Parity - Particle In A Box (Infinite Potential)

[catinabox]

I am trying to work out the wavefunctions for a particle in a box between -a/2 and a/2.I have already gone through the solution for a box between 0 and a and got the solution $$\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a} )$$So I can see that for -a/2 to a/2 I have $$\sqrt{\frac{2}{a}}sin(\frac{n\pi(x+\frac{a}{2})}{a})$$Which by some trig leads to $$\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})cos(\frac{n\pi}{2})+\sqrt{\frac{2}{a}}cos(\frac{n\pi x}{a})sin(\frac{n\pi}{2})$$Now i can see it differs for even and odd n as for even n $$sin(\frac{n\pi}{2})=0$$ for odd n $$cos(\frac{n\pi}{2})=0$$.

(NOT SURE WHATS HAPPENED WITH LATEX HERE :()Therefore even n leads to $$\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})cos(\frac{n\pi}{2})$$ odd n leads to $$\sqrt{\frac{2}{a}}cos(\frac{n\pi x}{a})sin(\frac{n\pi}{2})$$From research I have found that the wavefunction for n even is in fact just $$\sqrt{\frac{2}{a}}sin(\frac{n\pi x}{a})$$ and odd n just $$\sqrt{\frac{2}{a}}cos(\frac{n\pi x}{a})$$This is were I am confused because the $$cos(\frac{n\pi}{2})$$ for even n is positive or negative 1 and $$sin(\frac{n\pi}{2})$$ for odd n is positive or negative 1.Why is only the positive chosen, is this to do with normalistion?Any help is much appreciated.Thank you.

 

[bigevil]

n only has physical meaning for positive values because if you evaluate the energy eigenfunctions, you find that

$$E_n = \frac{n^2 \pi^2 \hbar^2}{2ma^2}$$

Mathematically, at least, it is unnecessary for n to be less than 0. Going by this equation, E0 is by definition the lowest possible energy state, or ground state.