Particle in a box, boundary co-ordinate change

[I_laff]

If you have a particle in a 1-d box with a finite potential when $0 < x < L$ and an infinite potential outside this region, then the normalised wavefunction used to describe said particle is $\psi (x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$.
However, if you had say instead a finite potential when $\frac{-L}{2} < x < \frac{L}{2}$ and an infinite potential outside this region, then wouldn't the wavefunction now change?

 

[I_laff]

Actually would the wavefunction be $\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})$ instead?

 

[hilbert2]

Yes, it would.

 

[Nugatory]

Actually would the wavefunction be $\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})$ instead?

Assuming you've done the math right (I haven't checked), yes. However, this change has no physical significance because the problem is still the same square well of width $L$. It's analogous to what happens if we decide to draw the prime meridian through something other than the Greenwich observatory - everything that is a function of longitude changes, but the surface of the Earth does not.

You can set the problem up either way, and in practice you will pick whichever way makes the algebra easiest for you.