Infinite square well centered at zero from -b to b

[cepheid]

Homework Statement

You don't need it verbatim. I'm just trying to solve for the eigenstates and eigenvalues of the Hamiltonian for a one-dimensional infinite square well, with a particle of mass M inside. I'm embarrassed to say it, but the question is throwing me off because the infinite well is centred at zero, ranging from -b < x < b rather than from 0 < x < a.

Homework Equations

$$\mathcal{H}\psi(x) = E\psi(x)$$

$$\mathcal{H} \equiv -\frac{\hbar^2}{2M}\frac{d^2 }{dx^2}\left(\right) + V()$$

$$V = 0 \ \ \mbox{inside the well.}$$

The Attempt at a Solution

Immediately from the ODE the solution is obviously:

$$\psi(x) = A \sin \left(\frac{\sqrt{2ME}}{\hbar}x\right) + B \cos \left(\frac{\sqrt{2ME}}{\hbar}x\right)$$

$$\psi(b) = \psi(-b) = 0$$

$$\Rightarrow A \sin \left(\frac{\sqrt{2ME}}{\hbar}b\right) + B \cos \left(\frac{\sqrt{2ME}}{\hbar}b\right) =0$$

$$\Rightarrow -A \sin \left(\frac{\sqrt{2ME}}{\hbar}b\right) + B \cos \left(\frac{\sqrt{2ME}}{\hbar}b\right) =0$$

I'm really not sure how to proceed. There is no value of b for which both the sine and cosine will be zero, suggesting that for some eigenstates, A is zero, and for others, B is zero. Griffiths even hints that even states are given by only cosines and odd states by sines, but I can't figure out how to arrive at this result systematically.

 

[Meir Achuz]

I recommend you use a variable z=x+b. Then the well will be between
z=0 and z=2b, with the efunction sin(n\pi z/2b).