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Demonstrating Propagator is Virtual

  1. Jan 26, 2014 #1
    Hello,

    I've been given a problem which is asking me to flesh out a calculation. The calculation is word for word like this;

    So I am being asked to work this out in detail, fill in all the steps. I thought that maybe I should try and show that the four momentum squared is not conserved at the vertex, i.e that the sum of the four momentum squared of the two electrons could not equal the four momentum squared of the photon and it was that violation that showed the particle was virtual;

    [itex]q^2=(p_e + p_e)^2[/itex]

    [itex]q^2=2p_e^2[/itex]​

    Where [itex]q^2[/itex] is the four momentum squared of the photon (i.e the mass of the photon) and [itex]p_e^2[/itex] is the four momentum squared of the electron, hence on one side of the vertex there are two electrons and on the other side there is a photon. The four momentum for the electron is [itex](E, \textbf{p})[/itex] so;

    [itex]q^2=2[E^2-(p_x^2+p_y^2+p_z^2)][/itex]

    [itex]=2(E^2-|\textbf{p}|^2) [/itex]

    [itex]q^2=2m^2[/itex]​

    Clearly this is going nowhere near the result result in the problem. Am I on the right track? I know I have probably made a dozen mistakes here as I am new to these sort of calculations. I would greatly appreciate if anyone could explain how to achieve the result above.

    Thanks
     
  2. jcsd
  3. Jan 27, 2014 #2
    Sorry, when I say

    What I mean is use the fact that four momentum is conserved at the vertex to show that the equation for energy mass and momentum does not apply to the propagator, hence virtual.
     
  4. Jan 27, 2014 #3

    strangerep

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    Science Advisor

    Something seems strange to me in the original wording. I just did a quick calculation, and found that ##E\Delta E - p\Delta p = 0## only if ##\Delta E^2 - \Delta p^2 = 0##, i.e., if the photon is on-shell.

    Are you sure you've got the original wording right, and complete?
     
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