Currently going through an old question which concerns a particle of mass $m$ on the interval $[-a,a]$ with potential $V=V_0$. In the question it says to show that the energy levels of the particle are $E=V_0+\frac{n^2\pi^2\hbar^2}{8ma^2}$, but after numerous tries and checks I keep getting $E=V_0+\frac{n^2\pi^2\hbar^2}{2ma^2}$, just want to check that the answer in the question isn't just wrong so I stop wasting my time trying to figure out where I went wrong. Thank you for your help!
WORKING:
Letting $k=\frac{(2m{E-V_0)^\frac{1}{2}}}{\hbar}$, I got the only non trivial solution to the SE is $Acos(kx)+Bsin(kx)$ and using the conditions that the wave function should be 0 at the endpoints, I got $ka=n\pi$, but the answer in the question works only if $ka=\frac{n\pi}{2}$ and im not sure why this should be the case.
The wavefunction must be assumed to be $0$ at the endpoints $\pm a$. Therefore $$\begin{cases}A\cos (ka)+B\sin (ka)=0\\ A\cos (-ka)+B\sin(-ka)=A\cos(ka)-B\sin(ka)=0 \end{cases} $$ Hence there are two classes of solutions:
odd solutions: $A=0$ and $ka=n\pi$, $n\in \mathbb{Z}$.
even solutions: $B=0$ and $ka=n\pi+\pi/2$, $n\in \mathbb{Z}$.
Combining odd and even solutions you have a solution for any $k_n=\frac{n\pi}{2}$, $n\in \mathbb{Z}$ and its associated energy is given by $$E_n=\frac{\hbar^2k_n^2}{2m}+V_0=\frac{n^2\pi^2\hbar^2}{8ma^2} +V_0,\qquad n\in \mathbb{N}$$