# Shifted infinite well

1. Sep 17, 2009

### E92M3

I can solve the infinite will with length a. But what happens when the coordination is shifted, namely:
$$V(x)=0,\frac{ -a}{4}<x<\frac{3a}{4}$$

I use the usual solution:
$$\psi=Asin(kx)+Bcos(kx)$$
Now when I apply the first boundary condition:
$$\psi(\frac{ -a}{4})=\psi(\frac{3a}{4})=0$$
I can't get rid of one of the terms (namely the B=0 like when the well was defined from 0 to a).
I know that A will be the same as if the well was defined from 0 to a and I can get k by applying the other boundary condition, but I can't do anything if I can't get rid of the cos term.

2. Sep 17, 2009

### Avodyne

If psi(x) is a solution to the Schrodinger equation with potential V(x), then psi(x-b) is a solution to the Schrodinger equation with potential V(x-b). So all you need to do is shift the solution you have by the correct amount.

You can also do it your way. Use psi(-a/4)=0 to solve for B in terms of A. Plug this into psi(3a/4)=0; every term will have an A in it, so cancel out A, since you know it isn't zero. Now you have an equation whose only unknown is k. You should be able to show that the solutions are k = pi n/a, where n=1,2,3,...

3. Sep 18, 2009

### E92M3

This is not working too well either.
Solving for B:
$$\psi(\frac{-a}{4})=Asin(k\frac{-a}{4})+Bcos(k\frac{-a}{4})=0$$
$$B=\frac{-Asin(k\frac{-a}{4})}{cos(k\frac{-a}{4})}=\frac{Asin(k\frac{a}{4})}{cos(k\frac{a}{4})}$$
Then:
$$\psi(\frac{3a}{4})=Asin(k\frac{3a}{4})+\frac{Asin(k\frac{a}{4})}{cos(k\frac{a}{4})}cos(k\frac{3a}{4})=0$$
$$\frac{sin(\frac{3ak}{4})}{cos(\frac{3ak}{4})}+\frac{sin(\frac{ak}{4})}{cos(\frac{ak}{4})}=0$$
$$\frac{sin(\frac{3ak}{4})cos(\frac{ak}{4})+cos(\frac{3ak}{4})sin(\frac{ak}{4})}{cos(\frac{3ak}{4})cos(\frac{ak}{4})}=0$$
$$sin(\frac{3ak}{4})cos(\frac{ak}{4})+cos(\frac{3ak}{4})sin(\frac{ak}{4})=0$$
$$sin(\frac{3ak}{4}+\frac{ak}{4})=sin(ak)=0$$
Then I indeed recover:
$$k=\frac{n\pi}{a}$$
Plugging it back in:
$$\psi(x)=Asin(\frac{n\pi}{a}x)+Bcos(\frac{n\pi}{a}x)$$
This is not what I expected and I am quite sure that it is wrong since if I plug in the limits, psi is non-zero at the end points.

Further more, this should be just the same as the infinite well with length a shifted over. So shouldn't the solution be the same but with a phase constant added? Namely:
$$\psi=Asin(\frac{n\pi}{a}x)\rightarrow\psi=Asin(\frac{n\pi}{a}(x+\frac{a}{4}))$$
I think that this is the answer, but that's kind of cheating since it is reverse engineered from another infinite well. I can't figure out why I can't do it top down.

Last edited: Sep 18, 2009