Bound states with energy below the bottom of the well?

1. V(x) V0 x −L/2 0 L/2 Bound states with energy below the bottom of the well? A few students asked whether the square, finite potential well could have a bound solution with energy below the bottom of the well. Could there be such a solution?

2. V(x) V0 x −L/2 0 L/2 Let’s try to find a solution with energy E < 0. Since the well is right-left symmetric, we can assume the solution will have definite parity. Inside: The parity-even solution is Acosh kx and the parity-odd solution is Asinh kx, where outside inside outside

3. V(x) V0 x −L/2 0 L/2 Outside: To be normalizable, the solution can only be A′e−k′|x|where k′ = outside inside outside

4. We consider symmetric (cosh) and anti-symmetric (sinh) solutions. For the symmetric solution, the continuity conditions at x = L/2 are together they imply k tanh kL/2 = –k′, which cannot be because k, tanh kL/2 and k′ are all positive! V0 outside inside outside x −L/2 0 L/2

5. For the anti-symmetric solution, the continuity conditions at x = L/2 are together they imply k coth kL/2 = –k′, which likewise cannot be because k, coth kL/2 and k′ are all positive! V0 outside inside outside x −L/2 0 L/2