# Momentum Operator

1. Dec 24, 2011

### hokhani

Consider a particle in one dimention. there is a dirac delta potential such as V=-a delat(x). The wave functions in two sides(left and right) are Aexp(kx) and Aexp(-kx) respectively. So if the momentum operator acts on the wave functions, would give two complex numbers while the momentum operator is Hermitian. what is the justification here?

2. Dec 24, 2011

### Bill_K

To see it more clearly, step away from the delta function for a moment and consider instead a finite square well potential, V(x) = -V0 for |x| < a and zero otherwise. Assuming the well is deep enough to have a bound state, the ground state wavefunction will be sinusoidal within the well and exponential without. Inside the well the kinetic energy p2/2m is positive and p is real. But outside the well is a classically forbidden region where the kinetic energy is negative and p is imaginary.

This is perfectly consistent with p being Hermitian, because the thing that must be real is not p itself, it's the matrix element or expectation value ∫ψ*pψ dx