My understanding is that a classical idealized particle, moving in one dimension, with momentum p and kinetic energy T comes into contact with an infinite step-function potential V, there will be an (instantaneous) elastic collision - the particle's momentum becomes -p, so its energy remains constant ( momentum is conserved by transferring 2p additional momentum to the barrier, but if its mass is taken to be infinite, this will not change its velocity/energy). Further, if I'm not mistaken, the same thing will happen with any finite V, where V > T. First: are both of those statements accurate? If so , I'm having trouble showing it in Hamiltonian form. As I understand it, the whole system, where the barrier is at x=0, should be describable by H = p2/2m - Vθ(x) Where θ(x) is the unit step function (x<0 θ(x)=0; x>=0 θ(x)=1), and V is a constant > p2/2m. So, Hamilton's equations give ∂H/∂p = dx/dt = p/m -∂H/∂x = dp/dt = -Vδ(x) Differentiating the first and setting it equal to the second gives us a Newtonian-style F=ma equation (F = dp/dt; a = d2x/dt2): m a= -Vδ(x) However, this can't be right, because as far as I can tell, the force at x=0 is proportional to V -- specifically, the total change in momentum between x-dx and x+dx is exactly V...while for an elastic collision, it would be -2p -- so I'm apparently missing something. Any ideas?