Schrodinger equation, potential energy function

[Cogswell]

A particle of mass m is in the state
[tex] \Psi (x,t) = A e^{-a[(mx^2 / \hbar ) + it]}[/tex]Find A
For what potential energy function V(x) does [tex]\Psi[/tex] satisfy the Schrodinger equation?


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

 

[DrClaude]

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.

 

[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.

 

[DrClaude]

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##.

 

[DrClaude]

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

 

[Cogswell]

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##.

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

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

 

[DrClaude]

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

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

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

 

[Cogswell]

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

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

And so that would just become:

[tex]\int_{- \infty}^{\infty} A A^*dx = 1[/tex]

??

 

[DrClaude]

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

 

[Cogswell]

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

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:

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

Which becomes:

[tex]\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[/tex]

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

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

 

[DrClaude]

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

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

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

 

[Cogswell]

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

 

[DrClaude]

Great, thank you!

You're welcome!

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

Not that I know of.