Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2019 / Paper 2 / Section A(ii) / 15 / iii / Solution

For $E>0$, write $\psi =A\sin kx+B\cos kx$ with $k^2=2mE/{\hslash }^2$. The condition at zero gives $B=0$; the condition at $L$ gives $kL=n\pi$, where $n=1,2,\dots$. The $n=0$ solution is identically zero and is not a state. Normalization gives $A^2{\int }_0^L{\sin }^2(n\pi x/L)dx=A^{2}L/2=1$. HenceChoosing positive $A$ fixes an irrelevant overall phase.