Assume that you have a one dimension box with infinite energy outside, and zero energy from 0 to L. Then my understanding of the Schrodinger equation is that the equation inside will be: -h^2/2m*d2/dx2ψ = ihd/dtψ And the energy eigenstates are given by ψ(x,t) = e-iwt*sin(kx) where k = n*π/L and w = k2h/2m and n = 1,2,3... Well, that's nice but what about if there is a positive potential energy U0 inside? -h^2/2m*d2/dx2ψ + U0ψ = ihd/dtψ What is the solution ψ(x,t) in explicit form? And how do you find the energy eigenstates? My guess at solving this would be to solve -h^2/2m (ψ)'' = (E-U)ψ as an ODE. I still get ψ = sin(kx), but I'm not sure whether this satisfies the boundary conditions or not. Also, I get that k2 = 2m(E-U)/h2. My naive assumption is that ψ(0,t) = ψ(L,t) = 0 since it can't exist pass an infinite potential wall and because then the function would be continuous, since ψ = 0 at all other points. If this is true, then energy eigenstates of the particle must be ψ(x) = Asin(nπ/L*x), where A is any arbitrary phase factor and n = 1,2,3... Thus, assuming ψ(x,t) = e-iwtψ(x) Thus we get h2k2/2m*ψ(x,t)+U*ψ(x,t) = hw*ψ(x,t) Factoring out, we get h2k2/2m + U = hw = E Thus w = hk2/2m + U/h = h/2m*(nπ/L)2 + U/h My interpretation of this is that given a certain n, the eigenstates "look" the same in complex phase space, but they are spinning around faster at a rate of U/h. Energy is quantized in this box to E = h/2m*(nπ/L)2 + U where n = 1,2,3... My main problem arises when U is negative, and chosen such that U = -h/2m*(π/L)2, thus leaving the total energy at the ground state as zero, and the wave function is "standing still" with w = 0. What's going on here? Even worse is when U is greater than E0, the angular frequency is negative, does that mean the wavefunction is spinning in the opposite direction as normal positive energy wavefunctions?