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Definition of effective QED coupling.

  1. Jun 2, 2013 #1
    (PS: this post was also posted at the quantum mechanics/field theory forum, but I did not get any replies there)

    Often when one speaks about the effective QED coupling one defines it as

    $$e = \frac{Z_2 Z_3^{1/2}}{Z_1} e_0 \ \ \ \ (*)$$

    when ##Z_1 = Z_2## by the Ward identity this turns out to be ##Z_3^{1/2}e_0## and some authors just define the coupling to be this right away.
    So why do some make a point that the effective coupling is really defined according to (*). In what sense is this the 'natural definition'?

    I have thought about it, and the best answer I have come up with is the following:
    For an effective coupling one wants a definition which captures as much information about the interaction as possible so that when one has a large effective coupling, one can also say that the probability for an interaction is large. This thus requires us to take in the effects from the propagator of the electron (Z_2), the propagator of the photon (Z_3) and the vertex function (Z_1). Since there are two electron propagators connected to each vertex one gets two factors of ##Z_2^{1/2}## while one gets just one factor of ##Z_3^{1/2}## from the photon propagator.

    Any insights would be appreciated!
  2. jcsd
  3. Jun 3, 2013 #2
    e is defined with that normalization so that at long distances the potential between two unit charges is ##e^2 / 4 \pi r##, in natural units. Srednicki discusses this a bit in his chapter 63.

    Your thoughts sound right. The Z-factors are there to compensate for things like the fact that, for example, ##\bar{\psi}(x)| 0 \rangle## is not a properly normalized one-particle state (rather, its projection onto the one-particle subspace of the Hilbert space does not have norm 1). To normalize it properly you have to multiply by a Z-factor. This carries over to the propagator of the ##\psi## field, which needs to be multiplied by a Z-factor to have the right normalization to describe propagation of an electron.

    You might consider a similar situation with no Ward identity, like a Yukawa theory with a fermion coupled to a scalar. Srednicki works out a lot in this theory. You could calculate the long-distance potential between two fermions. This should have the form ##V(r) = g^2 e^{-m r}/4 \pi r## where ##m## is the scalar mass and ##g## is some effective coupling. You can calculate ##V(r)## from the low-energy limit of tree-level scalar-scalar elastic scattering. In the Born approximation, the low-energy scattering amplitude is just the Fourier transform of the potential. Once you have ##V(r)## in the form above, you'll get a formula for ##g## in terms of the bare coupling, which will probably look exactly like the first formula in your post.
    Last edited: Jun 3, 2013
  4. Jun 5, 2013 #3
    I have actually read that from an older source,where z3 is defined as 1-c where c=α/3∏(lnλ2/m2) for the other there are factor (1-B) and (1+L) and you define e=(1-B)(1+L)√(1-c)e and just because of ward identity B=L,so up to order of α2 one has e=√(1-c)e as is described in every book.Now those B,c and L came from radiative correction to electron propagator,vacuum polarization and vertex correction respectively.It is just for the purpose of renormalization that this coupling is defined.As you may remember that renormalization in qed is carried out by a redefinition of charge and mass,
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