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What is the physical mass and coupling?

  1. Jun 22, 2005 #1
    I have a very basic question about exactly how to match the experimentally measured masses and coupling constants to the parameters in the lagrangian density in a given QFT. Let me specialize to a particular theory and perhaps people here can help me out.

    I've just computed the tadpole and self-energy diagram up to 1 loop in [tex]\phi^{3}[/tex] theory using dimensional regularization. If you do power counting you'll find these are the only divergent diagrams up to 1 loop. (I know [tex]\phi^{3}[/tex] is probably not a good QFT because the Hamiltonian is not bounded from below, but this is for practice.)

    [tex] \mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2} m^{2} \phi^{2} - \frac{\lambda \mu^{\epsilon}}{3!} \phi^{3} [/tex]

    I got, up to factors of 2 and minus signs (I can never get this right; if anyone could go thru the computation and tell me what they got that'll be great, because they don't match Cheng and Li's gauge theory problem book's solutions)

    1-point function:
    [tex] \tau \equiv \frac{i \lambda m^{2}}{32 \pi^{2}} \left( \frac{1}{\epsilon} - \gamma +1 + Log[ \frac{\mu}{m^{2}} ] \right) [/tex]

    (I'm not sure why I have [tex]Log[{\mu}/{m^{2}}][/tex] - the dimensions aren't right.)

    2-point function
    [tex] \Sigma[\lambda, m, p^{2}, \mu] \equiv \frac{i \lambda^{2} \mu^{2 \epsilon}}{16 \pi^{2}} \left( \frac{1}{\epsilon} - \gamma + Log[4 \pi] - \int_{0}^{1} Log[m^{2}-p^{2}x(1-x)] dx \right) [/tex]

    When I (actually Mathematica) did the remaining integral in the 2-point function I found

    [tex] -2 + 2 \kappa[m^{2}, p^{2}] ArcTan[\frac{1}{\kappa[m^{2}, p^{2}]}] + Log[m^{2}] [/tex]
    [tex] \kappa[m^{2}, p^{2}] \equiv \sqrt{\frac{4 m^{2}}{p^{2}}-1} [/tex]

    I know I need to introduce counterterms. Say I do the MS scheme and so the re-normalized 1- and 2-point functions are as above but without the [tex]1/{\epsilon}[/tex] term. Then what? How do I get the physical mass? Physical coupling constant?

    I think I should do

    \mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2} m^{2} \phi^{2} - \frac{\lambda \mu^{\epsilon}}{3!} \phi^{3} + \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi (Z_{\phi}-1) - \frac{1}{2} m^{2} \phi^{2}(Z_{m} Z_{\phi} - 1)- \frac{\lambda \mu^{\epsilon}}{3!} \phi^{3} (Z_{\lambda}Z_{\phi}^{3/2}-1)

    but I'm actually not sure if I need to introduce additional operators. For example I'm confused about how to deal with the [tex]p^{2}[/tex] dependence. I am probably doing something wrong? If I get a term linear in [tex]p^{2}[/tex] it would have allowed me to determine the wavefunction renormalization factor [tex]\phi_{0} = \phi_{R} \sqrt{Z_{\phi}}[/tex].

    As one can see I have lots of problems. Any help is very much appreciated.
  2. jcsd
  3. Jun 22, 2005 #2


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    It's too long ago I played with these things to jump into the calculations and help you out ; however I could try to give you a better overall picture (it might be that you understood this, in which case my blathering serves no purpose).

    It might be interesting to read the summary of the relevant chapter in Peskin and Schroeder I wrote about it:

    However, the basic point is that we have to DEFINE, by a physical process, what we mean by "physical mass" or "coupling constant". The physical mass is usually given by a pole in the propagator, because that corresponds to the mass of the "asymptotic" free field theory particles. For the coupling constant, you have to choose a definition. For instance, the charge of the electron is usually the electrostatic repulsion at "large distances", which is described by a limiting case of a scattering process. So, the quantity that you get from the PDG for, say, alpha, has been obtained under certain conditions, and these conditions are the ones you should try to calculate as a function of the bare parameters e0, m0, ... These choices of definition of physical quantities are called "renormalization conditions".

    This aspect becomes crucially important in QCD, where there is no natural, perturbative large-distance, low energy solution available.

    Last edited: Jun 22, 2005
  4. Jun 23, 2005 #3


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    Hey, good questions actually. I don't have the time to actually jump into the calculations (I did them years ago and thye have long since left my memory), but I strongly suspect there is an algebra problem, already in the one point correlation functions you listed. If you get wrong mass dimension terms, there is something fishy going on, especially if they persist through renormalization.

    Dont worry about factors of 2 and/or 4pi, no one gets them right, even textbooks often have them wrong, but negative signs are important.. Factors of i, sometimes are troubling as well (usually initially, and not so important afterwards).

    Just eyeballing those eqns, there are a few things that jump out, like the absence of readable pole terms, from experience they normally always are there and are more or less guarenteed to appear in textbook examples.
  5. Jun 23, 2005 #4
    I need to worry about every factor of -1's, [itex]\pi[/itex]'s and i's if I want to learn this stuff properly - this is my goal.

    By pole terms do you mean the [itex]1/{\epsilon}[/itex]'s? I'm doing dim reg, so my integrals live in [itex]4-2{\epsilon}[/itex] spacetime dimensions.
  6. Jun 23, 2005 #5


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    Would the dimensions of [tex]Log[{\mu}/{m^{2}}][/tex] be happier if you rewrote it as [tex]0.5 Log[{\mu}/m}][/tex]?

  7. Jun 23, 2005 #6
    Not really: it remains unhappy because [itex]Log[{\mu}/{m^{2}}] = \frac{1}{2} Log[{\sqrt{\mu}}/{m}][/itex], and so we're just exchanging [itex]mass/mass^{2}[/itex] for [itex]\sqrt{mass}/mass[/itex].
  8. Jun 24, 2005 #7
    This dimension issue is bugging me. I'll post my steps and if anyone cares to follow thru, let me know if you spot an error.

    We know that

    [tex] \mathcal{L} = \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2} m^{2} \phi^{2} - \frac{\lambda \mu^{\epsilon}}{3!} \phi^{3} [/tex]

    must be of mass dimensions [itex]4-2\epsilon[/itex] to make the action dimensionless, in the sloppy physicists' language where [itex]\hbar = c = 1[/itex]. Looking at the kinetic term we deduce the dimensions of [itex]\phi[/itex] must be [itex]1-\epsilon[/itex]. Therefore the dimensions of [itex]\lambda_{0}[/itex] must be [itex]4-2\epsilon-(3-3\epsilon) = 1+\epsilon[/itex]. So we introduce an arbitrary parameter [itex]\mu[/itex] of mass dimension and put [itex]\lambda_{0} = \lambda \mu^{\epsilon}[/itex].

    For the tadpole diagram we basically need to integrate over the propagator of a closed loop. Since there are 3 ways of contracting one [itex]\phi^{3}[/itex] with the external state we have

    [tex]\tau = \frac{-i \lambda \mu^{\epsilon}}{2} \int \frac{d^{4-2\epsilon}k}{(2 \pi)^{4-2\epsilon}} \frac{i}{k^{2}-m^{2}+i\delta}[/tex]
  9. Jun 24, 2005 #8
    Peskin (A.44) says

    [tex] \int \frac{d^{4-2\epsilon}l}{(2 \pi)^{4-2\epsilon}} \frac{1}{(l^{2}-\Delta)^{n}} = \frac{(-1)^{n}i \Gamma[n-(2-\epsilon)]}{(4 \pi)^{2-\epsilon} \Gamma[n]} \left( \frac{1}{\Delta} \right)^{n-(2-\epsilon)}[/tex]

    If we identify k = l, n = 1, and [itex]\Delta = m^{2}[/itex], we see that the tadpole diagram is proportional to

    [tex]m^{2} \left( \frac{\mu}{m^{2}} \right)^{\epsilon}[/tex]

    which is how the [itex]mass/mass^{2}[/itex] in the logarithm came about.
  10. Jun 25, 2005 #9


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    Up to here I followed, and I don't see any flaw. Care to specify how you got to the logarithm ?

  11. Jun 27, 2005 #10
    Given any quantity A and some small [itex]\epsilon << 1[/itex]. We have

    [itex]A^{\epsilon} = exp[ Log[ A^{\epsilon} ] ] = exp[ \epsilon Log[ A ] ] \approx 1 + \epsilon Log[ A ][/itex]

    where the last step just comes from Taylor expanding the exponential.
  12. Jun 28, 2005 #11
    By the way I now have fair confidence that this funny mass dependence is correct; Cheng and Li's problem book on gauge theory got it wrong and I asked the authors if it was a mistake and Li replied saying yes.
  13. Jun 29, 2005 #12


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    It is still funny that this logarithm with dimensions appears. Usually these things mean that some OTHER logarithm, with the opposite dimension, is absorbed in a "non-relevant constant" somewhere.
    I'm now thinking the following: the dimension of a diagram, does that make sense ? After all, they are quantum-mechanical AMPLITUDES, which should be dimensionless quantities (except of course if they are amplitude DENSITIES).
    So I was thinking, what happens when epsilon is not a small number, but when epsilon equals 1/2 (1+2 dimensional minkowski space) ?
    Now of course there is also the energy-momentum conserving delta function to turn the diagram expression into a genuine quantum mechanical amplitude.
    I didn't inquire into it, just some thoughts on how to get out of the riddle...

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