Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

QED renormalization in Peskin's

  1. Jul 20, 2015 #1
    I think I have found a mistake/wrong formulation at Peskin’s, when he discusses the renormalization of QED.

    In particular, he defines the 1PI of the electron’s self-energy on page 331 as: [itex] –i\Sigma( \displaystyle{\not}p ) [/itex] and the corresponding counterterm on page 332 as: [itex] i( \displaystyle{\not}p \delta_2 - \delta_m ) [/itex]. It is then logical to assume that the combined term can be written in 1-loop order as:
    [tex] –i\Sigma( \displaystyle{\not}p ) = –i
    \Sigma_2( \displaystyle{\not}p ) + i( \displaystyle{\not}p \delta_2-\delta_m ) [/tex]
    Then if one uses the on-shell renormalization conditions (pg. 332):
    [tex] \Sigma ( \displaystyle{\not}p = m) = 0 [/tex]
    [tex] \frac{d}{d\displaystyle{\not}p} \Sigma( \displaystyle{\not}p )\bigg|_{\displaystyle{\not}p = m} = 0
    [/tex]
    they yield the counterterms:
    [tex] \delta_2 = \frac{d}{d\displaystyle{\not}p }\Sigma_2( \displaystyle{\not}p ) \bigg|_{\displaystyle{\not}p = m} [/tex] and
    [tex] \delta_m = -\Sigma_2(m) +m \delta_2 [/tex]
    Peskin claims though that: [tex] \delta_m = -\Sigma_2(m) [/tex]
    I implore you to share any insights on the matter, cause this triviality is driving me crazy!
     
  2. jcsd
  3. Jul 20, 2015 #2

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're right, there's some confusion there. Recall that, if ##m = m_0 + \delta m##, then
    $$ \delta_m = Z_2 m_0 -m = Z_2(m - \delta m) -m = m \delta_2 - Z_2 \delta m.$$
    Therefore, you'll find that
    $$ \Sigma_2(m) = Z_2 \delta m \approx \delta m,$$
    where the last approximation is good at this order of perturbation theory.

    Alternatively, if you check the errata you'll find

    • p. 333: The extreme left-hand side of eq. (10.42) should be "delta_m - m delta_2" instead of "delta_m"
    which agrees with my result above.
     
  4. Jul 20, 2015 #3
    I've spent a day searching for alternative formulations. Totally forgot about the ERRATA, so I guess it was my fault. Thank you very much for the quick reply.
     
  5. Jul 21, 2015 #4
    Now that we are at it, I think they have the coefficients of the [itex]\delta[/itex] terms also wrong. For example the [itex]\Pi_2[/itex] diagram was calculated before the normalization and thus it has to contain the bare electric charge [itex]e_0[/itex] and not the renormalized charge [itex]e[/itex]. Another way to see this is the definition of the renormalized charge: [itex]e=Z^{1/2}_3e_0≈(1+\delta_3 /2)e_0[/itex]. It would make no sense for [itex]\delta_3[/itex] to have a [itex]e[/itex] coefficient as in (10.44) on page 333.
     
  6. Jul 21, 2015 #5

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The quantities on pg 333 are computed using the Lagrangian (10.38) (note the comment below Fig 10.4) so they depend on the renormalized coupling. Since the counterterms are introduced to cancel divergences of one-loop diagrams, they must include terms that depend on ##e^2##.

    As an aside, (10.37) defines the relationship between ##e## and ##e_0##, which involves ##Z_{1,2}## in addition to the factor you have above.
     
  7. Jul 21, 2015 #6
    If this is the case, then when I try to compute the correct QED beta function I get:

    [tex]\beta(e) = \mu \frac{d}{d\mu} e = \mu \frac{d}{d\mu} [(1+\frac{\delta_3}{2})e_0] = \frac{e_0 e^2}{12\pi^2} [/tex]

    and not the [itex]e_0^3[/itex] factor (or e^3 factor, I no longer know what is right).

    PS: In QED [itex]Z_1=Z_2[/itex], so those factors cancel out.
     
  8. Jul 21, 2015 #7

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think this is ok, since
    $$ e_0 = e + O(e^3),$$
    so
    $$\beta(e) = \frac{e^3}{12\pi^2} + O(e^5).$$
    The inaccuracies are shoved off to the next loop order. If we were computing the 2-loop beta function, we'd have to keep track of how the higher-order corrections to the one-loop coefficient propagate though to the next order.


    Oh right, thanks for the reminder!
     
  9. Jul 21, 2015 #8
    So what happens at higher loop orders? Do we get a [itex]e_0 ⋅ e^n[/itex] factor and then solve [itex]e_0[/itex] for [itex]e[/itex] and replace?

    What about the
    [tex] \beta(e) = \mu \frac{d}{d\mu} e \bigg|_{e_0} [/tex]
    equation (that I just noticed on pg. 417) ? Setting [itex]e_0=e[/itex] at any order would yield a convenient result. But then again, I don't see why the beta function should be defined like that from the derivation of the Callan-Symanzik equation.
     
    Last edited: Jul 21, 2015
  10. Jul 21, 2015 #9

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That expression doesn't mean set ##e=e_0##, it means take the derivative in such a way that ##\partial e_0/\partial M=0##. A more familiar notation might be
    $$\left( \frac{\partial e}{\partial M} \right)_{e_0,\Lambda}.$$
    To see that it works for QED, we need to incorporate dimensional transmutation into the relationship between the bare and renormalized coupling. I'm not familiar enough with P&S to know where it's gone in their expression, but you would have
    $$ e_0 = \mu^{\epsilon} \left( 1 + \frac{e^2}{24\pi^2} \frac{1}{\epsilon} + O(e^3) \right) e$$
    and can differentiate and solve this for the ##\beta## function in the ##\epsilon\rightarrow 0## limit.

    At higher orders I would want to compute the higher order corrections to either the equation I just wrote down, or use the Callen-Symanzik equation, since now we will probably have non-trivial contributions coming in from the anomalous dimension terms.
     
  11. Jul 22, 2015 #10
    So there is always one [itex]e_0[/itex] factor, which we trade for its expansion in [itex]e[/itex]. The [itex]\delta_3[/itex] contains only [itex]e[/itex] prefactors. In the end we get a beta-function depending only on [itex]e[/itex], which one can use to calculate the running coupling.
     
  12. Jul 22, 2015 #11

    fzero

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I think the ##e_0## factor will appear in certain expressions that you can use to compute the beta function, like the one you used above. The method I used above keeps ##e_0## off by itself and the method P&S use in Ch 12 uses only the renormalized variables. Whichever method you use, I believe that it is a theorem that the first coefficient in the beta function will always be the same (perhaps also the 2nd for QED). You might learn something about scheme dependence eventually, which is about how you treat the finite parts of the counterterms and how it can change the higher-order terms in the beta function. It's possible that having to trade expansions between bare and renormalized variables is effectively part of the scheme dependence at higher orders too, but I am not certain.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook