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(\f rac{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.