Simple Harmonic Oscillator - Schrodinger Equation

1. Nov 26, 2009

cpmiller

1. The problem statement, all variables and given/known data

One possible solution for the wave function ψn for the simple harmonic oscillator is

ψn = A (2*αx2 -1 ) e-αx2/2

where A is a constant. What is the value of the energy level En?

2. Relevant equations

The time independent Schrodinger wave equation

d2ψ / dx2 = (α2 * x2 - β )ψ

β = 2mE/ $$\hbar$$2

3. The attempt at a solution

So I took the solution that we were given ψn and found the second derivative of this. (I found A[3α + 3α2x2 -2α3x4]e-αx2/2, but I haven't triple checked my algebra yet, I'll do that later).

So I have
d2ψ / dx2 = (α2 * x2 - β )ψ

A[3α + 3α2x2 -2α3x4]e-αx2/2 = (α2 * x2 - β ) * A (2*αx2 -1 ) e-αx2/2

(In a nutshell I plugged the solution we were given into the time independent Schrodinger wave equation).

Then I did some tedious algebra, observed that my A's cancelled, and my exponentials cancelled, leaving me with

3α + 3α2 x2 - 2α3x4 = (α2 x2 - β) (2αx2 -1)

So some tedious algebra and polynomial division (oh and subsitituting β = 2mE/ $$\hbar$$2) yields

E = [2 α x2 - α - 4α/[2αx2 -1] ]

Which would all be well and good, except I have approximately 0 confidence in this answer.

I'm finding this material really confusing, and I don't really know what an answer should look like, is it okay to have α's in it? I don't know... It seems too convenient that my A's just cooperatively went away, that part just seems too good to be true.

Anyway, if anybody could give me any encouragement, or pointers towards another way of looking at this, I'd really appreciate it!

Thanks!

2. Nov 26, 2009

Feldoh

Well, I didn't check your math but it seems that your Hamiltonian does not include the potential energy for the harmonic oscillator.

I think you should get something proportional to $$\hbar \omega$$

3. Nov 26, 2009

cpmiller

Argh....Double and triple checking my post and I leave out obvious things...

α2 = m*k / $$\hbar$$2

The potential is k*x2 /2 so having the α in the Shrodinger equation should include the potential. I'm relatively certain that the Schrodinger equation that I gave above is correct, in that it is derived in the book (and I've rechecked my typing....)

However, you are right about the solution that I got at the end, I meant to type:

E = [hbar /2m] [2 α x2 - α - 4α/[2αx2 -1] ]

(For some reason it really wants that hbar to be a kappa... it shows up as hbar in my preview and then kappa when I submit the post.... mysterious, but beyond my comprehension... more about this problem to make me feel stupid )

So does my approach at solving this seem reasonable? I had another thought about how to solve it using the Hermite polynomials, such that my solution would be given in terms of A rather than alpha, but I don't know if that would be a reasonable approach...

Last edited: Nov 27, 2009
4. Nov 27, 2009

Feldoh

No your answer doesn't make sense. The energy should be position independent.

5. Nov 27, 2009

jdwood983

I think your energy eigenvalue doesn't make sense because you are solving the Schroedinger equation for a free particle when you are told that it is a possible solution to the simple harmonic oscillator. Your time-independent Schroedinger equation will be:

$$-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}+V(x)\Psi(x)=E\Psi(x)$$

You should plug in the trial wavefunction and see what you come to using this.

EDIT: Oh yeah, $V(x)=\frac{1}{2}m\omega^2x^2$.

Last edited: Nov 27, 2009
6. Nov 27, 2009

cpmiller

Feldoh,

Thanks for your response. I had a feeling my solution wasn't right and I couldn't figure out why... at least I've learned a way to know I'm wrong.

Jdwood,

I think that my Schroedinger equation is for the harmonic oscillator. I like your way of writing it a lot better than the way my book did it with the alphas and betas, I think I understand the meaning better now that I see what you did... but if you plug in the values for alpha and beta (and do a little bit of multiplication) you get:

- [hbar /2m] * d2ψ / dx2 = - (k/2) x2 + E * ψ

which is what you have, except with potential written in terms of k (because it's like a spring constant). I think maybe plugging into Schroedinger isn't the way to go :-(

7. Nov 27, 2009

jdwood983

Plugging into the Schroedinger equation is the only way to go. I see what your textbook does in trying to simplify the mathematics by using substitutions, but sometimes too many substitutions lead to not seeing the big picture.

So I went back to the beginning:

$$\Psi(x)=\left(2\alpha x^2-1\right)\exp\left[-\frac{\alpha x^2}{2}\right]$$

then the first derivative gives

$$\frac{\partial\Psi}{\partial x}=4\alpha x\exp\left[-\frac{\alpha x^2}{2}\right]-\alpha x\left(2\alpha x^2-1\right)\exp\left[-\frac{\alpha x^2}{2}\right]$$

and the second derivative

$$\frac{\partial^2\Psi}{\partial x^2}=\alpha^2x^2\left(2\alpha x^2-1\right)\exp\left[-\frac{\alpha x^2}{2}\right]-\alpha\left(2\alpha x^2-1\right)\exp\left[-\frac{\alpha x^2}{2}\right]-8\alpha^2x^2\exp\left[-\frac{\alpha x^2}{2}\right]+4\alpha\exp\left[[-\frac{\alpha x^2}{2}\right]$$

But this can be reduced to

$$\frac{\partial^2\Psi}{\partial x^2}=\left(2\alpha^3x^4-11\alpha^2x^2+5\alpha\right)\exp\left[-\frac{\alpha x^2}{2}\right]$$

So this is where it seems you went awry, your untriple-checked algebra! This should help.

EDIT: Whoops, I forgot the factor of $A$ in the math above. Each term should have that constant multiplied in front so that you can cancel it in the end.

8. Nov 27, 2009

cpmiller

Jdwoods,

Thanks so much for your help!

I really appreciate both the math checking and the encouragement that my way of approaching the problem wasn't totally off base, plus your writing of the Schroedinger equation helped me understand this better!

I think I'm going to have to learn LaTeX, your equations are a lot easier to follow than mine....

Thanks again,

Caroline

9. Nov 27, 2009

jdwood983

LaTeX is definitely the best way to write papers, journals, and (if your teacher allows it) homework. I've been using http://www.lyx.org/" [Broken] as my go-to editor for any LaTeX needs (completely free and usable in any operating system) for several years (I believe I first used version 1.4.something, they're on version 1.6.4 now)

The best way to learn LaTeX is to keep at it. When you post here, click the capital Sigma ($\Sigma$) to show the LaTeX Reference and use the TeX environment. It may take longer to post homework questions, but it'll go faster as you learn it.

Last edited by a moderator: May 4, 2017