# Homework Help: A gaussian wavefunction of the harmonic oscillator

1. Oct 12, 2014

### photomagnetic

1. The problem statement, all variables and given/known data
A particle of mass m in the harmonic oscillator potential V(x) = (mω2x2)/2 is described at time
t = 0 by the wavefunction

χ(x, t = 0) = 1/[(2πσ2)1/4] exp[-x2/(4σ2)]

What is <E> at time t?

2. Relevant equations

3. The attempt at a solution
<T>+<V>= <E>

I've found the expectation value of V, which is the easy part. just finding x^2 expectation value is enough.
for <T> the formula is pretty straight forward BUT the integral is getting out of hand.
How can I just simplify it?
There is a hint "Hint: Use generating function to calculate overlap integrals" but no matter what I try to do it doesn't look like the integral formula given by the hint.

ey2-(s-y)2= sigma [Hn(y)/n!] sn

Basically it should look like a hermite polynomials. but how =(

2. Oct 13, 2014

### Simon Bridge

After a few lines of <T> you end up with having to evaluate a gaussian integral right?
How would you normally do that?

3. Oct 13, 2014

### vela

Staff Emeritus
Please show us what you've done so far.

4. Oct 13, 2014

### photomagnetic

I've found <V>= S Y*(x,t) (mw2x2/2) Y(x,t) which was easy to integrate

for the energy part in the wave function I wrote down e-i[(p2/2m)+V(x)]

<T>= (-h2/2m) S Y* (a2Y /ax2Y)

I am trying to find this. I know that the wave function is the zeroth state's wave function. So <E> should be hbar.w/2

(sorry for my notation)

5. Oct 13, 2014

### vela

Staff Emeritus
6. Oct 13, 2014

### vela

Staff Emeritus
How is $\sigma$ in the wave function defined? Is it an arbitrary constant or does it have a specific value in this problem?

7. Oct 13, 2014

### photomagnetic

it is (mw/pi hbar)= 1/σ2
but I don't think I can use that, you need to find the answer w/o knowing σ. so that I will be able to show <E>= hbar w/2

if I use σ as it is, calculating that integral becomes a painful activity.

8. Oct 13, 2014

### vela

Staff Emeritus
I don't think you calculated $\langle V \rangle$ correctly. You may want to rethink your approach to this problem.

The idea behind using the generating function is that
$$\int e^{-2xt-t^2}f(x)\,dx = \int \left[\sum_{n=0}^\infty H_n(x) \frac{t^n}{n!}\right]f(x)\,dx = \sum_{n=0}^\infty \frac{t^n}{n!}\left[\int H_n(x) f(x)\,dx\right].$$ We evaluate the integral on the left, expand the result in powers of $t$, and identify it with the expression on the right.

For example, consider the integral
$$\int e^{-2xt-t^2}x^2 e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2} (1+2t^2),$$ which you can verify relatively easily. This result tells us that
\begin{eqnarray*}
\int H_0(x)x^2 e^{-x^2}\,dx &= \frac{\sqrt{\pi}}{2}, \\
\int H_2(x)x^2 e^{-x^2}\,dx &= \frac{\sqrt{\pi}}{2} 2\times 2! = 2\sqrt{\pi},
\end{eqnarray*} and all of the other integral vanish.

9. Oct 13, 2014

### photomagnetic

<YlHlY> this gives the expectation value of energy right?

then <V> gives hbarw/4 which is the answer for the expectation value of the potential energy.
I am 100% sure about that part.
well I'll see about the next part. thank anyway.

10. Oct 14, 2014

### vela

Staff Emeritus
If you don't know the value of $\sigma$, you can't say that $\psi(x,0)$ is an energy eigenstate, so how did you deal with the time dependence when trying to calculate $\langle \psi(x,t) \lvert V \rvert \psi(x,t) \rangle$?

11. Oct 22, 2014

### photomagnetic

for harmonic oscillator's wavefunctions.
just like stated in the question.
time doesn't have a role. after calculating the Cn's you see the <E(t)> is equal to <E(t=0)>
after doing that it's simple. I've found the rest as well. Thanks for the replies.