Why shifting the infinite well gives different wavefunctions

[ehrenfest]

Homework Statement

The time-independent Schrödinger equation solutions for an infinite well from 0 to a are of the form:

$$\psi_n(x) = \sqrt{2/a} \sin (n \pi x/ a)$$

If you move the well over so it is now from -a/2 to a/2, then you can replace x with x-a/2 and get the new equations right?

If I try to actually apply these new boundary conditions to the sin function, I get something different, however:$$\psi_n(x) = \sqrt{2/a} \sin (2 \pi n x/a)$$

How can these two techniques give different results?

Homework Equations

The Attempt at a Solution

 

[Dick]

Probably because you did it wrong. You managed to halve the period of a wavefunction by translating it. That's not supposed to happen, is it?

 

[ehrenfest]

I just applied the boundary conditions to sin(k_n * x) by plugging in (a/2, 0) or (-a/2, 0).

This yields k_n = 2 n pi/a?

 

[Meir Achuz]

For the symmetric well, there are also solutions cos(n\pi x), with n odd.

 

[ehrenfest]

But I get A cos (k_n a/2) + B sin(k_n a/2) = 0 and A cos (k_n -a/2) + B sin (-k_n a/2) = 0 implies that B/A = tan (k_n a/2) = -B/A which implies that B is 0?

 

[ehrenfest]

Maybe the problem is that I am assuming that the k_n is the same for both sine and cosine?

 

[Dick]

sin(n*pi*(x-a/2))=sin(n*pi*x-n*pi/2). Clearly this is +/- cos(n*pi*x).

 

[ehrenfest]

OK. The problem was that in post number 5 I divided by A which could have been 0.