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Dimensional analysis of the fermion mass renormalization

  1. Mar 29, 2013 #1
    In the textbook, usually the fermion mass renormalization is introduced as follows: the mass shift [itex]\delta m[/itex] must vanish when [itex]m_0=0[/itex]. The mass shift must therefore be proportional to [itex]m_0[/itex]. By dimensional analysis, it can only depend logarithmically on [itex]\Lambda[/itex] (the ultraviolet cutoff): [itex]\delta m \sim \ln \Lambda[/itex].

    Here I don't understand how the "dimensional analysis" is performed, can somebody explain it in detail? Thanks in advance.
     
  2. jcsd
  3. Mar 29, 2013 #2
    Take a look at e.g. the diagram (a) here, in QED: http://ej.iop.org/images/1751-8121/45/38/383001/Full/jpa374448f4_online.jpg

    This diagram will correct the mass of the electron. Schematically, this diagram has a value that looks like

    ##\int d^4 p \frac{\gamma^\mu p_\mu + m_e}{p^2 + m^2}\frac{1}{p^2}##

    The integrand has dimension 1/[mass]^3, so the diagram has a whole has dimension [mass]^1.

    Now, the integral superficially seems to be linearly divergent, like ##\int d^4 p (1/p^3)##. In that case the result would be proportional to ##\Lambda## times a dimensionless constant (since it must have dimensions of [mass]^1). However, looking a little closer we see that the term involving ##\gamma^\mu p_\mu## is odd in ##p## and therefore cancels out, leaving

    ##m_e \int d^4 p \frac{1}{p^2 + m^2}\frac{1}{p^2}##

    Now we can see that the integral is actually logarithmically divergent, so the diagram is proportional to ##m_e \ln \Lambda##. Here ##m_e## has units of [mass]^1 and ##\ln \Lambda## is dimensionless, so the diagram has units of [mass]^1, as required.

    The point of the symmetry+dimensional analysis argument is to argue that this same thing will happen generally. The diagram has to have units of [mass]^1, and must be proportional to ##m_e##. So whatever the diagram evaluates to, it has to look like ##m_e## times something dimensionless. The only dimensionless thing you can make out of ##\Lambda## is ##\ln \Lambda##, so at worst diagrams like this can only diverge like ##\ln \Lambda## and not, say, ##\Lambda^2##.
     
  4. Mar 29, 2013 #3
    Thanks a lot, you gave a very clear explanation. Now I understand it fully.
     
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