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Transition from QFT to Classical Physics using Path Integrals?

  1. Aug 18, 2009 #1
    I have a question in Srednicki's book regarding path integrals, but first I'll set it up so that no familiarity of the book is required to answer the question.

    The vacuum to vacuum transition amplitude for the photon field in the presence of a source is given by: [tex]<0|0>_J=\int \mathcal D A e^{iS+\int J_\mu A^\mu}[/tex] where [tex]S=\int -\frac{1}{4}F_{\mu \nu}F^{\mu \nu} d^4x[/tex]. The result is [tex]e^{i\int \int J^\mu(x)\Delta(x-y)_{\mu \nu} J^\nu(y)} [/tex]. If you rewrite J, the current density, in terms of the line integral [tex] e \oint dx^\mu [/tex], and choose a path for the line integral, then this vacuum to vacuum transition amplitude, being equal to [tex]e^{-iE_{0}t [/tex], should give you the energy of the situation described by your line integral. In this way by choosing the path where an electron and anti-electron are just sitting a small distance R apart for a long time T, you actually recover the Coloumb force from QFT, [tex]V(R)=-\frac{\alpha}{R}[/tex]. This is equation (82.22) Srednicki.

    My question is what does it mean to have the vacuum to vacuum amplitude in the presence of a source? If there is a source then is it really vacuum? Or does the vacuum mean it's a vacuum with respect to only photons, but you are allowed to have a fermion source [obviously J(x)]? If this is the case, then could you in principle use this same method but use the Dirac Lagrangian in place of the photon Lagrangian, and have the source J(x) be a photon source, and from this derive the energy between two photons? If one wants to derive Maxwell's equations (and not just Coloumb's law) from the path integral approach, how would you do this?

    Also, this is the first time I've ever seen QFT used to describe something in position space. Is this the standard way to check if QFT gives the same results as classical physics - vacuum expectation values in presence of sources?

    The coupling constant in QED gets big if the energy is high, and high energy is equivalent to short distances. Is this reflected in Coloumb's law with the 1/r^2 dependence? The beta function and renormalization seemed to have disappeared from this example, so this must mean that the 1/r^2 dependence is not due to changing coupling constants with energy?
  2. jcsd
  3. Aug 19, 2009 #2


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    Not when the source is "on". The idea is that the source is "off" (equal to zero) before some time t1 and after some later time t2, and that the initial state (before t1) is the vacuum. Then the computed amplitude is the amplitude to find the system in the vacuum state after time t2.
    No, for the same trick to work the source has to couple just to the field, so you would need a source like Srednicki's eta that couples to the Dirac field psi.
    Of course, QFT does not give the same results as classical physics; it's quantum! But just the fact that the quantum lagrangian is the same as the classical lagrangian means that, in suitable limits, the physics is the same.
    No, the 1/r^2 is a classical effect. Quantum effects effectively make the electron charge a function of r, and e^2 grows logarithmically with r (though this language is somewhat misleading; a more accurate statement is that there are quantum corrections to Couloumb scattering amplitudes that can be summarized by taking e^2 to be a function of r).
  4. Aug 19, 2009 #3
    I'm guessing there's no easy way to have [tex]\eta (x)[/tex] describe a photon, whereas it's easy to have [tex]J(x)[/tex] describe a fermion.

    I was hoping one could calculate the energy between two photons because I heard somewhere that QFT says that this vacuum energy should be infinite, which is troublesome because cosmology says it ought to be zero (the cosmological constant or something - it's been awhile since I looked at general relativity).

    At first I was amazed that Srednicki, in deriving (82.22), seem to have got a value for the self-energy. And indeed that term does come from integrating over the same line in the rectangular Wilson loop. But then I thought this was sweeping the self-energy under a rug, because the reason that this term is not infinite is because we assumed space-time is discrete via a lattice spacing, so of course there is no self-energy because a continuum doesn't exist! So in a way that's pretty cheap.

    I'm unconvinced of equation (82.18), or the "perimeter law". So that's basically saying that if you keep the aspect ratio of a certain Wilson loop the same (to maintain shape or c-tilda), then the value of vacuum expectation value will vary as the perimeter P. Just looking at this statement seems wrong. If you imagine two points infinitismally close together, and connect them with a rigid rod (so that their distance is constant), and move them along the loop, then I can see that the integral is proportional to the perimeter, because if you travel twice the distance on a path then that's twice the distance you integrate over a function whose value you insert the length of the rod. But clearly this doesn't work for a rod of finite length. It seems this type of problem would come up in self-inductance in classical E&M, but I can't find a book that shows this. I should probably check Jackson's book.
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