Problem: The particle in a 1D box [0, a] Eqs.: The general solution of the time-independent Schrödinger eq. may be written as ψ(x) = Acos(kx) + Bsin(kx), E = ħ2k2/2m. Imposing the boundary conditions ψ(0) = ψ(a) = 0 , we get immediately A = 0, ka = nπ (for any positive integer n). Using x' = x - a/2 , plus the boundary condition and elementary trigonometry, we also get ψ(x) = Bsin(kx) = Bcos(kx') , same eigenvalue. Everything is nice and dandy up to now, but, if I start from the box [-a/2, a/2] and impose boundary conditions ψ(-a/2) = ψ(a/2) = 0 , I seem to be getting B = 0, ka = nπ (but this time for odd n only) What's the reason for this inconsistency? Either in the first case we should take only odd n also, and nobody noticed it in QM textbooks (unlikely), or I'm making some silly mistake, but I can't point my finger where; I've checked the sol. to the boundary condition system many times, it seems fine to me.