Schrodinger equation, potential energy function

1. Apr 8, 2013

Cogswell

A particle of mass m is in the state
$$\Psi (x,t) = A e^{-a[(mx^2 / \hbar ) + it]}$$

Find A
For what potential energy function V(x) does $$\Psi$$ satisfy the Schrodinger equation?

Do I just re-arrange for A? (Sorry if I seem really dumb). I'm not really getting this.

2. Apr 8, 2013

Staff: Mentor

To find $A$ means to normalize the function. You want
$$\int_{-\infty}^{\infty} \left| \psi(x,t) \right|^2 dx = 1$$

For the second part, you need to plug the wave function in the Schrödinger equation to find $V(x)$. My guess is that you will need to use both the time-dependent and the time-independent Schrödinger equations.

3. Apr 8, 2013

Cogswell

Here's my attempt, which took over 2 hours >.<

I hope the final answers are right, but am I going about it the right way?
Sorry the photos are blurry.

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4. Apr 8, 2013

Staff: Mentor

For the normalization, you have to integrate $| \psi(x,t) |^2 = \psi(x,t)^* \psi(x,t)$. Time should disappear for the final result for $A$.

5. Apr 8, 2013

Staff: Mentor

I've check the calculation of $V(x)$ and you got it right!

6. Apr 8, 2013

Cogswell

Sorry the picture is really blurry. My final answer does have time in it:

$$A = (\dfrac{2ame^{4ait}}{\pi \hbar})^{1/4}$$

7. Apr 8, 2013

Staff: Mentor

And that is not correct. I think you forgot to take the complex conjugate.

8. Apr 8, 2013

Cogswell

If I do that wouldn't everything cancel out before I even integrate it?

$$\int_{- \infty}^{\infty} A A^* e^{-a[mx^2 / \hbar + it]} e^{a[mx^2 / \hbar + it]} dx = 1$$

And so that would just become:

$$\int_{- \infty}^{\infty} A A^*dx = 1$$

??

9. Apr 8, 2013

Staff: Mentor

No, because only the imaginary part changes sign:
$$\left( A e^{-a [m x^2/\hbar + i t]} \right)^* = A e^{-a [m x^2/\hbar - i t]}$$

10. Apr 8, 2013

Cogswell

Ah yes, that makes sense.

After a bit of working out again, I still think I'm missing something...
Even if I use $$\left( A e^{-a [m x^2/\hbar + i t]} \right)^* = A e^{-a [m x^2/\hbar - i t]}$$
my integral becomes:

$$\int_{-\infty}^{\infty} A e^{-a [m x^2/\hbar + i t]} A e^{-a [m x^2/\hbar - i t]} dx = 1$$

Which becomes:

$$\int_{-\infty}^{\infty} A^2 e^{-a m x^2/\hbar} e^{ -a i t} e^{-a m x^2/\hbar} e^{a i t} dx = 1$$

If I evaluate the integral, I get the final result of $$A = \displaystyle \left(\frac{2am}{ \pi \hbar}\right)^{1/4}$$

And then the time cancels out. So how would I get time left in the final result?

Last edited: Apr 8, 2013
11. Apr 9, 2013

Staff: Mentor

That is the correct answer. You might have misunderstood what I wrote earlier, but time should disappear from the final answer.

12. Apr 10, 2013

Cogswell

Great, thank you!
Is there a 'thanks' or 'like' button on these forums? I can't seem to find it.

13. Apr 10, 2013

Staff: Mentor

You're welcome!

Not that I know of.