Infinite Well Solutions: How Can Different Techniques Yield Contrasting Results?

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Homework Statement


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

[tex]\psi_n(x) = \sqrt{2/a} \sin (n \pi x/ a)[/tex]

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:[tex]\psi_n(x) = \sqrt{2/a} \sin (2 \pi n x/a)[/tex]

How can these two techniques give different results?

Homework Equations


The Attempt at a Solution

 
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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?
 
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?
 
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Maybe the problem is that I am assuming that the k_n is the same for both sine and cosine?
 
OK. The problem was that in post number 5 I divided by A which could have been 0.