Eigenstates of a Particle in a Box at x = a/2?

[Domnu]

What are the eigenstates of a particle in a box whose bounds are $$x = -a/2$$ and $$x = a/2$$?

Solution
Well, the eigenstates where $$x = 0, a$$ are just

$$\varphi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi x}{a}$$,

so why wouldn't the eigenstates just be

$$\varphi_n = \sqrt{\frac{2}{a}} \sin \frac{n \pi (x+a/2)}{a} = \sqrt{\frac{2}{a}}\sin \left(\frac{n \pi x}{a} + \frac{n \pi}{2}}\right)$$

?

 

[Avodyne]

No reason at all; those are the eigenstates!

 

[Domnu]

Yes, I just figured out the reason... I was slightly afraid of the negative sign that resulted in my answer for the Sin x solution, but it doesn't matter... negative signs don't have any effect on eigenstates.