Rewriting the propagator for the free particle as integral over E

[quasar_4]

Homework Statement

(This is all with respect to a free particle)
Show that the propagator $$U(t) = \int_{-\infty}^{\infty} |p><p| exp\left(\frac{-i E(p) t}{\hbar}\right) dp$$ can be rewritten as an integral over E and sum over the $$\pm$$ index as:

$$U(t) = \sum_{\alpha = \pm} \int_{0}^{\infty} \left(\frac{m}{\sqrt{2mE}}\right) |E, \alpha><E, \alpha| exp\left(\frac{-i E t}{\hbar}\right) dE$$.

Homework Equations

The Hamiltonian for a free particle is just H = P^2/2m (all kinetic energy) and the allowed values of momentum are plus or minus 2mE. So, for a given energy eigenvalue, there is a degenerate 2 dimensional subspace.

The Attempt at a Solution

Well... I can see that when we change to sum over E instead of P (or rather, integrate) that we have to change the limit of integration from 0 to infinity because the energy cannot be negative for a free particle. Also, it makes fine sense to me to sum over alpha because we're dealing with a degenerate subspace. What I can't figure out is the constant in front! Can someone help point me in the right direction?

 

[lanedance]

could the constant come from the intgeration variable change?

ie what is dE in terms of dp?