1D infinite square well and parity

  • Thread starter siifuthun
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  • #1
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I think I'm on the right track for this problem, but I'm not entirely sure.

Find the solutions to the one-dimensional infinite square well when the potential extends from -a/2 to +a/2 instead of 0 to +a. Is the potential invariant with respect to parity? Are the wave functions? Discuss the assignment off odd and even parity to the solution

So the wave functions should look the same as if it was an infinite square well going from 0 to +a, except it's going from -a/2 to a/2, so the parity should be exactly the same, right? V(x) = infinity at x< -a/2 and when x> a/2.

So I tried solving for the Schrodinger equation using:

http://img145.imageshack.us/img145/4013/03bp1.jpg [Broken]

And we know that at -a/2 and a/2, the wavefunction must be equal to 0. We also know that for energy level E_1, where n=2, there's a node right at x=0, so the wavefunction must also equal 0 at x = 0 for that case. So:

http://img174.imageshack.us/img174/4408/04ya7.jpg [Broken]

So we know from above that B=0, plugging that into equation we get:

http://img98.imageshack.us/img98/5520/05ep7.jpg [Broken]

Now here's my problem, when I try to solve this, I get:

http://img170.imageshack.us/img170/5769/06vs2.jpg [Broken]

Which is a problem, since it means A=0. Am I wrong in my assumption that the wavefunction for E_1 state is 0 at x=0?
 
Last edited by a moderator:

Answers and Replies

  • #2
10
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There might be a problem with the last statement of E_1 state is 0 at x=0
.
Why do you think that is so?
 
Last edited:
  • #3
10
0
Problem with integration

I finally managed to find out what the heck it was. There is a problem with the integration of the function. Try using http://integrals.wolfram.com/index.jsp. Very useful mathematical program.
 

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