compton effect and derive the expression for compton shift. Derviation of compton shift: Thus these two experiments/observations clearly demonstrate that light is not a continuous thing and it has a particle nature. 1.1.2. de Broglie hypothesis [1924] Based on the above observations and particularly influenced by the idea of Einstein that light has particle-like nature, Louis de Broglie, in 1924, in his PhD thesis, proposed that electrons and all matter also have wave-like properties. This is called wave-particle duality. 4

At that time, it was just a hypothesis but later on, in 1927, it was verified by Davisson and Germer in the famous double-slit experiment. According to de Broglie, the wavelength of a particle of mass m, moving with velocity v is given by λ=

h , mv

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where h is the Planck’s constant. There are several interesting things to note. • The wave associated with a material particle is called matterwave. • The wavelength of a matter-wave decreases with the increase in mass or velocity of the associated particle. • Since the value of h is very small (order of 10−34 in SI unit), wave-nature becomes significant only for a particle of very small mass and small velocity. • For a daily-life particles (dust particle, sand grains, car, human beings etc), wave-like character is negligible and hence does not affect their behavior significantly. However, it does not mean that these things don’t have wave-like properties. • For sub-atomic particles like electrons, protons, and neutrons, wave character is significant and it affects their behavior, which can be observed in different experiments. • If a sub-atomic particle moves at a very high speed, it loses its wave-nature to some extent. • It is very obvious that if a wave moves in a circle then the wave will survive only when the perimeter (circumference) of the circle equals an integer number of wavelengths i.e. 2πr = nλ, r being the radius of the circle. Otherwise, the wave will just disappear because of destructive interference [Fig. ??].

Figure 3: Wave on a circle If we assume the wave to be a matter wave then we can use Eq. ?? here, which gives nh mv nh =⇒ mvr = 2π 2πr =

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This is nothing but the Bohr’s postulate of stationary orbits. Thus, Bohr’s stationary orbit is a consequence of de Broglie’s wave-particle duality. 1.1.3. Electron as wave [1927] As we have seen above, light, which is not associated with any mass or volume i.e. particle-like properties, indeed has a particle nature. Now, we will see that electron, which has a definite 5

mass and volume, can also behave as a wave (as proposed by de Broglie). This can be demonstrated by the famous double-slit experiment. Surprisingly, the same experiment can be used to demonstrate that light behaves as a wave rather than a particle. So, double-slit experiment is really an amazing piece of work in science (It is not included in your chemistry syllabus.). You can watch this beautiful illustrative youtube-video on double-slit experiment (https://tinyurl.com/y8xsam6p). 1.2. Concept of operators: Let us consider some simple mathematical equations. • 1+2=3 • 3×2=6 d 2 x = 2x • dx

What are those symbols (+, ×) in the first two equations and what is d ? You never thought about this. Am I right? OK, the symthat dx d bols (+, ×) and dx are nothing but representation of some mathematical operation. I know, you knew this. But do you know, these are called mathematical operators or simply operators. Thus, an operator is nothing but a symbol representing some mathematical operations. In other words, operators are short-hand notation of some mathematical rules. For example, whenever we see the symbol + between two numbers or two variables, it says that the two numbers or the two variable should be added together obeying the mathematical rules of addition. Simid larly, when we see the symbol dx preceding something, it says that we should take a derivative of that something with respect to the variable x, following the rules of differentiation. 1.2.1. Eigenfunctions, eigenvalues, X Some operators after operating on a function return another d 2 d function. For example, dx x = 2x. Here, the operator dx oper2 ates on function x and returns a different function x. X It is also possible that an operator after operating on a function returns the same function multiplied by a constant. For d 5x example, dx e = 5ex . In this case, the function is called an eigenfunction of the operator and the constant is called eigenvalue of the function. d2 2 X Similarly, dx 2 (sink1 x + cosk1 x) = −k1 (sink1 x + cosk1 x). Thus, d2 2 (sink1 x+cosk1 x) is an eigenfunction of the operator dx 2 and −k1 is the corresponding eigenvalue. X But what does eigenfunctions and eigenvalues really mean? Later on, in this quantum chemistry course, you’ll come to know that operators basically represent some properties of the system and the functions represent some states of the systems. The eigenfunctions represent those states of the system in which the system can reside under certain conditions and when the system is in that state the value of the property (corresponding to that operator) will be equal to the eigenvalue. I hope this is quite confusing!! Am I right? OK, read it again, this time slowly and let me know if it makes sense. X Another attempt to understand the meaning of eigenfunction and eigenvalue. Let us suppose two conditions – (a) when you 6

are relaxing and (b) when you are jogging. Now, suppose there is an instrument that displays ”jogging” only when your heart beat is above 110 per minute. But when it is less than 110 beats per minute, it does not display anything. So,it is obvious that the instrument will give a display only when you are jogging. Thus, here your two conditions are two states. Your ”jogging” condition is an eigenfunction of the instrument (the operator) and the display is the eigenvalue. 1.2.2. Linear operators: Suppose, f and g are two functions, k is a constant and Aˆ is an operator. Now, if ˆ = k Af, ˆ Akf Operator does not operate on a constant. ˆ + g) = Af ˆ + Ag ˆ distributive property A(f

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then the operator Aˆ is called a linear operator. 1.2.3. Hermitian operator • An operator is said to be Hermitian if it satisfies the following rule, called turn-over rule Z Z   ∗ ˆ ∗ ∗ ˆ f Agdτ = A f gdτ, (7) where ∗ represents complex conjugate. Such operators are also called self-adjoint operator. • What is the meaning of self-adjoint? Adjoint is a terminology used (in this context) for a particular transposition of a matrix. When you transpose a matrix and takes complex conjugate of each element then the resultant matrix is called adjoint of the original matrix. For example, let a matrix   A11 A12 A= A21 A22 The adjoint of this matrix will be   ∗ A11 A∗22 † A = A∗12 A∗22 Thus, if 

then

1 + 2i 2 + 3i A= 4 − 5i 5 + 7i



  1 − 2i 4 + 5i A = 2 − 3i 5 − 7i †

• Thus, a Hermitian matrix is symmetric matrix will always  1  A= 2 3 is a Hermitian matrix.

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one for which A = A† . A real be Hermitian. For example  2 3 5 7 = A† 7 9

• Ok, the concept of hermitian or hermiticity is very much clear in the context of matrix. But how can we understand it in the context of operators? • Actually, later on, you’ll came to know that operators can be represented by a matrix and then Hermitian operators are always represented by real symmetric matrices. • According to above definition of Hermitian operator, Z Z   ∗ ∗ ∗ ˆ ˆ A f gdτ f Agdτ = Here, you can consider left hand side as Af g element of the R the ∗ ˆ matrix A i.e. Af g = f Agdτ . Similarly, the element Agf will R ˆ dτ . By definition, Agf is the transpose be given by, Agf = g ∗ Af of Af g . Now, if you take the complex conjugate of Agf then it   R R Aˆ∗ f ∗ gdτ =right hand side. becomes A∗ = g Aˆ∗ f ∗ dτ = gf

But, taking transpose followed by complex conjugation is the operation of making adjoint. Thus, right hand side of the above equation is an adjoint of the left hand side. Hence, the hermitian operator (or matrix) A is self-adjoint. • Now, you know, what is hermitian operator and why it is called self-adjoint? Let us now have a look at some important characteristics of a hermitian operator. Characteristics of a hermitian operator X The eigenfunctions of a hermitian operator can be complex or real but the eigenvalues are always real. Proof: Let Aˆ be a hermitian operator and f and a are the eigenfunction and eigenvalue respectively. Then ˆ = af Af Multiplying both sides (from left) by f ∗ and integrating the resultant, we can write Z Z ∗ ˆ f Af dτ = a f ∗ f dτ (8) Now, taking adjoint of the above equation, we get Z  ∗ Z ∗ ˆ Af f dτ = a f ∗ f dτ By definition, Thus,

R

ˆ dτ = f Af ∗

Z a ∗

(a − a )

Z

R

ˆ Af

∗

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f dτ (as Aˆ is hermitian.).

f ∗ f dτ = a∗

Z

f ∗ f dτ (10)

∗

f f dτ = 0.

R Since f ∗ f dτ 6= 0, this means a − a∗ = 0 =⇒ a = a∗ , which is possible only when a is real. Thus, the eigenvalue(s) of a hermitian operator are always real. 8

X Two eigenfunctions of a hermitian operator are always orthogonal to each other. Proof: Let f and g be two different eigenfunctions of a hermitian operator Aˆ and the corresponding eigenvalues be a and b respectively. Thus ˆ = af, Ag ˆ = bg Af From definition of hermitian operator, we can write Z Z  ∗ ˆ ˆ Af gdτ f ∗ Agdτ = Z Z f ∗ bgdτ = (af )∗ gdτ Z Z b f ∗ gdτ = a f ∗ gdτ, since a is real, a∗ = a Z (b − a) f ∗ gdτ = 0.

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Since, R ∗ f and g are two different functions, b 6= a. This means, f gdτ = 0. Thus, two different eigenfunctions of a hermitian operators are always orthogonal to each other. X Hermitian operators are represented by a real symmetric matrix. X The eigenfunctions of a hermitian operator form a ‘complete set’. That means, any arbitrary function can be expressed as a linear combination of eigenfunctions of hermitian operator. X Any observable (e.g. energy, momentum) is always represented by a hermitian operator. 1.2.4. Orthogonality and normalization of function Normalization and orthogonality are two important concepts arise when we deal with eigenfunctions in quantum mechanics. • A function f is said to be normalized if Z f ∗ f dτ = 1 • If a function is square integrable i.e if for a function (f ), Z f ∗ f dτ = finite over a space then it will be possible to normalize that function. • The meaning of normalization and more will be discussed later on. • Two functions f (x) and g(x) are said to be orthogonal to each other if they satisfy the relation Z f ∗ (x)g(x)dτ = 0 (12) • Orthogonality of two functions simply means that the two functions do not share a common space in the given interval (i.e. in the limit of the integration). • That means, if we plot the two functions in the same range of the variable x then the overlap between the graphs of the two functions will be zero. 9

• More about orthogonality will be discussed later on. 1.2.5. Commutation of operators X What to do when two operator appear as a product? Let us ˆ are two operators and f be a function. Then consider Aˆ and B ˆ means that at first B ˆ should operate on f and then on the AˆBf ˆ=√ ˆ , Aˆ should operate. For example, if Aˆ = d , B 2 result of Bf dx (i.e. square root operator) and f = x2 + 2x + 1. In this case, h i ˆ ˆ ˆ ˆ ABf = A Bf i d hp 2 2 ˆ operates on f = x + 2x + 1 at first, B dx (13) d ˆ Aˆ will work = [± (x + 1)] On result of Bf, dx = ±1 and h i ˆ ˆ ˆ ˆ B Af = B Af r d 2 2 = (x + 2x + 1) at first, Aˆ operates on f dx p ˆ will work on the result of Af ˆ . = 2 2(x + 1) B

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X The above example reflects a very important characteristic about the product of two operators. As we see above, the result of √ d operation of the product of dx and 2 operators on the same function depends upon the order of operations. This usually ˆ 6= B ˆ Af ˆ . happens i.e. AˆBf X However, for some pair of operators, result of operations of the products of the two operators may not depend upon the order ˆ =B ˆ Af ˆ . In such cases, the two operators of operations i.e. AˆBf are said to commute with each other. ˆ = k2 , that means operator Aˆ represents a X Let Aˆ = k1 x and B x ˆ represents multiplication multiplication by k1 x and operator B by kx2 , k1 and k2 being constants. Now, if the function f = g(x), then ˆ = k1 x × k2 × g(x) = k1 k2 g(x) AˆBf x

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and

ˆ Af ˆ = k2 × k1 x × g(x) = k1 k2 g(x) (16) B x Thus, the operators k1 x and kx2 commute with each other. h i ˆ ˆ ˆ−B ˆ Aˆ is called commutator of the X The quantity A, B = AˆB h i ˆ B ˆ = 0, the two operators commute two operators. when A, h i ˆ ˆ and when A, B 6= 0, the two operators do not commute. h i ˆ B ˆ = AˆB ˆ +B ˆ Aˆ is called anti-commutator of X The quantity A, +

the two operators. X So, now you know what a commutator is. But what does it really mean, when we say that two operators commute with each 10

other? If you go to the literal meaning of commute with each other, you will be surprised. It means, the two operators can commute – talk – or interact – with each other. It means, some sort of communication is possible between the two operators. But what kind of communication can be possible between two operators? X The answer is eigenfunctions. When two operators commute, they share the same set of eigenfunctions. That means, when two operators commute, both of them can extract information from a set of functions. X In other words, when two operators commute, an eigenfunction of one operator will also be an eigenfunction of the second operator and when the system is in such a state, both the properties – corresponding to the two operators – can be measured simultaneously and with any degree of precision. This is related to Heisenberg Uncertainty principle and will be discussed in more details there. For now, you should know that if two operators do not commute then you cannot measure the two properties simultaneously with any arbitrary degree of precision. X This also means that when two operates commute that means informations about the system i.e. about the states of the system can transfer through these operators. X This can also be put as when two operators commute that means the information about the system is not lost when the observables corresponding to these operators are measured. X Or like, two commuting operators behave as a transparent medium for the passage of information about the system and hence information is not lost during their measurement. 1.2.6. Fundamental commutator: Some of the fundamental commutation relationships in quantum mechanics are as follows • [ˆ x, yˆ] = [ˆ x, zˆ] = [ˆ y , zˆ] = 0, where xˆ, yˆ and zˆ are the position operators in three Cartesian coordinates directions. Proof: Position operators are simply multiplication by the coordinates (x, y or z). Thus, if f be a function then [ˆ x, yˆ] f = xˆyˆf − yˆxˆf = xyf − yxf = 0, x and y are numbers [ˆ x, zˆ] = [ˆ y , zˆ] = 0 can similarly be proved. • [ˆ px , pˆy ] = [ˆ px , pˆz ] = [ˆ py , pˆz ] = 0, where pˆx , pˆy and pˆz are the three Cartesian components of linear momentum operator, respectively. Proof: Linear momentum operator (e.g. in x−direction) is given as d pˆx = −i~ , dx √ h where i = −1 andh ~ = 2π . Let a function f = f (x, y). Then, [ˆ px , pˆy ] f (x, y) = pˆx pˆy f (x, y) − pˆy pˆx f (x, y) d d d d = −~2 f (x, y) + ~2 f (x, y) dx dy dy dx  2  2 d d = −~2 f (x, y) − f (x, y) dxdy dydx 11

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Note that f (x, y) is a state function. So, the term in parenthesis will vanish. Hence [ˆ px , pˆy ] = 0. Similarly, other commutators can be proved. • [ˆ y , pˆx ] = 0. Proof: Let f (x, y) be the function. Therefore, [ˆ y , pˆx ] f (x, y) = yˆpˆx f (x, y) − pˆx yˆf (x, y) d df (x, y) + i~ {yf (x, y)} = −i~y dx dx df (x, y) df (x, y) = −i~y + i~y dx dx =0

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• [ˆ x, pˆx ] = i~. Proof: Let f (x) be the function. [ˆ x, pˆx ] f (x) = xˆpˆx f (x) − pˆx xˆf (x) d d = −i~x f (x) + i~ {xf (x)} dx dx df (x) df (x) + i~x + i~f (x) = −i~x dx dx = i~f (x)

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Thus, [ˆ x, pˆx ] = i~ h i ˆ ˆ ˆ z , where L ˆ x, L ˆ y and L ˆ z are the three Cartesian • Lx , Ly = i~L components of angular momentum. Proof: The x, y and z-component of angular momentum operator is given by ˆ x = yˆpˆz − zˆpˆy L ˆ y = zˆpˆx − xˆpˆz L ˆ z = xˆpˆy − yˆpˆx L

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If f be the function, then i h ˆyf − L ˆyL ˆ xf ˆ xL ˆ ˆ Lx , Ly f = L = (ˆ y pˆz − zˆpˆy ) (ˆ z pˆx − xˆpˆz ) f − (ˆ z pˆx − xˆpˆz ) (ˆ y pˆz − zˆpˆy ) f   d d = −~2 y − z (zfx0 − xfz0 ) dz dy   
d d − z −x yfz0 − zfy0 dx dz  d d d d = −~2 y (zfx0 ) − y (xfz0 ) − z (zfx0 ) + z (xfz0 ) dz dz dy dy 

d d d d − z (yfz0 ) + z zfy0 + x (yfz0 ) − x zfy0 dx dx dz dz     00 0 00 00   = −~2  yzf yxf z 2f xz + yfx −  zz −  xy     00 00 2  0 00 00 0      zx + z   + zxf zyf f + xyf − xzf − xf  zy −   yx zz yz y    0  2 0 = −~ yfx − xfy = i~ [−ˆ y pˆx + xˆpˆy ] f ˆzf = i~L (21) 12

h

i ˆ ˆ ˆz Thus, Lx , Ly = i~L h i ˆ • Lx , yˆ = i~z. Proof: Let f be a function. h i ˆ ˆ x yˆf − yˆL ˆ xf Lx , yˆ f = L = (ˆ y pˆz − zˆpˆy ) yf − y (ˆ y pˆz − zˆpˆy ) f     d d d d yf − y −i~y + i~z f = −i~y + i~z dz dy dz dy   d(yf ) d(yf ) 2 0 0 = i~ −y +z + y fz − yzfy dz dy     2 0 0 2 0 0 = i~ − y fz − 0 + zyf  y + zf + y fz − yzf y  



= i~zf (22) h

i ˆ Thus, Lx , yˆ = i~z 1.3. Uncertainty relation: X Suppose, you want to determine the position of a particle. This is quite simple, you simply fix the origin of the coordinate system and with respect to that origin you can determine the position of the particle. X Suppose, you also want to measure velocity of the particle, which is now moving. This time the position of the particle will vary with time. X If the particle does not move with very hight velocity, it will be possible to tell both the position and velocity of the particle simultaneously to a very high degree of precision. X But, what if the particle is moving with a very hight velocity? In this situation there will be some uncertainty in determining both the position and velocity simultaneously. X Now, if the particle is a sub-atomic particle (or in general, say, a quantum particle) then you cannot determine both the position and velocity of the particle simultaneously to any arbitrary degrees of precision. X The reason is, because, as mentioned before, a quantum system does not precisely behave as a material particle – it has both particle as well as wave-like characters – you cannot determine the position of a wave (or could you!!). X This failure of measuring both the position and velocity (and hence momentum) of quantum systems simultaneously is a fundamental limit posed by the nature and was first put in the form of a principle by Werner Heisenberg in 1927. This is called Heisenberg uncertainty principle. X It states “It is impossible to determine both the exact position and exact momentum of a quantum system, at the same time, not even theoretically.” X If the uncertainty in determining the position of a quantum system is ∆x and uncertainty in determining its momentum is ∆px then 13

section no.

the product of the two uncertainties will always be greater than or at least equal to ~/2 i.e. ∆x∆px ≥

~ 2

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X You may notice that the product of the two uncertainties is of the order of Planck’s constant, which is 10−34 in SI unit. This is an extremely small quantity. X So, how much important is this uncertainty principle? Let’s have a quick look in two cases – (1) a quantum particle e.g. electron and (2) a macroscopic particle e.g. a bicycle. Importance of uncertainty principle: XX Let, an electron is moving with velocity 2.2 × 106 m per sec (This is approximate speed of an electron in Hydrogen atom) and the uncertainty in determining the velocity be 10%. This means uncertainty in velocity is ∆v = 2.2 × 105 m per sec. We know, the mass of electron is 9.31 × 10−31 kg. This gives, the uncertainty in momentum as ∆p = m∆v = 9.31 × 10−31 × 2.2 × 105 = 2.0482 × 10−25 kg.m / sec. Using Eq. ??, the uncertainty in position of the electron will be ≥ 6.626 × 10−34 /(4.0 × 3.14 × 2.0482 × 10−25 ) = 0.2075 × 10−9 m = 0.2075 nm. Remember, the dimension of an atom is 0.1–0.5nm. Thus, if the velocity of the electron will be determined with a certainty of 90% (i.e. 10% uncertainty) then the position of the electron determined simultaneously will be of the dimension of size of the atom i.e. the determined position will be highly uncertain. In fact, regarding the position you can just say that the electron is in the atom.’ XX Now, let us take the example of a bicycle. Let the mass of the bicycle be 8 kg and is moving with a velocity of 18 km/h (5 m/sec). Let the uncertainty in determining the velocity be 10% (same as we considered for the electron). This gives, ∆p = m∆v = 8×5×0.10 = 4 kg.m/sec. And hence ∆x ≥ 6.626 × 10−34 /(4.0 × 3.14 × 4) = 0.1318 × 10−31 m. This is an extremely small value and is absolutely negligible in comparison to the dimension of the bicycle. Thus if the velocity (or momentum) determined for the bike is 90% accurate then the position will be 100% accurate. That means both position and momentum of the bike can be determined simultaneously. XX Thus, we see that a small uncertainty in velocity can lead to very large degree of uncertainty in the position of a en electron (a quantum system), whereas in case of a bicycle (a macroscopic system) the same amount of uncertainty in velocity leads to practically no uncertainty in position. This demonstrates that uncertainty principle is extremely significant for quantum systems. More about uncertainty relationship: • So, Heisenberg uncertainty principle tells about the natural limit of simultaneous determination of position and momentum (or velocity) only. Why only position and momentum? • Is the principle also valid for other pairs of observables? • If yes, then what special characteristics do these observables possess to fit into uncertainty relationship? To answer these questions, we’ll have to go back a little bit. 14

- As mentioned previouslysection no.), in quantum mechanics, observables are represented by mathematical operators. - Some pair of operators commute with each other and some don’t. - If you remember, the two operators corresponding to position, xˆ and momentum pˆx do not commute with each other. Such pair of observables (operators), which do not commute are called canonically conjugate variables. - The uncertainty relationship holds only for those who do not commute i.e. canonically conjugate variables. Applications of Heisenberg uncertainty principle: • Non-existence of free electron in nucleus of an atom – This is an interesting example. Through this, we’ll prove that an electron does not exist in nucleus of an atom. – But before that let me tell why we are interested in this topic. – You know that radioactive nucleus emits three types of radiations α, β and γ. Among these three, β radiation consist of electrons. – Thus, it is really surprising that electrons come out of the nucleus. Does that mean nucleus of an atom contains electrons? In order to prove that electrons do not exist in nucleus, we proceed as follows. # The maximum kinetic energy of an electron emitted out of the nucleus (as β radiation) is 4MeV. # The rest mass of an electron is m0 = 9.1 × 10−31 kg. # The diameter of a nucleus is ≈ 2 × 10−14 m. # If the electron exists in nucleus and if one determines the position of electron then the maximum uncertainty in determining the position of the electron would be ∆x = 2 × 10−14 m. # Thus, the minimum uncertainty in momentum would be ~ 2∆x 6.626 × 10−34 J. sec = 4.0 × π × 2 × 10−14 m = 0.2638 × 10−20 kg. m. sec−1

∆p( min) =

# Thus, if the electron exists in nucleus, it will have a minimum momentum of pmin = 0.2638 × 10−20 kg. m. sec−1 # Hence, the minimum energy of the electron would be 2 Emin = p2 c2 + m20 c4

Emin Emin Emin


2
2
4 = 0.2638 × 10−20 × 3 × 108 + 9.1 × 10−31 3 × 108 = 0.62631 × 10−24 + 0.0067076 × 10−24 = 0.633 × 10−24 = 0.7956 × 10−12 J, [1 eV = 1.6 × 10−19 J] = 0.4973 × 107 eV = 4.973 MeV

# Thus, we see that if the electron has to stay in the nucleus, it’s energy would be greater than the maximum energy observed. Hence, electrons cannot exist in the nucleus of an atom. 15

(

• Size of H-atom • Energy of a harmonic oscillator • Energy of particle in a box from Heisenberg uncertainty principle (This will be discussed when we study the particle in a box problem – sec 1.5) 1.4. Basic Postulates of quantum mechanics (This was actually a topic on Sch¨odinger time-independent equation but I changed it for convenience.) Just like, the behavior of macroscopic particles is governed by some fundamental equations such as Newton’s laws of motion, the behavior of quantum mechanical systems can also be studied using a fundamental equation called Schr¨odinger’s equation. But before exploring Schr¨odinger’s equation let us discuss the basic postulates of quantum mechanics i.e. those things which are pre-assumed for developing the theory of quantum mechanics. The basic postulates of quantum mechanics and a brief discussion of each are as follows Postulate 1. The state of a quantum mechanical system is completely described by a mathematical function, Ψ(r, t), called wavefunction. Discussion on postulate 1: – The wavefunction is nothing but a mathematical function. If you think, you can realize that this is a very simple concept. Suppose, a circle – a very simple thing. All the properties of a circle can be described by a simple equation and that is x2 + y 2 − r2 = 0

–

–

– –

(24)

How did we reach at this equation? Simply by combining all the information about a circle. For example, a circle is a twodimensional symmetrical object. The circle has a boundary that is defined at equal distance from the origin. Thus, the shape of a circle depends upon two Cartesian coordinates x and y (twodimensional). The size of the circle is related to one variable – its radius. Thus, an equation that can describe everything about a circle should contain three things – the two Cartesian coordinates and the radius – related in such a way that the boundary of the circle lies at equal distance from the origin. Eq. ?? has all these information and hence we can say that the wavefunction for a circle is Ψcircle = x2 + y 2 − r2 It is obvious that the wavefunction should depend on the coordinates of each entity in the system. The state of a system can change with time, if a force is applied on it and hence the wavefunction should depend on time also. Thus, wavefunction is a function of {r} (coordinates of all the entities in the system) and t (time). The equation x2 + y 2 − r3 = 0 also depends upon x, y and r but it is not an acceptable equation for a circle. Similarly, just any function cannot be accepted as a wavefunction. For a function to be a valid wavefunction, it should satisfy certain conditions. A valid wavefunction is a single valued, continuous and quadratically integrable function. All the partial first derivatives of a valid wavefunction with respect to the coordinates of each entity in the system should also be continuous. 16

– A valid wavefunction should satisfy the boundary conditions for the system. – A valid wavefunction should be finite everywhere within the boundary of the system. Postulate 2. An observable (e.g. energy) is represented by a linear hermitian operator. Discussion on postulate 2: – Why should an observable be represented by an operator? – To determine the value of an observable one must perform some sort of activity/experiment involving the system and that’s what an operator does. An operate operates on a function and gives a result. – Suppose you want to determine the taste of a mango then you should eat (= activity/experiment) a mango. If you eat a banana, you can not determine the taste of a mango. – Similarly, if you operate an operator on a non-eigenfunction then you cannot get the property represented by that operator. Thus, the value of an observable is obtained only by operating the corresponding operator on an eigenfunction. – Why should the operator representing an observable should be hermitian? It’s simple. An observable is something, which is real and a real result is always obtained when the operator is hermitian. So the operator representing an observable should be hermitian. A non-hermitian operator does not represent an observable. Postulate 3. The wavefunction (Ψ) of a system evolves with time following the equation ˆ = i~ dΨ , (25) HΨ dt ˆ is the Hamiltonian or total energy operator of the system. where H Discussion on postulate 3: – When a system, in a particular state, be acted upon by a timevarying force (field), the system may switch from one state to another. This equation (Eq.??) represents that time evolution of the state of the system. – This is the time-dependent Schr¨odinger equation. One can easily extract the time-independent Schr¨odinger equation from it. – Note that there is only one Schr¨odinger equation – The timedependent one. Time-independent form is simply derived from the time-dependent ones (We’ll do it at some point.). ˆ As – Now, let us talk a little bit about Hamiltonian i.e. H. mentioned above, it is the total energy operator. – What does mean by ”Total energy operator”? It means it represents the total energy of the state of the system on which it is operating. – A system can have two types of energy – Kinetic and potential. – Apart from these two, a system can also have an energy due to interaction with the external applied field e.g. interaction with and external electric field or magnetic field. ˆ can be written as – Thus, H ˆ = Tˆ + Vˆconfiguration + Vˆinteraction , H (26) 17

section no.

where Tˆ is the kinetic energy operator, Vˆconfiguration is the potential energy operator due to the relative position of the particled in the system and Vˆinteraction is the potential energy operator due to the interaction of the system with the external applied field. – Later on, we’ll discuss more about these operators in particular, how do they look like? Postulate 4. In an experiment, measuring an observable (corresponding to the ˆ of the system, the only value that can be obtained will operator A) ˆ = aΨ, where be the eigenvalue (a) that satisfies the equation, AΨ Ψ is the wavefunction of the system in that state. Discussion on postulate 4: – As discussed previously, we know that when an operator operates on a function, the result can be another function (different from the original one) or the same function. – In an experiment (applying an operator), the system (wavefunction) does not change (you don’t end up with a banana while determining the color of a mango!!). This clearly says that the wavefunction should be an eigenfunction of the operator. And if that is the case then the observable will simply be the eigenvalue. – You can say that the state of a system changes when a timevarying field (time-dependent operator) is applied. Exactly. You are right. But, then we are talking about the time-evolution of state of the system and the properties calculated in that case is not the property of a particular state of the system rather in this case the property is a transition property. That means a property related to the transition of the system from one state to another. This can be explained by giving example of light absorption, for example. So, above, when we say observable that means the property of a particular state of the system. Postulate 5. The average value of an observable, corresponding to the operator ˆ for a system in state Ψ is given by A, +∞ R

hAi =

ˆ Ψ∗ AΨdτ

−∞ +∞ R −∞

(27) Ψ∗ Ψdτ

Before discussing this postulate, let us first get rid of the denominator in Eq. ??. We can do this by considering that the wavefunction +∞ R ∗ Ψ is normalized and hence Ψ Ψdτ = 1. This reduces Eq. ?? to −∞

Z+∞ ˆ hAi = Ψ∗ AΨdτ

(28)

−∞

Discussion on postulate 5: – This postulate basically reflects the statistical nature of quantum mechanics. – It actually differentiates two different cases 18

section no.

– (1) If a system is in a pure quantum state Ψ and if Ψ is an eigenfunction of the operator Aˆ then it is sure that a measurement of property A will always gives the eigenvalue a. If we repeat the measurement then also each time we get the value a. This is quite obvious!! Isn’t it?? Let go for the second case. – (2) If a system is in a mixed state. Mixed state simply means when the wavefunction of the system is a linear combination of several states. Let the system be in a mixed state of two states φ1 and φ2 and the total wavefunction of the system be Ψ = c1 φ1 + c2 φ2 . In this case, if we measure the property A then we can get the two different eigenvalues (corresponding to the functions φ1 and φ2 . Both of these functions are eigenˆ otherwise the total wavefunction will not be an functions of A, eigenfunction of this operator). The probability of obtaining the eigenvalue corresponding to φ1 will be c21 and that corresponding to φ2 will be c22 . Since the total probability of obtaining the eigenvalue should be 1 (a measurement always leads a result), c21 + c22 = 1. Postulate 6. The wavefunction of the system is antisymmetric with respect to the interchange of coordinates between any two fermions in the system. Discussion on postulate 6: – Suppose a system consist of two particles 1 and 2 having coordinates x1 and x2 respectively. Let, Ψ(x1 , x2 ) be the wavefunction of the system. On exchanging the coordinates of the particles i.e. particle 2 be placed to x1 and particle 1 be placed to x2 then the resultant wavefunction will be Ψ(x2 , x1 ). Then, the postulate simply says Ψ(x1 , x2 ) = −Ψ(x2 , x1 ) – This is a basic criteria for fermions and comes from Fermi-Dirac statistics. – Fermions are entities having half-integral spin. For example, electrons, protons and neutrons have a spin = 1/2 and hence are fermions. – Fermions obeys Pauli exclusion principle i.e. two fermions can not reside in the same quantum state in a system. 1.4.1. Kinetic and potential energy operators in Hamiltonian • In Eq. ??, we encounter the Hamiltonian, which is a total energy operator. ˆ consists of kinetic energy operator Tˆ • We have also seen that H and potential energy operators Vˆconfiguraiton and Vˆinteraction . • Let us discuss each of these operators in details. This is important because in next sections we’ll have to deal with Hamiltonian explicitly. • Kinetic energy operator: – To derive the expression of kinetic energy operator, we’ll start with the expression for the linear momentum operator. For convenience let’s first consider a one-dimensional case i.e. pˆx = −i~

d dx

– Kinetic energy of a system is related to the linear momentum 19

by

p2 T = 2m – Thus, the kinetic energy operator in one-dimension becomes 1 2 Tˆx = pˆ 2m x 2 1 d Tˆx = i~ 2m dx 2 2 ~ d Tˆx = − 2m dx2 In three-dimensions, it becomes  2  2 2 2 ~ ∂ ∂ ∂ Tˆ = − + + 2m ∂x2 ∂y 2 ∂z 2 ~2 2 ˆ T =− ∇, 2m

(29)

(30)

where ∇ is called nabla, its a short form of writing the partial first derivative with respect to Cartesian coordinates. To represent the partial second derivative with respect to the Cartesian coordinates, square of nabla i.e. ∇2 is used. Thus, now you know how does kinetic energy operator look like. • Potential energy operators – A system acquires potential energy from two sources (1) Internal interaction i.e. interaction between particles within the system (Vˆconfiguration ) and (2) due to interaction with externally applied field (Vˆinteraction ). – Vˆconfiguration represents the potential energy operator due to the internal interaction of the particles of the system, which depends upon the configuration of the system. – It is a function of Cartesian coordinates of each particle in the system i.e. Vˆconfiguration = V ({xi , yi , zi })

(31)

– Since Vˆconfiguration has no real mathematical operator, it is a multiplicative operator i.e. simple multiplication by a function of Cartesian coordinates of the particles of the system. – For example Vˆconfiguration , for a one-dimensional simple harmonic oscillator, is 21 kx2 , where k is the spring constant (or force constant) of the spring. – Now comes Vˆinteraction . It corresponds to interaction with the externally applied field. For example, if we apply a field that oscillates with time then Vˆinteraction would be say cosωt, where ω is the frequency of oscillation and t is the time. • In absence of any external field, the Hamiltonian will simply be ˆ = Tˆ + Vˆ H

(32)

2

ˆ = − ~ ∇2 + V ({xi , yi , zi }) H 2m 20

(33)

• If a system consist of particles, which do not interact with each other – independent particles – then the Hamiltonian will simply ˆ free = − ~2 ∇2 . be the kinetic energy operator i.e. H 2m 1.4.2. Sch¨odinger time-independent equation: As mentioned earlier, Eq. ?? is a fundamental equation of quantum mechanics. This is time-dependent Schr¨odinger equation. It represents the time-evolution of a quantum state of the system. However, in general, in absence of any external field the state of the system does not evolve over time and hence for such cases it is much convenient to use a time-independent version of Eq. ??. Time-independent Schr¨odinger equation can easily be obtained from Eq. ?? using separation of variable technique.1 . Let’s do it. So, we have d ˆ HΨ(r, t) = i~ Ψ(r, t) dt In separation of variable technique, we assume that the function Ψ(r, t) can be written as a product of two functions each is a function of one variable only. i.e. Ψ(r, t) = ψ(r)φ(t). Inserting this in timedependent Schr¨odinger’s equation, we can write d ˆ Hψ(r)φ(t) = i~ ψ(r)φ(t) dt dφ(t) ˆ Hψ(r)φ(t) = i~ψ(r) dt

(34)

ˆ is time-independent (remember there Now, since the Hamiltonian H is no time-dependent field in Eq. ??), we can write dφ(t) dt

(35)

1 ˆ 1 dφ(t) Hψ(r) = i~ ψ(r) φ(t) dt

(36)

ˆ φ(t)Hψ(r) = i~ψ(r) Dividing both sides by ψ(r)φ(t), we get

Thus, we have an equation (Eq. ??) whose left hand side is a function of spatial coordinates (i.e. r) only and right hand side is a function of time only. Thus, each side should be equal to a constant. Let this constant be represent by E, which later on can be shown to be the energy of the state. Thus, 1 dφ(t) =E φ(t) dt φ(t) = e−iEt/~ i~

(37)

and 1 ˆ Hψ(r) = E ψ(r) ˆ Hψ(r) = Eψ(r) This is time-independent Schr¨odinger equation. 1.4.3. Nature of Schr¨odinger’s equation 1

you can know more about separation of variables techniques at (https://tinyurl.com/7c7zkas)

21

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section no.

• In this section we’ll discuss the nature of the time-independent Schr¨odinger’s equation. ˆ Hψ(x) = Eψ(x)   ~2 2 − ∇ + V (x) ψ(x) = Eψ(x) 2m ~2 d2 ψ(x) = (E − V (x)) ψ(x) − 2m dx2 d2 ψ(x) 2m = 2 (V (x) − E) ψ(x) dx2 ~

(39)

• Thus, Schr¨odinger equation is a second order partial differential equation. • Note, in Eq. ?? full derivative is used not partial. This is because Eq. ?? is written for one-dimensional case. Since in onedimensional case there is no other variable a partial derivative simply becomes full derivate. • So, what does this second order partial differential equation says? • The second derivative of a function is related to the curvature of the function (https://tinyurl.com/y8l2l573). • As an approximation, when second derivative of a function is positive then the graph of the function will be concave up i.e. ^ • similarly, when the second derivative of a function is negative, the graph of the function will be concave down i.e. _ 1.4.4. Conditions imposed on the wavefunctions We have already discussed this while discussing postulate 1 in section 1.4. However, as a revision let’s collect all those conditions here. So, the conditions imposed on a wavefunction are • A valid wavefunction should be a single valued, continuous and quadratically integrable function. • All the partial first derivatives of a valid wavefunction with respect to the coordinates of each entity in the system should also be continuous. • A valid wavefunction should satisfy the boundary conditions for the system. • A valid wavefunction should be finite everywhere within the boundary of the system. • A valid wavefunction for an electronic system should be antisymmetric with respect to the exchange of any two electrons. 1.4.5. Probability interpretation of wave function • The most fundamental concept in quantum mechanics is wavefunction. • It contains all the information about the system in a particular state. • So, if you have access to the wavefunction of a particular state of the system that means you can tell everything about the system in that state. • This simply means that a wavefunction is loaded with enormous amount of information. 22

• So, how can you interpret a wavefunction? • What does a wavefunction really mean? • Max Born was the first to interpret a wavefunction. His interpretation was a statistical one. • A wavefunction can be a real, imaginary or a complex function. But, according to Born’s interpretation, the square modulus of a wavefunction (Ψ) i.e. |Ψ|2 is always a real number and it represents the probability density of detecting a particle at a particular place. • Thus, |Ψ|2 = Ψ∗ Ψ represents the probability density of detecting the particle. • The quantity Ψ∗ Ψdτ then represents the probability of finding the particle in volume element dτ . • Ψ is thus a probability distribution function • Since the particle must be present H ∗somewhere in it’s boundary, Hthe sum of all probabilities i.e. Ψ Ψdτ should be equal to one. represents taking integration over the whole boundary of the system. Z Ψ∗ Ψdτ = 1 (40) boundary

Orthogonality and normalization of wavefunctions: • A wavefunction satisfying Eq. ?? is called a normalized wavefunction. • A normalized wavefunction simply represents a confirmation that the particle is somewhere in the limits of integration (i.e. within the boundary of the system). • Two wavefunctions ψ and φ are said to be orthogonal when Z φ∗ ψdτ = 0 (41) • As mentioned earlier, orthogonality of wavefunctions represent that the two wavefunctions do not share any common space, when plotted on graph. • Orthogonality also represents a mutual exclusion of the two states of the system. That means a state does not carry any information about it’s orthogonal state. • So, if a system is in one state ψ then you can never measure the property corresponding to the orthogonal state φ. • In other words, using the wavefunction ψ one can never calculate the properties of corresponding to wavefunction φ. • The quantity Z φ∗ ψdτ

(42)

is called the overlap integral and it represents how much alike the two states are to each other? 1.5. Particle in a box: • So far, we have learned the basics of quantum mechanics. Now, it’s turn to see how can we apply these to some very simple systems. • These simple systems are ideal systems called model systems. 23

section no.

• A model system is designed in such a way that they represent the most simplified scenario of a real system. • One of such model systems is particle in a box model.

The following discussion will help us to understand this topic.

 What is this particle in a box model? – Particle in a box is a simple system. – We consider a one-dimensional region of certain length (say, L) along x-axis between x = 0 and x = L. – A constant potential is applied in the region 0 < x < L. For convenience this constant potential is considered to be 0. – In regions other than 0 < x < L, infinite potential is applied. – Finally, we put a quantum mechanical particle of mass m in the region 0 < x < L. – Thus, pictorially, particle in a 1D box would look like

Figure 4: Pictorial representation of particle in a 1D box  What real system does a particle in a box model represent? – Particle in a box actually represents an electron in a linear molecule. – Let us suppose a linear conjugated molecule. For example, buta1,3-diene (CH2 = CH − CH = CH2 ). – The particle inside the box in Fig. ?? is actually representative of the mobile electrons i.e. π-electrons in buta-1,3-diene. – The length, L of the box in Fig. ?? represents the length of the molecule i.e. distance between the two end C-atoms. – The boundary of the box in Fig. ?? represent the two ends of the molecule. – This is very obvious that the electrons in a molecule can not fall out of the molecule when it reaches the end as if some force forbids the electrons to fall out. This fact is represented by the infinite potential at the two end of the box in Fig. ??. – It is also reasonable to consider that the electrons within the molecule experiences a uniform (a constant) potential. This is represented by a constant (or for convenience, zero potential) within the box. – Thus, you see, particle in a box model can reasonably be used as a simplified model for a linear molecule with π-electrons i.e. linear conjugated molecules.  How can we apply quantum mechanical principles to study this model? 24

– In quantum mechanics, the basic equation is Schr¨odinger’s equation. – At first, we will write down the correct Hamiltonian for the system that appears in Schr¨odinger’s equation. – We will then solve the time-independent Schr¨odinger equation for our model. – Note, we are not interested in time-dependent problem as there is no time-dependent applied field. – After solving the time-independent Schr¨odinger’s equation, we will get the energy and wavefunction of different states of the system.  What real properties can we calculate? – By solving Schr¨odinger equation, we get the wavefunction and energy of different states of the system. – From wavefunctions, we can calculate any observable for our model system. – From energies of different states, we can calculate the energy difference between states, which in basically the excitation energy. – From excitation energy, we can calculate the wavelength of light that can be used for causing a particular transition. – It is important to keep in mind that particle in a 1D box is a simplified model and hence we cannot expect it to reproduce the experimental values of excitation energy or excitation wavelength or other properties. – However, it can definitely be used to reproduce the experimental trend in a property. For example, the variation of excitation wavelength with the increase in length of linear conjugated molecules. 1.5.1. Setting up of Schr¨odinger equation for 1D box and solution Now, you know, what a particle in a 1D box model is and why and what are we going to do with this model. So, let’s move on • Hamiltonian for the particle in a 1D box model According to Eq. ??, the Hamiltonian in one-dimension is given by 2 2 ˆ = − ~ d + V (x) H 2m dx2 Since potential for a particle in a 1D box is infinite everywhere except in the region 0 < x < L, the particle cannot exist outside this region. Therefore, we should only consider the Hamiltonian in this region (0 < x < L), which would be ~2 d2 ˆ H=− 2m dx2

(43)

• Solving Schr¨odinger’s equation for particle in a 1D box: – As mentioned earlier, the particle does not exist outside the region 0 < x < L. Therefore, there is no meaning of wavefunction outside this region. – If Ψ(x) and E be the wavefunction and energy of the particle inside the box, the Schr¨odinger’s equation inside the box 25

section no.

would simply be

– –

– –

–

~2 d2 − Ψ(x) = EΨ(x), 0 < x < L 2m dx2 2mE d2 (44) Ψ(x) + Ψ(x) = 0, 0 < x < L dx2 ~2 √ d2 2mE 2 Ψ(x) + k Ψ(x) = 0, k = , 0<x
(46)

The constant can be obtained by normalizing the wavefunction. But, this we’ll do a bit later. Let us first apply the second boundary condition. Boundary condition 2: At x = L, Ψ(x) = Ψ(x = L) = 0. Putting this condition in Eq. ??. Bsin(kL) = 0

(47)

Since, B 6= 0 (otherwise the wavefunction Eq. ??, will become zero inside the box). Thus, Eq. ?? gives sin(kL) = 0 =⇒ kL = nπ,

(48)

where n = 0, 1, 2, 3, . . . . It is important to note that n = 0 is also not allowed, because it will also make the wavefunction zero inside the box. This suggests that the states of a particle in a 1D box are quantized. n = 1 corresponds to ground state, n = 2 is the first excited state and so on. Putting the expression for k in Eq. ?? gives √ 2mE L = nπ, n = 1, 2, 3, . . . ~ (49) n2 ~2 π 2 E= 2mL2 Thus, the expression for the energy of a particle in a 1D box is n2 h2 E= 8mL2 26

(50)

– Now, let’s find out the constant B in Eq. ??. The condition for normalization of wavefunction for particle in a 1D box is ZL

Ψ(x)2 dx = 1

0

ZL

B 2 sin2

nπx dx = 1 L

0

B2 2

 ZL  2nπx 1 − cos dx = 1 L 0

Since, n is an integer, the result of integration of cos 2nπx L will always be zero. So B 2L =1 2 r B=

2 L

This makes the wavefunction (Eq. ??), for a particle in a 1D box as r 2 nπx (51) Ψ(x) = sin L L – The wavefunction and energy for the ground state i.e. the state corresponding to n = 1 are r Ψgs (x) =

2 πx sin L L (52)

h2 Egs = 8mL2 • More fun with particle in a box model So, now, we know both the wavefunction and energy of different states of a particle in a 1D box. Now, based on our previous learnings, let us do some more fun with these wavefunction and energies. – The first thing we can do is to see how the wavefunction varies along x-axis. We can also check the variation of probability density. For this, we can directly plot Ψ(x) and |Ψ(x)|2 . The plots are shown in Fig. ??. For this figure, we have considered a box of length 2˚ A. There are few interesting things to note in Fig. ??. i. Both the wavefunction and probability density are always zero at the two ends of the box i.e. at x = 0 and x = L. ii. Nodes are the points where both the wavefunction and the probability density have zero value. That means the particle never occupy these points. 27

Probability density n = 1

Variation of PIB 1D n = 1 wavefunctions with x 1

1 n=1

0.7

0.6 0.5

2

0.7

0.6

1 (x)|

0.8

0.4

0.5 0.4

0.3

0.3

0.2

0.2

0.1

0.1 0

0 0.5

1 1.5 ˚)2 wavefunctions with x Variation of PIB 1D n x(= A

0

2

0.6

0.8

0.4

0.7

0.2 0 0.2

2

1 0.9

0.6

2 (x)|

1 0.8

0.5 0.4

0.4

0.3

0.6

0.2

0.8

0.5

1

1.5

2

1.5

2

1.5

2

˚ x(density A) Probability n=2

|

2 (x)

0

n=2

0.1

n=2

1

0 0

0.5 1 1.5 A3) wavefunctions with x Variation of PIB 1D nx(=˚

2

0

0.8

0.4

0.7 1

0.2

2

0.6

0.6

3 (x)|

1 0.9

0.5

|

1 0.8

0

0.5

1 ˚ A) Probabilityx(density n=3

0.2 3 (x)

n=1

0.9

0.8

|

1 (x)

0.9

0.4

4

0.3

0.4 0.6

0.2

0.8

n=3

0.1

n=3

1

0 0

0.5

1

1.5

2

0

0.5

1

x(˚ A)

x(˚ A)

Figure 5: wavefunctions and probability density for particle in a 1D box 2

5

iii. Within the box, there is no node for ground state wavefunction. iv. For the first exited state i.e. n = 2, there is one node within the box and for the second excited state i.e. n = 3, there are two nodes within the box. v. In general, for the mth excited state, there exist m nodes within the box. Note, for mth excited state, n = m + 1. 3

6

Weird quantum world Did you notice something weird? No! Lets read points ii and iv again. If the particle never exists at the nodes and there exist nodes within the box then how the particle moves from one side to the other side of the box? Can you explain this? It’s really very hard to realize it. We can simply say that this is the nature of a quantum particle. Since the energy and other properties of a quantum particle is not continuous (that means these are quantized), the movement of such particles and hence their spatial existence are also not continuous but discrete. The nodes can be considered as the forbidden regions for the particle. When we quantize the energy of a particle, we do so with it’s existence too. Numerical illustration of quantum weirdness: ∗ To have more feelings about quantum weirdness, let us do some calculations. Fig. ?? shows that there is a node at x = 1 for first excited state of a particle in a 1D box. ∗ So, to detect that the particle goes from left side of the 28

node to the right side without passing through the node, let us find out the probability of finding the particle in a small volume element slightly left to the node (between x = 0.9˚ A and x = 0.99˚ A) and a small volume element slightly right to the node (between x = 1.01˚ A and x = 1.1˚ A). ∗ In section 1.4.5, we have seen that the probability of finding a particle between two points x = x1 and x = x2 is given by x=x Z 2 Ψ∗ (x)Ψ(x)dx x=x1

∗ The wavefunction of the particle in first excited state is Ψn=2 (x) = sin(πx). [Note, length of the box is L = 2˚ A and for first excited state n = 2.] ∗ Thus, the probability of finding the particle slightly to the left of the node i.e. between x = 0.9˚ A and x = 0.99˚ A is x=x Z 2

Z0.99 Ψ∗ (x)Ψ(x)dx = sin2 (πx)dx = 0.00322

x=x1

0.9

Similarly, the probability of finding the particle slightly to the right of the node i.e. between x = 1.01˚ A and x = 1.1˚ A is x=x Z 2

Ψ∗ (x)Ψ(x)dx =

x=x1

Z1.1

sin2 (πx)dx = 0.00322

1.01

∗ Thus, we see that there is a finite probability of finding the particle immediately to the left or to the right of the node. But is zero at the node. This confirms that the particle moves from one side to the other side of node without ever existing at the node.

29

Another quantum weirdness There is another weird thing in Fig. ??. ∗ Let us have a look at the probability density plot for ground state. ∗ This plot tells us, how probable is it to find the particle at a particular point in the box. ∗ As the figure suggests, the probability of finding the particle is maximum at the center of the box. ∗ It seems that the particle is trying to avoid the boundary (as if they fear of something at the boundary). As we move towards the boundary, the probability decreases. ∗ This is weird. Classically, the probability of finding the particle should be equal at any point in the box. ∗ This is another weird nature of quantum particles and is completely because of their quantum mechanical nature. OK, we have talked enough about weirdness of a particle in a 1D box. Let us now discuss some other interesting features. – Orthogonality of two states: ∗ We know from previous sections that two orthogonal states do not share any common information. ∗ So, it will be fun to know, which two states of particle in a 1D box are orthogonal to each other? ∗ For this purpose let us two consider two different states of the system, say Ψm (x) and Ψn (x) states and evaluate the overlap integral between the two. ∗ The overlap integral between the two states is given by (From Eq. ??) x=L x=L Z Z mπx nπx sin x Ψm (x)Ψn (x)dx = sin L L x=0

x=0 x=L Z

x=L Z 1 (m − n)πx 1 (m + n)πx = cos dx − cos dx 2 L 2 L x=0 x=0  L (m − n)πx L (m + n)πx L sin − sin = 2(m − n)π L 2(m + n)π L 0 L L = sin((m − n)π) − sin((m + n)π) 2(m − n)π 2(m + n)π

Since both m and n are integers, each term on right hand side will be zero. Therefore, for any two different states of

30

a particle in 1D box, x=L Z Ψm (x)Ψn (x)dx = 0 x=0

Thus, the states of a particle in a 1D box are orthogonal to each other. – Studying physically observable quantity of real systems ∗ As mentioned before, particle in a 1D box is a simplified model and hence as such it cannot reproduce an experimental result. However, it can be used to study an experimental trend. ∗ In this section, we will see whether or not our particle in a 1D box model can reproduce the experimental trend of variation of excitation wavelength in linear conjugated molecules. ∗ For particle in a 1D box, the excitation energy for a transition from mth state to nth state is given by, ∆Emn

h2 n2 h2 m2 h2 2 2 − = (n −m ) (53) = En −Em = 8mL2 8mL2 8mL2

The wavelength (λmn ) corresponding to excitation energy ∆Emn is given by hc ∆Emn = , (54) λmn where c is the speed of light in vacuum. Combining Eqs. ?? and ??, we get 8mcL2 λmn = 2 (55) (n − m2 )h

Thus, for a transition from ground state (m = 1) to the first excited state (n = 2), we can write λ12

8mcL2 = 3h

(56)

Since, m, c and h are fundamental constants, Eq. ?? reflects that the excitation wavelength should increase with the increase in length of the box (or length of linear conjugated alkenes) ∗ An experimental study of UV-Visible spectra of several linear conjugated alkenes shows the following results Molecule Wavelength Ethylene 163 Butadiene 217 Hexatriene 268 Octatetraene 304 Decapentaene 328

31

∗ It is clear that experimentally the excitation wavelength of linear conjugated alkenes gradually increases with the increase in chain-length, which is also predicted by particle in a 1D box model. ∗ Thus, we see that particle in a 1D box model can be used to predict an experimental trend in excitation wavelength in linear conjugated alkenes. Actual comparison Problem: Buta-1,3-diene is a conjugated molecule. It has a structure CH2 = CH − CH = CH2 . If effective length of the molecule be 4˚ A then by using particle in a 1D box model, calculate the excitation energy for transition from ground state to the first excited state in buta-1,3-diene. Compare your result with the experimental value of 217nm. Solution: Putting the values of m = 9.1 × 10−31 kg, h = 6.626 × 10−34 J.Sec, c = 3.0 × 108 m and L = 4 × 10−10 m in Eq. ??, we get λ12 λ12 λ12

8 × 9.1 × 10−31 kg × 3.0 × 108 m.sec−1 2 L = 6.626 × 10−34 J.Sec = 32.961 × 1011 (4 × 10−10 )2 m, (L is in m.) = 527.4 × 10−9 m = 527.4nm

This wavelength is much larger than the experimental value. Thus, because of very simple nature, particle in a 1D box cannot really reproduce the experimental value. But as shown in previous section, it can be used to predict the experimental trend. 1.5.2. Free particle A concept closely related to our particle in a 1D box is free particle. As the name suggest, a free particle is on which is not acted upon by a field. Thus, if you remove the two infinite potentials at the two points (i.e. at x = 0 and x = L) in a particle in a 1D box then it turns to be a free particle. • Schrødinger equation, wavefunction and energy of a free particle – A free particle has the same Schrødinger equation as particle in 1D box i.e. Eq ?? except that there is no limit on x. – Thus, for a free particle, p 2 2mEfp d 2 Ψ (x) + k Ψ (x) = 0, k = (57) fp fp fp fp dx2 ~ The subscript “fp” is used to represent that the wavefunction and the energy are for a free particle. – Eq. ?? has the same solution, as the particle in a 1D box i.e. Ψfp (x) = C1 eikfp x + C2 e−ikfp x . – Note, there is no limit on x that means there is no boundary

32

condition. The quantity Z+∞ Ψ∗fp (x)Ψfp (x)dx

(58)

−∞

is infinite and hence the free particle wavefunctions are not normalizable. – If you remember, we get the energy expression for particle in a 1D box by normalizing the corresponding wavefunction. Since for a free particle, we cannot normalize the wavefunction, an explicit energy expression cannot be obtained. – However, we can have some idea about the energy if We can put one obvious condition on Ψfp (x). The obvious condition is that Ψfp (x) should be finite everywhere, even when x → ±∞. – Now, if that is the case then Efp cannot be negative. Because, Efp < 0 =⇒ kfp = imaginary =⇒ Ψfp (x) becomes infinity as x → +∞. – Thus, energy of a free particle should always be positive. • Comparison of eigenfunctions and eigenvalues of particle in a 1D box and a free particle – The eigenfunctions for particle in a 1D box is a sine function but that of a free particle is a linear combination of sine and cosine functions. – The eigenfunctions of particle in a 1D box is normalizable but that of a free particle is not. – The eigenvalues i.e energy for a particle in a 1D box varies inversely with the square of length of the box but no such condition can be derived for a free particle, because there is no boundary. – The energy of particle in a 1D box is quantized (depends on the quantum number n) but that of a free particle is not, in fact there is no energy expression for a free particle. 1.5.3. Properties of PIB wavefunctions (normalization, orthogonality, probability distribution) These topics have already been discussed previously. 1.5.4. Expectation values of x, x2 , px , and p2x for a particle in a 1D box and their significance in relation to the uncertainty principle • As mentioned in postulate 5 of quantum mechanics, the average value (also called expectation value) of a quantity A corresponding to the operator Aˆ in a normalized state Ψ is given by Z ˆ hAi = Ψ∗ AΨdτ, (59) all space

“all space” represents the boundary condition for the system. • Using Eq. ??, let us find out the expectation values of some important quantities like x, x2 , px and p2x . • But before doing this you must know what these quantities are and why should you do this? 33

• hxi and hpx i are the expectation values of the position and linear momentum of the particle. In simple words, they represent the average position and average linear momentum of the particle. • Similarly, hx2 i and hp2x i are the average of the square of the position and momentum of the particle. • These quantities are basically related to the standard deviation of position and linear momentum i.e. ∆x and ∆px terms appeared in the expression for Heisenberg Uncertainty principle. • The said relationship is given by p (60) ∆A = hA2 i − hAi2 • So, if we know these four expectation value, we can check if Uncertainty principle is valid for the particle or not. Now you know what these four expectation values are and why we want to calculate these values for particle in a 1D box. So, let’s do it. • Calculating hxi for particle in a 1D box: hxi =

ZL

Ψ∗n (x)ˆ xΨn (x)dx

0

ZL r =

 nπx  2 sin x L L

r

 nπx  2 sin dx L L

0

1 = L

 ZL  2nπx 1 − cos xdx L 0

 ZL  2nπx x − xcos dx L 0  2   L  1 x L 2nπx L2 2nπx − 2 2 cos = − xsin L 2 2nπ L 4n π L 0 L hxi = 2 (61) 1 = L

Thus, the average position of the particle is at the middle of the box. This can also be predicted from the ground state probability density plot in Fig. ??.

34

• Calculating hx2 i for particle in a 1D box: hx2 i =

ZL

Ψ∗n (x)ˆ x2 Ψn (x)dx

0

ZL r =

2 sin L

r

 nπx 

2

x

L

 nπx  2 sin dx L L

0

  ZL  2nπx 1 2nπx 1 − cos x2 dx = x2 − x2 cos dx L L L 0 0    3   nπx  L 2 2nπx 2L3 2nπ 2xL2 1 x + x sin − 3 3 sin + 2 2 cos = L 3 2nπ L 4n π L 8n π L 2 2 L L + 2 2 hx2 i = 3 2n π (62) 1 = L

ZL 

• Calculating hpx i for particle in a 1D box: hpx i =

ZL

Ψ∗n (x)ˆ px Ψn (x)dx

0

ZL r =

2 sin L

 nπx   L

0

2inπ~ =− 2 L

ZL sin

0 L Z

d −i~ dx

 nπx  L

cos

r

 nπx  2 sin dx L L

 nπx  L

dx

(63)

 2nπx sin dx L 0   L inπ~ L 2nπx =− 2 × cos L 2nπ L 0 hpx i = 0 inπ~ =− 2 L



We see that the expectation value of linear momentum for a particle in a 1D box is zero. This is obvious, because since we did not put any restriction on the direction of movement of the particle, it can move both in positive (towards x = L) or in negative direction (toward x = 0). On average, these two opposite linear momenta cancel out each other.

35

• Calculating hp2x i for particle in a 1D box: hp2x i =

ZL

Ψ∗n (x)ˆ p2x Ψn (x)dx

0

ZL r = 0

r  nπx    nπx  2 2 d 2 2 sin −~ sin dx L L dx2 L L

2 2 2

2~ n π = L3 ~2 n2 π 2 = L3

ZL

2

sin

 nπx  L

dx

0 ZL 

(64)

 2nπx 1 − cos dx L 0  L 2 2 2 ~nπ L 2nπx = x− sin L3 2nπ L 0  2 nπ~ = L

• Since, we now have all the four expectation values hxi, hx2 i, hpx i and hp2x i, let’s calculate the uncertainty in position and momentum of the particle in a 1D box. • Calculating ∆x p ∆x = hx2 i − hxi2 r L2 L2 L2 = + 2 2− 2n π 4 r3 1 1 =L + 2 2 12 2n π • Calculating ∆px p hp2x i − hpx i2 s 2 nπ~ = −0 L nπ~ = L

∆px =

• Calculating ∆x∆px r

1 1 nπ~ + 2 2× 12 2n π L r 1 1 = nπ~ + 2 2 12 2n π

∆x∆px = L

Even for minimum value of n i.e. n = 1 r 1 1 ~ + 2 = 1.14~ > ∆x∆px = π~ 12 2π 2 Thus, for any state of a particle in a 1D box, ∆x∆px > Heisenberg uncertainty principle is valid. 36

~ 2

i.e.

1.5.5. Extension of PIB to 2D and 3D • Particle in a 1D box is a simplified model for linear conjugated molecule. • However, a molecule is not one dimensional. • So, a better model would be a particle in a 2D or 3D box. • That’s why, we need an extension of our model to 2D and 3D. • This is very simple. We don’t even need to solve the Schrødinger’s equation in 2D or 3D. • Why? The reason is explained in the following. Wavefunction of particle in a 2D and 3D box • We have seen that the eigenfunction of Tˆx operator (which is basically the hamiltonian for particle q in a 1D box along x-axis) is a function of x variable only ( L2 sin nπx L ). • In stead of considering the box along x-axis, if we consider it along y-axis then the hamiltonian will be Tˆy and the eigenfunction will be a function of y variable only. • Thus, the eigenfunction of Tˆx behaves as a constant, when it is acted upon by Tˆy and that of Tˆy behave as a constant for Tˆx operator. • Hence, a product of eigenfunctions of the two operators will serve as an eigenfunction for both the operators. • So, the solution for particle q in a 2D box q (along x and y axes) can be written as

nx πx 2 Lx sin Lx

×

ny πy 2 Ly sin Ly

• Similarly, for 3D will be q q q box, the solution ny πy nx πx nz πz 2 2 2 Lx sin Lx × Ly sin Ly × Lz sin Lz

• Lx , Ly , and Lz are the length of the box along the three Cartesian axes and nx , ny , and nz are the respective quantum numbers. • OK, we have got the wavefunction for particle in a 2D and 3D boxes very easily. What will be the eigenvalues then? Engery of particle in a 2D or 3D box • This is also very simple. The eigenvalue will simply be the sum of the eigenvalues of the Tˆi (i ≡ x, y, z) operators. • Let us verify it. • Let Ψx , Ψy , and Ψz be the eigenfunctions of the operators Tˆx , Tˆy , and Tˆz with eigenvalues Ex , Ey , and Ez respectively. Thus   ˆ ˆ ˆ Tx + Ty + Tz Ψx Ψy Ψz = Tˆx Ψx Ψy Ψz + Tˆy Ψx Ψy Ψz + Tˆz Ψx Ψy Ψz = Ψy Ψz Tˆx Ψx + Ψx Ψz Tˆy Ψy + Ψx Ψy Tˆz Ψz = Ex Ψy Ψz Ψx + Ey Ψx Ψz Ψy + Ez Ψx Ψy Ψz = (Ex + Ey + Ez ) Ψx Ψy Ψz • Thus, we see, (1) the wavefunction for a particle in a 2D or 3D box will simply be a product of the wavefunctions for particle in a 1D box in different directions and (2) the corresponding 37

eigenvalue will be the sum of the eigenvalues for the particle in a 1D box in different directions. Wavefunction and energy – particle in a 2D and 3D box In 2D box For rectangular box (Lx = 6 Ly ) s 22 nx πx ny πy Ψnx ,ny = sin sin Lx Ly Lx Ly ! n2x n2y h2 Enx ,ny = + L2x L2y 8m For square box (Lx = Ly = L) r 22 nx πx ny πy sin sin Ψnx ,ny = L2 L L 2
h Enx ,ny = n2x + n2y 8mL2

(65)

(66)

In 3D box For cuboid box (Lx = 6 Ly 6= Lz ) s 23 nx πx ny πy nz πz Ψnx ,ny ,nz = sin sin sin Lx Ly Lz Lx Ly Lz ! n2z h2 n2x n2y + + Enx ,ny ,nz = L2x L2y L2z 8m For cubic box (Lx = Ly = Lz = L) r 23 nx πx ny πy nz πz Ψnx ,ny ,nz = sin sin sin L3 L L L 2
h Enx ,ny ,nz = n2x + n2y + n2z 8mL2

(67)

(68)

1.5.6. Concept of degeneracy • Two different states having same energy are called degenerate states. • In particle in a 1D box, it is obvious from Eq. ?? that no two states can have the same energy – because no two states can have same value for the quantum number n. • However, from Eqs. ??, ??, ?? and ?? one can easily see that particle in a 2D or in a 3D box can have same energies in two different states – because all the three quantum numbers nx , ny , and nz can have an integer value (1, 2, 3, . . . ) • As an example, Let us consider the energies of nx = 1, ny = 2 (in short (1, 2)) and nx = 2, ny = 1 (in short (2, 1)) states of a particle in a square box (Eq. ??). Both of these states have an 38

2

• • • •

• •

5h energy of 8mL 2 and hence are degenerate. Similarly, in a cubic 3D box, the states, for example (1, 1, 2), (1, 2, 1), (2, 1, are also degenerate. These degeneracies arise merely because of permutations in values of the quantum numbers. However, Eqs. ?? and ?? indicate that additional degeneracies are also possible due to ratio of the quantum number to the length of the sides of the box in different directions. For example, for a rectangular box with Lx : Ly = 2 : 1, the states (2, 5) and (10, 1) are degenerate. Check if, for this box, the pair of states (1, 4), (7, 2) and (1, 6), (9, 4) are degenerate or not? Similar, if the box is such that Lx : Ly = 1 : 2 then the states (2, 2) and (1, 4) will be degenerate. Check if, for this box, the pair of states (1, 6), (3, 2) and (1, 8), (4, 2) are degenerate or not? Degeneracy basically reflects the symmetry in a system. The more symmetrical a system is the more is the number of degenerate levels it can have.

Not in syllabus: Particle in a 1D box with a non-uniform potential • So far, we have considered a uniform potential within the box (for our convenience, we have considered it to be zero, V = 0). • What will happen if the particle is subjected to a non-uniform potential inside the box? In this section, we will address this question. • In simple words, applying a potential inside the box simply means applying a field. • For example, suppose you are riding your bicycle on a flat smooth road without any pit or speed-breaker or a turn or traffic. This means you are in a uniform potential (no disturbance). • Now suppose, there are speed-breakers at different places on the road or there is traffic on the road. • The speed-breakers or the traffic puts extra conditions on your cycling. You become more cautious during cycling. This reflects that you are cycling in a non-uniform condition ≡ non-uniform potential. • So, applying a non-uniform potential simply means creating trouble for the movement of the particle. • This means, in a non-uniform potential there will be extra restriction(s) on the particle. • Let us start with a very simple potential – a sinusoidal potential sinx, between 0 < x < L. For simplicity, we consider L = π 2. Quantum Chemistry II 39

2.1. Simple Harmonic Oscillator 2.1.1. What is an oscillator, a harmonic oscillator and a simple harmonic oscillator? We can start our discussion with the word ‘Oscillator’. An oscillator is simply anything that undergoes a periodic motion that means it goes forth and back about a fixed position. The fixed position is also called the equilibrium position and is nothing but the position of the system when it is not acted upon by any external force. For example, if you hit a hanging fruit softly (so, that it does not fall from the branch) then it starts oscillating. A harmonic oscillator is any system which when displaced from its equilibrium position experiences a restoring force that is directly proportional to the displacement. For example, let us consider a spring. If you press a spring a tension is generated in it and when you release the pressure the spring immediately relaxes and jumps in a direction opposite to the direction of application of force. Another Spring at equilibrium

Displacement 𝒙 Stretched spring

Applied force

Restoring force (𝑭)

example of a harmonic oscillator is a pendulum. If you drag the bob of a pendulum from its equilibrium position and release it then the bob will move towards its equilibrium position.

Displacement in this direction 𝑥

Equilibrium position

F) e( c r o g F ction n i r e sto dir Re this

Bob

in

A simple harmonic oscillator is a harmonic oscillator that is not acted upon by an external force. That means, you displace the harmonic oscillator from its equilibrium position and release it freely without applying any extra force. 2.1.2. Why do we study simple harmonic oscillator in quantum mechanics? What real system does it represent? We know that in a molecule, the atoms are bonded with each other. 40

These bonds do not remain static. They always oscillates. The nature of these oscillations can be very simple (as in the case of di-atomic molecules) or very complex (as in case of molecules with large number of atoms). In order to study these oscillations (also called bond vibration) the problem of simple harmonic oscillator serves as the simplest model. Compressed

Equilibrium position

Stretched

2.1.3. What problem are we going to solve? Let us first define our system. Let our system consists of a mass m attached to a wall with a weightless spring of equilibrium length l along x−axis, as shown in figure ??. Now, if the spring is stretched 𝒙=𝟎 Spring at equilibrium

𝒙=𝒙

m1

𝒍 Displacement 𝒙 m1

Stretched spring

Restoring force (𝑭)

Figure 6: Our one-dimensional simple harmonic oscillator. by moving the mass by a distance x and then released freely, the system will start exhibiting simple harmonic oscillations about x = 0. Let the frequency of oscillation be ν.2 Our target is to study the system, for which, we will (i) write down the Hamiltonian for the system and (ii) setup the Schr¨odinger equation for it. (iii) we are not going to solve it3 but we will directly study the characteristic features of the energies and wavefunctions of different states of our simple harmonic oscillator. 2.1.4. Hamiltonian for a simple harmonic oscillator in one dimension In a simple harmonic oscillator, the restoring force is directly proportional to the displacement from equilibrium position. If F be the restoring force and x be the displacement from equilibrium position, 2

As mentioned above, simple harmonic oscillation/motion is a periodic motion that means the system moves from one side of equilibrium position to the other side. When the system starts from one side (say, right side of x = 0) it crosses the equilibrium position and goes to the other side (left side of x = 0), then again when it returns back to its previous side (left side of x = 0) then it is said that one oscillation is completed. The number of oscillations completed by the system in one second is called frequency of oscillation. It has a unit of sec−1 or Hertz (Hz). 3 This is a little bit complicated problem. That’s why we’ll not solve the Schr¨odinger equation for simple harmonic oscillator.

41

then F ∝x F = −kx,

(69)

where k is the proportionality constant and is also called force constant. This force constant is related to the mass of the system and frequency of oscillation. The relation between force constant (k), linear frequency of oscillation (ν) and mass (m) of the system is given by r 1 k ν= 2π m r (70) k ω= , ω = 2πν = Angular frequency of oscillation. m We know that the Hamiltonian is given as ˆ = Tˆ + Vˆ , H

(71)

where Tˆ and Vˆ are the kinetic and potential energy operators respectively. The kinetic energy operator has a universal expression ~2 d2 ˆ T =− 2m dx2 But the potential energy operator depends on the system under study. For simple harmonic oscillator, the potential energy operator has to be obtained from the restoring force (the only expression we know so far). We know, for a conservative force i.e. a force which depends only upon the position coordinates, potential energy is given by F =− V =−

dV dx Z F dx

Since, the restoring force in a simple harmonic oscillator is also conservative in nature, the potential energy operator can be written as Zx 1 (72) Vˆ = − (−kx)dx = kx2 2 0

Thus, the Hamiltonian for a simple harmonic oscillator is ~2 d2 1 2 ˆ H=− + kx 2m dx2 2 ~2 d2 1 ˆ H=− + mω 2 x2 2 2m dx 2

42

(73)

mass m replaced by reduced mass µ Note that, if the system consists of more than one particle then the mass (m) should be replaced by the reduced mass (µ) of the system. For example, if our system consists of two masses m1 and m2 attached with each other by the spring then the reduced mass (µ) of the system will be 1 1 1 = + µ m1 m2 m1 m2 µ= m1 + m2 The concept of reduced mass is used to reduce a many-body problem to a single body problem. 2.1.5. Setting up of the time-independent Schr¨odinger equation The general time-independent Schr¨odinger equation in one-dimension is given by ˆ v (x) = Ev ψ(x), Hψ (74) ˆ is the Hamiltonian of the system and ψv (x) and Ev are the where H wavefunction and energy of v th state of the system. As this model system is used to study vibrations in molecules, here the quantum number is represented by v. Thus, for a 1D simple harmonic oscillator, we can write   1 ~2 d2 + mω 2 x2 ψv (x) = Ev ψ(x) (75) − 2m dx2 2 2.1.6. Solution of the time-independent Schr¨odinger equation for one-dimensional simple harmonic oscillator. Wavefunctions of simple harmonic oscillator: The general solution of Eq. ?? is given by 1

ψn (x) = Nv · Hv (αx) · e− 2 α where

r Nv =

2 2

x

, v = 0, 1, 2, . . .

α 2n n!π 1/2

(76)

(77)

is the normalization constant, r α=

mω ~

and Hv (αx) is a polynomial expression called Hermite polynomial.4 Note that the vibrational quantum number v can have any positive integer values including 0. The state with v = 0 is called the ground state, v = 1 is the first excited state and so on. The value of Hermite polynomial for v = 0 and v = 1 states are H0 (αx) = 1; 4

H1 (αx) = 2αx

The general expression for Hermite polynomial is given by v

2

Hv (y) = (−1) · ey ·

43

dn −y2 e dy n

(78)

Hermite polynomials in Eq. ?? Problem: From the general expression of Hermite polynomial, 2

Hv (y) = (−1)v · ey ·

dn −y2 e dy n

(79)

prove that Eq. ?? is true. Solution: For v = 0, Eq. ?? becomes d0 −y2 H0 (y) = (−1) · e · 0 e dy 0

y2

d0 dy 0 simply −y 2

e

represents no-derivative. That means the function remains intact. Thus, 2

2

H0 (y) = ey · e−y = 1 or, H0 (αx) = 1. Now, for v = 2, Eq. ?? becomes d 2 2 H1 (y) = (−1)1 · ey · e−y dy   2 y y2 = −e × −2ye = 2y As y = αx, H1 (αx) = 2αx Wavefunctions for ground and first excited states. Using the above informations about the normalization constants (Eq. ??) and Hermite polynomials (Eq. ??), fill the missing parts in the following to derive the ground state (v = 0) and first excited state (v = 1) wavefunctions for simple harmonic oscillator. Wavefunction for ground state (v = 0):

ψ0 (x) = r ψ0 (x) =

·

α − 12 α2 x2 · e π 1/2

1

· e− 2 α

2 2

x

(80)

Wavefunction for first excited state (v = 1):

ψ1 (x) = r ψ1 (x) =

·

1

· e− 2 α

2α3 − 12 α2 x2 · x · e π 1/2

2 2

x

(81)

Variation of ψ0 (x) and ψ1 (x) with x. We know from the previous discussions that the wavefunction represents the probability amplitude and square of the wavefunction represents the probability density. Let us see how does these two quantities for ground and first excited states of simple harmonic 44

oscillator vary with the displacement x. ����� ��������

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Figure 7: Variation of ψ0 (x), ψ1 (x), ψ02 (x) and ψ12 (x). Characteristics of the plots • Number of nodes in ψ0 (x) and ψ1 (x) There is no node in the ground state but there is one in the first excited state. This can easily be predicted also from the expressions of the wavefunctions (Eqs ??, ??). The node in first excited state occurs at x = 0 i.e. at equilibrium position of the particle. • Most probable position of the particle in ground state The most probable position of finding the particle in ground state is at x = 0 i.e. at the equilibrium position of the particle. This is counter-intuitive. Since the particle is exhibiting simple harmonic motion, classically, the kinetic energy of the particle is maximum at x = 0 and hence the particle spends least time at the equilibrium. Similarly, since the kinetic energy of particle is minimum at the two extreme points, the particle is supposed to spend maximum time at the two extreme points. However, this is true only for classical case. In quantum mechanics, things are different and the most probable position of finding the particle at the equilibrium is a quantum mechanical characteristic of the system. Show mathematically that the most probable position of finding the particle exhibiting simple harmonic motion in ground state is x = 0. The most probable position of finding the particle exhibiting simple harmonic motion in ground state is obtained in two steps – (A) by taking derivative of ψ02 (x) with respect to x and setting the same to zero and then (B) finding the value of x obtained in step A for which the second derivative of ψ02 (x) with respect to x is negative.

45

Step A: d 2 ψ (x) = 0 dx 0 r 2 d α − 1 α 2 x2 e 2 =0 dx π 1/2 α d −α2 x2 e =0 π 1/2 dx
α 2 −α2 x2 × −2α x × e =0 π 1/2 −xe

−α2 x2

= 0,

2α3 as, 1/2 6= 0 π

This equation has three roots – (i) x = 0 and (ii & iii) x = ±∞. Step B:  o d2 2 2α3 n −α2 x2 2 −α2 x2 ψ (x) = − 1/2 e + x −2α xe dx2 0 π
2α3 2 2 = − 1/2 1 − 2α2 x2 e−α x π Now, d2 2 2α3 At x = 0, 2 ψ0 (x) = − 1/2 < 0 dx π 2 d At x = ±∞, 2 ψ02 (x) = 0 dx Thus, the probability of finding the particle will be maximum at x = 0. • Most probable position of the particle in first excited state Figure ?? clearly shows that the probability of finding the particle in its first excited state is maximum at two equidistant points on the two sides of the equilibrium position. Show mathematically that there are two most probable positions of finding the particle exhibiting simple harmonic motion in its first excited state. We’ll follow the same steps as we did for ground state. Step A:

d dx

r

d 2 ψ (x) = 0 dx 1 !

2α3 − 1 α2 x2 xe 2 π 1/2

2

=0

2α3 d 2 −α2 x2 xe =0 π 1/2 dx  o 2α3 n −α2 x2 2 2 −α2 x2 × e × (2x) + x × −2α xe =0 π 1/2
2 2 x 1 − α2 x2 e−α x = 0, 46

4α3 as, 1/2 6= 0 π

This equation has five roots – (i) x = 0, (ii & iii) x = ± α1 and (iv & v) x = ±∞. Step B: n 2 2
−α2 x2
d2 2 4α3 h 2 2 −α x 2 ψ (x) = 1 − α x e + x e 0 − 2α x dx2 1 π 1/2 oi
 2 2 2 −α2 x2 + 1−α x −2α xe
4α3 2 2 = 1/2 1 − 5α2 x2 + 2α4 x4 e−α x π

Now, d2 2 4α3 At x = 0, ψ (x) = 1/2 > 0 Minima dx2 1 π d2 2 ψ (x) = 0 Inflection point At x = ±∞, dx2 1 1 d2 2 8α3 and at x = ± , ψ (x) = − 1/2 < 0 Maxima α dx2 1 π e Thus, the particle in first excited state has two most probable positions equidistant from the equilibrium position. Energies of different states of simple harmonic oscillator: The energy of the v th state of a simple harmonic oscillator is   1 Ev = v + ~ω (82) 2 The energy of ground state (v = 0) of a simple harmonic state is E0 = 21 ~ω. Thus, we see that the energy of the lowest state (v = 0) of a simple harmonic oscillator is 21 ~ω and is not zero. Zero-point energy The ground state energy of a systems is also called the zeropoint energy. This is the lowest energy a system can have. This is the energy of the system at absolute zero temperature. Classically, any system at absolute zero is assumed to be at absolute rest. All the vibrations or motions in a system is assumed to cease at absolute zero. However, for quantum mechanical systems this is not true. Even at absolute zero a quantum mechanical system fluctuates. The existence of non-zero zero-point energy is consistent with the Heisenberg uncertainty principle. Problem: Using the definition of Uncertainty in linear momentum, show that the lowest energy of a simple harmonic oscillator is not zero. Hint: We need to calculate the uncertainty in momentum i.e. p ∆px = hp2x i − hpx i2 for the ground state.

47

Solution: So, let us calculate the required quantities. Z+∞ hp2x i = ψ0 (x)ˆ p2x ψ0 (x)dx −∞ Z+∞r

= −∞

 2 r 1 2 2 α d α −2α x − 12 α2 x2 · e · e −i~ dx dx π 1/2 π 1/2

α~2 = − 1/2 π α~2 = − 1/2 π =−

Z+∞ 2 − 21 α2 x2 − 21 α2 x2 d e e dx dx2 −∞ Z+∞

1

e− 2 α

−∞ +∞ 3 2 Z

α~ π 1/2

−∞

2 2

x


1 2 2 α 2 α 2 x2 − 1 e− 2 α x


2 2 α2 x2 − 1 e−α x dx

 +∞ +∞ Z Z α3 ~2 2 2 2 2 = − 1/2 α2 x2 e−α x dx − e−α x dx π −∞  −∞+∞  +∞ Z Z α3 ~2  2 2 2 2 2 = − 1/2 2α x2 e−α x dx − 2 e−α x dx π 0  √ √ 0 3 2 π π α~ = − 1/2 2α2 × 3 − 2 × 4α 2α π √  √ α3 ~2 π π = − 1/2 − 2α α π 1 = α2 ~2 2 

Z+∞ hpx i = ψ0 (x)ˆ px ψ0 (x)dx −∞ Z+∞r

= −∞

 r 1 2 2 d α α −2α x − 21 α2 x2 −i~ dx · e · e dx π 1/2 π 1/2

iα~ = − 1/2 π =−

iα~ π 1/2

iα3 ~ = 1/2 π

Z+∞ 1 2 2 d 1 2 2 e− 2 α x e− 2 α x dx dx −∞ Z+∞

1

e− 2 α

−∞ +∞ Z

xe−α

2 2

x

2 2

−∞

x

1

e− 2 α

2 2

x

× (−α2 x)dx

dx

= 0 The integrand is an odd function of x.

48

Thus, uncertainty in momentum r p ∆px = hp2x i − hpx i2 =

α~ α2 ~2 −0= √ 2 2

If this is the minimum uncertainty in linear momentum, then the minimum energy will be Emin

(∆px )2 = 2m

Therefore, Emin

1 (∆px )2 = = 2m 2m



α~ √ 2

2

α2 ~2 = 4m

Since, r α=

mω ~

1 =⇒ Emin = ~ω > 0 4 Thus, the zero-point energy of a simple harmonic oscillator is not zero.

Heisenberg Uncertainty principle for the simple harmonic oscillato Let us verify the Heisenberg Uncertainty principle for the ground state of a particle exhibiting simple harmonic motion. For this, we need to derive the expression for uncertainty in position in linear momentum of the particle. We already know from previous derivation, uncertainty in linear momentum of a particle exhibiting simple harmonic motion is ∆px = √1 α~ 2 Now, let us calculate the uncertainty in position i.e. p ∆x = hx2 i − hxi2 hx2 i =

Z+∞r −∞

α = 1/2 π =

2α π 1/2

α − 1 α 2 x2 2 e 2 x π 1/2

r

α − 1 α 2 x2 e 2 dx π 1/2

Z+∞ 2 2 x2 e−α x dx −∞ Z+∞

x2 e−α

2 2

x

dx

0

π 1/2 2α = 1/2 × Using standard value of the integral 4α3 π 1 = 2 2α

49

Now, hxi =

Z+∞r −∞

α = 1/2 π = 0,

α − 1 α 2 x2 e 2 x π 1/2

r

α − 1 α 2 x2 e 2 dx π 1/2

Z+∞ 2 2 xe−α x dx −∞

as the integrand is and odd function of x.

Therefore, uncertainty in position becomes r 1 1 √ ∆x = − 0 = 2α2 α 2 Finally, the product of ∆px and ∆x is 1 1 1 ∆x∆px = √ α~ × √ = ~ 2 α 2 2 Thus, Heisenberg uncertainty principle is valid for a particle exhibiting simple harmonic motion. Vibrational energy states are equi-spaced The gap between any two neighboring vibrational levels, say between v = v and v 0 = v + 1 is given by     1 1 0 ~ω − v + ~ω ∆Ev0 −v = v + 2 2     1 1 = v+1+ ~ω − v + ~ω 2 2 = ~ω Thus, the gap between any two neighboring vibrational states is independent of the vibrational quantum number of either states. It depends only upon the frequency of vibration. Thus, the vibrational energy levels of a simple harmonic oscillator are equi-spaced. Classical and quantum mechanical turning points: As discussed previously, during simple harmonic motion, the particle moves back and forth about the equilibrium position. At equilibrium position, the kinetic energy is maximum and it gradually decreases as the particle moves towards the extreme points. At the extreme points, kinetic energy becomes zero and the total energy becomes equal to the potential energy. These extreme points are called turning points (because from this point the particle return towards the equilibrium position). Thus, the classical turning point is characterized as a point, where total energy of the system is equal to its potential energy. 1 At classical turning point: E = kx2ctp 2r 2E xctp = ± k 50

Classically, the particle exhibiting simple harmonic motion should not go beyond this point. Classical turning point for an electron Assuming quantum mechanical ground state energy, calculate the xctp for a simple harmonic oscillator. Also assume that the angular q frequency of oscillation ω = stant k = 1 unit). r xctp =

2E0 = k

r

1 m

(this corresponds to a force con-

s ~ω = k

~ × k

r

k = m



~2 km

1/4

Putting m = 9.1 × 10−31 kg, h = 6.626 × 10−34 J.sec, k = 1 N.m−1 , we get xctp = ±3.3257 × 10−10 m. Classical turning point is not the quantum mechanical turning point Let us check the status of the particle beyond xctp , when we treat it quantum mechanically. To do so, we should check whether the particle exist or not beyond xctp . That means, we need to evaluate the wavefunction of our oscillator at some point beyond xctp say, at xctp + δx (where δx is a very very small but positive distance). Let us do this for the ground state wavefunction. r r α − 1 α2 (xctp +δx)2 α − 12 α2 (x2ctp +2xctp δx+δx2 ) 2 e e ψ0 (x = xctp + δx) = = π 1/2 π 1/2 r i α h − 1 α2 x2ctp −α2 xctp δx 2 ×e , = e π 1/2 (δx2 , being very small is neglected.) q q p mω 2E0 1 k Since, xctp = ± k , E0 = 2 ~ω, α = ~ and ω = m we can write

1 2 2 1 mω 2E0 1 mω ~ω 1 mω 2 1 α xctp = × × = × × = × = 2 2 ~ r k 2 ~ kr 2 k 2 mω 2E0 ~ω k mω mω 3/2 2 α xctp δx = ± × × δx = ± × × δx = ± √ δx = ± √ ~ k ~ k m k~ Thus, r ψ0 (x = xctp + δx) =

k α ± √m~ δx − 21 × e × e 6= 0 1/2 π

Thus, we see that beyond the classical turning points, the ground state wavefunction is not zero. In principle, one can verify it for the other states too. Therefore, the classical turning point is not the quantum mechanical turning point. The quantum mechanical oscillator goes beyond the classical turning point. This is also called the tunneling of the quantum mechanical oscillator into classically forbidden region.. Quantum mechanical turning point: In principle, there is always a finite probability of finding the particle anywhere between 51

x = +∞ and x = −∞. So, there is no as such quantum mechanical turning point of the simple harmonic oscillator. This is also obvious from the expression of wavefunction in ground state. Since, the wavefunction consist of an exponentially decaying function, it vanishes only at x = ±∞. This is another weird example of a quantum mechanical system. However, for practical purpose we can define a quantum mechanical turning point by arbitrarily setting a very small threshold value for the probability density (say, Pth ). Thus, we can assume that if the probability of finding the particle is less than Pth then the particle does not go beyond the corresponding position (xqtp ). Value of xqtp in ground state: Let us consider a particle exhibiting a simple harmonic motion be in its ground state. The quantum mechanical turning point (xqtp ) corresponding to the threshold probability density of Pth can be obtained from |ψ0 (xqtp )|2 = Pth 2 r α − 1 α2 x2qtp e 2 = Pth π 1/2 α −α2 x2qtp e = Pth π 1/2 Thus xqtp

1 =± α

s



ln

α √

 (83)

Pth π

Example: If an electron q exhibits simple harmonic motion with 1 angular frequency ω = m (this corresponds to a force constant k = 1 unit) then find xqtp corresponding to the threshold probability density Pth = 10−6 . Compare this with corresponding xctp . s r r  m 1/4 mω m 1 α= = × = 2 ~ ~ m ~ Thus, 

xqtp

2

~ =± m

1/4 s

m1/4 √ ln Pth ~π 



Using m = 9.1 × 10−31 kg, h = 6.626 × 10−34 J.sec, we get, xqtp = ±1.9694 × 10−9 m. The corresponding value of xctp is ±3.3257 × 10−10 m. Thus xqtp > xctp . That means, quantum mechanical oscillator goes beyond the classical turning point. Probability of finding the electron beyond the classical limit, when it is in ground state. For this, we first need to find a relation between xctp and α for the ground state.

52

Relationship between xctp and α: for the ground state s r r r  2 1/4 2E0 ~ω ~ k ~ = = × = xctp = k k k m mk s r r  1/4 mω m k mk = × = α= ~ ~ m ~2 1 Thus, xctp = , α =⇒ αxctp = 1 Now, the probability of finding the ground state electron within the classical limit i.e. within x = +xctp and x = −xctp is given by +x Z ctp

|ψ0 (x)|2 dx =

−xctp

+x Z ctpr

α − 1 α 2 x2 e 2 π 1/2

2

α dx = 1/2 π

−xctp

+x Z ctp

e−α

2 2

x

dx

−xctp

This is a standard definite integral given by +x Z ctp

e−α

−xctp

2 2

x

  (αxctp )3 (αxctp )5 (αxctp )7 2 αxctp − + − + ... dx = α 3 10 42

Using the relation αxctp = 1, we get +x Z ctp −xctp

  1 1 1 − + . . . = 0.8427 |ψ0 (x)|2 dx = 1/2 1 − + 3 10 42 π 2

Finally, the probability of finding the ground state electron outside the classical limit will be 1 − 0.8427 = 0.1573. Thus, in ground state, there is ≈ 16% chance that the electron will be found in the classically forbidden region. Probability of finding the electron beyond the classical limit, when it is in first excited state. As before, we first need to find a relation between xctp and α for the first excited state. Relationship between xctp and α: for the first excited state s r r r  2 1/4 9~ 2E1 3~ω 3~ k xctp = = = × = k k k m mk s r r 1/4  mω m k mk α= = × = ~ ~ m ~2 √ 3 Thus, xctp = , √α =⇒ αxctp = 3

53

Now, the probability of finding the first excited state electron within the classical limit i.e. within x = +xctp and x = −xctp is given by +x Z ctp −xctp

|ψ1 (x)|2 dx =

+x Z ctp

r

2α3 − 1 α2 x2 xe 2 π 1/2

!2

2α dx = 1/2 π

+x Z ctp

α2 x2 e−α

−xctp

−xctp

After evaluating the integral and using the relation αxctp = the first excited state), we get

√

2 2

x

dx

3 (for

+x Z ctp −xctp

|ψ1 (x)|2 dx = 0.8886

Therefore, the probability of finding the particle within the classical region in first excited state is 0.8886. Therefore, the probability of finding the electron outside the classical region in first excited state will be 1 − 0.8886 = 0.1114. Thus, in first excited state, there is more than 11% chance that the electron will be found in the classically forbidden region. Note: The probability of finding the particle in classical region in first excited state (0.8886) is more than that in ground state (0.8427) When does a quantum mechanical simple harmonic oscillator behave as a classical simple harmonic oscillator? As we have seen above, the probability of finding a simple harmonic oscillator in the classical region increases on going from the ground state to the first excited state. This probability will keep on increasing as one goes to the higher and higher excited states. That means, as one goes to higher excited state, the simple harmonic oscillator tends to avoid the classically forbidden region and tends to stay within the classical region. In other words, the higher the vibration state of the oscillator the more classically it behaves. From this, one can easily conclude that for a very large value of v (vibrational quantum number) the quantum mechanical simple harmonic oscillator will behave as a classical simple harmonic oscillator. Why there is a finite probability of finding the oscillator in classically forbidden region? You can get a hint of this weird nature from particle in a 1D box model. We know that in particle in a 1D box, the potential at the two walls and outside of the 1D box is infinite. However, no such condition is imposed on a simple harmonic oscillator. Because of the infinite potential at the two walls and outside of the box, it is not possible to solve the corresponding Schr¨odinger equation outside the box and hence there is no wavefunction of the particle outside the box.   ~2 d2 + Vout ψout (x) = Eout ψout (x) − 2m dx2   ~2 d2 − + ∞ ψout (x) = Eout ψout (x), as Vout = ∞ 2m dx2 54

That means, the particle does not exist outside the box. But in case of simple harmonic oscillator, since no such condition is there, the particle tunnels into the classically forbidden region. If you consider a finite potential at the two walls and outside the 1D box then also you can see the tunneling of the particle through the walls of the box. Tunneling is a very important quantum mechanical phenomena. Many chemical reactions are now know which occurs through quantum mechanical tunneling effect.

55

Hydrogen atom So far we have studied the basic principles and concepts in quantum mechanics and applied those to a very simple model (particle in a box) and a little bit complex model (simple harmonic oscillator). Now, let us apply our knowledge of quantum mechanics for studying a real but relatively simpler system – a H-atom in absence of any externally applied field. Description of the problem we are going to solve: A H-atom consists of a positively charged nucleus of mass M and a negatively charged electron of mass m. The electron is in a state of continuous motion around the nucleus. In absence of any externally applied field, the only force the electron experiences is the electrostatic force of attraction (i.e. Coulombic force)5 due to the nucleus. e2 , =− 4π0 r2

Fne where

q r = (xe − xn )2 + (ye − yn )2 + (ze − zn )2

(84)

(85)

is the distance between the electron and the nucleus and 0 is the permittivity of vacuum. (xe , ye , ze ) and (xn , yn , zn ) are the Cartesian coordinates of the electron and nucleus respectively. 𝑥" , 𝑦" , 𝑧" 𝑀 𝑟 𝑥& , 𝑦& , 𝑧& 𝑚 Electron

Nucleus

Figure 8: Pictorial representation of H-atom Our target is to study the H-atom. That means to calculate the energies and wavefunctions of different quantum mechanical states of H-atom. In order to do so, we will proceed as before – (A) At first, we will find out the expression of Hamiltonian for the system, (B) then we’ll set up the corresponding Schr¨odinger equation, and (C) finally solve the Schr¨odinger equation. (A) Hamiltonian for H-atom: We know, the Hamiltonian consists of kinetic energy and potential energy operators. ˆ = Tˆ + Vˆ H (86) There are few things to consider here. (i) Kinetic energy operator (Tˆ) corresponds to the kinetic energy of each 5

The Coulombic force between two charges Q1 and Q2 separated by a distance r is given by FQ1 Q2 =

Q1 Q2 4π0 r2

56

particle present in the system. Since, H-atom contains one electron and one nucleus, the kinetic energy operator will be the sum of kinetic energy operator for the electron (Tˆe ) and for the nucleus (Tˆn ). Thus, Tˆ = Tˆe + Tˆn

(87)

(ii) Since, a H-atom is a three dimensional system, the kinetic energy operator of electron and nucleus will be  2  2 2 2 ~ ∂ ∂ ∂ ~2 2 ˆ Te = − + + =− ∇ 2m ∂x2e ∂ye2 ∂ze2 2me e  2  (88) 2 2 2 2 ~ ∂ ∂ ∂ ~ 2 =− ∇ , Tˆn = − + + 2M ∂x2n ∂yn2 ∂zn2 2mn n where xe , ye , ze represent the Cartesian coordinates of the electron and xn , yn , zn are those for the nucleus. m and M are the masses of the electron and the nucleus respectively. The symbol ∇ is called   ‘Nabla’ and ∂ ∂ ∂ is the short form of writing the ‘Del operator’ i.e. ∇ = ∂x + ∂y + ∂z .   ∂ ∂ ∂ 2 Similarly, ∇ = ∂x2 + ∂y2 + ∂z 2 (iii) The potential energy of the system can be obtained from Eq. ??. We know, Fne = − dVdrne . This gives, e2 ˆ V = Vne = − 4π0 r

(89)

(iv) Thus, the Hamiltonian of H-atom becomes
~2 e2 2 2 ˆ H=− ∇ e + ∇n − 2me 4π0 r

(90)

(v) It is much easier to deal with one-body problem rather than a twobody problem. Any two-body problem can be simplified to a one-body problem by introducing the concept of reduced mass (µ) and center of mass. H-atom system can also be simplified in this way. The reduced mass of H-atom is given by µ=

mM . m+M

By using this, the Hamiltonian (Eq. ??) of H-atom becomes   2 ∂ ∂ ∂ e2 ~ ˆ + + − , H=− 2µ ∂x2 ∂y 2 ∂z 2 4π0 r

(91)

where (x, y, z) are given by x = |xe − xn | y = |ye − yn | z = |ze − zn | The kinetic energy part in H-atom Hamiltonian (Eq. ??) represents the relative motion of electron and nucleus together. odinger equation for the H-atom in Cartesian coordinates: (B) Schr¨ 57

If ψn (x, y, z) and En be the wavefunction and energy of nth state in H-atom then Schr¨odinger equation can be written as # " 2 2 e ~ p ψn (x, y, z) = En ψn (x, y, z), − ∇2 − 2µ 4π0 x2 + y 2 + z 2  2  2 2 ∂ ∂ ∂ + + where ∇2 = ∂x2 ∂y 2 ∂z 2 (92) Converting Schr¨ odinger equation in Cartesian coordinates coordinates to Spherical polar coordinates:6 Why we need this conversion? The Schr¨odinger equation (Eq. ??) for H-atom contains three variables (x, y, z) and hence is a multivariate equation. The first step of solving this equation involves the application of separation of variable technique. That means at first we need to break the Eq. ?? into equations containing only one variable. Without doing this, it is not possible to solve this equation. But,  the Schr¨odinger equation containspthe potential  2since e energy term 4π , which contains the term r given by x2 + y 2 + z 2 , 0r it is not possible to separate the three variables in this equation. One way to separate the variables in Eq. ?? is the conversion of the whole equation in Spherical polar coordinates (r, θ, φ). Since, in spherical polar coordinates the potential energy term is a single variable term, the problem associated with it is automatically solved. That’s why we first need to convert the Schr¨odinger equation (Eq. ??) from Cartesian coordinates to Spherical polar coordinates. Spherical polar Coordinates and its relation with Cartesian coordinates: The Spherical polar coordinates of a point represents the location of a point on the surface of a sphere (hence the name “spherical”). The spherical polar coordinates are given by r, θ and φ, where r is the distance of the point from the center of the sphere (the origin) that means r is the radius of the sphere, θ is the angle made by the vector ~r with the z-axis and φ is the angle with x-axis made by the projection of r on xy− plane. The relationships between Cartesian coordinates (x, y, z) and Spherical polar coordinates (r, θ, φ) are given by x = rsinθcosφ y = rsinθsinφ z = rcosθ

(93)

Conversion of Schr¨ odinger equation (Eq. ??) to spherical polar coordinates: 6

Sometimes, people use the name polar Coordinates in place of Spherical polar coordinates. Actually, the term polar coordinates is used for 2D systems only and hence it has only two variables (r, θ). Spherical polar coordinates is applicable to 3D systems. Since, in our case, we are dealing with a 3D system we should use the term Spherical polar coordinates.

58

z !, #, $

Y Figure 9: Relation between Spherical polar and Cartesian coordinates The Schr¨odinger equation for H-atom is given by Eq. ??. It is obvious that we just need to convert the ∇2 into Spherical polar coordinates because the potential energy part is already expressed in terms of r. Now, for any function f , we can write ∂ 2f ∂ 2f ∂ 2f ∇f= + + ∂x2 ∂y 2 ∂z 2 2

Furthermore, ∂f ∂r ∂f ∂θ ∂f ∂φ ∂f = + + ∂x ∂r ∂x ∂θ ∂x ∂φ ∂x ∂f ∂f ∂r ∂f ∂θ ∂f ∂φ = + + ∂y ∂r ∂y ∂θ ∂y ∂φ ∂y ∂f ∂r ∂f ∂θ ∂f ∂φ ∂f = + + ∂z ∂r ∂z ∂θ ∂z ∂φ ∂z From Eq. ??, we can write p y z r = x2 + y 2 + z 2 , cosθ = p and, tanφ = . x x2 + y 2 + z 2

(94)

(95)

Derivation of x-component of ∇2 f : From Eq. ??, differentiating r, θ and φ w.r.t. x we get ∂r ∂θ 1 ∂φ sinφ = sinθcosφ; = cosθcosφ; =− ∂x ∂x r ∂x rsinθ Putting Eq. ?? into x-component of Eq. ??, ∂f ∂f 1 ∂f sinφ ∂f = sinθcosφ + cosθcosφ − ∂x ∂r r ∂θ rsinθ ∂φ

(96)

(97)

Differentiating Eq. ?? w.r.t. x ∂ 2f ∂ 2 f ∂r ∂f ∂φ ∂f ∂θ = sinθcosφ + sinθ (−sinφ) + cosφ cosθ ∂x2 ∂r2 ∂x ∂r ∂x ∂r ∂x 2 ∂ f ∂θ 1 ∂f ∂φ 1 ∂f ∂θ 1 + cosθcosφ 2 + cosθ (−sinφ) + cosφ (−sinθ) r ∂θ ∂x  r ∂θ ∂x r ∂θ ∂x 2 ∂f 1 ∂r sinφ ∂ f ∂φ 1 ∂f ∂φ + cosθcosφ − 2 − − cosφ ∂θ r ∂x rsinθ ∂φ2 ∂x rsinθ ∂φ ∂x     sinφ ∂f cosθ ∂θ sinφ ∂f 1 ∂r − − 2 − − 2 r ∂φ r ∂x sin θ ∂x sinθ ∂φ (98) 59

Again, using the expressions from Eq. ??, we get 2 sin2 φ ∂f cos2 θcos2 φ ∂f cos2 θcos2 φ ∂ 2 f ∂ 2f 2 2 ∂ f = sin θcos φ 2 + + + ∂x2 ∂r r ∂r r ∂r r2 ∂θ2 cosθsin2 φ ∂f sinθcosθcos2 φ ∂f sinθcosθcos2 φ ∂f + 2 − − r sinθ ∂θ r2 ∂θ r2 ∂θ 2 2 2 sin φ ∂ f sinφcosφ ∂f cos θsinφcosφ ∂f sinφcosφ ∂f + 2 2 + 2 2 + + ∂φ r2 ∂φ r sin θ ∂φ2 r sin θ ∂φ r2 sin2 θ (99)

Derivation of y-component of ∇2 f : From Eq. ??, differentiating r, θ and φ w.r.t. y we get ∂θ 1 ∂φ cosφ ∂r = sinθsinφ; = cosθsinφ; = ∂y ∂y r ∂y rsinθ

(100)

Putting Eq. ?? into y-component of Eq. ??, ∂f 1 ∂f cosφ ∂f ∂f = sinθsinφ + cosθsinφ + ∂y ∂r r ∂θ rsinθ ∂φ

(101)

Differentiating Eq. ?? w.r.t. y ∂ 2 f ∂r ∂f ∂φ ∂f ∂θ ∂ 2f = sinθsinφ 2 + sinθ (cosφ) + sinφ cosθ 2 ∂y ∂r ∂y ∂r ∂y ∂r ∂y 1 ∂ 2 f ∂θ 1 ∂f ∂φ 1 ∂f ∂θ + cosθsinφ 2 + cosθ (cosφ) + sinφ (−sinθ) r ∂θ ∂y r ∂θ ∂y r ∂θ ∂y   2 ∂f 1 ∂r cosφ ∂ f ∂φ 1 ∂f ∂φ + cosθsinφ − 2 + + (−sinφ) ∂θ r ∂y rsinθ ∂φ2 ∂y rsinθ ∂φ ∂y     cosφ ∂f cosθ ∂θ cosφ ∂f 1 ∂r − 2 + − 2 + r ∂φ sinθ ∂φ r ∂y sin θ ∂y (102) Again, using the expressions from Eq. ??, we get 2 ∂ 2f cos2 φ ∂f cos2 θsin2 φ ∂f 2 2 ∂ f = sin θsin φ 2 + + ∂y 2 ∂r r ∂r r ∂r cos2 θsin2 φ ∂ 2 f cosθcos2 φ ∂f sinθcosθsin2 φ ∂f + + − r2 ∂θ2 r2 sinθ ∂θ r2 ∂θ 2 2 2 cos φ ∂ f sinφcosφ ∂f sinθcosθsin φ ∂f − + − 2 2 r2 ∂θ r2 sin2 θ ∂φ2 r sin θ ∂φ 2 cos θsinφcosφ ∂f sinφcosφ ∂f − − ∂φ r2 ∂φ r2 sin2 θ

(103)

2 ∂ 2f cos2 φ ∂f cos2 θsin2 φ ∂f cos2 θsin2 φ ∂ 2 f 2 2 ∂ f = sin θsin φ 2 + + + ∂y 2 ∂r r ∂r r ∂r r2 ∂θ2 cosθcos2 φ ∂f 2sinθcosθsin2 φ ∂f cos2 φ ∂ 2 f sinφcosφ ∂f + − + − 2 2 r2 sinθ ∂θ r2 ∂θ r2 sin2 θ ∂φ2 r sin θ ∂φ 2 cos θsinφcosφ ∂f sinφcosφ ∂f − − ∂φ r2 ∂φ r2 sin2 θ (104)

60

Derivation of z-component of ∇2 f : From Eq. ??, differentiating r, θ and φ w.r.t. z we get ∂θ sinθ ∂φ ∂r = cosθ; =− ; =0 ∂z ∂z r ∂z Putting Eq. ?? into z-component of Eq. ??, ∂f ∂f sinθ ∂f = cosθ − ∂z ∂r r ∂θ Differentiating Eq. ?? w.r.t. z

(105)

(106)

∂ 2f ∂ 2 f ∂r ∂f ∂θ sinθ ∂ 2 f ∂θ 1 ∂f ∂θ = cosθ − sinθ − − cosθ ∂z 2 ∂r2∂z ∂r ∂z r ∂θ2 ∂z r ∂θ ∂z ∂f 1 ∂r − sinθ − 2 ∂θ r ∂z

(107)

Again, using the expressions from Eq. ??, we get 2 sin2 θ ∂f sin2 θ ∂ 2 f sinθcosθ ∂f sinθcosθ ∂f ∂ 2f 2 ∂ f = cos θ 2 + + 2 + + ∂z 2 ∂r r ∂r r ∂θ2 r2 ∂θ r2 ∂θ (108) 2 ∂ 2f sin2 θ ∂f sin2 θ ∂ 2 f 2sinθcosθ ∂f 2 ∂ f = cos θ 2 + + 2 + ∂z 2 ∂r r ∂r r ∂θ2 r2 ∂θ

(109)

Final expression of ∇2 in Spherical polar coordinates: Adding Eqs. ??, ?? and ??, we get ∂ 2 f 2 ∂f 1 ∂ 2f cosθ ∂f 1 ∂ 2f ∇f= 2 + + + + ∂r r ∂r r2 ∂θ2 r2 sinθ ∂θ r2 sin2 θ ∂φ2 2

(110)

which can be written in the final form as 1 ∂ ∇2 = 2 r ∂r



∂ r2 ∂r



1 ∂ + 2 r sinθ ∂θ



∂ sinθ ∂θ

 +

1 ∂2 r2 sin2 θ ∂φ2

(111)

Hamiltonian and Schr¨ odinger equation for H-atom, in Spherical polar coordinates: From Eqs. ?? and ??, we can write down the Hamiltonian for H-atom in Spherical polar coordinates.  2       2 2 ~ 1 ∂ ∂ 1 ∂ ∂ 1 ∂ e 2 ˆ = − H r + 2 sinθ + 2 2 − 2µ r2 ∂r ∂r r sinθ ∂θ ∂θ 4π0 r r sin θ ∂φ2 and ˆ n (r, θ, φ) = En ψn (r, θ, φ) Hψ (112)

61

Note that the wavefunction is now written with variables (r, θ, φ). Separation of variable technique: As mentioned before, the main reason of converting the Schr¨odinger equation for H-atom in Spherical polar coordinates is to make it suitable for separating the variables in it, so that we can solve it. An equation containing only one variable is easy to solve than an equation containing many variables. Separation of variables is a technique by which one can break a multivariate equation into simple equations containing only one variable. Eq. ?? has three variables (r, θ, φ) so, our next step is to separate these three variables and obtain three different equations in r, θ and φ separately. Before we proceed, let us have a brief description of steps involved in separation of variable technique. Steps involved in separation of variable technique: The general steps involved in separation of variable technique are – (i) Write down the solution of the equation as a product of functions of only one variable. A more in-depth discussion on it is given later on. In particular, how the emergence of quantum number is related to it. Anyway, in this step, The wavefunction is written as a product of three functions each depending on one and only one coordinate i.e. ψ(r, θ, φ) = R(r) × Θ(θ) × Φ(φ). (ii) Rearrange the main equation in such a way that the left and right hand sides of the equation depends only one one variable. In Cartesian coordinates, the Eq. ?? cannot be written in this way because of the potential energy term that contains √ 2 1 2 2 . However, x +y +z

we can do this with Eq. ??. (iii) If Step (ii) is finished successfully, one can write each side of the equation as a constant. This is because each side depends only on one variable and thus two different quantities depending on two different things will be equal if and only if both equals to the same constant. Applying separation of variable technique on Eq. ?? Since there are three variables in Eq. ??, we will separate the variables in two steps. (A) At first we will separate the radial part (R(r), which depends only on r) and the angular part (Y (θ, φ), which depends one θ and φ). (B) Then we will separate the θ and the φ parts from Y (θ, φ). (A) Separation of radial (r) and angular (θ, φ) parts of Eq. ?? We will follow the three steps of separation of variable technique (as described above). But before that let us first rewrite the Eq. ?? in a more convenient way. Thus, 62

     2  2 ∂ψ 1 ∂ ∂ψ 1 2µ ∂ ψ e n n n r2 + 2 sinθ + 2 2 + + En ψn = 0 ∂r r sinθ ∂θ ∂θ r sin θ ∂φ2 ~2 4π0 r (113) Note, in the the above equation, for clarity we have omitted the variables in ψ. 1 ∂ r2 ∂r

Step (i): Write down the solution of the equation as a product of functions of only one variable. We can write the wavefunction ψ(r, θ, φ) as a product of two functions – one depending only on r (R(r)) and the other (Y (θ, φ)) depending one θ and φ. ψ(r, θ, φ) = R(r) × Y (θ, φ) or, ψ = R × Y

(114)

Step (ii): Write down the Eq ?? in such a way that the left and right hand sides of the equation depends only one one variable.

Putting Eq. ?? in Eq. ??.      2  1 ∂ ∂RY 1 ∂ 2 RY 2µ e 1 ∂ 2 ∂RY r + 2 sinθ + 2 2 + 2 + E RY r2 ∂r ∂r r sinθ ∂θ ∂θ ~ 4π0 r r sin θ ∂φ2 Since R depends only on r and Y depends only on (θ, φ), we can write      2  2 Y ∂ ∂R R ∂ ∂Y R ∂ Y 2µ e r2 + 2 sinθ + 2 2 + + E RY = 0 2 r ∂r ∂r r sinθ ∂θ ∂θ r sin θ ∂φ2 ~2 4π0 r 2

r Multiplying both sides by RY , we get       2 1 ∂ 1 ∂ ∂Y 1 ∂ 2 Y 2µr2 e 2 ∂R r + sinθ + + 2 +E =0 R ∂r ∂r Y sinθ ∂θ ∂θ ~ 4π0 r Y sin2 θ ∂φ2

Writing all the r-dependent parts on left hand side and Y -dependent parts on the right hand side.    2    2 1 ∂ ∂R 2µr e 1 ∂ ∂Y 1 ∂ 2Y r2 + 2 sinθ − +E =− R ∂r ∂r ~ 4π0 r Y sinθ ∂θ ∂θ Y sin2 θ ∂φ2 (115) Step (iii): Write each side of the Eq. ?? as a constant. Since left hand side of this equation depends only on r and the right hand side depends only on angles (θ, φ), each should be equal to a constant (say, A). Thus, the radial part of Schr¨odinger equation for H-atom is    2  1 d 2µr2 e 2 dR r + 2 +E =A R dr dr ~ 4π0 r or, d dr



dR r2 dr



2µr2 + 2 ~



 e2 + E R − AR = 0. 4π0 r

63

(116)

Note the change of partial derivative to full derivative in Eq. ??. This is because there is only one variable in this equation. (B) Separating the θ and the φ parts: The angular part of Schr¨odinger equation for H-atom is   1 ∂ ∂Y 1 ∂ 2Y − sinθ − =A Y sinθ ∂θ ∂θ Y sin2 θ ∂φ2 or, 1 ∂ sinθ ∂θ

  ∂Y 1 ∂ 2Y sinθ + + AY = 0 ∂θ sin2 θ ∂φ2

(117)

Note: Here we still use the partial derivatives, because it contains two variables θ and φ. Separating the polar (θ) and azimuthal (φ) parts of Eq. ?? For separating the polar and azimuthal parts of Eq. ??, we will proceed in the same way, as before. Step (i): Write down the solution of the Eq. ?? as a product of functions of only one variable. The solution of Eq. ?? can be written as Y (θ, φ) = Θ(θ) × Φ(φ) or, Y = ΘΦ

(118)

Step (ii): Write down the Eq. ?? in such a way that the left and right hand sides of the equation depends only one one variable. Putting Eq. ?? in Eq. ??,   1 ∂ ∂ΘΦ 1 ∂ 2 ΘΦ sinθ + + AΘΦ = 0 sinθ ∂θ ∂θ sin2 θ ∂φ2 Since Θ depends only on θ and Φ depends only on φ, we can write   Φ ∂ ∂Θ Θ ∂ 2Φ + AΘΦ = 0 sinθ + sinθ ∂θ ∂θ sin2 θ ∂φ2 2

θ Multiplying both sides by sin ΘΦ ,   sinθ ∂ ∂Θ 1 ∂ 2Φ sinθ + + Asin2 θ = 0 2 Θ ∂θ ∂θ Φ ∂φ   ∂Θ sinθ ∂ 1 ∂ 2Φ 2 sinθ + Asin θ = − Θ ∂θ ∂θ Φ ∂φ2

Step (iii): Write each side of the Eq. ?? as a constant. 64

(119)

As before, since left hand side of Eq. ?? depends only on θ and right hand side depends only on φ, we can write each of them equal to a constant (say, B). Thus, the polar part of Schr¨odinger equation is   dΘ sinθ d sinθ + Asin2 θ = B Θ dθ dθ or  
d dΘ sinθ sinθ + Asin2 θ − B Θ = 0 (120) dθ dθ and the azimuthal part of Schr¨odinger equation is 1 d2 Φ =B − Φ dφ2 or, d2 Φ + BΦ = 0 dφ2

(121)

Solving the azimuthal equation (Eq. ??) of H-atom The azimuthal equation (Eq. ??) is a standard second order differential equation. It has a general solution of Φ = ceimφ ,

(122)

where m = B 1/2 . φ is called the azimuthal angle and its values can vary from 0 to 2π. It is obvious from Fig. ?? that after tracing an angle of 2π we reach at the same point. Thus, the value of Φ should also be the same at φ = 0 and φ = 2π. Therefore, Φ(φ = 0) = Φ(φ = 2π) c = ce2imπ e2imπ = 1 cos2mπ + isin2mπ = 1,

Using, eix = cosx + isinx

Since the right hand side is a real number and left hand side is a complex quantity, the equality will be satisfied if and only if the imaginary part equals zero i.e. sin2mπ = 0 This is possible if and only if m be a positive or negative integer including zero. Thus, m = 0, ±1, ±2, ±3, . . . m is called the magnetic quantum number. Determining the normalization constant c in Eq. ??:

65

c in the azimuthal wavefunction (Eq. ??) can be determined by applying the condition of normalization. Thus, Z2π

Φ∗ Φdφ = 1

0

Z2π

c2 e−2imφ e2imφ dφ = 1

0

c2

Z2π dφ = 1 0

2

c × 2π = 1

1 c= √ 2π

Thus, the azimuthal wavefunction becomes 1 Φm = √ eimφ 2π

(123)

The subscript m is added to represent the dependence of Φ on m. Solution of the θ-part or polar equation i.e. Eq. ?? of H-atom The general solution of Eq. ?? is the associated Legendre’s polynomial (Plm (cosθ)), given by For positive m: " Plm>0 (cosθ) = sinm θ c0

∞ X c2k k=0

c0

cos2k θ + c1

∞ X c2k+1 k=1

c1

# cos2k+1 θ , (124)

where ck+2 =

(k + m)(k + m + 1) − A ck , and A = l(l + 1) (k + 1)(k + 2))

For negative m: Plm<0 (cosθ) = (−1)|m|

(l − |m|)! m>0 P (cosθ) (l + |m|)! l

(125)

Thus, the normalized solution of Eq. ?? is Θl,m = Nl,m Plm (cosθ),

Nl,m = Normalization constant (126)

The subscripts l and m are added to show the dependence of Θ on these two quantities. l is an integer and is called azimuthal quantum number. The value of l controls the value of magnetic quantum number, m. For each value of l there are 2l + 1 allowed values of m given by −l, −(l − 1), . . . , 0, . . . , (l − 1), l. The restrictions on the values of l and m is for ensuring the convergence of the series in Eq. ??. The values of c0 and c1 are also chosen for each set of (l, m) in such a way that the 66

convergence of the series is ensured. The expression of Plm (cosθ) for some selected sets of (l, m) are – l ↓, m → 0 1 2

-2 2 1 8 sin θ

-1 1 2 sinθ 1 2 sinθcosθ

1 2

0 1 2 1 cosθ −sinθ
2 3cos θ − 1 −3sinθcosθ 3sin2 θ

Solution of the radial equation i.e. Eq. ?? of H-atom The general solution of radial equation (Eq. ??) of H-atom is given by  Rn,l (r) = where, ck+1

2r na0

l X ∞

 ck

k=0

2r na0

k

k+l+1−n = ck (k + 1)(k + 2l + 2)

r

e− na0 (127)

In Eq. ??

µαc e2 a0 = and α = (128) ~ 4π0 ~c µ = reduced mass, α is called fine structure constant, c = speed of light h is called reduced Planck’s constant, e = charge of in vacuum and ~ = 2π an electron and 0 = permittivity of vacuum. All these quantities are in SI unit and are constants. Therefore, a0 is also a constant. It is called Bohr’s radius and it’s value is 0.529177˚ A. n in Eq. ?? is a positive integer (6= 0) and is called Principal quantum number. The value of n controls the value of azimuthal quantum number l. For each n, l has values from 0 to (n − 1). This restriction on the values of l is imposed for ensuring convergence of the series in Eq. ?? Energy expressions The solution of radial part also gives the expression for energy (En )7 . µe4 En = − 2 2 2 80 n h

(129)

Why there are quantum number? The Step (i) of separation of variable technique involves the writing of the wavefunction as a product of three functions, each depending on one variable only. As such it is not a simple task. We cannot always write any multivariate function as a product of one-variable functions. However, we know that a function can be written as a series expansion. This property can be used to express a multivariate function as a product of single-variable functions. Suppose, there be a function f (r, θ) = sin(rθ). 7

ˆ n = En ψn (See, Eq. ??) This is the energy appeared in Schr¨ odinger equation: Hψ

67

How can you write this function as f (r, θ) = g(r)h(θ)? It’s a difficult task. However, if you write, sin(rθ) as a series expansion, then it becomes simple. Thus, ∞

X X (rθ)2k+1 (rθ)3 (rθ)5 (−1)k gk (r)hk (θ) + −··· = = sin(rθ) = rθ − 3! 5! (2k + 1)! k=0 k (130) Thus, we see that although it may not be possible to write a multivariate function as a product of single variable functions yet it is possible to write the same as a sum of products of single-variable functions. Note that the example given above is a simple case because we knew the function. In case of Schr¨odinger equation, we do not know the form of the wavefunction (ψ(r, θ, φ)), but still we expressed it as a product of R(r), Θ(θ) and Φ(φ). That’s why, later on when one solves the individual equations (Eqs. ??, ?? and ??), one gets the solutions as series expansions. Since these series represent the part of wavefunctions i.e. a real system, it must converge to satisfy the real situations and boundary conditions. This requirement of convergence of the series puts restrictions on different quantities like A and B in Eqs. ?? and ??. These restrictions emerges in the form of quantum numbers. There is one equation corresponding to each variable in the wavefunction and each equation gives rise to one quantum number. That’s why there are three quantum numbers. But we know four quantum numbers! Yes, that’s true. There is the fourth quantum number – spin quantum number. It emerges when one considers the relativistic effect. Since the electrons in an atom moves with a speed comparable to the speed of light one should always consider relativistic effect in the description of electronic structure. Relativistic effect is out of the scope of this book. Summary important points to remember: • Three quantum numbers are appeared from the solution of Schr¨odinger equation for H-atom. • Principal quantum number (n) is obtained from the solution of radial (r) part. It is always a positive integer (6= 0). Thus, n = 1, 2, 3, . . . • Azimuthal quantum number (l) is obtained from the solution of polar (θ) part. Its value is controlled by the value of n. For each n, l = 0, 1, 2, . . . , (n − 1). • Magnetic quantum number (m) is obtained from the solution of azimuthal (φ) part. Its value is controlled by the value of l. For each l, m = −l, −(l − 1), . . . , 0, . . . , (l − 1), l. • The conditions on the values of n, l and m comes directly from the solution of the equations. For example, – n should always be a positive integer. A negative integer value of n is not allowed as this will make the radial wavefunction Rn,l (Eq. ??) exponentially increasing and thus at r = ∞ the wavefunction will become ∞, which is not allowed. Similarly, n = 0 will also make Rn,l = ∞ for any value of r. 68

• • •

•

– The value of l is controlled by the value of n in such a way that the recursion series in Eq. ?? is converged. – The value of m is controlled by the value of l for the same reason i.e. to allow the recursion series in Eq. ?? to converge. The allowed values of m as positive or negative integers including zero is already discussed while solving the azimuthal equation. Thus, each quantum mechanical state is characterized by these three quantum numbers (nlm). There is a fourth quantum number called spin quantum number, which arises due to relativistic effect and it takes the values of ± 12 for each value of m. The first state or the ground state of H-atom is characterized by n = 1, l = 0, m = 0 or simply (100). Since, n = 1 allows only one value of l i.e. l = 0, which in turn allows only one value of m i.e. m = 0, the ground state has a degeneracy level of 1. That means, the ground state is non-degenerate. For the first excited state, n = 2, which allows two values of l i.e. l = 0, 1. Now, l = 0 allows only one value of m i.e. m = 0 and l = 1 allows three values of m i.e. m = −1, 0, +1. Thus, the first excited state is four fold degenerate, as shown below n 1 2 2 2 2

l m n, l, m 0 0 1, 0, 0 0 0 2, 0, 0 1 -1 2, 1, −1 1 0 2, 1, 0 1 1 2, 1, 1

• Thus, degeneracy of ground state = 1, degeneracy of first excited state = 4. In general, any state designated by the Principal quantum number n has a degenearcy of = n2 . • The dependence of energy (Eq. ??) on n2 also reflects the level of degeneracy of the state. • Since energy (E) appeared only in the radial part of Schr¨odinger equation, it depends only on the principal quantum number. However, inherently it takes into account the number of values of l and m also. • Since the energy depends only on n, the states are primarily designated by the value of n. Thus, n = 1 is called K-shell (which is the ground state), n = 2 is L-shell, n = 3 is M-shell and so on. • Since each value of n corresponds to a range of l values, each value of l is called a sub-shell. Thus l = 0 is called s-subshell, l = 1 is p-subshell, l = 2 is d-subshell, l = 3 is f-subshell and so on. • Furthermore, since each value of l corresponds to (2l + 1) number of m values, each subshell is further divided into (2l + 1) number of orbitals. Thus, each s-subshell contains 1 orbital, each p-subshell contains 3 orbitals, each d-subshell contains 5 orbitals, each f -subshell contains 7 orbitals and so on. • The orbitals within a subshell are further named according to their m values and/or structural features. For example, the three porbitals can be named as p+1 , p0 and p−1 . Later on, we’ll see that 69

the p-orbital are actually named as px , py and pz . Similarly, the five d-orbitals are named as dxy , dxxz , dyz , dx2 −y2 and dz 2 . • Since there is only one orbital in a s-shell, it is not given separate name. Thus, the names s-subshell and s-orbital are synonymous. Origin of orbital’s name The names s, p, d, f, g, . . . are derived from spectroscopic terms. s stands for sharp, p for principal, d for diffuse and f for fundamental. These names (sharp, principal, diffuse, fundamental ) were originated long before quantum mechanics was developed. When the electronic structure was not known to the researchers they classified the spectral lines according to their pattern in the spectra. Later on, when the idea of quantum numbers become clear the name of the orbitals were given on the basis of the corresponding pattern of the spectral lines associated with those orbitals. The expression of radial (Rnl ) and angular (Y (θ, φ) = Θl,m × Φm ) parts and total wavefunctions (ψnlm ) for the first two states in H-atom Note that the radial and angular parts and hence the total wavefunctions given in Table ?? are not normalized. We’ll consider the normalization in the next section. Table 1: Radial, angular and total wavefunctions of first two states in H-atom n l m Rn,l Y (θ, φ) ψnlm Name 1 0

0

2 0

0

2 1

0

R −r/a0 e N10

R (2 − N20

r −r/2a0 a0 )e

R −r/2a0 re N21

2 1 +1

R −r/2a0 N21 re

2 1 −1

R −r/2a0 N21 re

A N00

A N00

A cosθ N10

N100 e−r/a0

N200 (2 −

r −r/2a0 a0 )e

N210 re−r/2a0 cosθ

A −N11 sinθe±iφ −N211 re−r/2a0 sinθe±iφ A N11 sinθe±iφ

N211 re−r/2a0 sinθe±iφ

1s

2s

2p

2p 2p

The normalization constants for the radial, angular and toR A tal wavefunctions are represented respectively by Nnl , Nlm R A and Nnlm (= Nnl × Nlm ). Note: in last row, the factor 12 in angular part is absorbed into the normalization constant. Normalization of Rn,l and Y (θ, φ) = Yl,m (θ, φ) for the first two states of H-atom: Before we proceed, let us discuss a little bit about normalization in Spherical polar coordinates. The normalization of a wavefunction (say, ψ(r, θ, φ)) requires to solve the following equation to find the value of the normalization constant N . 70

rmax Z θmax φ Z Zmax N 2 ψ ∗ (r, θ, φ)ψ(r, θ, φ)dτ = 1, rmin θmin φmin

where (rmin , θmin , φmin ) and (rmax , θmax , φmax ) are respectively the lower and upper limits of (r, θ, φ). The integration should be carried over all the variables present in the wavefunction taking into account the proper limits and the volume element (dτ ). The limits of r, θ and φ are rmin , rmax ≡ 0, ∞ θmin , θmax ≡ 0, π φmin , φmax ≡ 0, 2π Care should be taken to consider the volume element dτ . In Cartesian coordinates, the volume element is simply dxdydz. dτ in Cartesian coordinates: dxdydz But in Spherical polar coordinates it is not that simple. dτ in Spherical polar coordinates: r2 sinθdrdθdφ This expression for volume element in Spherical polar coordinates is derived in Appendix. Thus, to normalize different parts of wavefunctions, we need to solve the following equations Z∞ r part only:

∗ Rn,l Rn,l r2 dr = 1

(131)

Θ∗l,m Θl,m sinθdθ = 1

(132)

Φ∗m Φm dφ = 1

(133)

0

Zπ θ part only: 0 Z2π

φ part only: 0

Zπ Z2π Yl,m i.e. Angular part:

∗ Yl,m Yl,m sinθ dθ dφ = 1

0 0 Z∞ Zπ Z2π

whole wavefunction: 0

0

(134)

∗ ψn,l,m ψn,l,m r2 sinθdrdθdφ = 1

0

(135) Now we can proceed to normalize the different parts of the wavefunctions.

71

Normalizing R1,0 Z∞

∗ R1,0 R1,0 r2 dr = 1

R −r/a0 R1,0 = N10 e

0

Z∞

R N10


2

0 R N10

Z∞


2

e−2r/a0 × r2 dr = 1 (136) r2 e−2r/a0 dr = 1

0

2 a30 R = 1 =⇒ N10 = 3/2 × 4 a0


R 2 N10 Normalizing Y0,0 Zπ Z2π

∗ Y0,0 Y0,0 sinθdθdφ = 1

0 π Z Z2π

A Y0,0 = N00

0

0

A N00


2

sinθdθdφ = 1 (137)

0


A 2

Zπ

N00

Z2π sinθdθ

0 A N00


2

dφ = 1 0

1 A × 2 × 2π = 1 =⇒ N00 = √ 2 π

Combining the above two normalization constant, we get 1 N100 = p 3 . πa0

(138)

Thus, the complete expression of ground state wavefunction becomes 1 ψ100 = p 3 e−r/a0 . πa0

(139)

Normalizing R2,0 Z∞

∗ R2,0 R2,0 r2 dr = 1,

R R2,0 = N2,0

0

Z∞


R 2

N20 0



r 2− a0

2



r 2− a0



e−r/2a0

e−r/a0 × r2 dr = 1

 ∞  Z Z∞ Z∞  
4 1 R 2 2 −r/a0 3 −r/a0 4 −r/a0 N20 4 r e dr − r e dr + 2 r e dr = 1   a0 a0 0 0 0  
1 1 4 2 R R N20 4 × 2a30 − × 6a40 + 2 × 24a50 = 1 =⇒ N20 =p 3 a0 a0 8a0 (140) 72

Since the wavefunction of first excited state ψ200 contains the same angular part (Y00 ) as the ground state, we can write the total wavefunction of first excited state of H-atom as   1 r ψ200 = p 2− e−r/2a0 (141) 3 a0 32πa0 Normalizing R2,1 Z∞

∗ R2,1 R2,1 r2 dr = 1,

R R2,1 = N2,1 re−r/2a0

0

Z∞

R N21


2

0 R N21


2

Z∞

r2 e−r/a0 × r2 dr = 1 (142) r4 e−r/a0 dr = 1

0 R N21


2

1 R × 24a50 = 1 =⇒ N21 =p 5 24a0

Normalizing Y1,0 Zπ Z2π

∗ Y1,0 Y1,0 sinθdθdφ = 1

0 π Z Z2π

A Y1,0 = N10 cosθ

0

0

A N10


2

cos2 θsinθdθdφ = 1 (143)

0 A N10


2

Zπ

cos2 θsinθdθ

0


A 2

N10

Z2π dφ = 1 0

2 A × × 2π = 1 =⇒ N10 = 3

r

3 4π

Combining the normalization constants for Y10 and R2,1 , the total wavefunciton of the (2, 1, 0) state of H-atom becomes 1 ψ210 = p r cosθ e−r/2a0 5 32πa0

73

(144)

Normalizing Y1,±1 Zπ Z2π

∗ Y1,±1 Y1,±1 sinθdθdφ = 1

0

0 π Z Z2π 0

A N11


2

A Y1,±1 = ∓N11 sinθe±iφ

sin2 θe∓iφ e±iφ sinθdθdφ = 1 (145)

0 A N11


2

Zπ

sin3 θdθ

0


A 2

N11

Z2π dφ = 1 0

4 A = × × 2π = 1 =⇒ N11 3

r

3 8π

Thus, the (211) wavefunction becomes 1



ψ2,1,±1 = ∓ p r sinθ e 64πa50

±iφ− 2ar



0

(146)

Radial and angular distribution: From the probabilistic interpretation of wavefunction, we know that the ψ(r, θ, φ)ψ ∗ (r, θ, φ) represents the probability density of the electron at point (r, θ, φ). When we multiply this probability density with a volume element dτ , then ψ(r, θ, φ)ψ ∗ (r, θ, φ)dτ becomes the probability of finding the particle in a small sphere around the point (r, θ, φ). Note: We set the sum of all such probabilities within all space i.e. the total probability to 1 for normalizing the wavefunction. Thus, if P (r, θ, φ) be the probability density at the point (r, θ, φ), we can write P (r, θ, φ)dτ = ψ(r, θ, φ)ψ ∗ (r, θ, φ) r2 sinθ dr dθ dφ

(147)

Since the wavefunction has two parts – radial and angular, there exist both the radial and angular distributions. Thus, the radial and angular probability densities are given as ∗ P (r) = Rn,l Rn,l ∗ P (θ, φ) = Yl,m Yl,m

(148)

∗ The quantity 4πr2 Rn,l Rn,l is called radial distribution function. It describes the probability of finding the electron in a small volume element between r and r + dr in all directions in an orbital. Note: since there is no directional preference, the volume element simply becomes 4πr2 (see the footnote 8 ). The angular probability density describes the shape of the space around the nucleus where the probability of finding the electron is maximum. That means, it describes the shape of the orbitals.

8

If τ be the volume of a sphere, then τ=

4 2 πr =⇒ dτ = 4πr2 dr 3

74

Plot of radial distribution function in H-atom Spherical harmonics, concept of orbitals, Real and imaginary wavefunctions and shape of the orbitals: Spherical Harmonics: The solutions of angular part of the wavefunctions i.e. Ylm (θ, φ) are called spherical harmonics. Spherical harmonics are basically some special functions described on the surface of a sphere. Why sphere? Because, θ and φ are two spherical polar coordinates and spherical polar coordinates represent the location of a point on a sphere. Later on, you’ll see how can one obtain the shape of different orbitals from the spherical harmonics. In an atom, the spherical harmonics describe the shape of the space where the electrons can move. That means spherical harmonics describe the shape of different orbitals. Since, for any s-orbital (l = 0), the spherical harmonics is a constant (Y00 = 2√1 π ), it indicates a spherically symmetrical shape. Thus, an s-orbital is spherical in shape. ���� ���� ���� ���� �

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Figure 10: visualizing s-orbital You can plot the spherical harmonics to visualize it (as in Fig. ??). This can be done by many softwares. In Appendix is given a step by step method of how to plot spherical harmonics using GNUPLOT (a free software). For any p-subshell (l = 1), there are three spherical harmonics corresponding to three p-orbitals. The spherical q harmonics of a p-orbital with 3 cosθ). This is called pz orm = 0 is a real function (Eq. ??, Y10 = 4π bital. The name comes from the functional form (Remember, z = rcosθ in Eq. ??). When the spherical harmonics is plotted, it looks like a dumb-bell with the lobes along the z-axis. This justifies the nomenclature of l = 1, m = 0 orbital as pz . The other two p-orbitals corresponding to m = ±1 have complex spherical harmonics (Eq. ??). These orbitals can be termed as p+1 and p−1 . 2.1.1. energy expressions (without derivation), degeneracy 2.1.2. Hydrogenic wave functions up to n = 2 (expression only) 75

Figure 11: visualizing pz -orbital 2.1.3. real wavefunction 2.1.4. concept of orbials and shapes of s and p orbitals

76

3. University questions: 2

2017: 1) Verify that the wavefunction Ae−Bx /2 is an eigenfunction of the p simple harmonic oscillator (in one dimension) Hamiltonian. Here B = 2π (mk)/h (terms have their usual significance). Find the expression for the eigenvalue. Solution: The Hamiltonian for a simple harmonic oscillator is given by h2 d2 1 2 ˆ H=− 2 + kx , 8π m dx2 2 where k is the force constant. Let us apply this hamiltonian on the given 2 wavefunction Ψ(x) = Ae−Bx /2 . h2 d2 Ψ(x) 1 2 ˆ HΨ(x) = − 2 + kx Ψ(x) 8π m dx2 2 2 2 d −Bx2 /2 1 2 −Bx2 /2 h e + kx Ae = −A 2 8π m dx2 2 n o 1 2 h d 2 −Bx2 /2 = −A 2 −Bxe + kx2 Ae−Bx /2 8π m dx 2  o 1 2 n  h 2 −Bx2 /2 −Bx2 /2 = AB 2 x −Bxe +e + kx2 Ae−Bx /2 8π m 2 2 2 h 1 h 2 2 2 = −AB 2 2 x2 e−Bx /2 + AB 2 e−Bx /2 + kx2 Ae−Bx /2 8π m 8π m 2 p Since B = 2π (mk)/h =⇒ B 2 = 4π 2 mk/h2 . Putting these in the above equation, √ 2 2 h 4π mk 2π mk h2 −Bx2 /2 1 2 −Bx2 /2 2 2 −Bx /2 ˆ HΨ(x) = −A 2 xe +A e + kx Ae 2m 8π m h2 h 8π 2 r 1 h k −Bx2 /2 1 2 −Bx2 /2 2 e + kx Ae = −A kx2 e−Bx /2 + A 2 4π m 2 r h k −Bx2 /2 = Ae 4π m r 1 k 2 = ~ωAe−Bx /2 , where, ω = = angular frequency of oscillation. 2 m Thus, the given wavefunction is an eigenfunction of a one-dimensional q simple harmonic oscillator. The corresponding eigenvalue is 21 ~ω, where ω = mk . 2) From the eigenvalue obtained above, make an estimate of the positions of the Classical Turning points x1 and x2 i.e. values of x after which the motion reverses its direction. Use proper arguments. Solution: The classical turning point is the point, where the kinetic energy of the simple harmonic oscillator becomes zero and the whole energy (i.e. the eigenvalue obtained above) becomes equal to the potential energy (i.e. 12 kx2 ). So, at classical turning point, 1 1 ~ω = kx2 2 2r ~ω x=± k Thus, r ~ω x1 = + rk ~ω x2 = − k 77

 3/2

1 3) The 1s wavefunction for the H-atom is Ψ = e−r/a0 , where a0 is a0 the Bohr radius. Calculate the probability of finding the electron within a distance a0 from the nucleus. Given √1 π

Z∞

xn e−bx dx =

n! bn+1

0

; (n > −1)

Solution: The probability of finding a particle (having wavefunction Ψ(r)) between the two points r = r1 and r = r2 is given by Zr2 P = r1 √1 π

Ψ∗ (r)Ψ(r) × 4πr2 dr

 3/2

1 In our case, Ψ(r) = e−r/a0 and r1 = 0, r2 = a0 . Therefore, the a0 probability of finding the electron within the Bohr radius is

Za0 P =

1 √ π



0

4 = 3 a0

Za0

1 a0

3/2

−r/a0

e

1 ×√ π



1 a0

3/2

e−r/a0 × 4πr2 dr

r2 e−2r/a0 dr

0

 a0 Z Z  Z d 4 2 e−2r/a0 dr − r2 e−2r/a0 dr dr = 3 r a0 dr   a0 0 Z    a0 a0 4 = 3 r2 − e−2r/a0 − 2r − e−2r/a0 dr a0 2 2 0  a0 Z 4 1 = 2 − r2 e−2r/a0 + re−2r/a0 dr a0 2 0   a0 Z Z  Z 1 2 −2r/a0 d 4 + r e−2r/a0 dr − r e−2r/a0 dr dr = 2 − r e a0 2 dr 0  a0 2 4 1 a0 a = 2 − r2 e−2r/a0 − r e−2r/a0 − 0 e−2r/a0 a0 2 2 4 0  2  a0 −2 a20 −2 a20 −2 a20 4 = 2 − e − e − e +0+0+ a0 2 2 4 4 = 1 − 5e−2 = 0.3233 Thus, the probability of finding an electron in 1s orbital of a H-atom within Bohr radius is 0.3233. 4) For the 1s wavefunction, find hri, the average distance of the electron from the nucleus. Solution: The average distance of an electron in 1s orbital is hri =

Z∞

1 √ π

0

4 = 3 a0

Z∞



1 a0

3/2

1 e−r/a0 × r × √ π

r3 e−2r/a0 dr

0

78



1 a0

3/2

e−r/a0 × 4πr2 dr

Using the standard integration Z∞

xn e−bx dx =

0

n! bn+1

; (n > −1)

we can write 4 hri = 3 a0

Z∞

r3 e−2r/a0 dr

0

3! 4 × a30 16a−4 0 3 = a0 2 =

5) The 2s wavefunction for the H-atom is Ψ2s = N (2 − ar0 )e−r/2a0 , where N is a constant, r is the distance from the nucleus and a0 is the Bohr radius. Find the distance from the nucleus, in terms of a0 , at which the radial probability density shows maxima. Solution: The radial probability density for the 2s wavefunction of H-atom is given by 4πr2 Ψ∗2s Ψ2s . The distance at which it is maximum, can be obtained by taking its derivative with respect to r and setting the result to zero. Thus,
d 4πr2 Ψ∗2s Ψ2s = 0 dr   d r −r/2a0 r −r/2a0 2 4π × N (2 − )e r × N (2 − )e =0 dr a0 a0 (  ) 2 r d r2 2 − e−r/a0 = 0 4πN 2 dr a0 ( )  2 2 d r 2r − e−r/a0 = 0, as 4πN2 6= 0 dr a0       2 2 r2 1 −r/a0 2r r − e 2− =0 2r − + e−r/a0 × 2 2r − a0 a0 a0 a0    r2 2r r2 4r −r/a0 2r − − + 2 +4− e =0 a0 a0 a0 a0  
r 1 r 2− × 2 r2 − 6ra0 + 4a20 e−r/a0 = 0 a0 a0 This gives five roots as (i) r = 0, (ii) 2 − ar0 = 0 =⇒ r = 2a0 (iii) e−r/a0 = 0 =⇒ r = ∞ √ (iv) and (v) r2 − 6ra0 + 4a20 = 0 =⇒ r = (3 ± 5)a0 = +5.236a0 , −0.764a0 Among these five, (i) r = 0, (ii) r = 20 , (iii) r = ∞ and (v) r = −0.764a0 are not valid. Because, (i) r = 0 indicates nucleus and at nucleus probability density cannot be maximum. (ii) r = 2a0 indicates a node, i.e. at this point the wavefunction Ψ2s becomes zero and hence at this point the probability density cannot be maximum. (iii) r = ∞ indicates infinite distance from nucleus. At this point too, the electron density cannot be maximum. (v) r = −0.764a0 is a negative distance and hence is invalid. 79

Therefore, we are left with only one option i.e. r = 5.236a0 . This is the point at which the probability density will be maximum. 6) Justify the existence of a non-zero zero point energy in case of a quantum mechanical oscillator in the light of Heisenberg Uncertainty principle. Solution: For a one-dimensional quantum mechanical simple harmonic os−Bx2 /2 cillator, the ground state wavefunction is given by Ae , where B = √ 2π mk/h. The expectation value of linear momentum hpx i will be   Z+∞ d 2 2 hpx i = Ae−Bx /2 dx Ae−Bx /2 −i~ dx −∞

Z+∞ 2 = i~A2 B xe−Bx dx −∞

The integrand is an odd function and hence the value of the integral will be zero. Thus, hpx i = 0. Now, we need the expectation value of p2x i.e. hp2x i.   Z+∞ 2 −Bx2 /2 2 −Bx2 /2 2 d Ae dx hpx i = Ae −~ dx2 −∞

Z+∞
2 = −A2 ~2 e−Bx B 2 x2 − B dx −∞

Z+∞ Z+∞ 2 2 = −A2 B 2 ~2 x2 e−Bx dx + A2 B~2 e−Bx dx −∞

r √ π π = −A2 B 2 ~2 √ + A2 B~2 B √ 2B B √ 2 2 Bπ 2 2 = −A ~ + A ~ Bπ 2 √ 1 = A2 ~2 Bπ 2

−∞

Thus, the uncertainty in linear momentum (∆px ) would be p ∆px = hp2x i − hpx i2 r 1 2 2√ = A ~ Bπ − 0 2 1/4  Bπ = A~ 4 If this uncertainty in linear momentum be taken as the minimum uncertainty then the minimum energy (Emin ) of the oscillator would be, ∆p2x 2m √ A2 ~2 Bπ = 4m Since, none of the quantity on the right hand side is zero, the minimum energy of the quantum mechanical oscillator is not zero. Thus, the oscillator has a non-zero zero-point energy and is consistent with the Heisenberg uncertainty principle. Emin =

80

2016: 1) For the photoelectric effect in sodium metal the following results were obtained where Kmax is the maximum kinetic energy of emitted electrons and 1012 Kmax /erg 3.41 1.95 λ/˚ A 3125 4047 λ is the wavelength of the incident radiation. Calculate Planck’s constant h and the ‘work function’ for sodium. Solution: The relation between the maximum kinetic energy (Kmax ) of the photoelectrons emitted during a photoelectric experiment, wavelength (λ) of radiation used and the work function (W0 ) is given by Kmax =

hc − W0 , λ

where h and c are Planck’s constant and speed of light in vacuum respectively. Given, when λ = 3125 × 10−10 m, Kmax = 3.41 × 105 J. and when λ = 4047 × 10−10 m, Kmax = 1.95 × 105 J. Thus, h × 3 × 1010 3.41 × 10 = − W0 3125 × 10−10 h × 3 × 1010 5 1.95 × 10 = − W0 4047 × 10−10 5

Note all the quantities are used in MKS unit, hence the value of h and W0 obtained from these two equations will be in J.sec and J respectively. Subtracting the above two equations, we get   1 1 5 20 − (3.41 − 1.95) × 10 = h × 3 × 10 3125 4047 1.46 × 105 × 4047 × 3125 h= 922 × 3 × 1020 h = 6.675 × 10−12 J.sec Putting this value of h in one of the above equations we get, 6.675 × 10−12 × 3 × 1010 − W0 3.41 × 10 = 3125 × 10−10 W0 = 6.408 × 105 + 3.41 × 105 = 9.818 × 105 J 5

ˆ is a linear operator and that M ˆ f1 = bf1 and M ˆ f2 = bf2 2) Given that M prove that C1 f1 + C2 f2 where C1 and C2 are constants, is an eigenfunction of ˆ with eigenvalue b. M Solution: Given ˆ f1 = bf1 M ˆ f2 = bf2 M ˆ is a linear operator. Thus, we can write M ˆ (C1 f1 + C2 f2 ) = M ˆ C 1 f1 + M ˆ C 2 f2 M ˆ f 1 + C2 M ˆ f2 Since M ˆ is a linear operator. = C1 M = C1 bf1 + C2 bf2 Using the above eigenvalue equations. = b (C1 f1 + C2 f2 ) 81

ˆ Thus, we see that if f1 and f2 be the eigenfunctions of a linear operator M with the same eigenvalue (b), then the linear combination (C1 f1 + C2 f2 ) will ˆ with the same eigenvalue (b). also be an eigenfunction of M 3) Estimate the minimum uncertainty in the x-component of the velocity of an electron whose position is measured to an uncertainty 1.0 × 10−11 m. Solution: The required minimum uncertainty in velocity is ~ 2∆px h = 4πme ∆vx

∆x =

6.626 × 10−34 J.Sec = 4 × 3.14 × 9.1 × 10−31 kg × 1.0 × 10−11 m = 5.797 × 106 m

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4) A function that becomes infinite at a point must not be quadratically integrable. Justify or criticize. Solution: Quadratic integration of a function Ψ means evaluating the value R ∗ of Ψ Ψdτ over a certain limit (usually the boundary of the system). The limit of integration is very important here. If the function has finite values within the limit of integration then the function will be quadratically integrable but if the function has infinite value at any point within the limit of integration then the function will not be quadratically integrable within that limit. And if a function is not quadratically integrable withing the boundary of the system then the function cannot be used as a valid wavefunction for the system. If the boundary of the system is from −∞ to +∞ and the function becomes infinite at a point then it is definite that the function is not quadratically integrable. However, if the function becomes infinite only at a point beyond the boundary of the system then the function will be quadratically integrable and hence can serve as a valid wavefunction for the system. 5) Sketch Ψ and |Ψ2 | for the n = 3 state for a particle in a 1D box (0 ≤ x ≤ L). Solution: For a particle in a 1D box, the wavefunction of n = 3 state is given by r 2 3πx sin Ψ= L L 2 The required plots of Ψ and |Ψ | are

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6) Identify the locations of nodes in the wavefunction with n = 4 for a particle in a 1D box. Solution: The wavefunction for n = 4 state of a particle in a 1D box is given by r 2 4πx sin Ψ(x) = L L The nodes appear at the points where the wavefunction is zero. The above function will be zero at points for which 4x/L = an integer or 0. Thus for 82

n = 4 state for a particle in a 1D box, nodes appear at x = 0, L4 , L2 , 3L 4 and L. The first and last points i.e. x = 0 and x = L are common to any state of the particle in a 1D box. 7) Which of the functions 6cos4x, 5x3 , 3e−5x , ln2x, sin3x are eigenfunction of d2 the operator dx 2 ? For each eigenfunction, state the eigenvalue. Solution: Let us check one by one d2 6cos4x = −16 × 6cos4x dx2 d2 3 5x = 30x dx2 d2 −5x 3e = 25 × 3e−5x 2 dx d2 1 ln2x = − dx2 2x2 2 d sin3x = −9sin3x dx2 Thus out of the five given funcitons, three of them i.e. 6cos4x, 3e−5x and d2 sin3x are eigenfunctions of the operator dx 2 with eigenvalues of −16, 25 and −9 respectively. 8) For a particle in a 2D rectangular box with sides of lengths L and 2L, show that (2,2) and (1,4) states are ‘accidentlly’ degenerate. Solution: In a 2D rectangular box, a degeneracy, which occurs due a combination of symmetry of the box and the quantum numbers is called ‘accidental degeneracy’. The energy of a particle in a 2D rectangular box in a quantum state (nx , ny ) is given by ! 2 2 2 n h nx y + 2 E(nx , ny ) = 2 8m Lx Ly Given Lx = L, Ly = 2L. Thus h2 E(nx , ny ) = 8m

n2y n2x + L2 4L2

!

Thus for the two states (2,2) and (1,4), the energies are   h2 4 4 5h2 E(2, 2) = + = 8m L2 4L2 8mL2   h2 1 16 5h2 E(1, 4) = + = 8m L2 4L2 8mL2 Therefore, the two states (2,2) and (1,4) are accidentally degenerate for a particle in a 2D rectangular box with lengths L and 2L. 9) Find an expression for hp2x i in the case of n = 2 state of a particle in a 1D box, without performing an explicit differentiation. Find the numerical value of hp2x i for a box of length 2.0˚ A. Solution: Since for a particle in a 1D box, the energy is solely kinetic, the expectation value of p2x can be written as hp2x i = 2mEn , 83

where m is the mass of the particle and En is the energy of the nth state. For 4h2 n = 2, in a box of length L, we know E2 = 8mL 2 . Therefore, we can write hp2x i

4h2 = 2m 8mL2 h2 = 2 L

For a box of length L = 2.0˚ A, hp2x i

(6.626 × 10−34 )2 2 = J .sec2 /m2 −10 2 (2.0 × 10 ) = 1.0976 × 10−47 Kg2 m2 sec−2

10) Show that a Bohr orbit for a hydrogen-like atom has a circumference which is an integral multiple of de Broglie wavelength of the electron moving around the nucleus. Solution: The Bohr’s theory demands that an electron exist only in those orbits around the nucleus, where the angular momentum of the electron is an integral multiple of h/2π. i.e. mvr =

nh , 2π

where m, v, and r are mass of the electron, velocity of the electron and radius of the orbit respectively. n is the principal quantum number. Now, according to de Broglie’s hypothesis, relation between the wavelength of the wave associated with the moving electron having momentum mv is given by mv =

h λ

Combining the two equations, we can write h h r=n λ 2π 2πr = nλ Thus, the circumference of the Bohr orbit should be an integral multiple of de Broglie wavelength. 11) The energy of a particle of mass m in a 1D box of length l is measured. What are the possible values that can result from the measurement if at the time of measurement, the state function is 1 30 2 i)Ψ = x(l − x) for 0 ≤ x ≤ l l5   21 2 3πx ii)Ψ = sin for 0 ≤ x ≤ l l l 

Solution: In case of (i):

84

hEi =

Zl 

30 l5

 21

0

15h2 = 2 5 2π ml

Zl 0

h2 d2 x(l − x) − 2 8π m dx2 



30 l5

 21 x(l − x)dx

x(l − x)dx

15h2 x2 l x3 − = 2 5 2π ml 2 3 5h2 = 2 2 4π ml In case of (ii): 

l 0

    12 Z l   12 2 3πx 3πx h2 d2 2 hEi = sin sin dx − 2 l l 8π m dx2 l l 0

2

=

9h 8ml3

Zl

2sin2

3πx dx l

0

9h2 = 8ml3

 Zl  6πx 1 − cos dx l 0

2

9h 8ml2 12) Find the expectation value of kinetic energy of the harmonic oscillator in the ground state. Given  α 1/4 αx2 e− 2 , Ψg = π where α is a constant. Solution: =

  Z∞  1/4 2 α h2 d2  α 1/4 − αx2 − αx2 hT i = e − 2 e 2 dx π 8π m dx2 π −∞

h2 =− 2 8π m

r

h2 =− 2 8π m

r

α π

Z∞ −∞



e−

αx2 2

Z∞


αx2 α2 x2 − α e− 2 dx Z∞



α 2 2 2 α x2 e−αx dx − α e−αx dx π −∞  −∞∞  r Z Z∞ 2 α 2 h 2 2 2α x2 e−αx dx − 2α e−αx dx =− 2 8π m π 0 r  √0 r  2 h α 2 π 1 π =− 2 α −α 3/2 4π m π 2 α 4α r r r   h2 α α π α π =− 2 − 4π m π 4 α 2 α αh2 = 16π 2 m 85

13) The force constant of a harmonic oscillator of reduced mass 1.5 × 10−27 kg is 10Nm−1 . Calculate the zero point energy. Solution: Zero-point energy of a harmonic oscillator is given by s 1 k , EZP = h 2 µ where h = Planck’s constant, k = force constant and µ = reduced mass. Putting the given values of k and m and standard value of h in the above equation, we can write s 1 10 N.m−1 −34 EZP = × 6.626 × 10 J.Sec × 2 1.5 × 10−27 kg = 2.7049 × 10−20 J

14) Find the most probable value of r for 1s electron in a H-atom. Solution: The wavefunction for an electron in 1s orbital in H-atom is given by  3/2 1 1 e−r/a0 Ψ1s = √ π a0 Therefore, the most probable value of r will be obtained by differentiating the radial probability density 4πr2 Ψ∗1s Ψ1s with respect to r and setting it to zero. ( )  3/2  3/2 d 1 1 1 1 0= 4πr2 × √ e−r/a0 × √ e−r/a0 dr π a0 π a0 n o 4 d r2 e−2r/a0 0= 3 a0 dr   8 r 0 = 3r 1 − e−2r/a0 a0 a0

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It has three roots. (i) r = 0, this indicates nucleus. The probability of finding electron cannot be maximum in nucleus. So, this is not a valid root. (ii) e−2r/a0 =⇒ r = ∞. At infinite distance too, the probability of finding the electron cannot be maximum. So, this is also not the required point. (iii) (1 − ar0 ) =⇒ r = a0 . This represents a distance at Bohr radius distance and is the required most probable value of r. 15) Plot Ψ1s and Ψ2s vs r. Solution:

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16) Show that Ψ0 and Ψ1 for harmonic oscillator are orthogonal to each other. Solution: Ψ0 and Ψ1 for harmonic oscillator are given as  α 1/4 2 Ψ0 = e−αx /2 π  α 1/4 √ 2 Ψ1 = 2αxe−αx /2 π 86

In order to show that these two states are orthogonal to each other, the overlap integral between the two should be zero. r Z∞ Z∞  1/4   2 1/4 √ 2α α α 2 2 2 xe−αx dx e−αx /2 × 2αxe−αx /2 dx = π π π −∞

−∞

The integrand is an odd function of x and hence the value of the definite integral will be zero. Thus, we see, that the overlap integral between Ψ0 and Ψ1 states of a harmonic oscillator is zero. Hence these two states are orthogonal to each other. 17) Show that the degree of degeneracy of a H-atom bound state energy level is n2 (spin consideration being omitted), n =principal quantum number. Solution: In H-atom, the Hamiltonian commutes with the orbital angular ˆ 2 and one of its components (say Lz ). Since, in our case, we momentum L are not considering the spin, the states are designated by values of three quantum numbers n, l, and ml . For a given value of n, all the states with l = 0, .., (n − 1) have the same energy. Similarly, for a given value of n, l all the (2l + 1) states having ml = −l, . . . , +l have the same energies. Therefore, n−1 P degeneracy is (2l + 1) = 1 + 3 + 5 + · · · + (2n − 1) = n2 . 0

2015: 1) Determine, citing reasons, whether the following function is acceptable as well behaved wavefunction or not within the indicated interval: e−x cosx (0, ∞). Solution: e−x is a continuous decaying function and cosx is an oscillating function. Both are single valued, continuous, finite and square integrable functions within the interval (0, ∞). Therefore, their product e−x cosx is also single valued, continuous, finite and square integrable within that interval. Therefore, e−x cosx is an acceptable and well behaved wavefunction. 2) Write down an expression for ‘Compton Shift’ and indicate therein the parameter ‘Compton Wavelength’. What is the value of scattering angle for which the Compton shift becomes maximum? Solution: Compton shift is give by ∆λ =

h (1 − cosθ) , me c

where ∆λ = change in wavelength due to Compton effect – also called Compton shift, h = Planck’s constant, me = mass of the electron, c = speed of light in vacuum and θ is the scattering angle. In the above equation, mhe c is called Compton wavelength. To find the scattering angle for which the Compton shift becomes maximum, one needs to take derivative of ∆λ with respect to θ and set the result to zero. Therefore,

d dθ



d ∆λ = 0 dθ 

h (1 − cosθ) = 0 me c h sinθ = 0 me c θ = 0◦ or 180◦

When θ = 0◦ , ∆λ = 0 and when θ = 180◦ , ∆λ = 2 mhe c . Thus, the Compton shift becomes maximum at 180◦ scattering angle. 87

3) Heisenberg uncertainty principle demands that the zero-point energy of a Harmonic oscillator be non-zero. Explain. Solution: Please see question 6 of year 2017. 4) Evaluate the commutator: [ˆ xn , pˆx ]. Solution: [ˆ xn , pˆx ] ψ(x) = xˆn pˆx ψ(x) − pˆx xˆn ψ(x)   d d n ψ(x) + i~ {xx ψ(x)} = x −i~ dx dx dxn n dψ(x) n dψ(x) = −i~x + i~x + i~ψ(x) dx dx dx n−1 = i~nx ψ(x) Thus, [ˆ xn , pˆx ] = i~nxn−1 5) Photoelectric effect gives a strong evidence in favor of particle nature of light. Account for the statement. Solution: Photoelectric effect is the emission of electrons from a metal surface when the surface is illuminated with a light of frequency, greater than some minimum frequency (characteristics of the metal). It has been observed that the number of electrons emitted from the metal surface depends only on the frequency of the incident light and not on the intensity. This observation is very striking and is in direct contradiction of the wave nature of light. If light behaves as a wave then the 6) Find the de Broglie wavelength of electrons that have been accelerated from rest through a potential difference of 1kV. Solution: The potential different used for accelerating the electron is 1 kV = 1000 V. Therefore, the kinetic energy of the electron is 1000 eV = 1000 × 1.602 × 10−19 J. p The momentum of the electron = 2 × 9.1 × 10−31 kg × 1.602 × 10−16 J = 1.707 × 10−23 kg.m.sec−1 . −34 h So, the de Broglie wavelength of the electron will be λ = mv = 6.626×10 1.707×10−23 = −5 0.6055 × 10 m 7) Show that the wavefunctions corresponding to different eigenvalues of a Hermitian operator are orthogonal. Solution: Let ψ and φ be the wavefunctions corresponding to respectively ˆ the different eigenvalues a and b of a hermitian operator O. ˆ = aψ Oψ ˆ = bφ Oφ ˆ is hermitian, we can write Since O I I  ∗ ∗ ˆ ˆ ψ Oφdτ = φ Oψ dτ I I ∗ ψ aφdτ = φ (bψ)∗ dτ I (a − b)ψ ∗ φdτ = 0, (a, b are real) Since a 6= b,

I

ψ ∗ φdτ = 0 88

That means the two wavefunctions are orthogonal to each other. d2 8) If csinax is an eigenfunction of dx 2 , calculate the eigenvalue. Solution: d2 csinax = −ca2 sinax 2 dx Therefore, the eigenvalue is −a2 . iEt 9) Explain whether the function e ~ Ψ(x) represents a stationary state or not. Solution: The expectation value of any property hPˆ i of the system in the iEt state e ~ Ψ(x) is given by I n o∗ iEt iEt ˆ e ~ Ψ(x) Pˆ e ~ Ψ(x)dτ hP i = I −iEt iEt = e ~ Ψ∗ (x)Pˆ e ~ Ψ(x)dτ Note that the operator Pˆ is time-independent. Therefore, I −iEt iEt hPˆ i = e ~ e ~ Ψ∗ (x)Pˆ Ψ(x)dτ I = Ψ∗ (x)Pˆ Ψ(x)dτ, iEt

which is independent of time. Thus, the function e ~ Ψ(x) represents a stationary state. 14h2 10) Calculate the degeneracy of the level having an energy of 8ma 2 for a particle of mass m confined in a cubic box of edge length a. Solution: The energy of a particle of mass m confined in a cubic box of edge length a, is given by E(nx , ny , nz ) =

n2x

+

n2y

+

n2z


h2 8ma2

2

14h The given energy is 8ma 2 . This energy can be obtained for the following combinations of (nx , ny , nz ): (1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,2,1),(3,1,2). That means the state is 6-fold degenerate. 11) For what molecular phenomenon does the simple harmonic oscillator serve as model? Mention its limitation in this respect. Solution: Simple harmonic oscillator serves as a model for molecular vibrations i.e. to model the vibration of bond in a molecule. This is applicable only for those vibrations which is harmonic that means only those vibrations which occur at the bottom of the potential energy surface. In other words, vibrations corresponding to the lower vibrational state. If there is anharmonicity in the vibration then simple harmonic oscillator cannot serve as a good model. 12) The radial wavefunction for the 1s orbital of a H-atom is given by

R10 (r) =

2 3/2 a0

e−r/a0

Calculate the average radius of the 1s orbital. Given Z∞

xn e−ax dx =

0

89

n! an+1

Solution: hri =

Z∞

R10 (r)rR10 (r)r2 dr

0 Z∞

e−r/a0 r 3/2

a0

0

4 = 3 a0

2

2

=

Z∞

3/2 a0

e−r/a0 r2 dr

r3 e−2r/a0 dr

0

4 3!a40 × a30 16 3 = a0 2 =

13) Find out hpx i for a harmonic oscillator in its ground state. Given Ψg =

 a 1/4 π

2

e

− ax2

.

Solution: Please see question 6 of year 2017. 14) Zero point energy of a simple harmonic oscillator does not violate Heisenberg uncertainty principle. Justify. Solution: Please see question 6 of year 2017. 15) Three 2p orbitals of H-atom are p1 = f1 (r)rcosθ p2 = f2 (r)rsinθcosφ p3 = f3 (r)rsinθsinφ where the terms have their usual significance. Identify them as 2px , 2py and 2pz with reasons. Solution: The relation between Cartesian and Polar coordinates is

z !, #, $

Y x = rsinθcosφ y = rsinθsinφ z = rcosθ, Comparing the angular part of the given three 2p orbitals with the above relations, one can easily identify that p1 represents py orbital, p2 represents px orbital and p3 represent pz orbital. 90

16) Define radial distribution function (RDF) and give the plot of RDF corresponding to the following wavefunction against r/a0 . Given Ψ = A(2a0 − r)e−r/2a0 Solution: Radial distribution function is defined as a function that describes the variation of probability density with distance. For a quantum mechanical systems in state Ψ(r), the radial distribution function is given by 4πr2 Ψ∗ (r)Ψ(r). For Ψ(r) = A(2a0 − r)e−r/2a0 , radial distribution function will be 4πA2 (2a0 r − r2 )2 e−r/a0

��������

A plot of this function against r/a0 is

����

h

i d2 ˆ ˆ ˆ 2014: 1) Evaluate the commutator A, B , where Aˆ = dx 2 and B = x. Solution:  2  h i ˆ B ˆ ψ(x) = d , x ψ(x) A, dx2 d2 d2 = 2 {xψ(x)} − x 2 ψ(x) dx dx d2 ψ(x) d2 ψ(x) =x + ψ(x) × 0 − x dx2 dx2 =0 2) The Hamiltonian operator for a system is given by ~2 d2 ˆ H=− . 2m dx2 If it is operated on a wavefunction Ψ = 2l sin nπx l find dispersion in energy 2 2 2 given by the expression σE = hE i − hEi .

91

Solution: hEi =

Zl 0

  ~2 d2 2 nπx 2 nπx sin − sin dx l l 2m dx2 l l

2~2 n2 π 2 = ml4

Zl

sin2

nπx dx l

0

h2n2 = 4ml4

Zl

  2nπx 1 − cos dx l 0  l 2 h2n l 2nπx = x− sin 4ml4 2nπ l 0   2 h2n l = l− sin2nπ − 0 + 0 4ml4 2nπ n2 h2 = 4ml3 and 2

hE i =

Zl 0

 2 ~2 d2 2 nπx 2 nπx sin − sin dx l l 2m dx2 l l

4~4 = 4m2 l2

Zl sin

nπx d4 nπx sin dx l dx4 l

0

4~4 n4 π 4 = 4m2 l6 h4 n4 = 32m2 l6

Zl

0 Zl 0

sin2

nπx dx l

  2nπx 1 − cos dx l

This is the same integral asappeared above. h4 n4 = 32m2 l5 So, the dispersion in energy will be σE = hE 2 i − hEi2  2 2 2 h4 n4 nh = − 32m2 l5 4ml3 n4 h4 (l − 2) = 32m2 l6 3) Find whether the following wavefunctions are well-behaved wavefunctions or not (i) e−x (−∞, ∞), (ii) xe−x (0, ∞). Solution: (i) e−x becomes infinite at x = −∞. That means, it is not quadratically integrable in the given interval. So, it is not a well-behaved wavefunction. (ii) xe−x is a continuous function. It has also finite values in the indicated 92

interval. it is also quadratically integrable.  ∞  Z Z Z∞ Z 
d 2 −2x −x 2 2 −2x x e dx dx xe dx = x e dx − dx 0 0  ∞ 1 −2x 2 = − e (2x + 2x + 1) 4 0 1 = 4 So, xe−x is a well-behaved wavefunction. 4) A particle is trapped in a 1D box of length L. Apply de Broglie’s hypothesis to obtain energy levels. Solution: For a particle to exist in a 1D box of length L, L should be an integral multiple of the de Broglie wavelength of the particle. i.e. L = nλ. h (de Broglie relation). Let there be a constant potential (V ) inside But λ = mv mv 2 the box. Then the energy of the particle can be written as E = 2 + V =⇒ p mv = (E − V )2m Therefore, nh mv n2 h2 n2 h2 2 L = 2 2= mv 2m(E − V ) n2 h2 E= +V 2mL2 5) Find the expression of permitted energies when (i) an electron is in a box of 0.01 nm length, (ii) a 10 gm marble is in a box of 10 cm length. Mass of electron = 9.1 × 10−31 kg. Solution: The energy of a particle is quantized. According to the particle in a 1D box model, (i) the expression of permitted energy of an electron in a box of length 0.01 nm is given by L=

Eelectron = =

n2 h2 8me l2 6.626 × 10−34


2

n2

8 × 9.1 × 10−31 × (0.01 × 10−9 )2 = n2 × 0.603 × 10−15 J

J

(ii) the expression of permitted energy of a 10 g marble in a box of 10cm length is given by Emarble

n2 h2 = 8mmarble l2
2 6.626 × 10−34 n2

8 × 10 × 10−3 × (10 × 10−2 )2 = n2 × 0.549 × 10−63 J

J

6) An excited atom gives up excess energy by emitting a photon of characteristic frequency. The average period that elapses between the excitation of the atom and the time it radiates is 1.0 × 10−8 sec. Find the inherent uncertainty in the frequency of the photon. Solution: According to Heisenberg uncertainty principle, we can write ∆E × τ ≥ 93

h , 4π

where ∆E is the uncertainty in the energy of the photon emitted and τ is the average time that elapses between the excitation of the atom and the time it radiates. Given, τ = 1.0 × 10−8 sec. Using the relation ∆E = h∆ν, where ∆ν is the uncertainty in frequency of the photon. We can write ∆ν ≥

1 4πτ

1 sec−1 −8 4 × 3.14 × 1.0 × 10 ∆ν ≥ 7.962 × 106 sec−1

∆ν ≥

7) Show that in a rectangular box with sides L1 = L and L2 = 2L, there is an accidental degeneracy between the states nx = 1, ny = 4 and nx = 2, ny = 2. Solution: Please see question 8 of year 2016. 8) Given that the work function of chromium is 4.40 eV, calculate the kinetic energy of electrons emitted from a chromium surface that is radiated with UV radiation of wavelength 200 nm. Solution: Einstein’s equation of photoelectric effect is given by: T = hc λ −W , where T = kinetic energy of the electron emitted, λ = wavelength of light used, W = work function of the metal, h and c are the Planck’s constant and speed of light in vacuum respectively. Given, λ = 200 × 10−9 m, W = 4.40eV = 4.40 × 1.602 × 10−19 J. Thus we can write 6.626 × 10−34 J.Sec × 3 × 108 m.sec−1 − 4.40 × 1.602 × 10−19 J T = −9 200 × 10 m −19 = 993.9 × 10 − 7.0488 × 10−19 J = 986.851 × 10−19 J

9) Regard a chain of N conjugated carbon atoms, bond length RCC , as forming a box of length L = (N − 1)RCC . Find the allowed energies. Estimate the wavelength of the lowest energy transition. Solution: The required energy is n2 h2 En = 8mL2 n2 h2 = 2 8m(N − 1)2 RCC For estimating the wavelength of the the lowest energy transition, we should calculate the wavelength for the energy different E2 − E1 . Thus 22 h2 12 h2 − 2 2 8m(N − 1)2 RCC 8m(N − 1)2 RCC 3h2 = 2 8m(N − 1)2 RCC

E2 − E1 =

Thus the required wavelength is 2 hc 8hcm(N − 1)2 RCC λ= = E2 − E1 3h2

10) An electron is travelling at 1/4th the speed of light. Calculate the de Broglie wavelength taking into account the relativistic change in mass. Solution: For a body moving with velocity v, its mass is given by m0 M=q , v2 1 − c2 94

where m0 is the rest mass of the body i.e. it’s mass when it is at rest v = 0. Now, if an electron of rest mass me is moving with a velocity v = 41 c, then the relativistic mass of the electron will become me M=q 1 1 − 16 4me =√ 15

Therefore, the required de Broglie wavelength will be h M √v 15h = me c

λ=

11) Find and plot the radial distribution function for the following hydrogenic orbital Ψ = Arcosθe−r/2a0 , where A and a0 are constants. Solution: The radial part of the given wavefunction is R = Are−r/2a0 . Therefore, the radial distribution function for the given hydrogenic orbital will be  2 2 2 2 −r/2a0 |R| × 4πr = 4πr Are = 4πA2 r4 e−r/a0

������

The required plot is

�

12) Normalize the wavefunction of a harmonic oscillator in the v = 1 level, 2 ψ1 = Axe−ax /2 . Given, r Z∞ 1 π 2 −ax2 xe dx = 4a a 0

Solution: Z+∞ −∞

Axe

−ax2 /2

2

dx = 1

Z+∞ 2 A2 x2 e−ax dx = 1 −∞ Z+∞

A2 × 2

2

x2 e−ax dx = 1

0

1 2A 4a 2

95

r

π =1 a  3 1/4 4a A= π

13) The wavefunction for 2pz orbital of H-atom is ψ2pz = p

1 32πa50

re−r/2a0 cosθ,

where a0 =Bohr radius. Find the average value of the electron-nucleus sepaR∞ n −bx n! . ration in the 2pz orbital of H-atom. Given x e dx = bn+1 0

Solution: Z∞ Zπ Z2π 1 1 −r/2a0 p p hri = re cosθ × r × re−r/2a0 cosθ × r2 sinθdrdθdφ 5 5 32πa0 32πa0 0

0

0

1 = 32πa50

Z∞

r5 e−r/2a0 dr

0

Zπ

sinθcos2 θdθ

0



3

cos θ 1 6 6 × 5! × 2 a × − 0 32πa50 3 = 320a0

π

=

0

Z2π dφ 0

× 2π

14) For a harmonic oscillator consisting of a particle of mass 1.33 × 10−25 kg., the difference in adjacent energy levels is 4.82 × 10−21 J. Calculate the force constant of the oscillator. Solution: The energy difference between two adjacent states of a harmonic oscillator consisting of mass m and oscillating with frequency ω, in v th state is given by r k Thus, ∆Ev,v+1 = hω, where ω = m 2 m∆Ev,v+1 k= h2 Given, m = 1.33 × 10−25 kg and ∆Ev,v+1 = 4.82 × 10−21 J. Therefore, 1.33 × 10−25 × (4.82 × 10−21 )2 (6.626 × 10−34 )2 = 7.037 kg.sec−2

k=

15) The wavefunction for the state of lowest energy of 1D harmonic√ oscillator 2 is ψ = Ae−Bx , where A is the normalization constant and B q = 2hkµ . Show

that the total energy E of the lowest state is 21 hω, where ω = µk . Solution: Please see question 1 of year 2017. 16) ‘p-orbitals are not designated as p0 , p+1 , p−1 but as px , py , pz .’ Explain. Solution: The px , py and pz representation is based on the orientation of the lobes of the three p orbitals along the three Cartesian axes, whereas the p0 , p+1 and p−1 representation is based on the value of magnetic quantum number. It has been found that the wavefunction corresponding to p0 is a real function but those corresponding to p+1 and p−1 are imaginary functions. In order to convert the imaginary functions into real functions, linear combinations of p+1 and p−1 are taken, which results in two functions, whose spherical harmonics, when plotted, gives lobes along x and y axes. These two linear combinations are therefore named as px and py respectively. The spherical harmonics of p0 function has lobes along the z axis and hence is called pz orbital. 96

2013: 1) Determine whether each of the following functions is acceptable or not as a state function over the indicated intervals: (i) e−|x| (−∞, ∞), (ii) cos−1 x[−1, 1] Solution: (i) The function e−|x| is not differentiable at x = 0 and hence is not an acceptable as a state function. (ii) The function cos−1 x in the interval of [−1, 1] is continuous and finite. It is also quadratically integrable. Z+1 −1

cos−1 x × cos−1 xdx

1 dx =⇒ dx = −sinzdz. Therefore, Let z = cos−1 x =⇒ dz = − √1−x 2

Z+1 −1

cos−1 x × cos−1 xdx = −

x=+1 Z

z 2 sinzdz

x=−1

 x=+1 Z d z 2 sinzdz dz = − z 2 sinzdz − dz x=−1  x=+1 Z = − −z 2 cosz + 2 zcoszdz 

Z 

Z

x=−1



= − −z 2 cosz + 2z

Z

coszdz − 2

Z 

d z dz

Z



x=+1

coszdz dz x=−1

 x=+1 = − −z 2 cosz + 2zsinz + 2cosz x=−1 h ix=+1 p
2 −1 −1 2 = x cos x − 2 1 − x cos x − 2x

x=−1

2

2

= 0 − 0 − 2 − π + 0 − 2 = −(π + 4) Hence cos−1 x, [−1, 1] is an acceptable state function. 2) Prove that the functions Ψn (x) = (2a)−1/2 eiπnx/a , n = 0, ±1, ±2, ... over the interval −a ≤ x ≤ a are each normalized and mutually orthogonal. Solution: (i) Normalization: Z+a Z+a 1 2iπnx |Ψn (x)|2 dx = e a dx 2a −a −a 1 a h 2iπnx i+a = × e a −a 2a 2iπn
1 e2iπn − e−2iπn = 4inπ 1 = sin2nπ = 0 2nπ

97

problem

(ii) For orthogonality: Z+a Z+a iπmx iπmx 1 e a e a dx Ψm (x)Ψn (x)dx = 2a −a

=

1 2a

−a Z+a

e

iπ(m+n)x a

dx

−a

h iπ(m+n)x i+a a 1 × e a = −a 2a iπ(m + n)   a iπ(m+n) −iπ(m+n) e −e = iπ(m + n) 2a sin{(m + n)π} = π(m + n) = 0 as both m and n are integers. Thus the functions are orthogonal to each other but are not normalized. 3) Benzene may be regarded as a square box of 4˚ A edge length containing 6π electrons. Find the expression for the energy of the ground state of the system. Calculate the minimum energy required to promote one electron to the lowest unoccupied energy level. Solution: The energy of a particle in a 2D square box of length L is given by
n2x + n2y h2 E(nx , ny ) = 8mL2 Since there are six electrons, the first three energy levels will be occupied by them. Therefore, the lowest unoccupied energy level is the fourth energy level. The quantum numbers corresponding to first four energy levels are (nx = 1, ny = 1), (nx = 1, ny = 2), (nx = 2, ny = 1), and (nx = 2, ny = 2). Thus, the minimum energy required to promote one electron to the lowest unoccupied energy level corresponds to the energy different between the levels (nx = 1, ny = 2) and (nx = 2, ny = 2). Therefore,

12 + 22 h2 22 + 22 h2 − E(2, 2) − E(1, 2) = 8mL2 8mL2 3h2 = . 8mL2 Given L = 4˚ A= 4.0 × 10−10 m, m = 9.1 × 10−31 kg. 3(6.626 × 10−34 )2 E(2, 2) − E(1, 2) = J 8 × 9.1 × 10−31 × (4.0 × 10−10 )2 = 11.31 × 10−19 J 4) Prove that a Hermitian operator produces eigenvalues that are real. Solution: Let Aˆ be a hermitian operator and ψ be a normalized eigenfunction of Aˆ with eigenvalue a. Then we’ll have to prove that a is real. From the

98

definition of hermitian operator, we can write, I I  ∗ ∗ ˆ ˆ ψ Aψdτ = ψ Aψ dτ I I ∗ ψ aψdτ = ψ(aψ)∗ dτ I I ∗ ∗ a ψ ψdτ = a ψψ ∗ dτ I ∗ (a − a ) ψ ∗ ψdτ = 0 H Since ψ is normalized, ψ ∗ ψdτ 6= 0. This means a − a∗ = 0 =⇒ a = a∗ , which is possible only when a is real. 5) Explain what you understand by Compton effect. Determine the scattering angle for which the shift in wavelength would be maximum. Solution: Compton effect is the change in wavelength of an x-ray when it is scattered by an electron. The amount by which wavelength changes is called Compton shift. For second part, please see question 2 of year 2015. 6) Evaluate hpx i for a particle confined in a 1D box. Solution: For a particle in a 1D box of length L (along x-axis, between x = 0 q

to x = L), the wavefunction is given by

hpx i =

ZL r 0

=−

2 nπx L sin L

 r h d 2 nπx 2 nπx sin −i sin dx L L 2π dx L L

ih nπ × πL L

inh =− 2 2L

ZL

ZL sin

sin

nπx nπx cos dx L L

0

2nπx dx L

0

ih = (cos2nπ − cos0) 4πL =0

�

��

7) Draw the Ψ and Ψ2 plots as a function of x for the particle in a 1D box, (0 ≤ x ≤ L) in the state n = 2. Determine the slopes of Ψ and Ψ2 at x = 0, L/2. Solution: Slope of ψ and ψ 2 at x = 0, L/2

�

�

99

d ψ dx d ψ dx

√ 2π 2 2 2π 2πx = √ = × cos L L L L L x=0 r √ 2π 2 2πx 2 2π = =− √ × cos L L L L L x=L/2 2 2πx 2πx 2π = × 2 × sin =0 × cos × L L L L x=0 2 2πx 2πx 2π = × 2 × sin =0 × cos × L L L L x=L/2 r

x=0

x=L/2

d 2 ψ dx x=0 d 2 ψ dx

x=L/2

8) The kinetic energy of an electron trapped in a cubical box of edge-length ‘a’ is given as E = 14h2 /8ma2 . How many degenerate states do you expect? Solution: Please see question 10 of year 2015. 9) Consider a particle with quantum number ‘n’ moving in a 1D box of length ’l’. (i) Determine the probability of finding the particle in the region, 0 ≤ x ≤ l/4. (ii) For what value of ‘n’, the above determined probability is a maximum. Solution:(i) Probability of finding the particle in the region, 0 ≤ x ≤ l/4 is given by

P (0 ≤ x ≤ l/4) =

Zl/4r 0

2 nπx sin × l l

r

2 nπx sin dx l l

 Zl/4 2nπx 1 − cos dx l 0  l/4 1 l 2nπx = x− sin l 2nπ l 0   1 l l nπ = − sin −0+0 l 4 2nπ 2 1 1 nπ = − sin 4 2nπ 2 1 = l

(ii) The value of sine function lies between 0 and 1. So, the above probability will be maximum when sin nπ 2 = 0, which is possible for all the even values of n i.e. n = 2, 4, 6, ...
1/4 −αx2 /2 10) Justify that the harmonic oscillator wavefunctions ψ0 (x) = απ e  3 1/4 2 and ψ1 (x) = 4απ xe−αx /2 are orthogonal. Solution: Please see question 16 of year 2016. 11) The radial wavefunction for the 1s orbital of a hydrogen atom is given by R10 (r) =

2 3/2 a0

e−r/a0

Calculate the probability that an electron described by a hydrogen atomic 1s wavefunction will be found within one Bohr radius (a0 ) of the nucleus.

100

Solution: Za0 0

4 |R10 (r)|2 r2 dr = 3 a0

Za0

r2 e−2r/a0 dr

0

 a0 Z Z  Z d 4 2 e−2r/a0 dr − r2 e−2r/a0 dr dr = 3 r a0 dr   a0 0 Z    4 a0 a0 = 3 r2 − e−2r/a0 − 2r − e−2r/a0 dr a0 2 2 0  a0 Z 1 4 = 2 − r2 e−2r/a0 + re−2r/a0 dr a0 2 0   a0 Z Z  Z 4 1 2 −2r/a0 d −2r/a0 −2r/a0 = 2 − r e +r e dr − r e dr dr a0 2 dr 0  a0 2 1 a0 a 4 = 2 − r2 e−2r/a0 − r e−2r/a0 − 0 e−2r/a0 a0 2 2 4 0  2  2 2 2 a0 −2 a0 −2 a0 −2 a0 4 = 2 − e − e − e +0+0+ a0 2 2 4 4 = 1 − 5e−2 = 0.3233

12) What is the zero-point energy of a particle executing simple harmonic oscillations? Solution: Zero-point energy of a particle executing simple harmonic oscillations is 21 hω, where h = Planck’s constant and ω = frequency at which the oscillator oscillates. 13) Calculate the most probable radius (rmp ), at which an electron will be found when it occupies a 1s orbital of a hydrogenic atom of atomic number Z.  3/2 1 Z ψ1s = √ e−zr/a0 π a0 Solution: The wavefunction for an electron in 1s orbital in a hydrogenic atom is given by  3/2 1 Z Ψ1s = √ e−r/a0 π a0 Therefore, the most probable value of r will be obtained by differentiating the radial probability density 4πr2 Ψ∗1s Ψ1s with respect to r and setting it to zero. ( )  3/2  3/2 1 Z 1 Z d 4πr2 × √ 0= e−r/a0 × √ e−r/a0 dr π a0 π a0 4Z 3 d n 2 −2r/a0 o r e 0= 3 a0 dr   8Z 3 r 0= 3 r 1− e−2r/a0 a0 a0 It has three roots. (i) r = 0, this indicates nucleus. The probability of finding electron cannot be maximum in nucleus. So, this is not a valid root. (ii) e−2r/a0 =⇒ r = ∞. At infinite distance too, the probability of finding the electron cannot be maximum. So, this is also not the required point. (iii) (1 − ar0 ) =⇒ r = a0 . This represents a distance equal to Bohr radius 101

and is the required most probable value of r. 14) Calculate the frequency of radiation emitted when a harmonic oscillator of frequency 6.0 × 1013 Sec−1 goes from v = 8 to v = 7 level. Solution: The energy gap between two adjacent states of a harmonic oscillator is given by ∆E = hω, where h = Planck’s constant and ω = frequency of oscillation. If ν be the frequency of the light emitted due to this change of state, then ∆E = hν. Therefore, the frequency of light emitted will be the same as the frequency of oscillation of the oscillator i.e 6.0 × 1013 Sec−1 . 15) The radial part of the 2s and 2p orbitals of hydrogen atom are given (in a.u.) as: R20 = N1 (2 − r)e−r/2 and R21 = N2 re−r/2 , where N1 and N2 are constants. Depict graphically the plots of the radial distribution functions against r and comment. Solution: The required plot is

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(i) The figure clearly shows that 2s orbital has one node at r = 2, whereas 2p orbital has no node and 2s also has larger electron density near the nucleus than 2p orbital. (ii) The electron density in 2p orbital also spans into the region of 2s orbital, which reflects that 2s electron does not screen the 2p orbitals completely. (iii) At large distance, the probability density decreases must faster for the 2p orbital than 2s orbital. 2012: 1) Consider a particle of mass m confined in an infinitely deep potential well, where V = 0 for |x| < L: The wavefunction is of the form Ψ(x) = Asinkx + Bcoskx. Apply boundary condition on the wavefunction to deduce that k = nπ 2L , n = 0, 1, 2, . . . with A = 0 for odd n, and B = 0 for even n. Solution: The given wavefunction is Ψ(x) = Asinkx + Bcoskx The boundary conditions are (i) when x = L, Ψ(x) = 0 and (ii) when x = −L, Ψ(x) = 0. Application of these conditions on Ψ(x) gives: 0 = AsinkL + BcoskL Condition (i) 0 = −AsinkL + BcoskL Condition (ii) From these two equations, we get AsinkL = 0, On subtracting the above two equations BcoskL = 0, On Adding the above two equations It is obvious that both A and B cannot be zero – as it will make the whole wavefunction zero. So, The above two equations says that – (i) If A = 0, B 6= 0 and hence coskL = 0 =⇒ kL = nπ 2 , when n is odd. (ii) If B = 0, A 6== 0 and hence sinkL = 0 =⇒ kL = nπ, when n is any 102

integer. The second part can be modified as kL = nπ 2 , when n is even. 2) Determine, citing reasons, whether each of the following functions is acceptable or not as a wavefunction over the indicated intervals: sin−1 x, [−1, 1] and 1/x[0.∞]. Solution: (i) The function sin−1 x in the interval of [−1, 1] is continuous and finite. It is also quadratically integrable. Z+1 −1

Let z = sin−1 x =⇒ dz = Z+1 −1

sin−1 x × sin−1 xdx =

sin−1 x × sin−1 xdx

√ 1 dx 1−x2

=⇒ dx = coszdz. Therefore,

x=+1 Z

z 2 coszdz

x=−1

 Z  x=+1 Z  Z d = z 2 coszdz − z 2 coszdz dz dz x=−1  x=+1 Z = z 2 sinz − 2 zsinzdz x=−1

  x=+1 Z Z  Z d = z 2 sinz − 2z sinzdz + 2 z sinzzdz dz dz x=−1  2 x=+1 = z sinz + 2zcosz − 2sinz x=−1 h ix=+1 p
2 −1 −1 2 = x sin x + 2 1 − x sin x − 2x x=−1

2

2

=

2

π π π +0−2+ −0−2= −4 4 4 2

Hence sin−1 x, [−1, 1] is an acceptable state function. (ii) 1/x, [0, ∞] becomes ∞ at x = 0. So, it is not an acceptable wavefunction. 3) The de Broglie wavelength of the electron moving in the first orbit of the hydrogen atom (Bohr model) is equal to the circumference of the orbit. Justify. Solution: Please see question 10 of year 2016. 4) The term ‘state’ and ‘energy level’ are not synonymous in quantum me9h2 chanics. For the particle in a cubic box, consider E ≤ 8ma 2 . How many states and levels lie within this range? Solution: For a particle of mass m in a cubic box of length, the expression for energy is given by E=

n2x

+

n2y

+


h2 , 8mL2

n2z +

where nx , ny and nz are the three principal quantum numbers along the three 9h2 Cartesian coordinate axes. For E ≤ 8ma 2 , the following combinations of nx , ny and nz are possible – (1,1,1),(1,2,2),(2,1,2),(2,2,1). So, there are four energy levels and two states. The two states are non-degenerate ground state (1,2,2) and three-fold degenerate first excited state {(1,2,2),(2,1,2),(2,2,1)}. 5) Light of wavelength 552 nm or greater will not eject photoelectrons from a potassium surface. What is the work function (in eV) of potassium?

103

Solution: The required work function (W0 ) is 6.626 × 10−34 J.Sec × 3.0 × 108 m.Sec−1 552 × 10−9 m = 3.601 × 10−19 J

W0 =

6) Evaluate the expectation value of momentum of a particle described by 2 the wavefunction e−ax , where x ranges from −∞ to ∞. Solution:   Z+∞ d h 2 2 e−ax dx hpx i = e−ax −i 2π dx −∞

ih =− 2π =

Z+∞   −ax2 −ax2 e × −2axe dx

−∞ +∞ Z

iah π

2

xe−2ax dx

−∞

= 0 The integrand is an odd function of the variable. d is hermitian. 7) Define a hermitian operator. Confirm that the operator, ~i dx Solution: (i) Hermitian operator: A hermitian operator is one which obeys the turn-over rule. That means, if f and g be two functions and Aˆ be a hermitian operator then I I  ∗ ∗ ˆ ˆ f Agdτ = g Af dτ.

In other words, an operator is said to be hermitian if it is self-adjoint i.e. ˆ where ∗ represents taking complex conjugate. Aˆ∗ = A, (ii) Let f (x) and g(x) be the two functions. Then, For the given operator d Aˆ = ~i dx , we can write, Z+∞ Z+∞ ~ d ~ f ∗ (x) g(x)dx = f ∗ (x)dg(x) i dx i −∞ −∞   +∞ Z Z  ∗ Z ~ ∗ df (x) = f (x) dg(x) − dg(x) dx i dx −∞  +∞ Z ∗ ~ ∗ df (x) = f (x)g(x) − g(x)dx i dx −∞  +∞ Z ~ ∗ = f (x)g(x) − g(x)df ∗ (x) i −∞ The first term will vanish at the two limits because the wavefunction for a confined particle vanishes at very large distance. Thus Z+∞ Z+∞ ~ d ~ g(x)df ∗ (x) f ∗ (x) g(x)dx = − i dx i −∞

−∞

104

On analogy to the first step we take, we can write Z+∞ Z+∞ Z+∞ ~ ~ d ~ d ∗ g(x)dx = − f (x) f ∗ (x) g(x)df ∗ (x) = − g(x) i dx i i dx

−∞

−∞

−∞ Z+∞

∗ ~ d f (x) dx g(x) i dx

=



−∞

Thus,  ∗ Z+∞ Z+∞ ~ d d ~ g(x) g(x)dx = f (x) dx f ∗ (x) i dx i dx −∞

−∞

d Hence, ~i dx is hermitian. 8) Two operators A and B have simultaneous eigenfunction with different eigenvalues. Show that A and B commute. ˆ with Solution: Let, ψ be an eigenfunction of the two operators Aˆ and B eigenvalues a and b respectively and a 6= b. Thus we can write

ˆ = aψ Aψ ˆ = bψ Bψ Therefore, we can write h

i   ˆ ˆ ˆ ˆ ˆ ˆ A, B ψ = AB − B A ψ ˆ −B ˆ Aψ ˆ = AˆBψ ˆ − Baψ ˆ = Abψ

ˆ − aBψ ˆ = bAψ = baψ − abψ =0

Thus, if two operators A and B have simultaneous eigenfunction with different eigenvalues then they will commute with each other. 9) For a particle of mass m in a cubic box of length L with zero potential inside and infinite potential on the walls and outside,show that if the cube is distorted in Z direction only by an amount ∆L and if ∆L is small, then the energy changes approximately by n2z h2 ∆E = − ∆L 4mL3 State explicitly the assumption you make in deriving this result. Solution: The energy of a particle of mass m in a cubic box of length L with zero potential inside and infinite potential on the walls and outside is given by
h2 2 2 2 E (nx , ny , nz ) = nx + ny + nz 8mL2 If the box is distorted in Z direction by an amount of ∆L, then the energy of the elongated box can be written as E∆L (nx , ny , nz ) = n2x + n2y 105


h2 n2z h2 + 8mL2 8m(L + ∆L)2

Here, it is assumed that the change in length is very small, so that we can still consider that the box to be cubic. Therefore, change in energy will be ∆E = E (nx , ny , nz ) − E∆L (nx , ny , nz )
h2
h2 n2z h2 2 2 − n + n + = n2x + n2y + n2z x y 8mL2 8mL2 8m(L + ∆L)2 n2z h2 n2z h2 = − 8mL2 8m(L + ∆L)2   n2z h2 1 1 = − 8m L2 (L + ∆L)2 n2z h2 (L + ∆L)2 − L2 = × 8m L2 (L + ∆L)2 n2 h2 2L∆L + ∆L2 = z × 2 8m L (L + ∆L)2 Since, the change in length is very small, we can neglect ∆L2 and L+∆L ∼ L. Thus n2z h2 2L∆L × ∆E = 8m L4 2 2 nh = z 3 ∆L 4mL 2 2 2

10) For a particle in a 1D box of length L, hpi = 0 and hp2 i = n Lπ 2~ . Find the uncertainty in momentum and estimate the zero point energy of the particle. Solution: The required uncertainty is given by p ∆p = hp2 i − hpi2 r n2 π 2 ~2 = −0 L2 nπ~ = L nh = 2L If this is the minimum uncertainty in momentum, then the minimum energy i.e. zero-point energy (ZPE) of the particle will be EZPE

∆p2 = 2m n2 h2 = 8mL2

2011: 1) What is Compton effect? Use a properly labelled diagram to illustrate your answer. Define Compton wavelength and determine its value. Solution: Compton effect is the change in wavelength of an x-ray when it is scattered by an electron. The amount by which wavelength changes is called Compton shift. The Compton shift is given by ∆λ = λf − λi =

h (1 − cosθ) , me c

The value of Compton shift for a scattering angle of 90◦ i.e the quantity is called Compton wavelength. Putting the value of h, me and c, we get 106

h me c

d re tt e a Sc Incident photon

n to o ph

𝜆$

𝜃 𝜆#

Electron at rest

Recoil electron

−34

J.Sec −11 Compton wavelength = 9.1×106.626×10 m −31 kg×3.0×108 m.Sec−1 = 0.2427 × 10 d 2) Examine which of the following operators are linear: dx , SQR (i.e. squaring) Solution: If f and g are two functions and c1 and c2 are two constants then df dg d d (i) dx {c1 f + c2 g} = c1 dx + c2 dx . Thus, dx is a linear operator. 2 2 2 2 2 (ii) (c1 f + c2 g) = c1 f + c2 g + 2c1 c2 f g 6= c21 f 2 + c22 g 2 . Thus, SQR is not a linear operator.  d d + x, dx −x . 3) Evaluate the commutator dx Solution:         d d d d d d + x, −x ψ = +x −x ψ− −x +x ψ dx dx dx dx dx dx       d dψ d dψ = +x − xψ − −x + xψ dx dx dx dx d dψ dψ d d dψ dψ d = +x − (xψ) − x2 ψ − +x − (xψ) + dx dx dx dx dx dx dx dx dψ d = 2x − 2 (xψ) dx dx dψ dψ = 2x − 2x − 2ψ dx dx = 2ψ   d d =⇒ + x, −x =2 dx dx

4) Determine whether the following functions are acceptable over the indicated intervals: ex (−∞, 0), tanθ(0, π/2). Solution: (i) ex is a continuous and finite function in the given interval

107

(∞, 0). It is also quadratically integrable, as shown below. Z0 −∞

ex × ex dx =

Z0

e2x dx

−∞

1  2x 0 e −∞ 2
1 0 e − e−∞ = 2 1 = (1 − 0) 2 1 = 2 =

Thus, ex , (−∞, 0) is an acceptable function over the indicated interval. (ii) The function tanθ in the interval (0, π/2) becomes infinite at θ = π2 . So, tanθ in the interval (0, π/2) is not an acceptable function. 5) The speed of a certain proton is 4.5 × 105 m/sec. If the uncertainty in its momentum is to be reduced to 0.01 percent, what uncertainty in its location must be tolerated? Given, mass of a proton = 1.673 × 10−27 kg Solution: Let the initial uncertainty in position and momentum of the proton be ∆x and ∆px respectively. According to Heisenberg uncertainty principle, h . we can write: ∆px ∆x ≥ 4π If we reduce the uncertainty in momentum then corresponding uncertainty in position can be increased such that Heisenberg uncertainty principle holds. Let us consider that if the uncertainty in momentum be reduced by 0.01% then the corresponding increase in uncertainty be w%. Therefore, new unx certainty in momentum = ∆px − 0.01∆p = 0.9999∆px and new uncertainty 100 (w+100)∆x . Again, applying Heisenberg uncertainty in position = ∆x + w∆x 100 = 100 principle, we can write (w + 100)∆x h ≥ 100 4π 0.9999(w + 100) h h ≥ 100 4π 4π 0.9999(w + 100) ≥1 100 100 w≥ − 100 0.9999 w ≥ 0.01

0.9999∆px ×

Thus, if the uncertainty in momentum be reduced by 0.01% then the corresponding uncertainty in position can be increased to 0.01%. 6) A certain system is described by a Hamiltonian operator d2 ˆ H = 2 + x2 dx 2 ˆ determine the eigen(i) Show that Ψ = Axe−x /2 is an eigenfunction of H; value. (ii) Determine A, so that the function is normalized.

Z∞

x2 e−ax

2

0

108

1 dx = 4

r

π a3

Solution: (i)  2 d 2 −x2 /2 ˆ = HΨ + x Axe dx2  d  2 −x2 /2 2 −x2 /2 −x e +e + Ax3 e−x /2 =A dx
2 2 2 = A 1 − x2 e−x /2 (−x) + Ae−x /2 (0 − 2x) + Ax3 e−x /2 

= −Axe−x

2

/2

+ Ax3 e−x

2

2

= 2Ax3 e−x /2 − 3Axe−x = (2x2 − 3)Ψ

/2

2

/2

− 2Axe−x

2

/2

+ Ax3 e−x

2

/2

ˆ and the eigenvalue is (2x2 − 3). Thus Ψ is an eigenfunction of H (ii) Z∞

Ψ2 dx = 1

0

Z∞



Axe

−x2 /2

2

dx = 1

0

A2

Z∞

2

x2 e−x dx = 1

0

A2

1√ π=1 4 A=±

2 π 1/4

7) Draw the Ψ and Ψ2 plots as a function of x for the article in a 1D box (0 ≤ x ≤ L): in the state n = 2. Determine the slopes of Ψ and Ψ2 at x = 0, L/2. Prove that the maximum and minimum for Ψ occur at L/4 and 3L/4 respectively. Solution: Part (i) and (ii): Please see question 7 of year 2013. q Part (iii): Ψ for n = 2 state for a particle in a 1D box of length L is, At extreme points of Ψ, r d 2 2πx sin =0 dx L L r 2 2π 2πx × cos =0 L L L 2πx cos =0 L 2πx (2m + 1)π = , m = 0, 1, 2, ... L 2 L x = (2m + 1) 4

2 2πx L sin L .

Since x > L is not allowed, only two values of m i.e. 0 and 1 are valid. Since, the second derivative of Ψ will always be negative (derivative of cosine function is negative of sine function), maxima will occur at L/4 and at 3L/4. 8) An electron in a 3D rectangular box with dimensions 5˚ A, 3˚ A and 6˚ A makes a radiative transition from the lowest excited state to the ground state. 109

Calculate the frequency of the photon emitted. Solution: Ground state of a particle i a 3D box corresponds to (1,1,1) state and the lowest excited state corresponds to (1,1,2) state. Further, the energy of the particle in (nx , ny , nz ) state is given by ! 2 2 2 n nx nz h2 y E(nx , ny , nz ) = + + L2x L2y L2z 8m ˚, Ly = 3˚ Given Lx = 5A A, Lz = 6˚ A0.0455. Therefore, the required frequency of the photon emitted is given by E(1, 1, 2) − E(1, 1, 1)  h    h 1 4 1 1 1 1 = + + − + + 8m L2x L2y L2z L2x L2y L2z 3 × 6.626 × 10−34 J.Sec 3h = = 8mL2z 8 × 9.1 × 10−31 kg × (6 × 10−10 m)2 = 7.58 × 1014 Sec−1

ν=

9) Find the most likely location of a particle in a 1D box of length ‘L’ in the ground state. Solution: The most likely position of a particle in a 1D box can be obtained by setting the first derivative of probability density i.e. |Ψ(x)|2 to zero. Thus, for ground state   d 2 2 πx sin =0 dx L L 2 πx πx π × 2sin × cos × =0 L L L L 2πx sin =0 L 2πx =⇒ = mπ, where m = 0, ±1, ±2, ±3, ... L mL x= 2 m = 0 gives x = 0 i.e one of the wall of the box and since, at the wall, potential is infinite, the particle cannot exist there. Furthermore, since x cannot be larger than or equal to the length of the box, only m = +1 is the allowed value for m. Thus, the most likely location of a particle in a 1D box is x = L/2 i.e at the center of the box. 10) For a H-atom identify the following orbitals with given radial parts as (i) N2 β2 re−α2 r and (ii) N3 re−α3 r , where Ni , βi , αi are constants. Explain your answer using the plots of the orbitals. Solution:

weird

2010: 1) Examine whether the following are acceptable wavefuncitons: (i) sin−1 x [−1 ≤ x ≤ 1 (ii) 1/x[0 < x < ∞]. Solution: Please see question 2 of year 2012 2) The de Broglie wavelength of the electron moving in the first orbit of the hydrogen atom (Bohr orbit) is equal to the circumference of the orbit. Justify. Solution: Please see question 3 of year 2012 3) Zero point energy of a harmonic oscillator is a consequence of uncertainty 110

principle. Explain. Solution: Please see question 6 of year 2017 4) Verify whether the squaring operator, ( )2 is linear or not. Solution: Please see question 2 of year 2011 5) Find the value of the commutator [ˆ x, pˆx ]. Solution: [ˆ x, pˆx ] ψ(x) = xˆpˆx ψ − pˆx xˆ d dψ + i~ (xψ) = −i~x dx dx dψ dψ = −i~x + i~x + i~ψ dx dx = i~ψ =⇒ [ˆ x, pˆx ] = i~ 6) Find the radial distribution function for the following hydrogenic orbital, Ψ = Arcosθe−r/2a0 , where A and a0 are constants. Solution: Please see question 11 of year 2014 ˆ then 7) “If φ1 and φ2 are degenerate eigenfunctions of a linear operator A, a linear combination of the eigenfunctions is also an eigenfunction of the operator with the same eigenvalue.” Prove the statement. Solution: Please see question 2 of year 2016 8) For a particle in a 1D box, show that Ψm and Ψm−1 are orthogonal to each other. Solution: ZL

ZL r Ψm Ψm−1 dx =

0

2 mπx sin L L

r

2 (m − 1)πx sin dx L L

0

1 = L

ZL 2sin

mπx (m − 1)πx sin dx L L

0

ZL 

 πx (2m − 1)πx cos − cos dx L L 0     1 L L L L = sinπ − sin0 − sin(2m − 1)π + sin0 L π π (2m − 1)π (2m + 1) =0 1 = L

Thus, the two wavefunctions in questions are orthogonal to each other. 9) For a particle of mass ‘m’, confined in a 1D box of length ‘L’ find the value of ∆px , where ∆p2x = hp2x i − hpx i2 and hpx i is zero. Hence determine the minimum possible value of ∆x for the third level.

111

Solution: For a particle in a 1D box r ZL r 2 2 nπx nπx hp2x i = sin pˆ2x sin dx L L L L 0

2 = L

ZL 0

 2 d nπx nπx −i~ dx sin sin L dx L

2~2 =− L

ZL

 2 2 nπx nπ nπx sin × − 2 × sin dx L L L

0

n2 ~2 π 2 = L3

ZL

2sin2

nπx dx L

0

 ZL  2nπx 1 − cos dx L 0  L 2 2 2 n~π L 2nπx n2 ~2 π 2 n2 h2 = x− sin = = L3 2nπ L 0 L2 4L2 n2 ~2 π 2 = L3

Therefore, p hp2 i − hpx i2 r x n2 h2 −0 = 4L2 nh = 2L According to Heisenberg uncertainty principle, we can write ∆px =

h 4π nh h ∆x ≥ 2L 4π L = 2nπ

∆px ∆x ≥

∆xmin

L Thus, for n = 3, ∆xmin = 6π . 10) What is the frequency of the light absorbed when a CO molecule (assumed to be a rigid rotor) makes a transition from the rotational ground state to the first excited state. Given, bond length of CO = 112.8 pm and 1 a.m.u. = 1.66 × 10−27 kg. Solution: The energy of J th level of a rigid rotor is given bye

EJ =

h2 J(J + 1), 8π 2 µr2

where µ = reduced mass, r = length and other terms have their usual significance. For CO, mC mO µ= mC + mO 12 × 1.66 × 10−27 × 16 × 1.66 × 10−27 = (12 + 16) × 1.66 × 10−27 = 6.857 × 1.66 × 10−27 kg 112

Given r = 112.8 × 10−12 m. Therefore, the frequency of light absorbed when a CO molecule makes a transition from the rotational ground state (J = 0) to the first excited state (J = 1) is E1 − E0 h h = 2 2 ×2 8π µrCO

ν=

6.626 × 10−34 = Sec−1 −27 −12 −12 16.0 × 3.14 × 3.14 × 6.857 × 1.66 × 10 × 112.8 × 10 × 112.8 × 10 = 2.9 × 1010 Sec−1 11) What is an orbital? what does the following diagram represent for a pz orbital? Solution: An orbital is a three-dimensional space around the

+ +

- -

nucleus, where the probability of finding an electron is maximum. The given diagram represents the spherical harmonics for the pz orbital. The signs ‘+’ and ‘-’ represents the two phases of the wavefunction of pz orbitals and hence it shows that the two lobes of pz orbital have different phases. 2009: 1) State Heisenberg’s uncertainty principle for position and momentum. The speed of an electron in x-direction is measured as vx = 300m/sec with an uncertainty of 0.01%. What would be uncertainties in position ∆x and ∆y, when position is measured simultaneously with speed? Comment on your result. Given that me = 9.1 × 10−31 kg. x- and y-directions are mutually perpendicular. Solution: (i) Heisenberg uncertainty principle It states that “It is impossible to determine both the exact position and exact momentum of a quantum system, at the same time, not even theoretically.” (ii) The relation between the uncertainty in linear momentum and position of a particle along x-axis is given by ∆px ∆x ≥

h 4π

Given, 0.01 × 300 100 = 0.01m.sec−1 =⇒ ∆px = m∆vx = 9.1 × 10−31 kg × 0.01m.sec−1 ∆vx =

= 9.1 × 10−33 kg.m.sec−1

113

Thus, ∆x ≥

h 4π∆px

6.626 × 10−34 J.sec ≥ 4 × 3.14 × 9.1 × 10−33 kg.m.sec−1 ≥ 0.0579 m Since the two operators yˆ and pˆx commute with each other because the two directions x and y are independent of each other (orthogonal), there exist no uncertainty relation between position along y axis and momentum along x axis. Thus, whatever may be the uncertainty in px , the measurement of particles position along y axis will not be affected and hence the corresponding uncertainty in y will be zero.   2) Evaluate the commutator xˆ2 , pˆx Solution:  2  xˆ , pˆx ψ = xˆ2 pˆx ψ − pˆx xˆ2 ψ d dψ + i~ (x2 ψ) = −i~x2 dx dx dψ dψ = −i~x2 + i~x2 + i~ψ × 2x dx dx = 2ix~ψ  2  xˆ , pˆx = 2ix~ 3) (i) What is Compton effect? (ii) Determine the scattering angle, for which the shift in wavelength would be maximum. (iii) Find the maximum kinetic energy of recoil electrons when x-rays of wavelength 12.0pm are scattered from a target. Solution: For part (i) and (ii), please see question 2 of year 2015. (iii) The kinetic energy of recoil electron in Compton scattering is given by   1 1 K.E = hc − , λ λ0 where λ and λ0 are the wavelength of the incident and scattered photons respectively. We know, λ0 − λ =

h (1 − cosθ) , m0 c

where θ is the angle of scattering, m0 is the rest mass of the electron. So, the kinetic energy of recoil electron will be maximum when cosθ = −1 =⇒ λ0 − λ = m2h0 c . Thus the maximum kinetic energy of the recoil electron is hc(λ0 − λ) K.E = λ0 λ hc × m2h0 c  = 2h m0 c + λ λ = =

2h c

2h2
+ m0 λ λ

2×6.626×10−34 3×108

2 × 6.626 × 10−34


2


+ 9.1 × 10−31 × 12 × 10−12 × 12 × 10−12

87.8077 × 10−68 = = 4.771 × 10−15 J 184.044 × 10−54 114

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4) Explain why p-orbitals are often represented as px , py and pz , instead of p0 , p+1 and p−1 . Solution: Please see question 16 of year 2014. 2 ˆ = − d22 + x2 . Give 5) Determine the eigenvalue when Ψ(x) = Axe−x /2 and H dx the schematic plot of Ψ(x) and |Ψ(x)|2 versus displacement in x-direction. Solution: For part (i) please see question 6(i) of year 2011. (ii)

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6) Calculate the probability of finding a free particle confined in a 1D box of length L within the region from L/4 to 3L/4. Solution: The required probability is 3L/4 Z

r

P =

2 nπx sin L L

!2 dx

L/4

1 = L

3L/4 Z 

 2nπx 1 − cos dx L

L/4

 3L/4 1 L 2nπx = x− sin L 2nπ L L/4 3 1 3nπ 1 1 nπ = − sin − + sin 4 2nπ  2 4 2nπ 2 1 1 nπ 3nπ = + sin − sin 2 2nπ 2 2 7) What is the concept associated with de Broglie hypothesis? Starting from this hypothesis, arrive at the energy expression for a free particle confined in a 1D box. Solution: The concept associated with the de Broglie hypothesis is waveparticle duality. That means all the materials also have wave-like properties. The wavelength of the wave associated with a particle of mass having momentum p is given by h λ= p For a particle confined in a 1D box of length L, the length of the box should be an integral multiple of the wavelength of the associated wave. Otherwise, the particle will be annihilated due to the destructive interference of the associated waves. Thus, L = nλ h L = n Using the above relation. p If there is a fixed potential V inside the box and E be the energy of the 115

particle then, p =

p 2m(E − V ). Thus nh L=p 2m(E − V ) n2 h2 2 L = 2m(E − V ) n2 h2 E= +V 2mL2

8) Hydrogen atoms 1s wavefunction is given by, 1 Ψ=√ π



1 a0

1/2

e−r/a0 .

Determine the most probable value of r in this state. Solution: Please see question 14 of year 2016. 2008: 1) Determine stating reasons whether each of the following functions is acceptable or not as a state function over the indicated intervals: (i) e−|x| (−∞ ≤ x ≤ ∞), (ii) sinx(0 ≤ x ≤ ∞). Solution: (i) Please see question 1 of year 2013. (ii) sinx is periodic function. It’s values lies between -1 and +1. In the range (0 ≤ x ≤ ∞) it is finite, continuous, single-valued and square integrable. So, it is an acceptable state function over the indicated interval. 2) An electron is confined in a 1D box of length L. What should be the length of the box to make its zero-point energy equal to its rest mass energy (me c2 )? Explain the result in terms of the Compton wavelength. Solution: The energy of an electron in a 1D box of length l is given by n2 h2 E= 8me L2 If this energy equals the rest mass energy of the electron then, n2 h2 me c = 8me L2 h n √ L= me c 2 2 2

Thus, if the length of the box is √n2 times the Compton wavelength of an electron then the energy will hbe equal mass of the electron. i to the2 rest 2 ˆ . H ˆ = ~ d 2 + V (x) 3) Evaluate the commutator xˆ, H 2m dx Solution: h i ˆ ˆ −H ˆ xˆψ xˆ, H ψ = xˆHψ  2 2   2 2  ~ d ~ d =x + V (x) ψ − + V (x) xψ 2m dx2 2m dx2 ~2 x d2 ψ ~2 d2     = + xV (xψ) −  xV (x)ψ  (x)ψ −  2 2 2m dx 2m dx 2 2 2 2 ~ x d ψ ~ x d ψ ~2 ψ d2 x = − − 2m dx2 2m dx2 2m dx2 =0

116

αx2

4) Show that Ψ = ( απ )1/4 e− 2 is an√energy eigenfunction of the simple harmonic oscillator in 1D. Given, α = ~kµ . Terms have their usual significance. Solution: The hamiltonian of the simple harmonic oscillator in 1D is s 2 2 ˆ = − ~ d + 1 µω 2 x2 , where ω = k and k is the force constant H 2µ dx2 2 µ Thus,

  2 2 ~ d 1 α 1/4 − αx2 2 2 ˆ = − HΨ + µω x e 2 2µ dx2 2 π    2αx − αx2 1 2 2  α 1/4 − αx2 ~2  α 1/4 d 2 − e + µω x × =− × × e 2 2µ π dx 2 2 π     αx2 ~2  α 1/4 2αx − αx2 1 2 2  α 1/4 − αx − =− × e 2 + e 2 + µω x × × (−α) × x − e 2 2µ π 2 2 π  2 2 2  ~2 α 1 2 2  α 1/4 − αx2 ~αx + + µω x = − e 2 2µ 2µ 2 π 

Putting the expressions for α and ω, s ( )  α 1/4 αx2 2 kx ~ k 1 2 ˆ HΨ = − + + kx e− 2 2 2 µ 2 π s 1 k = ~ ψ 2 µ ˆ for a simple harmonic oscillator and the Thus, Ψ is an eigenfunction of H q eigenvalue is 21 ~ µk . 5) The lifetime of an excited state is 109 sec. Calculate the uncertainty in the energy of this state. Solution: According to Heisenberg uncertainty principle, the uncertainty in energy (∆E) of a state and its lifetime (τ ) is given by h 4π 6.626 × 10−34 ∆E ≥ J 4 × 3.14 × 109 ∆E ≥ 0.5275 × 10−25 J

∆Eτ ≥

6) Apply de Broglie’s hypothesis to arrive at the Bohr’s quantization rule. Solution: Please see question 3 of year 2012. 7) Benzene may be regarded as a 4.0˚ Asquare box containing 6 pi electrons. Find the expression for energy of the ground state of the system. Calculate the minimum energy required to promote one electron to the lowest unoccupied energy level. Solution: Please see question 3 of year h 2013. i ˆ ˆ ˆ ˆ 8) Two operators A, B are related as A, B = 0. How accurately can one ˆ for a system in eigenstate measure the physical property corresponding to A, ˆ Justify your answer. Assume the eigenstate as non-degenerate. of B? h i ˆ with eigenvalue b. Since A, ˆ B ˆ = Solution: Let ΨB be an eigenfunction of B 117

ˆ −B ˆ Aˆ = 0 =⇒ AˆB ˆ=B ˆ A. ˆ Thus, if the system is in ΨB , 0, this means AˆB we can write ˆ AΨ ˆ B = AˆBΨ ˆ B B ˆ B = AbΨ ˆ B = b × AΨ ˆ B is an eigenfunction of B, which is possible only when This indicates AΨ ˆ ˆ AΨ h B= i constant × ΨB , which means ΨB is also an eigenfunction of A. So, if ˆ B ˆ = 0, then one can measure the physical property corresponding to Aˆ A, ˆ with 100% accuracy for a system in eigenstate of B. 9) A normalized wavefunction F (φ) = Aeimφ is defined in the range 0 ≤ φ ≤ 2π. Find A. Using the conditions for acceptability of a wavefunction, obtain the allowed values that the quantum number ‘m’ may take up. Solution: Applying normalization condition, Z2π

F ∗ (φ)F (φ)dφ = 1

0

Z2π

A∗ e−imφ Aeimφ dφ = 1

0

|A|2

Z2π dφ = 1 0

|A|2 2φ = 1

1 |A| = √ 2π

Since, this is an azimuthal equation after each 2π rotation the function should return to its initial value. Thus, the condition of acceptability for the given wavefunction is F (φ) = F (φ + 2π). This is possible only when m = 0, ±1, ±2, · · · . 10) Define radial distribution function and state its significance. Plot radial distribution function corresponding to the following wavefunction against r/a0 : Ψ = A(2a0 − r)e−r/2a0 . Terms have their usual significance. Solution: Please see question 16 of year 2015. 2007: 1) (i) Show that hpi is zero for all the stationary states of a particle in a 1D box of length L. (ii) Does it mean that the particle is at rest? Explain. Solution: For part (i) Please see question 6 of year 2013. (ii) The zero expectation value of linear momentum, hpi means the average value of linear momentum of the particle is zero. It does not mean that the particle is at rest. Particle in a box has equal probability of moving towards the left or right sided wall and hence has equal values of left and right momentum. So, on average, the positive and negative (left and right values) momentum cancel out each other and the average becomes zero. 2) Determine whether each of the following functions is normalizable or not over the indicated intervals: (i) ex (0, ∞), (ii) sinx(0, 2π). Normalize the function that can be normalized. 118

Solution: (i)

Z∞ 0

 ∞ ex × ex dx = 2 e2x 0 = ∞

So, ex is not normalizable in the interval (0, ∞) (ii) Z2π 0

1 sinx × sinxdx = 2

Z2π 0

(1 − cos2x) dx

2π 1 1 x − sin2x = 2 2 0 1 = [2π − 0 − 0 + 0] 2 1 = 2 

So, the function sinx is normalizable in the interval (0, 2π). 3) A photon of momentum 5.33 × 10−22 g.cm.sec−1 is required to excite the electron in a H-atom from the ground state to the first excited state. Use this information to find the ionization energy of the H-atom. h Solution: Since E = hc λ and from de Broglie relation we know p = λ . This means E = pc. Thus energy of the photon is 5.33 × 10−22 × 3 × 1010 erg. This is the energy gap between the ground and first excited states. The energy required to excite an electron from mth state to nth state, in H-atom, is given by   1 1 me4 Emn = 2 2 − 80 h n2 m2 For ground state (m = 1) to first excited state (n = 2) transition, in H-atom, is given by 3 me4 E12 = − × 2 2 4 80 h Now, ionization energy (IE) will be a transition from ground state (m = 1) to n = ∞ state. Thus me4 (0 − 1) 820 h2 4 IE = E12 , Using the above relation. 3 4 IE = × 5.33 × 10−22 × 3 × 1010 erg = 21.32 × 10−12 erg. 3

E1∞ =

4) For the ground state of the H-atom, find the most probable distance of the electron from the nucleus, give Ψ1s =

1 3/2 π 1/2 a0

e−r/a0

Solution: Please see question 14 of year 2014. 5) Express mathematically the Planck’s distribution law for black-body radiation and show that under a particular condition it reduces to the RayleighJeans law. 119

Solution: Planck’s distribution law for black-body radiation is given by, 1 2hc2 , Bλ = 5 hc λ e λkT − 1 hc

where terms have their usual meanings. In longer wavelength region, e λkT ∼ hc . Thus, 1 + λkT 2hc2 1 Bλ = 5 hc λ 1 + λkT −1 2hc2 λkT = 5 × λ hc 2ckT , which is Rayleigh-Jeans law. = λ4 6) Show that, in a rectangular box with infinitely high potential walls and of sides a = L and b = 2L, there is a degeneracy between the states (1,4) and (2,2). Write down the wavefunction for the state (1,0) and comment on the result. Solution: Please see question 8 of year 2016. 7) How can you use kinetic energy of photo-electron vs frequency plot in photoelectric experiments to distinguish between two metals. Solution: The relation between kinetic energy (T ) of photo-electron and the frequency (ν) of light used is T = hν − W0 , where h = Planck’s constant and W0 = work function of the metal. So a plot of T vs nu will be a straight line having slope h (which is a constant) and an y-intercept of −W0 . Since work function W0 is a characteristics of a metal, two different metals will have different values of W0 i.e. different values of y-intercept. Thus, for two metals the plot will look like

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8) The work function for metallic Cesium is 2.14 eV. Calculate the kinetic energy of the electrons ejected by light of wavelength 300 nm. Solution:Einstein’s equation of photoelectric effect is given by: T = hc λ −W, where T = kinetic energy of the electron emitted, λ = wavelength of light used, W = work function of the metal, h and c are the Planck’s constant and speed of light in vacuum respectively. Given, λ = 300 × 10−9 m, W = 2.14eV = 2.14 × 1.602 × 10−19 J. Thus we can write 6.626 × 10−34 J.Sec × 3 × 108 m.sec−1 − 2.14 × 1.602 × 10−19 J −9 300 × 10 m −19 = 6.626 × 10 − 3.428 × 10−19 J = 3.198 × 10−19 J

T =

120

h i   ˆ ˆ ˆ 9) Given [ˆ x, pˆx ] = i~, evaluate A and xˆ, A , where xˆ, pˆ2x = A. Solution: Given, [ˆ x, pˆx ] = i~ =⇒ xˆpˆx − pˆx xˆ = i~   A = xˆ, pˆ2x = xˆpˆ2x − pˆ2x xˆ = xˆpˆx pˆx − pˆx pˆx xˆ = xˆpˆx pˆx − pˆx (ˆ xpˆx − i~) = xˆpˆx pˆx − pˆx xˆpˆx + i~ˆ px = (ˆ xpˆx − pˆx xˆ) pˆx + i~ˆ px = i~ˆ px + i~ˆ px = 2i~ˆ px Thus, h i ˆx xˆ, Aˆ = xˆAˆ − Aˆ = 2i~ˆ xpˆx − 2i~ˆ px xˆ = 2i~ (ˆ xpˆx − pˆx xˆ) = 2i~ × i~ = −2~2 10) Explain whether “squaring” is a linear operation. Solution: Please see question 2 of year 2011. 2006: 1) If Ψ(x) = N sin nπx L for 0 ≤ x ≤ L, find N, where N is the normalization constant. Solution: ZL 

nπx 2 N sin dx = 1 L

0

 ZL  2nπx 1 − cos dx = 1 L 0  L 2 L 2nπx N x− sin =1 2 2nπ L 0 N2 (L − 0 − 0 + 0) = 1 2 r

N2 2

N=

2 L

2) State, with reasons, whether the functions Ψ(φ) = eiφ/2 and Ψ(x) = sin−1 x are acceptable wavefunctions. Solution: (i) Within the interval (0, 2π), Ψ(φ) is a finite, continuous and single valued. It is also square integrable. Z2π

Ψ∗ (φ)Ψ(φ)dφ =

0

Z2π

e−iφ/2 eiφ/2 dφ

0 Z2π

=

dφ = 2π 0

121

Hence, Ψ(φ) is an acceptable wavefunction. (ii) Please see question 2 of year 2012. 3) A micro-particle can be found anywhere with equal probability within a length L. Find roughly its uncertainty in momentum. If the latter is the minimum possible value of its momentum, find also the minimum kinetic energy. Solution: Since the probability of finding the particle at any point within the length L is equal, the uncertainty in position will be ∆x = L. Therefore, h . the uncertainty in momentum will be ∆p = 2πL Now, if the above uncertainty in momentum be minimum, then the minimum kinetic energy will be h2 (∆p)2 = 2 K.E.(min) = 2m 8π mL2 4) The peak in the sun’s emitted energy occurs at about 480 nm. Estimate the temperature of its surface assuming as a black-body like emitter. The second radiation constant, C2 = hc k = 1.439cm.K) Solution: 2 5) A state Ψ = N e−ax satisfies an eigenvalue equation HΨ = EΨ with d2 2 H = − dx 2 + x . Find possible values of a and eigenvalue E. 2 Solution: Since Ψ = N e−ax satisfies the eigenvalue equation HΨ = EΨ. This means Ψ is normalized. Thus Z+∞ 2 N 2 e−2ax dx = 1 −∞

N2

r

π =1 2a  1/4 2a N= π

Thus, a = π2 N 4 . Now, putting the expression of Ψ in the eigenvalue equation. ˆ EΨ = HΨ    1/4 2a d2 2 2 = − 2 +x e−ax dx π  1/4 o  2a 1/4 2a d n 2 2 =− −2axe−ax + x2 e−ax π dx π  1/4   2a 2 −ax2 2 −ax2 a e =2 − 2ax e + N x2 e−ax π  1/4
2a 2 e−ax = 2a − 4a2 x2 + x2 π The eigenvalue should not depend upon the position. Therefore, −4a2 + 1 = 0 =⇒ a = ± 12 and hence E = ±1. 6) How is the de Broglie hypothesis used to obtain Bohr’s quantization rule for H-atom? Solution: Please see h i question 10 of year 2016. 2 d 7) Evaluate x2 , dx without using [x, px ] = i~. 2 122

Solution:   2 2
d d2 2 2d ψ 2 x , 2 ψ=x − x ψ dx dx2 dx2 d2 ψ d2 ψ = x2 2 − x2 2 − 2ψ dx dx = −2ψ   d2 =⇒ x2 , 2 = −2 dx 8) Explain whether the functions Ψ, −Ψ and 2iΨ represent the same state, given that Ψ is real. Solution: The expectation value of the property (P) of a system is given by, H ∗ Ψ Pˆ Ψdτ hP i = H ∗ Ψ Ψdτ If the wavefunction Ψ is multiplied by a constant k (real or imaginary) then the property will not change. Thus H H H ∗ k ∗ k Ψ∗ Pˆ Ψdτ (kΨ)∗ Pˆ kΨdτ Ψ Pˆ Ψdτ H H H = ∗ = hP i = (kΨ)∗ kΨdτ Ψ∗ Ψdτ k k Ψ∗ Ψdτ If k = −1, the wavefunction is −Ψ and if k = 2i, the wavefunction is 2iΨ. Thus, all the three wavefunctions Ψ, −Ψ and 2iΨ represent the same state. 9) Calculate hx2 i for a particle in a 1D box with 0 ≤ x ≤ L and comment on its value as n → ∞. Solution: hx2 i =

ZL r 0

2 nπx sin × x2 × L L

r

2 nπx sin dx L L

 ZL  2nπx 1 − cos x2 dx L 0  L  L 1 2xL2 2nπx x2 L 2nπx L3 2nπx 1 x3 − sin − cos + 3 3 cos = L 3 0 L 4n2 π 2 L 2nπ L 4n π L 0 L2 L2 L2 L2 = −0+ cos2nπ − 3 3 cos2nπ + 0 − 0 + 3 3 3 2nπ  4n π  4n π 2 2 2 2 L L L L = + 3 3+ − 3 3 cos2nπ 3 4n π 2nπ 4n π 1 = L

When n → ∞, all the terms except the first one will → 0. Thus as n → ∞, 2 hx2 i → L3 . 10) Identify the following hydrogenic orbitals with given radial parts as (i) N1 (β2 − r)e−α2 r and (ii) N2 e−α3 r , where Ni , αi and βi are constants. Explain your answer using plots of the orbitals. Solution: Radial distribution plots of the two given wavefunctions are shown below. Since the wavefunction (i) N1 (β2 − r)e−α2 r has one node (except at r = 0 and r = ∞, it represents a 2s orbital. Similarly, since the wavefunction (ii) N2 e−α3 r has no node (except at r = 0 and r = ∞, it represents a 1s orbital.

123

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2005: 1) Indicate and explain the salient features of the photoelectric effect. Solution: Please see relevant section of the text (Section: ??). 2) Evaluate [ˆ xn , pˆx ] Solution: Please see question 4 of year 2015. 3) Show that Ψi (x) and Ψj (x) representing the wavefunctions corresponding to two different sates of a particle confined in a 1D box are orthogonal. Solution: ZL

ZL r Ψi Ψj dx =

0

2 iπx sin L L

r

2 jπx sin dx L L

0

1 = L

ZL 2sin

iπx jπx sin dx L L

0

 ZL  (i − j)πx (i + j)πx cos − cos dx L L 0    1 L L L L = sinπ − sin0 − sin(i + j)π + sin0 L (i − j)π (i − j)π (i + j)π (i + j) = 0, as both i and j are integers.

1 = L

Thus, the two states are orthogonal to each other. 4) Write down time independent Schr¨odinger equation of a 1D harmonic oscillator. Comment on its zero-point energy. Solution: Time independent Schr¨odinger equation of a 1D harmonic oscillator is   1 2 1 ~2 d2 − + kx Ψ(x) = (v + )~ωΨ(x), 2m dx2 2 2 where k =force constant, v = vibrational quantum number and ω is angular frequency of oscillation. The zero-point energy of the oscillator corresponds to the value of energy for the quantum number v = 0. Thus, zero-point energy is 12 ~ω and non-zero. 5) From the probability density of 1s electron, show that the most probable distance of the electron from the nucleus is equal to Bohr radius. Given Ψ1s =

1 1/2 (πa30 )

e−r/a0

Solution: Please see question 14 of year 2016. 6) Find the de Broglie wavelength of electrons that have been accelerated from 124

rest through a potential difference of 1.00 KV. h = 6.626 × 10−34 J.Sec; me = 9.11 × 10−31 kg. Solution: de Broglie relation is λ=

h , p

where p = momentum of the electron. If the electron is accelerated from rest through a potential difference of 1.00kV, then the enegy of the electron will be = 1.602 × 10−19 × 1000 J. √ 2mE The momentum of the electron will be p = √ = 2 × 9.11 × 10−31 × 1.602 × 10−19 × 1000 = 1.708 × 10−23 kg.m.sec−1 Therefore, the de Broglie wavelength of the electron will be 6.626×10−34 −11 m λ = 1.708×10 −23 = 3.879 × 10 7) What is meant by degeneracy? Find the quantum numbers associated with the energy level 17h2 /8ma2 of a particle in a cubical box and hence indicate the degree of degeneracy. Solution: When two different states corresponding to different values of quantum numbers, possess the same energy, the two states are called degenerate and this phenomenon is called degeneracy. The quantum numbers associated with the energy level 17h2 /8ma2 of a particle in a cubical box are : (nx = 2, ny = 2, nz = 3), (nx = 2, ny = 3, nz = 2), (nx = 3, ny = 2, nz = 2). So, degree of degeneracy is 3. 8) Solve the Schr¨odinger equation for a particle in a 1D box and find the expressions for the wavefunction and energy. Solution: Please see the relevant text (section: ??). 9) Calculate the frequency of radiation emitted when a linear harmonic oscillator of frequency 6.0 × 1013 sec−1 goes from n = 2 to n = 1 state. Solution: The frequency of radiation emitted by a linear harmonic oscillator (of frequency ω) when it goes from state v 0 to v is given by   1 1 1 0 (v + )hω − (v + )hω = (v 0 − v)ω ν= h 2 2 In our case, v 0 = 2 and v = 1. Thus ν = (v 0 − v)ω = ω = 6.0 × 1013 sec−1 10) Calculate the probability that an electron in the ground state of a H-atom will be found within one Bohr radius of the nucleus. Given Ψ1s =

1 1/2 (πa30 )

e−r/a0

Solution: Please see question 3 of year 2017. 2004: 1) Sketch the energy distribution curves of the radiation emitted from a black body at two different temperatures and point out two characteristic features of the energy distribution. Solution: Please see relevant section in the text. text to be added 2) Derive the expression for the energy of a particle in a 1D box using the de Broglie formula. Solution: Please see question 7 of year 2009. 3) Find the average position hxi of a particle constrained to lie in a 1D box of length ‘a’ and with infinitely high potential walls in the lowest state E1 . 125

Would this location be a good place to seek the particle if it is in the next energy level E2 ? Given Znπ n2 π 2 2 zsin zdz = 4 0

Solution: The average position hxi of a particle constrained to lie in a 1D box of length ‘a’ and with infinitely high potential walls in any state n is given by hxi =

Za r

r 2 nπx 2 nπx sin x sin dx a a a a

0

 Za  2nπx x 1 − cos dx a 0 2 a a2 2nπx ax 2nπx 1 x − 2 2 cos − sin = a 2 4n π a 2nπ a 0  2  2 a a2 1 a − 2 2 cos2nπ − 0 − 0 + 2 2 + 0 = a 2 4n π 4n π a a a = − 2 2 cos2nπ + 2 2 2 4n π 4n π 1 = a

For ground state n = 1 and cos2π = 1 and hence hxi = a2 . Thus, in ground state the average position of a particle in a 1D box is at the middle of the box. In the next energy level, n = 2, cos4π = 1 and hence hxi = a2 . Thus, the average position of the particle in the next energy level (E2 ) is also at the middle of the box. 4) Determine, citing reasons, whether each of the following functions is acceptable or not as a wavefunction over the indicated intervals: (i) 1/x[0, ∞], (ii) sin−1 x[−1, 1], (iii) e−x cosx[0, ∞]. Solution:For (i) and (ii) please see question 2 of year 2012. For (iii), please see question 1 of year 2015. 5) Calculate the zero-point energy of a harmonic oscillator consisting of a particle of mass 5.16 × 10−26 kg and force constant 285N.m−1 . Solution: The zero-point energy of a harmonic oscillator consisting of a particle of mass m and force constant k is given by r 1 k EZPE = ~ 2 m In our case, m = 5.16 × 10−26 kg, k = 285N.m−1 . Thus, r 1 6.626 × 10−34 285 × EZPE = × 2 2 × 3.14 5.16 × 10−26 = 3.9207 × 10−21 J R d2 2 6) Classify the following operators as linear or non-linear: dx , ( ) , ( )dx, exp 2 d2 Solution: If φ and ψ be two functions and k be a constant then, (i) dx 2 (kφ + 2 d2 φ d2 ψ d kψ) = k dx2 + k dx2 . Thus, dx 2 is a linear operator. (ii) Please see question R2 or year 2011. R R R (iii) (kφ + kψ)dx = k φdx + k ψdx. So, ( )dx is a linear operator. (iv) exp(kφ + kψ) = exp(kφ) × exp(kψ) 6= k exp φ + k exp(ψ). So, exp is a non-linear operator. 126

  7) Evaluate the commutator xˆ, pˆ2x Solution: Please see question 9 of year 2007. 8) Estimate the wavelength of light absorbed when a π electron of butadiene is excited from the highest occupied energy level to the lowest vacant energy level. For the sake of simplicity assume that the pi electrons of butadiene move in a 1D box of length 7.0˚ A, me = 9.1 × 10−28 kg. Solution: The required wavelength can be obtained from the relation Elowest unoccupied level −Ehighest occupied level = hc λ . For butadiene, highest occupied level corresponds to n = 2 level and lowest unoccupied level corresponds to n = 3 level. Thus, required wavelength is hc E3 − E2 = λ 2 h hc (9 − 4) = 2 8mL λ 8mcL2 λ= 5h
2 8 × 9.1 × 10−31 × 3 × 108 × 7.0 × 10−10 m λ= 5 × 6.626 × 10−34 λ = 323.018 × 10−9 m 9) For a simple harmonic oscillator: hxi = 0, hpi = 0, hx2 i = h(mk)−1/2 (v + 1/2) and hp2 i = h(mk)1/2 (v + 1/2), the terms have their usual significance. Show that a simple harmonic oscillator obeys the uncertainty principle by computing ∆x and ∆p. Solution: p ∆x = hx2 i − hxi2 s h = √ (v + 1/2) − 0 mk s h = √ (v + 1/2) mk p ∆p = hp2 i − hpi2 q = h(mk)1/2 (v + 1/2) − 0 q = h(mk)1/2 (v + 1/2)

Thus, s

q h ∆x∆p = √ (v + 1/2) × h(mk)1/2 (v + 1/2) mk 1 = (v + )h 2 h The minimum value of v is zero. So, ∆x∆p ≥ h or, ∆x∆p ≥ 4π . Thus, Heisenberg uncertainty principle is obeyed by simple harmonic oscillator. 10) For the 1s state of the H-atom Ψ1s = b0 e−r/a0 . (i) Find the normalization constant b0 . (ii) Specify the values of n, l and m of 1s electron. (iii) Evaluate the probability density for a 1s electron at the nucleus. Given Z∞ n! xn e−qx dx = n+1 , n > −1, q > 0 q 0

127

Solution: (i) Z+∞ |Ψ1s |2 4πr2 dr = 1 0

Z+∞ 4π|b0 |2 r2 e−2r/a0 dr = 1 0

4π|b0 |2

2!a30 =1 8 1 |b0 | = p 3 πa0

(ii) For 1s electron, n = 1, l = 0, m = 0. (iii) Probability density at the nucleus is given by 1 1 −2r/a0 ∗ = e Ψ1s Ψ1s |r=0 = πa30 πa30 r=0 2003: 1) Indicate and explain the salient features of the photoelectric effect. Solution: Please see relevant text (section ??). 2) State the conditions of “acceptability of wave function” in quantum mechanics with explanation. Illustrate the function e−x is acceptable wavefunction within the range (0, ∞). Solution: A wavefunction is acceptable if it satisfies the following conditions: (i) It is finite, single valued and continuous function within the boundary of the system. (ii) It’s partial first derivatives with respect to the Cartesian coordinates of each particle of the system should be continuous. (iii) It should be quadratically integrable. (iv) It should satisfy the boundary condition of the system. (v) For fermions, it should be antisymmetric with respect to the exchange of position of any two fermions in the system. e−x is a single valued, continuous and finite function in the range (0, ∞). The derivative of e−x with respect to x is −e−x , which is also continuous over the indicated interval. The function is also square integrable, as shown below Z∞ 0

e−2x dx = −


1 −∞ 1 1 1  −2x ∞ 0 e = − e − e = − (0 − 1) = 0 2 2 2 2

Hence, e−x is an acceptable wavefunction within the range (0, ∞). 3) If Ψ1 and Ψ2 are two eigenfunctions with same eigenvalue E, prove that any linear combination C1 Ψ1 +C2 Ψ2 is also an eigenfunction with same eigenvalue. Solution: Please see question 2 of year 2016. 4) The eigenfunction for a particle  1/2 2 mπx Ψm (x) = sin , m = 1, 2, 3, · · · L L where x is defined within 0 ≤ x ≤ L. Show that these eigenfunctions are orthogonal. Solution: Please see question 3 of year 2005. 5) The wavefunction for the state of lowest energy of a 1D harmonic oscillator 128

2

1 (µk)1/2 . The is Ψ = Ae−Bx , where A is a normalization constant and B = 2h potential energy for the oscillator, V (x) = 21 kx2 . Write the Schr¨odinger equation for the system and qhence show that the total energy E of the lowest 1 k h state is 12 hν, where ν = 2π µ and ~ = 2π . Solution: The Hamiltonian for a simple harmonic oscillator is

h2 d2 1 ˆ H=− 2 + kx2 , 2 8π m dx 2 where k is the force constant. Let us apply this hamiltonian on the given 2 wavefunction Ψ(x) = Ae−Bx . h2 d2 Ψ(x) 1 2 ˆ HΨ(x) = − 2 + kx Ψ(x) 8π µ dx2 2 2 2 h d −Bx2 1 2 −Bx2 e + kx Ae = −A 2 8π µ dx2 2 n o 1 2 h d 2 −Bx2 = −A 2 −2Bxe + kx2 Ae−Bx 8π µ dx 2  o 1 2 n  h 2 −Bx2 −Bx2 = 2AB 2 x −2Bxe +e + kx2 Ae−Bx 8π µ 2 2 2 h h 1 2 2 2 = −4AB 2 2 x2 e−Bx + 2AB 2 e−Bx + kx2 Ae−Bx 8π µ 8π µ 2 p Since B = (µk)/2~ =⇒ B 2 = µπ 2 k/h2 . Putting these in the above equation, √ 2 2 π µk π h µk h2 −Bx2 1 2 −Bx2 2 2 −Bx ˆ xe + 2A e + kx Ae HΨ(x) = −4A 2 8π µ h2 h 8π 2 µ 2 s 1 h k −Bx2 1 2 −Bx2 2 = −A kx2 e−Bx + A e + kx Ae 2 4π µ 2 s h k −Bx2 = Ae 4π µ s 1 1 k 2 = frequency of oscillation. = hνAe−Bx , where, ν = 2 2π µ q 1 1 Thus, the total energy E of the lowest state is 2 hν, where ν = 2π mk . 6) (i) Write the Schr¨odinger equation for a particle in 1D box of length L having V = 0 inside the box and V = ∞ at the walls and outside the box. Find the expression for the energy and wavefunction of the particle. (ii) Why is the value of n = 0 of the quantum number not permitted? (iii) Find the average value of px , hpx i for n = 1 state. Comment on your result. (iv) If the particle is in its ground state, what is the probability of finding the particle in the range L/3 to 2L/3? Solution: (i) Let, the box lies between x = 0 and x = L along the x−axis and m be the mass of the particle. Then, the required Schr¨odinger equation will be h2 d2 ψ = Eψ, 0 < x < L − 2 8π m dx2 This is a standard second order differential equation of the form d2 ψ + k 2 ψ = 0, 2 dx 129

where k =

q

8π 2 mE h2 .

It has a standard solution, given by ψ = Acoskx + Bsinkx

A and B are constants, which can be determined by applying boundary conditions. The boundary conditions are (i)At x = 0, ψ = 0 And (ii)At x = L, ψ = 0 These boundary conditions give, 0=A 0 = AcoskL + BsinkL Since, B cannot be equal to zero (otherwise there will not be any wavefunction). So, sinkL = 0 =⇒ kL = nπ, where n is an integer. Putting the expression for k, we get r 8π 2 mE L = nπ h2 8π 2 mE 2 L = n2 π 2 2 h n2 h2 E= 8mL2 This is the required expression for energy. (ii) Note that, the value of n cannot be zero, as this will gives a ψ = 0. q (iii) The wavefunction of the particle in ground state is ψ = L2 sin πx L . Therefore, the probability of finding the particle in the range L/3 to 2L/3 is given by 2L/3 Z

L/3

r

2 πx sin L L

!2

1 dx = L

2L/3 Z 

 2πx 1 − cos dx L

L/3

 2L/3 1 L 2πx = x − sin L 2π L L/3   1 2L L 4π L L 2π = − sin − + sin L 3 2π 3 3 2π 3 √ √ 1 3 3 1 + ) = 0.609 = + ( 3 2π 2 2 ˚(with zero potential inside and 7) A cubic box with each side measuring 10A infinite potential outside) contains a system of 4 quantum particles. Find the degeneracy of the lowest energy state of the system. Solution: The energy of a particle in a cubic box is given by E(nx , ny , nz ) =

(n2x

130

+

n2y

+

n2z )

h2 8mL2

The lowest energy level corresponds to the quantum number (1, 1, 1). This energy level is non-degenerate and hence has degeneracy = 1. 8) The wavefunction for a 1s orbital for a H-atom is Ψ1s =

1 1/2 (πa30 )

e−r/a0

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where a0 = Bohr radius. Depict graphically the plots of (i) Ψ against r and (ii) 4πr2 Ψ2 against r. Explain the graph. Find the most probable value of r in the SI unit. Solution: The plot (i) Ψ against r and (ii) 4πr2 Ψ2 against r are shown below.

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Ψ vs r plot represents that the wavefunction gradually decreases with the increase in distance from the nucleus. After a certain distance the wavefunction becomes almost zero. 4πr2 Ψ2 is the radial distribution function and it represents the probability of finding the electron anywhere on the surface of a sphere located at a distance r from the nucleus. We see that the plot at first increases with the increase in distance and it reaches the maximum point at r = a0 and then with the further increase in distance from the nucleus the plot falls rapidly. This plot represents that the most probable location of an electron in 1s orbital is at the Bohr radius distance = 0.529 × 10−10 m. 2002: 1) Write down the (i) Stefan-Boltzmann law, (ii) Wien’s displacement law indicating the quantities involved. Solution: (i) Stefan-Boltzmann law: The total energy emitted from a blackbody is proportional to the fourth power of thermodynamic temperature of the surface. Thus, E = σT 4 , where E = total heat energy emitted, σ = Stefan’s constant and T = temperature of blackbody in Kelvin. (ii) Wien’s displacement law: The radiant energy emitted by a blackbody at a particular temperature is maximum for a wavelength that is inversely proportional to the temperature of the blackbody. Thus, λmax ∝ T1 , where λmax represents the wavelength at which the radiant energy emitted by a blackbody at temperature T is maximum. 2) Determine, citing reasons, whether each of the following functions is acceptable or not as a wavefunction over the indicated intervals: (i) e−x (0, ∞), (ii) e−x (−∞, ∞), (iii) sin−1 x(−1, 1), (iv) e−|x| (−∞, ∞). Solution: (i) Please see question 2 of year 2002. (ii) e−x becomes infinite at x = −∞. Thus, it is not an acceptable wavefunction in the interval (−∞, ∞) (iii) Please see question 1 of year 2015. (iv) Please see question 1 of year 2013. 3) The wavefunction of a particle of mass m which moves in one dimension between x = a and x = b is Ψ = Ax , where A is a normalization constant. (i) ab calculate A, (ii) show that x¯ = a−b ln ab , where x¯ represents the average value 131

of x. Solution: (i) Applying normalization condition, Zb

A | |2 dx = 1 x a  b 1 |A|2 − =1 x a   1 1 2 |A| − + =1 b a r |A| =

ab b−a

(ii) Calculating x¯ Zb x¯ =

Ψ∗ xΨdx

a

Zb r = a

ab = b−a

ab 1 ×x× b − ax Zb

r

ab 1 dx b − ax

1 dx x

a

ab [lnx]ba b−a ab = (lnb − lna) b−a ab b = ln b−a a =

4) Calculate the de Broglie wavelength of an electron travelling at 1% of the speed of light. (me = 9.1 × 10−28 g). Solution: The required de Broglie wavelength is λ = mhe v , where me = 9.1 × 10−31 kg, v = 0.01 × 3 × 108 m.sec−1 and h = 6.626 × 10−34 J.sec. Putting all these values, we get 6.626 × 10−34 λ= 9.1 × 10−31 × 0.01 × 3 × 108 = 0.2427 × 10−9 m 5) Calculate the frequency of radiation emitted when a linear harmonic oscillator of frequency 6.0 × 1013 sec−1 goes from v = 8 to v = 7 state. Solution: Please see question 14 of year 2013. 6) For a particle of mass m in a cubic box of length L with zero potential inside and infinite potential on the walls and outside, (i) write down the expressions for the wavefunctions and the energy. (ii) What is the degeneracy of the level that has an energy three times that of the lowest level? Solution: (i) Wavefunctions: r 23 nx πx ny πy nz πz sin sin sin Ψ(nx , ny , nz ) = L3 L L L 132

Energies: E(nx , ny , nz ) =

n2x

+

n2y

+

n2z


h2 8mL2 2

3h (ii) The lowest energy level has energy E(1, 1, 1) = 8mL 2 . The energy of the 9h2 level which has three times the energy of lowest energy level is E = 8mL 2. This energy corresponds to the following combinations of quantum numbers: (1, 2, 2), (2, 1, 2), (2, 2, 1). Thus, this energy level is three fold degenerate. 7) Sketch |Ψ|2 for n = 1 and n = 2 states of a particle in a 1D box of length L and indicate the most likely locations of the particle in the state n = 1 and n = 2. Solution:

most probable location for n=2

most probable location for n=1

most probable location for n=2

8) Consider a particle constrained to move in a 2D box. Determine [ˆ x, pˆy ] and interpret the result. Solution: The wavefunction for a particle constrained in a 2D box is given by 2 nx πx ny πy Ψ(x, y) = p sin sin Lx Ly Lx Ly Thus, [ˆ x, pˆy ]Ψ(x, y) = xˆpˆy Ψ(x, y) − pˆy xˆΨ(x, y)

( ) 2 nx πx ny πy d 2 nx πx ny πy d sin + i~ xp sin = −i~x p sin sin dy Lx Ly Lx Ly dy Lx Ly Lx Ly   2i~ nx πx ny π ny πy nx πx ny π ny πy =p −xsin × sin + xsin × sin Lx Ly Ly Lx Ly Ly Lx Ly =0

Thus, for a particle constrained to move in a 2D box, [ˆ x, pˆy ] = 0. The two operators commute. This means the position of the particle along x−axis and it’s momentum along y−axis can be determined simultaneously and accurately. This is obvious. Because the two Cartesian axes x and y are orthogonal to each other and hence independent. This makes the properties along each axes independent of each other. That’s why, all the one-dimensional operators along different Cartesian axes commute with each other. d2 9) Is the function Ae−ax an eigenfunction of the operator dx 2 ? If so, what is the eigenvalue? Solution: d d2 −ax Ae = A (−ae−ax ) = a2 Ae−ax 2 dx dx 2

d 2 Thus, Ae−ax is an eigenfunction of the operator dx 2 . The eigenvalue is a . 10) The radial wavefunction for 2s orbital of a H-atom is given by the following expression:   r r R2,0 = N 2 − e− 2a0 a0

133

where N is a constant. (i) Determine the number and location of node(s) in the 2s wavefunction. (ii) Write down the expression for the radial distribution function of a 2s electron and sketch the radial distribution curve. Solution: (i) The function   r r e− 2a0 R2,0 = N 2 − a0

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is zero at r = 2a0 and r = −∞. Since r = −∞ is not considered as a node, the 2s orbital has one node.  2 r r 2 2 2 2 (ii) Radial distribution function of a 2s orbital 4πr R2,0 = 4πN r 2 − a0 e− a0 . The corresponding sketch is

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2001: 1) What is photoelectric effect? How can the important features be explained? Solution: Please see relevant section in the text (section ??). 2) Calculate the minimum uncertainty in position for an automobile of mass 500kg moving with speed 50 ± 0.001km.hr−1 . Comment on the result. h = 6.627 × 10−34 J.Sec. Solution: The required minimum uncertainty in position is given by h 4πm∆v 6.626 × 10−34 = 4 × 3.14 × 500 × 0.001×1000 3600 −34 = 3.798 × 10 m.

∆xmin =

3) Verify that the operator ∇2 is linear. Solution: ∂2 ∂2 ∂2 2 ∇ = 2+ 2+ 2 ∂x ∂y ∂z If f and g be two functions of the Cartesian coordinates x, y, z and k be a

134

constant, then  ∂2 ∂2 ∂2 ∇ {k(f + g)} = + + {k(f + g)} ∂x2 ∂y 2 ∂z 2 ∂2 ∂2 ∂2 = 2 k(f + g) + 2 k(f + g) + 2 k(f + g) ∂x ∂y ∂z 2 2 ∂ ∂ ∂2 = k 2 (f + g) + k 2 (f + g) + k 2 (f + g) ∂x ∂y ∂z ∂2 ∂2 ∂2 ∂2 ∂2 ∂2 = k 2f + k 2g + k 2f + k 2g + k 2f + k 2g ∂x ∂x ∂y ∂y ∂z ∂z 2 2 2 2 2 ∂ ∂ ∂ ∂ ∂2 ∂ = k 2f + k 2f + k 2f + k 2g + k 2g + k 2g ∂x ∂y ∂z ∂x ∂y ∂z  2    ∂ ∂2 ∂2 ∂2 ∂2 ∂2 =k + + f + k + + g ∂x2 ∂y 2 ∂z 2 ∂x2 ∂y 2 ∂z 2 = k∇2 f + k∇2 g 2



Thus, ∇2 is a linear operator. 4) State the postulate regarding dynamical variables of a particle in quantum mechanics. From the definition of linear harmonic oscillator, obtain the expression for the potential energy and hence construct the Hamiltonian operator or the same. Solution: There are three postulates regarding dynamical variables in quantum mechanics. (i) Every dynamical variable is represented by a corresponding hermitian operator. (ii) In an experiment, measuring a dynamical variable (A) of the system, the ˆ = aψ, only value that can be obtained will be the eigenvalue that satisfies Aψ where Aˆ = hermitian operator corresponding to the dynamical variable (A), ψ = wavefunction of the system and a = eigenvalue. (iii) If a number of experiments are performed to measure the value of a dynamical variable (A), then the average or the expectation value of the dynamical variable associated with the operator Aˆ is given by +∞ R

hAi =

ˆ ψ ∗ Aψdτ

−∞ +∞ R −∞

ψ ∗ ψdτ

A linear harmonic oscillator is one which oscillates under a potential of 12 kx2 , where k = force constant. The Hamiltonian operator of a linear harmonic oscillator is given by h2 d2 1 2 ˆ H=− 2 + kx , 8π µ dx2 2 where µ = reduced mass of the particle exhibiting harmonic oscillation. 5) What are the characteristic features of linear harmonic oscillator obeying quantum mechanical laws. Solution: 6) Solve the Schr¨odinger wave equation for a particle in a one dimensional box and find the expression for eigenvalue and eigenfunction. Why is the value of n = 0 of the quantum number not permitted? Solution: Please see question 6 of year 2003. 7) What is normalization? Why is it necessary? What is its significance? 135

Solution: If ψ be the wavefunction of a particle in a particular state then R ∗ ψ ψdτ represents the net probability of finding the particle anywhere all space

in space. In principle, this probability should be equal to 1 (because the probability should be somewhere). Even if ψ is a valid wavefunction of the system, it may not give a total probability of 1. Normalization is the technique of scaling the wavefunction such that the total probability of finding the particle anywhere in space becomes 1. Normalization does not change the wavefunction, as in this process the wavefunction is multiplied by a constant quantity. Normalization also rectifies the dimension (or unit) of the wavefunction. Without normalization, any property calculated using the unnormalized wavefunction will miss the exact value by a constant factor and the calculated property will also suffer a dimensional error (unit related error). 8) Calculate the expectation value hxi for a particle in a 1D box of width L and infinitely high potential at the walls with origin at the one end of the box. Solution: Please see question 3 of year 2004. 9) The wavefunction for a 1s orbital for a H-atom is 1 Ψ1s = p 3 e−r/a0 , πa0

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where a0 is the Bohr radius. Depict graphically the plots of (i) Ψ against r, (ii) Ψ2 against r and (iii) 4πrr Ψ2 against r. Explain with the help of the above equation why the plots are different. Solution: The required plots are

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1 Ψ1s = p 3 e−r/a0 πa0 1 Ψ21s = 3 e−2r/a0 πa0 4 4πr2 Ψ21s = 3 r2 e−2r/a0 a0 From the above three equations, it is obvious that Ψ1s is an exponentially decaying function and so is reflected by its plot. Ψ21s is also an exponentially decaying function but has larger exponent than Ψ1s and hence it falls off rapidly as shown in figure. Finally, 4πr2 Ψ21s is the radial distribution function 136

and is a product of a parabolic function (r2 ) and an exponentially decaying function. At smaller value of r the parabolic function dominates over the exponential decay and hence the plot increases with the increase in r, at a certain value the increasing trend of r2 is balanced by the exponential decay, where the curve reaches a the peak. After that, the exponential decay starts dominating the parabolic r2 and the hence the curve falls off till it becomes zero at infinity. 2000: 1) Define a perfect black body. Write three characteristic features of intensity distribution of radiation emitted by a black body. Solution: (i) A perfect blackbody is a substance which absorbs all the radiations incident on it and when it is heated it emits radiations of all wavelengths. (ii) Three characteristic features of intensity distribution of radiation emitted by a black body: (a) The intensity distribution of radiation emitted by a blackbody is independent of the material constitution of the blackbody and depends only on its temperature. (b) At any temperature, the radiant energy emitted at a particular wavelength increases with the increase in wavelength and reaches a peak. The wavelength corresponding to this peak is inversely proportional to the temperature of the blackbody. With further increase in wavelength (at the same temperature) the radiant energy emitted by the blackbody decreases. (c) With the increase in temperature, the wavelength corresponding to the peak in the blackbody radiation plot shifts to the lower value i.e. a blue shift is observed. 2) If Ψ1 and Ψ2 are eigenfunctions of linear operator Aˆ with the same eigenvalues a, show that any linear combination of Ψ1 and Ψ2 will also be an eigenfunction of Aˆ with the same eigenvalue. Solution: Please see question 2 of year 2016. 3) Write down the Schr¨odinger equation for the time independent case choosing the 1D simple harmonic oscillator as an example. Solution: The time independent Schr¨odinger equation for a 1D simple harmonic oscillator is   1 2 2 h2 d2 + µω x ψ(x) = Eψ(x), − 2 8π µ dx2 2 q where µ = reduced mass of the system, ω = µk is the angular frequency, k = force constant and E = energy of the 1D simple harmonic oscillator in state ψ(x). 4) Of the three functions Ψ1 , Ψ2 and Ψ3 , choose the acceptable wavefunction citing proper reasons. Normalise the acceptable wavefunction. A1 , A2 and A3 are real constants: Ψ1 = A1 sinx, [0 ≤ x ≤ 2π] Ψ2 = A2 sin−1 y, [−1 ≤ y ≤ 1] Ψ3 = A3 exp(z), [−∞ ≤ z ≤ ∞] Solution: For Ψ1 , please see question 1 of year 2008. For Ψ2 , please see question 2 of year 2012. Ψ3 = A3 exp(z), [−∞ ≤ z ≤ ∞] becomes infinite for z = ∞. Thus it is not an acceptable wavefunction. 5) From the probability density of 1s electron show that the most probable 137

distance of electron from the nucleus is equal to the Bohr radius. Given 1 Ψ1s = p 3 e−r/a0 , πa0 where a0 is the Bohr radius. Solution: Please see question 14 of year 2016. 6) Find the value of the commutator (ˆ xpˆx − pˆx xˆ) of the quantum mechanical operators xˆ and pˆx . Solution: Please see question 5 of year 2010. 7) For the particle in a box, in one dimension, with infinitely high potential at the walls, write down the expression for energy and wavefunction. Show that here wavefunction Ψ1 (n = 1) and Ψ2 (n = 2) are orthogonal to each other. Solution: The expressions for energy and wavefunction of particle in a 1D box are r 2 nπx sin ψ(x) = L L 2 2 nh , L = Length of the box. E= 8mL2 (ii) ZL

ZL r Ψ1 Ψ2 dx =

0

2 πx sin L L

r

2 2πx sin dx L L

0

1 = L

ZL 2sin

2πx πx sin dx L L

0

ZL 

 πx 3πx cos − cos dx L L 0     1 L L L L = sinπ − sin0 − sin3π + sin0 L π 1π 3π 3 = 0. 1 = L

Thus, the wavefunctions Ψ1 (n = 1) and Ψ2 (n = 2) are orthogonal to each other. 8) An electron is confined in a 1D box of side 2˚ A. Calculate the frequency of the photon which can produce an excitation from n = 1 to n = 2. Given me = 9.1 × 10−28 g. Solution: If ν be the frequency of the photon that can produce the excitation from n = 1 to n = 2 then 4h2 h2 3h2 hν = E2 − E1 = − = 8mL2 8mL2 8mL2 3 × 6.626 × 10−34 ν= = 6.826 × 1015 sec−1 . −31 −10 2 8 × 9.1 × 10 × (2 × 10 ) 9) Show that eikx is an eigenfunction of the operator pˆx . What is the eigen value? Solution: d pˆx eikx = −i~ eikx = −i~ × ikeikx = k~eikx dx 138

Thus, eikx is an eigenfunction of pˆx and the eigenvalue is k~. 10) For a particle in a cubical box, write down the energy values for the condition: nx + ny + nz = 4 and indicate the level of degeneracy, if any. Solution: nx , ny , nz can have only integer value (6= 0). Thus, nx +ny +nz = 4 will be possible only for (nx = 1, ny = 1, nz = 2), (nx = 1, ny = 2, nz = 1) and (nx = 2, ny = 1, nz = 1). Thus, the energy value will be equal to h2 (n2x + n2y + n2z ) 8mL 2 . The level of degeneracy is 3. 11) Write down the Schr¨odinger equation for H-atom. What is an orbital? Draw schematically the 2s orbital of H-atom. Solution: The Schr¨odinger equation for H-atom   2   h2 ∂2 ∂2 e2 ∂ − 2 + + − ψ(x, y, z) = Eψ(x, y, z), 8π µ ∂x2 ∂y 2 ∂z 2 4π0 r where r = nucleus-electron separation distance, µ = reduced mass. An orbital is a three-dimensional space around the nucleus, where the probability of finding electron is maximum.

Figure 12: 2s orbital in H-atom 1999: 1) What is the kinetic energy of an electron whose de Broglie wavelength is 10−10 m? Solution: The momentum of the electron with de Broglie wavelength of 10−10 m is h 6.626 × 10−34 p= = = 6.626 × 10−24 kg.m.sec−1 −10 λ 10 Therefore, the kinetic energy of the electron will be p2 (6.626 × 10−24 )2 = = 2.412 × 10−17 J −31 2m 2 × 9.1 × 10 2) Explain the phenomenon of photoelectric effect in the light of Einstein’s theory. Solution: See the relevant section of the text (section ??). 3) Sodium has an ionization energy of 5.12 eV. Calculate the enegy of photoelectrons, when light of wavelength 200 nm shines on it. Work function of Sodium = 2.3 eV. Solution: When a metal having work function W0 is irradiated with a light of wavelength λ, the energy of photoelectrons (E) emitted is hc − W0 λ 6.626 × 10−34 × 3 × 108 = − 2.3 × 1.602 × 10−19 −9 200 × 10 = 9.939 × 10−19 − 3.6846 × 10−19 = 6.2544 × 10−19 J = 3.9041eV

E=

139