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Origin of Feynman Propogator.

  1. Apr 26, 2005 #1
    Ok - so essentially the Feynman propogator is a Green function solved for a point source - i.e a delta function in 3 dims.

    My question is more mathematical.

    When you solve for the Green Function you assume

    (id-m)G(x,x') = delta(x-x') (roughly)

    Then solving this going through the standard process (Fourier Transforming RHS, performing integral etc) you come up with the Green function aka the Feynman propogator.

    Now my question is this:

    What is the original justification (for ALL Green function proplems actually) that the RHS is a delta function?? Why/How do you assume this and what is the physics meaning??
     
  2. jcsd
  3. Apr 26, 2005 #2

    dextercioby

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    Evaluation of the generation functional for the unconnected Green functions comprizes the inverse of the kinetic operator.This inverse of the kinetic operator is found from the inverse condition (in 4D euclidean space)

    [tex] \int d^{4}\bar{z} \ A^{-1}\left(\bar{x},\bar{z}\right)A\left(\bar{z},\bar{y}\right)=\delta^{4}\left(\bar{x}-\bar{y}\right) [/tex]

    ,which is the general case...

    Daniel.
     
  4. Apr 26, 2005 #3

    ZapperZ

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    One can only justify this for a non-interacting system. For example, for a fermionic system, this is only valid for a free electron gas that is non-interacting.

    In an interacting system, you have self-energy consideration that has to be included in the derivation of the propagator. This comes in via both the real part and the imaginary part of the self-energy AFTER you solve for the integration around the appropriate poles (see, for example, http://w3.physics.uiuc.edu/~efradkin/phys561/gf.pdf). When you do this, the delta functions will "broaden" due to the self-interaction terms.

    So those delta functions are only for special cases.

    Zz.
     
  5. Apr 26, 2005 #4
    To the OP : it's essential that you understand the 'adiabatic switching on' of some interaction. This system, used on P.13 of the above link is crucial in understanding how the Feynman propagator can be derived in the case of an interacting many particles-system. Do you know what this technique means ? The word 'adiabatic' is crucial here. Remember that it primarily serves the purpose of approaching the real physical reality.

    A very well known system that copes with the difficulties associated with self-energy-terms (ie perturbation theory in QED), are the famous quasi-particles (a bit like the concept of effective mass in solid state physics). If you want a clear understanding of what they mean, just look in my journal. I have written some entries on this stuff. They are also essential in the many-particle-physics.

    Read the effective field theory entry :

    https://www.physicsforums.com/journal.php?s=&action=view&journalid=13790&perpage=10&page=11
    regards
    marlon
     
    Last edited: Apr 26, 2005
  6. Apr 26, 2005 #5

    dextercioby

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    Apparently each & everyone has a different story to tell...Even the OP.

    Daniel.
     
  7. Apr 26, 2005 #6
    There are some links here that I plan to read through later today - so that you everybody for explanations and links.

    dexter - sorry, but I didn't understand your answer - I think you are light years ahead of me right now.

    I think I'll try and repharase the question :

    You initially SOLVE some operator G(x,x') by IMPOSING that the RHS is equal a delta function.

    AFTER solving you use the G(x,x') for a whole host of other things that (as far as I can see) don't have any relation to the delta function.

    So...how can you justify creating an operator G that was originally specific to the delta function - and then use it to solve a load of other crazy stuff???


    Sorry if I am being dumb here (and for using caps - but they are just to emphasize the main words in my question).

    Thanks in advance.

    Richard

    ...PS sorry if the answer is crystal clear in one of the links too, I will read them later!
     
  8. Apr 27, 2005 #7

    vanesch

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    Well, I'd agree with Dexter here. You just want to find the INVERSE operation of "applying the differential operator".
    You have in general an equation to solve D y(x) = r(x)
    where D is a linear differential operator.
    A way to solve this is to solve first the problem D h(x) = delta(x)
    and then to put y(x) = integral h(x) r(x) dx

    Consider a similar problem in linear algebra:
    You have an equation to solve: A x = b

    A way to solve it (not numerically the smartest way, ok) is:

    Solve first A B = 1 (eg, calculate the inverse matrix of A and call it B),
    and then do x = B b.

    cheers,
    Patrick.
     
  9. Apr 27, 2005 #8

    dextercioby

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    Yep,there's a smart analogy with linear algebra,however,Green's method is really simple and useful.

    Daniel.
     
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