2006-7 Module 113 - Lecture 7

Module 113- Quantum theory and atomic spectroscopy LECTURE 7 – Quantization and hydrogen 17. Why we (sometimes) find quantised energy levels We saw earlier that for the case of a free particle there was a general solution to the Schrödinger equation for that particular system. This is often the case. Unfortunately, this means that mathematically there are often an infinite number of solutions to the Schrödinger equation! Well mathematically many solutions may be possible, but maybe physically they are not. We will call these extra criteria physical constraints on the acceptable wavefunctions. The result from the physical interpretations of the wavefunction that we considered in the last section: 1) The curvature gives you the kinetic energy 2) Ψ ( x ) Ψ ( x ) is the probability density ∗

Since Ψ ( x ) Ψ ( x ) has a probabilistic interpretation, it must satisfy certain physical requirements. For example, ∗

an obvious one is the total probability for finding a particle somewhere must be unity. ∗

∫ Ψ (x ) Ψ (x )dx

= 1

(7.1)

allspace

The notation “all space” just means here that we integrate over all possible values of x. Wavefunctions that satisfy the above equation are said to be normalised. Even if the integral equals some constant other than 1, which we can denote by

1 , we can multiply the N2

wavefunction by N to normalise the wavefunction (make its integral over space equal one). Wavefunctions that can be normalised are said to be normalized. Those that cannot are unacceptable. Often the wavefunction is such that the integral diverges (get larger as x increases) and these functions will clearly not normalise. Consequently, we find that a wavefunction MUST be finite everywhere (even if it is zero at some locations, that is okay) Consider the two functions below

These are both unacceptable because they manage to have two different values at the same position x, which contradicts the Born interpretation. In addition, no gaps can appear in the wavefunction because the value of Ψ ( x ) * Ψ ( x ) would be ill defined. In addition, the fact that a quantum object cannot have an infinite kinetic energy requires that there must be a finite value for the wavefunction curvature. These restrictions are normally summarised as follows:

A wavefunction is acceptable only if it (a) is continuous (b) has a continuous slope (c) is single valued and (d) is finite. These points become particularly important when there are so-called boundary conditions in place. These are limitations or restrictions on a wavefunction that depend on a particular physical situation in question. The standing wave considered earlier, like a plucked string, is a good example. A combination of the boundary conditions and the restrictions imposed by the physical interpretation of the wavefunction is usually so restrictive that acceptable solutions of the Schrödinger equation do not in general exist for arbitrary values of the energy E. An example of this is a particle in a potential box with very high walls. The particle is assumed not to have enough kinetic energy to escape the box. We don’t need to do the full mathematics, we just need to appreciate how the quantization seeps in. At the walls of the box, we assume that each wavefunction must equal zero, otherwise the quantum object can exist outside the confines of the box. We can see that only certain wavefunctions can obey these restrictions and that they have very different wavelengths and hence energies: quantization!!!

What the allowed energies are we shall discover by solving the Schrödinger equation for a particular system and selecting only those solutions that confirm to the restrictions. One such system is the hydrogen atom. This is the simplest atomic system, possessing one electron and a central nucleus and what’s more it displays the important features of more complicated atoms, namely that its electron can always be described by the four quantum numbers n, l, ml and ms (whether the electron is excited or not).

18. The hydrogen atom and the meaning of n, l and ml In matter of fact, the hydrogen atom is one of the few instances where the Schrödinger equation can be solved exactly. However, that does not mean it is easy to solve- without a background in pure mathematics to degree level, forget it. Its solution draws on the work of some of the greatest mathematicians of the 18th and 19th centuries. We don’t really need to solve it ourselves: we just need to understand roughly what is going on.

First we must set up the Hamiltonian. The hydrogen atom contains a proton of charge +q, and an electron of charge –q. The proton mass, m p , is roughly 1836 times greater than the mass of an electron me . We must include BOTH in the Schrödinger equation. Also, a hydrogen atom is a three dimensional quantum mechanical system, and therefore we have to use a three-dimensional version of the kinetic energy operator

−

h2 2m

⎛ ∂2 ∂2 ∂2 ⎞ h2 ⎜⎜ 2 + ⎟ + = − ∇2 2 2 ⎟ 2m ∂y ∂z ⎠ ⎝ ∂x

(7.2)

∂2 ∂2 ∂2 + + is known as the Laplacian operator. Finally, we must also ∂z 2 ∂y 2 ∂x 2 include the fact that there is a Coulombic attraction between the proton and electron, manifest as a potential energy V (r ) , where r is the separation between the proton and the electron. From electrostatics we have an expression for the Coulombic attraction where ∇ 2 =

V (r ) =

(− q )(+ q ) 4πε 0 r

= −

q2 4πε 0 r

(7.3)

where is the vacuum permittivity = 8.854 x 10-12 J-1 C2 m-1. The full Schrödinger equation for the hydrogen atom is therefore

h2 h2 q2 2 2 ∇p − ∇e − Hˆ = − 2m p 2me 4πε 0 r

The hydrogen Schrödinger equation in full is: 2 2 2 ⎞ ⎛ ⎜ − h ∇ p 2 − h ∇ e 2 − q ⎟ Ψ = EΨ ⎜ 2m 2me 4πε 0 r ⎟⎠ p ⎝

(7.4)

Doesn’t look like much, but the solution is very hard work and a person who can solve this is like a dog that speaks Norwegian- very rare. However, we’ll let others do the work for us and concentrate only on the solution. A significant simplification is made by noting that since the proton is so much heavier than the electron, its motion will be much smaller and hence we can place it at the centre of our co-ordinate system. If we do this, we immediately see that the proton/electron system has spherical symmetry, because the Coulomb potential only depends on the distance of separation between the electron and the proton. Mathematicians exploit this fact by transforming the equation from Cartesian co-ordinates to spherical ones. The three new co-ordinates are (r, θ and φ ).

The trick Schrödinger (who was a mathematical genius really) was to assume that the final wavefunction was just a product of three separate wavefunctions that each depended on one of the variables r, θ and φ . Ψn ,l ,ml (r , θ , φ ) = Rn ,l ( r )Θ l , ml (θ )Φ ml (φ )

(7.5)

Then we find the hydrogen Schrödinger equation splits into three separate equations, one of which depends on only the angle φ , another on the angle θ and finally one that just depends on r. The third contains both θ and φ but if you first solve the equation with angle φ you can then solve this one. The key point is that there are restrictions on the three individual wavefunctions, imposed by boundary conditions, which results in the quantization of energy. By solving the Schrödinger three equations in turn we actually retrieve the three quantum numbers n, l and ml . I will now explain where one of these boundary conditions comes from. Motion on a ring Let us consider a particle confined to move on a ring. We assume that its wavefunction only depends on the angle φ and that it possesses only the energy from its motion. We find that the appropriate Schrödinger equation again has the general form d2 Φ = −a 2 Φ 2 dφ

(7.6)

where Φ is the wavefunction (which is a function of φ ) and a is just a constant. This equation is very reminiscent of the one earlier for a free atom: d2 Ψ = −a 2 Ψ 2 dx but with Ψ replaced by Φ and x replaced by φ . The solution is therefore very familiar Φ (φ ) = N m e aφ = N m e imφ

(7.7)

where im = a , m is a real number and N m is the normalisation constant to be determined. However, there is one critical difference between this and the case of a free particle. For a free particle, there was no boundary condition: this time there is. Think of the motion along co-ordinate as rather like a wheel on a bicycle- after one complete rotation of the wheel (which is 2π radians in math speak), the wheel is back to its original position. For our wavefunction, it means that Φ(φ + 2π ) must be identical to Φ(φ ) . Applying this boundary condition to our proposed wavefunction e imφ = e im (φ + 2π ) = e imφ e im 2π

(7.8)

This can only be true if e im 2π = 1 and this turns can only be true if m2π is an integer multiple of 2πthat is, m may only assume integer values.

If we simply re-label m as ml , we can see that the boundary condition is the reason why only integer values of ml are allowed. We have just proved that ml can only have integer quantum numbers- easy!! As you’ve probably guessed, we can find the remaining quantum numbers from the two remaining Schrödinger equations. This is not something we will do this year!

Combining the two angular wavefunctions we obtain what is called a spherical harmonic: Yl , ml (θ , φ ) = Θ l , ml (θ )Φ ml (φ )

(7.9)

Yl , ml (θ , φ ) is the eigenfunction for an angular momentum operator. This is not very surprising,

since the angular co-ordinates define they behaviour of both Θ l , ml (θ ) and Φ ml (φ ) .

These

wavefunctions depend on two quantum numbers: l and ml. In fact, the eigenfunctions Yl , ml (θ , φ ) are exactly the same as those for a particle trapped on the surface of a sphere (Prof Atkins does this example in his textbook)! Below are the wavefunctions Y2, ml (θ , φ ) and yes you’re right these are just the d-orbitals you’ve see a hundred times before! The spherical harmonics determine the shape of the hydrogen wavefunction. Now, we must beware: the wavefunctions Yl , ml (θ , φ ) represent the angular part of the complete hydrogen wavefunction, hence a hydrogen atom with n = 2 and l = 1 will not have precisely the same appearance as one with n = 3, l = 1, because these have different principle quantum numbers n which we discover by solving the last of our three equations for the wavefunction Rnl (r ) .

The solution, Rnl (r ) , of the radial equation is a function of r and also depends on the value of n and l. We saw that we can force l to take only integer values if we have a suitable boundary condition. What type of boundary condition can force n to take only integer values? Well, it has to do with the Coulombic potential energy which we can model as a well with a proton at the centre:

This kind of boundary condition is very common, and is idealised in the so-called particle in a box model we saw at the beginning of the lecture. Basically, think back to our standing wave analogy: the walls of the potential are like the fixed points of the string. They set the boundary condition that forces only discrete values of energy to be supported by the Coulomb potential. The different discrete values are labelled by the quantum number n. What do these radial wavefunctions “look” like? Well, they depend on r and are parameterised by n and l (that means that there appearance is affected by the values these quantum numbers take).

Now before we finish I have to mention a little more mathematics but it is important: true solutions (eigenfunctions) of the must be orthogonal to one another. This means if we take two wavefunctions, multiply them together and integrate over all space then the result of the integration is a big fat zero. All the radial wavefunctions with the same l are orthogonal, as are all angular momentum wavefunctions with the same n and this means that if we leave the electrons in an atom alone, they will happily stay with their original wavefunctions and will not go hopping about from one wavefunction to the next unless an external force is applied. When we find two are more wavefunctions with the same energy (degenerate) that are not orthogonal, we must combine them together to form a true orthogonal pair: more on this in the next lecture. Note that none of this has explained the fourth quantum number ms. As the Hamiltonian we used doesn’t include spin, it is perhaps not surprising that we get no information on this quantum number. A slightly more complex equation- the Dirac equation – does include spin but it is not as easy to use as the Schrödinger equation. We will revisit this next year. Good. Now we see that solving the hydrogen provides the angular and radial wavefunctions for the hydrogen system. However, hydrogen is not a typical atom: for example, we know that in many-electron atoms states with the same n but different l are non-degenerate. It is natural to ask: •

Can we solve the Schrödinger equation for a many electron atom?

Unfortunately, we discover the answer is no! Key points in lecture 7 • •

• • •

The three quantum numbers: n, the principle quantum number; l , the orbital angular momentum; and ml the magnetic orbital quantum number are found by solving the Schrödinger equation for the hydrogen atom. The solution of the hydrogen atom is most easily found using spherical co-ordinates to describe the electron. When this is done, the equation is split into three separate equations, two of which are the same as for a particle moving on the surface of a sphere while the third depends only on the radial co-ordinate. The angular part of the wavefunction is described by special wavefunctions called spherical harmonics which correspond to the picture of “orbitals” chemists use. The radial wavefunctions are different for identical l but different n and is the reason for rows in the periodic table. The hydrogen wavefunctions are orthogonal, ensuring they are unique mathematical solutions.