Schrödinger equation energy level

Question

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.

Note that the function $\cos (\pi x/2a)$ is a solution. – Hamed Apr 12 '18 at 18:32 — Hamed, Apr 12 '18 at 18:32

score 2 · Accepted Answer · answered Apr 12 '18 at 18:37

2

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 $$E_n=\frac{\hbar^2k_n^2}{2m}+V_0=\frac{n^2\pi^2\hbar^2}{8ma^2} +V_0,\qquad n\in \mathbb{N}$$

answered Apr 12 '18 at 18:37

Lorenzo Q

5,269

Thank you!! I was completely disregarding the even solutions for some reason! – Derren Apr 12 '18 at 18:48
@Lorenzo Why does it necessarily imply that $k_n=n\pi/2$? – UnsinkableSam Aug 09 '19 at 18:22

Schrödinger equation energy level

1 Answers1