Quantum - Harmonic Oscillator

Quantum Explanations – Issue I - Quantum Mechanical Harmonic Oscillator Author: Ryan Jadrich Last Update: December 1, 2007

Introduction: Your every day Harmonic oscillator is a fairly simple problem to describe and model by simple easily solvable differential equations. On the other hand the quantum mechanical harmonic oscillator is at a much higher degree of difficulty in terms of understanding it and solving the necessary differential equations to model it. Most texts will simply give the ordinary harmonic oscillator differential equation and then move onto the differential equation describing that of the quantum version. Finally they will state that the solution is too difficult to work out and they will only give you the answer. This is lazy and if you can understand quantum mechanics you can probably understand the math behind the solutions. This is only one of the many occasions for this in quantum mechanics and other issues will cover different topics in all of the gruesome detail that anyone could ever want. Explanation: Part I. – First off we will consider that of a harmonic oscillator composed of two masses attached by a spring because it turns out that to a reasonable approximation the bonds holding molecules act in a nature like that of a spring. Figure I. – The two mass spring scenario

If the spring is a normal spring then this system will be described by two separate differential equations, one for the first mass and one for the second mass. Since the force on the separate masses only depends on the relative displacement from the rest length l 0 the equations for the two masses are exactly the same except the negative sign. This is due to the fact that the force on one mass will always be opposite that of the other mass. d 2 x1  k ( x 2  x1  l 0 ) dt 2 d 2 x2 m2 2   k ( x 2  x1  l 0 ) dt m1

(1)

Copyright © Ryan Jadrich 2007

(2)

-1-

Since the motion is governed only by that of the relative displacement between the masses (atoms) we want to get a differential equation describing that of the relative distance. This can be done by subtracting equation one from equation two and the result of this is given below. Note that in the derivative only the x variables are needed as in differentiation the constant l 0 would just disappear. d 2 ( x 2  x1 ) 1 1  k (  )( x 2  x1  l 0 ) 2 m1 m2 dt d 2x 1 1  k (  )x 2 m1 m2 dt

(3)

(4)

The next step is to take the messy expression involving the masses and define something referred to as the reduced mass which is represented by  which makes the expression a little neater. 1 1 1    m1 m2



m1m2 m1  m2

d 2x 1  k ( ) x 2  dt



d 2x  kx  0 dt 2

(5)

By golly that looks just like the differential equation describing the motion of a ordinary harmonic oscillator. So from any differential equations course the solution to this equation can easily be shown to be in its most general form as: x  c1 cos(t )  c 2 sin(t )

 v

k  1 1  2 2

k 

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Part II. – Here comes the cool stuff where we apply this to a quantum system where really tiny stuff don’t behave the way they should. The best place to start of is stating that in order to describe this system the use of the Schrödinger equation is needed and is stated below in its one dimensional form. 

 2 d 2  V ( x)  E 2m dx 2

(6)

For those who want a review the V(x) is the potential on a particle, the E is the energy which is taken to be a constant, the  is the wave function for a particle which describes not is absolute position but rather describes its probability of being found at a location as a continuous function of x. finally the m is the mass of the particle and the  is the reduced plank constant. The next part is to substitute the various conditions for our reduced mass system undergoing simple harmonic motion. In this case we have to substitute the spring potential energy function for V(x) and  for the mass and finally we will have our differential equation describing a quantum mechanical harmonic oscillator. This is given below with all substitutions. m 1 2 kx 2  2 d 2 1 2   kx   E 2  dx 2 2

V ( x) 

( 7)

This equation is a real pain in the neck to solve. The good thing though is that it certainly looks similar to another differential equation that we do have a solution for which is referred to as Hermite’s differential equation. The comparison of equation 7 in a rewritten form (8) with Hermite’s diffq (9) is given below for your visual inspection.  2mE kx 2   ' ' 2  2   0     y ' '2 xy '2my  0

(8)

(9)

From the similarities we can assume that (8) will have solutions at least similar to that of (9) which we know and will show you how to solve. The solution to (9) is not in the form of a nice equation but rather a series of terminating polynomials that stem from a power series solution.

Copyright © Ryan Jadrich 2007

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Our first step in solving Hermite’s differential equation is substituting in the general form of a power series which is shown below. 

y   cn x n

(10)

0

Since the first and second derivatives also appear in the equation we will be needing those which are easily found by differentiation of (10). The summation indexes are each shifted up by one for each derivative because the first and corresponding second terms will be zero. 

y '   c n nx n 1

(11)

1



y ' '   c n n(n  1) x n  2

(12)

2

Now that we have our necessary derivatives we can plug them into Hermite’s differential (9) equation and in the second term we can multiply through by x and in all terms move any constant coefficients into the summation. 





 cn n(n  1) x n2  2 x cn nx n1  2m cn x n  0 2

1



 c n(n  1) x n

2

n2

0





1

0

  (2c n n) x n   (2mc n ) x n  0

(13)

Now we have a problem. We want to add up the necessary series and form one unique series that will lead to our solution. The problem is that the equations do not start with the same powers of x, so we have to take out the necessary amount of terms from each. For this example we will want all of the summations to start at a power of 1. This means we will have to take out one term from equation (13) in the first and last summations so that they start at one like the middle summation. The adjusted equation is given below. 





3

1

1

2c 2   c n n(n  1) x n  2   (2c n n) x n  2mc0   (2mc n ) x n  0

Copyright © Ryan Jadrich 2007

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Now that we have the summations all beginning with the same power of x we have to use what is called a k substitution to get the summations to not only begin with the same power but with the same index number. In the first summation we set k=n-2. The second summation k=n, and the third k=n again. Then we change all n’s into their corresponding k forms including the summation starting value. If you are unsure of this method look up adding together power series and you should find a really simple explanation of this valuable technique. 





1

1

1

2c 2   c k  2 (k  2)(k  1) x k   (2c k k ) x k  2mc0   (2mc k ) x k  0

(15)

Now our equation not only starts with the same power but with the same index and now our power series can be added together into one big conglomerate as given in equation (16) below. 

2c 2  2mc0   [c k  2 (k  2)(k  1)  2c k k  2mc k ]x k  0

(16)

1

From (16) we get two separate equations that must be equal to zero in order for the expression to be true. The constant out front must sum to zero, and the constants inside the brackets inside the summation must equal zero as this must hold true for all x. 2c 2  2mc0  0 c 2   mc0

(17)

c k  2 (k  2)(k  1)  2c k k  2mc k  0 ck  2 

2(k  m) c k , k  1,2,3,... (k  2)(k  1)

Copyright © Ryan Jadrich 2007

(18)

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From equations (17) and (18) we have what is referred to as a recursion relation which means we will be able to get a series in terms of the unknown constants c0 and c1 . The first values of c are given below. c0  c0 , c1  c1 c 2   mc0 c3 

2(1  m) 2(1  m) c1  c1 (1  2)(1  1) 3!

c4 

2(2  m) 2 2 (2  m)( m) c2  c0 (2  2)(2  1) 4!

c5 

2(3  m) 2 2 (3  m)(1  m) c3  c1 (3  2)(3  1) 5!

c6 

2(4  m) 2 3 (4  m)(2  m)( m) c4  c0 (4  2)(4  1) 6!

c7 

2(5  m) 2 3 (5  m)(3  m)(1  m) c5  c1 (5  2)(5  1) 7!

Now that we have are coefficients to equation (10) we can plug in and get our general solution to Hermite’s differential equation. 

y   cn x n 0

  2 2 (2  m)( m) 4 2 3 (4  m)(2  m)( m) 6 y  c0 1  ( m) x 2  x  x  ... (19) 4! 6!     2(1  m) 3 2 2 (3  m)(1  m) 5 2 3 (5  m)(3  m)(1  m) 6 x  x  x  ...  c1  x 1  3! 5! 7!   This general expression can be written as two separate solutions that we will use based on whether or not the number m is and even or odd integer.   2 2 (2  m)( m) 4 2 3 (4  m)(2  m)( m) 6 2 y  c0 1  ( m) x  x  x  ... 4! 6!  

(20)

  2(1  m) 3 2 2 (3  m)(1  m) 5 2 3 (5  m)(3  m)(1  m) 7 y  c1  x1  x  x  x  ... 3! 5! 7!  

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(21)

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It can be easily seen if the integer m is even then (20) will terminate with the highest power equal to that even integer, and the opposite is true for (21). So if m is an even integer we take (20) as our solution and if m is odd we take (21) as our solution. Given below are the first few solutions for various integers of m and then immediately next to the solution is a scaled version of the polynomial which is called the Hermite polynomials. This is done to make them look pretty really, and this can be done as any constant multiple times the polynomials will be solutions to (9). m  0,

y  1,

H0 1

m  1,

y  x,

H1  2x

m  2,

y  1  2x 2 ,

m  3,

y  x

m  4,

y  1  4x 2 

H 2  22 x 2  2

4 3 x , 3!

H 3  2 3 x 3  12 x

2 2 (8) 4 x , 4!

H 4  2 4 x 4  48 x 2  12

Now before we move onto applying this result to the quantum mechanical harmonic oscillator we are going state two formulas with out proof. The first is a relationship that can be used to generate the Hermite Polynomials as shown above using: H n  (1) n e x

2

d n  x2 e dx n

( A)

The next relationship relates the Hermite polynomials of different order, d H n  2nH n 1 dx

( B)

Part III. – Finally we will solve the Schrödinger equation for a simple harmonic oscillator. This is the really fun stuff. Lets bring back are two equations which are again stated below.  2mE kx 2   ' ' 2  2   0     y ' '2 xy '2my  0

(8)

(9)

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We now know the solutions to equation (9) in terms of the Hermite polynomials. Since (8) and (9) look so much alike it might be possible to solve (8) with some form of Hermite polynomials. Let’s try. The first major problem is that Hermite polynomials are not orthogonal to one another unless a weighing function is brought into play. This is demonstrated below. 

H

n

H m dx  0



With the appropriate weighing function though the polynomials will be orthogonal as demonstrated below. 

H

H m e  x dx  0 2

n



(22) 

H

H n e  x dx  2 n n!  2

n



Now to prove these relationships we will be using relationship (A) for our Hermite polynomial of order n in (22). Doing so we get, 

(1)

n

 H n H me

x2





dx   H m 

d n  x2 e dx dx n

Then after integration by parts once we are left with, 

  d n  x2 d n 1  x 2  d n1  x 2   H e dx  H e  H '  m dx n  m dx n1 e dx  m dx n 1       

It is easy to se that the portion outside of the integral is equal to zero because of the weight function which when differentiated any number of times will go to zero at both limits of infinity. Now we can use Relationship (B) in our new equation to get, 



(1) n  H n H m e  x dx    H m ' 2







 2(1)1 m  H m1 

d n 1  x 2 e dx dx n 1

n 1

d x2 e dx dx n 1

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Now it is clear that if we perform integration by parts m times we will be left with, 

(1)

n

 H n H me

 x2





dx  2 (1) (m!)  H 0 m

m



d n m  x 2 e dx dx n m

mn

(C )

Now if m=n the last expression (C) will be, 



m m m x x  H m H m e dx 2 (1) (m!)  e dx 2

2





 2 (1) (m!)  2m

m

 2 m (m!)  Which proves the second part of (22) except in terms of m. In the case that m does not equal n then (C) can be integrated one more time to obtain, 

(1)

n

H

n

H me



x2



d nm  x 2 dx  2 (1) (m!)  H 0 n m e dx dx  m

2 d n m 1  2 (1) (m!) n m 1 e  x dx

m

mn

m



0

m



This proves the first part of (22). Now we continue on to solving our equation. Since the solutions of (8) must be orthogonal to one another it seems that we should define a new function based on (22) which includes the Hermite polynomial and a portion of the weighing function. This is defined as: Z n  H nex

2

/2

(23)

Then by rearranging (23) we get: H n  Z ne x

2

/2

(24)

Since (24) is a solution to (9) we will take all necessary derivatives of (24) and substitute the results into (9) to get a new differential equation. H n  Z ne x

2

H n ' Zn 'ex

/2 2

H n ' Z n ''e x

/2 2

/2

 Z n xe x

2

/2

 Z n ' xe x

2

/2

 Z n ' xe x

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2

/2

 Z n [e x

2

/2

 x2e x

2

/2

]

-9-

Now that we have all necessary derivatives we can just plug them all into Hermites differential equation and get a new differential equation. H n  Z ne x

2

H n ' Zn 'ex

/2 2

/2

H n '' Zn ''e x

2

 Z n xe x

/2

/2

 Z n ' xe x

y ' '2 xy '2my  0

Z ' ' e  Z ' xe  0  2mZ e x2 / 2

n

2

n

2

/2

 Z n ' xe x

2

/2

 Z n [e x

2

/2

 x 2e x

2

/2

]

(9) x2 / 2

 Z n ' xe x

2

x /2

2

/2

 Z n [e x

2

/2

 x 2e x

2

/2

 

]  2x Z n ' e x

2

/2

 Z n xe x

2

/2



(25)

n

After simplifying (25) a bit and rearranging terms we will end up with an equation that look strikingly similar to that of equation (8) which we are trying to solve. The comparison is given below. Z n ' ' ((2m  1)  x 2 ) Z n  0  2mE kx 2   ' ' 2  2   0    

(26)

(8)

The only issue with our equations is that in (8) we are lacking a coefficient of one in front of the x squared term. If we could only make (26) look identical to (8) we would have our solutions and be able to determine or energy relationship. To do this we need to define a new wave function in terms of a new variable that will satisfy (26).

 ( p)   ( x) p  x

(27)

(28)

 ( x)   ( p /  )   ( p)

Copyright © Ryan Jadrich 2007

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With are new expressions we will need the necessary second derivative of are new function to substitute into (8). This is worked out below. d ( x) dx d ( p )   dx dp dp d   ' [ p /  ]  ' dp d ' dx d '    (1 /  )  dx dp dp   ' ' (1 /  ) 2   ' '   ' '   2 ' '

(29)

Our next step is to substitute equations (27), (28), and (29) into (8) to get an equation in terms of our new wave function.  2mE kp 2   2  ' ' 2  2 2   0       2mE kp 2   ' ' 2 2  4 2   0     

(30)

Now that we have this extra constant multiple which can by the way we defined it to be anything we have an extra degree of freedom in the manipulation of our differential equation. We will choose it so that it makes the constant in front of the p squared term equal to one.

k 1  4 2 mk 2  

(31)

Then after plugging in alpha (31) we will get the differential equation below.  2E  ' '  

 m  p 2   0 k 

(32)

Now comparing (32) with (26) we shockingly see that they are identical.  2E m   ' '  p 2   0   k  2 Z n ' '((2m  1)  x ) Z n  0

(32)

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Therefore equation (32) has the same solutions as in (26) which is the product of a Hermite polynomial and a portion of the necessary weighing function to make different polynomials orthogonal to one another. If this were not true then the wave function would not approach zero at infinity and the probability of finding a particle in a localized region would be zero and this can’t be. So are solutions are given below: Z n ( p )  H n ( p )e  p

2

/2

Z n ( x)  H n (x)e 

2 2

x /2

(33)

( k )1 / 4   1/ 2 

The next bit is about the factor needed to normalize the solutions. This normalized wave function is stated below,

 norm  (2  m / 2  1 / 4 (m!) 1 / 2 ) Z n ( x)  (2  m / 2  1 / 4 (m!) 1 / 2 ) H n (x)e 

2 2

x /2

(34)

To prove this we will use the expression that we already proved which states that, 

H

H m e  x dx  2 m (m!)  2

m



Now using the Z functions which are our solutions that are stated in (33) we will attempt to normalize these functions. 



A 2  Z m Z m dx  A 2  H m (x)e  

2 2

x /2

H m (x)e 

2 2

x /2

dx  1





A

2

  H m (x) H m (x)e

2 2

x

dx  1



 A 2 (2 m )(m!)   1  A

1 (2 m )(m!) 

 A  (2 m / 2  1 / 4 (m!) 1 / 2 )

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Which is precisely the normalization coefficient that we had hoped for? With this normalization, 



 norm

2

1



And this is exactly what we wanted. So all in all our solutions are given by (34), and these give the probability amplitudes for a quantum mechanical harmonic oscillator. The last bit refers to the energy levels of the quantum mechanical harmonic oscillator. The allowed energy Eigen-values can be figured out by comparing equation (32) and (26) and by noting that m must be an integer to get this Hermite polynomial like solutions. 2E 

  2m  1 k

k 1 E  ( n  )  2 1 E  (n  ) 2

(35)

From (35) it can easily be seen that the energy can only take on certain values as is expected of a quantum system.

Sources: Silbey, Robert J., Robert A. Alberty, and Moungi G. Bawendi. Physical Chemistry. 4th ed. Wiley, 2005. Pahikkala, J.. "Hermite polynomials." PlanetMath.org. October 10 2006 . PlanetMath.org. 1 Dec 2007 . Aung, Pye, Phyo. "Application of Hermite Polynomials in the Quantum Simple Harmonic Oscillator." Physical Chemistry I Legacy Project

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