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Rewriting the propagator for the free particle as integral over E

  1. Nov 18, 2009 #1
    1. The problem statement, all variables and given/known data

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

    [tex] 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 [/tex].

    2. Relevant 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.

    3. 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?
     
  2. jcsd
  3. Nov 19, 2009 #2

    lanedance

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    Homework Helper

    could the constant come from the intgeration variable change?

    ie what is dE in terms of dp?
     
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