# Renormalization question in phi^4 theory

1. Dec 26, 2014

### copernicus1

I'm studying renormalization and I have a question about part of a textbook. In P&S at the top of p.324 they show the divergent amplitudes of phi^4 theory, and they say that the two-point vertex (which has superficial degree of divergence D=2 according to the formula they derive) will have a term proportional to $\Lambda^2$ and also terms proportional to $p^2\log\Lambda$. My question is, how do they know these terms just from D=2? That would seem to give the $\Lambda^2$ term, but how do they know there is no term proportional to, for example, $p\Lambda$?

2. Dec 27, 2014

### vanhees71

Renormalization theory depends on two things: The "power counting" getting you the superficial degree of divergence, which is in $phi^4$ theory in 1+3 space-time dimensions $D=4-E$, where $E$ is the number of external (amputated) legs of the proper (one-particle irreducible) diagrams that have to be renormalized and contribute counter terms to the bare Lagrangian.

Now there is Weinberg's power-counting theorem that states that after all subdivergences are regularized the overall power counting is indeed a polynomial up to the order $D$ and some powers of logarithms. Of course there can never be something like $\ln \Lambda$. This is an expression that doesn't make mathematical sense at all since you wrote down a logarithm with a dimensionful argument. It's one of the really bad weaknesses of Peskin & Schroeder (which otherwise is a marvelous textbook, when you take out the trivial typos): It gives expressions with dimensionful arguments of logarithms, and this even in the chapter on the Wilsonian ideas about renormalization, where everything is about the scaling introduced by the regulator and where it is of utmost importance to keep the dimensions within the logarithms correct.

The second ingredient in renormalization theory are, as in general for all quantum theory and most importantly in relativistic quantum field theory, symmetries. $\phi^4$ theory has only the Poincare symmetry and the "field-reflection symmetry", i.e., the symmetry under the discrete transformation $\phi \rightarrow -\phi$. The latter symmetry implies that all proper vertex functions with an odd number of (truncated) external legs must vanish. Thus the only superficially divergent diagrams are those with $E=0$ (vacuum fluctuations), $E=2$ (self-energies), and $E=4$ (four-particle vertex). The vacuum-fluctuation diagrams play no role as long as you deal with vacuum QFT only. They become important as parts of the partition sum in the canonical ensemble of relativistic quantum many-body theory (QFT at finite temperatures). The self-energy divergences get renormalized by the wave-function renormalization and mass renormalization, and finally the four-point vertex function gets renormalized by coupling-constant counter terms.

Now, it is good to preserve as many symmetries as you can. The cutoff-regularization is very bad, because it already spoils Poincare covariance, but now to evaluate loop diagrams, it's anyway more convenient (to a certain extent) to Wick rotate and work in Euclidean QFT and then do the appropriate analytical continuation to the physical real-time quantities (be warned that this latter step is everything else than trivial, and for more complicated theories it's more convenient to use other regularization procedures or no regularization procedure at all and just subtract the counter terms on the level of the loop integrals a la BPHZ; for a purely scalar $\phi^4$ theory the Wick rotation is fine, as far as I know since for vertex functions of a scalar theory the analytic continuation isn't that complicated). So after Wick rotation you can just take a cutoff in Euclidean four-momentum space, i.e., you don't integrate over the entire $\mathbb{R}^4$ for each loop momentum but only over a solid sphere of radius $\Lambda$, i.e., you choose the cutoff as four-dimensional radius in momentum space, and thus the theory stays Lorentz invariant (or more accurately stated invariant under four-dimensional rotations SO(4) in the Wick rotated Euclidean formulation). Note, however that you loose translation invariance in momentum space as long as the cutoff is finite.

Now it's clear that the two-point functions can only depend on the external four-momentum $p$. There are no other invariant building blocks than the "metric" $g_{\mu \nu}$ (or $\delta_{\mu \nu}$ in the Euclidean version of the theory) in $\phi^2$. So your regularized (!!!) integral must be an SO(4) scalar due to the rotational symmetry, which is kept intact by the Euclidean cutoff-regularization procedure. Now with $p$ and $\delta_{\mu \nu}$ you can build only $p^4$ as a scalar, and thus the two-point function must be a function of $p^2$. That's why there cannot be linear term in $p$. Due to Weinberg's power counting theorem, for $|p| \rightarrow \infty$, $\Lambda \rightarrow \infty$ you have the possible terms
$$A(\Lambda) p^2+B_1(\Lambda) m^2+B_2(\Lambda) \Lambda^2, \quad C(\Lambda) \ln(p^2/\Lambda^2), \quad D(\Lambda) p^2 \ln(p^2/\Lambda^2),$$
where the first three are renormalized by wave-function and mass renormalization counterterms.

The bad guys are the remaining terms, but there's the key theorem of renormalization theory by Bogoliubov, Parasiuk, Hepp, and Zimmermann which tells you that these divergent terms are there because of subdivergences of multi-loop diagrams, i.e., after subtracting the subdivergences with the corresponding counter terms of lower loop order the remainders of terms of this form are finite in the limit $\Lambda \rightarrow \infty$. Also there is no problem with overlapping divergences (Zimmermann's forrest formula).

For a pretty complete analysis of these issues, see my QFT manuscript:

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Let's check these arguments on the most simple one-loop diagram, the tadpole-diagram contribution to the self-energy. This is somewhat special, because after truncating the external legs it's effectively only a one-point function and thus cannot depend on the external momentum at all. Thus we exspect the the general form
$$\Sigma=B_1(\Lambda) m^2 + B_2(\Lambda) \Lambda^2$$
Now indeed the Euclidean loop integral is
$$\Sigma =C \int_0^{\Lambda} \mathrm{d} P \frac{P^3}{P^2+m^2},$$
where $C$ is a finite constant given by the 3-sphere hypersurface area and a factor $1/(2 \pi)^4$. This doesn't matter for our renormalzation problem. The integral now gives
$$\Sigma=\frac{C}{2} \left [\Lambda^2 + m^2 \ln \left (\frac{m^2}{m^2+\Lambda^2} \right ) \right].$$
This is indeed of the form predicted by Weinberg's theorem. Here you don't get any wave-function renormalization term, which would be $\propto p^2$, because of the special property that we deal effectively with a one-point function. The first term gives rise to a mass counterterm $\delta m^2=-C \Lambda^2/2+\mathrm{finite constant}$, and the 2nd one to a mass-scale counter term $m^2 \delta Z_m=- C m^2 \ln [m^2/(m^2+\Lambda^2)]/2+\text{finite constant}\times m^2$.

We distinguish here these two kinds of mass counter terms, because we also like to consider the limit $m \rightarrow 0$. In the latter case the log-term vanishes, and we'd only have a term $\propto \Lambda^2$ with a coefficient that is finite as a function of $\Lambda$ (here it doesn't depend on $\Lambda$ at all) and also finite for $m^2 \rightarrow 0$ (it doesn't depend on $m$ either in this case). It's always possible to choose a "mass-independent renormalization scheme" such that there are no infrared divergences introduced by renormalization (as would be the case when doing an on-shell renormalization scheme). You just have to choose the renormalization conditions for the logarithmically divergent pieces at a finite mass scale $m=M$, e.g., (leaving out the irrelevant vacuum divergences):
$$\Sigma(p^2=0,m^2=0;M)=0 \quad (\text{quadratically divergent piece, mass counter term}),$$
$$\partial_{p^2} \Sigma(p^2=0,m^2=M^2;M)=0 \quad (\text{logarithmically divergent piece, wave-function counter term}),$$
$$\partial_{m^2} \Sigma(p^2=0,m^2=M^2;M)=0 \quad (\text{logarithmically divergent piece, mass-scale counter term}),$$
$$\Gamma^{(4)}(s=0,t=0,u=0,m^2=M^2;M)=0 \quad (\text{logarithmically divergent piece, coupling-constant counter term}).$$
In the parantheses we've given the over-all degree of divergence and the type of counter term to renormalize the divergent piece. For our tadpole contribution to the self-energy this leads to the renormalized quantities.

In our case we get in this renormalization scheme
$$\Sigma_{\text{ren}}=\lim_{\Lambda \rightarrow \infty} \left [\Sigma-\Sigma_{m^2 \rightarrow 0}-m^2 \partial_m^2 \Sigma_{m^2=M^2} \right ] = \lim_{\Lambda \rightarrow \infty} \frac{m^2}{2} \left [\ln \left (\frac{m^2(M^2+\Lambda^2)}{M^2(m^2+\Lambda^2)} \right )-\frac{\Lambda^2}{\Lambda^2+M^2} \right]= \frac{m^2}{2} \left [\ln \left (\frac{m^2}{M^2} \right )-1 \right ].$$
As we see, this is finite in the limit $m^2 \rightarrow 0$, so that there are indeed no IR divergences added due to renormalization. At finite $m$ the renormalized self-energy depends on an arbitrary mass scale, $M$, which is the renormalization scale in this scheme.

Last edited: Dec 27, 2014
3. Dec 27, 2014

### copernicus1

Wow thanks for your thorough reply. There are one or two things I am still fuzzy on:

1) When you write that we can construct $p^4$ but then the two-point function is a function of $p^2$, how did you get from $p^4$ to $p^2$?

2) How do you know (or how did Peskin and Schroeder know) that the two-point function in $\phi^4$ theory doesn't have a term like $B_1(\Lambda)m^2$?

4. Dec 27, 2014

### vanhees71

1) was a typo, which I've corrected now.

2) But I said, this term should be there in the dimensionsl analysis, and indeed in the one-loop example it's there. So everything isonsistent.

5. Dec 28, 2014

### copernicus1

Ah ok. Thanks for your help. One last question. When you say that you can construct $p^2$ from the available quantities (i.e. from $\delta_{\mu\nu}$ and $p_\mu$), I see that you can construct $p^2=p^\mu p^\nu\delta_{\mu\nu}$, but the magnitude of $p^\mu$ is also invariant under SO(4) rotations in Euclidean 4-space, and this can be constructed from $p=\sqrt{p^\mu p^\nu\delta_{\mu\nu}}$. Why is this not allowed? Are we not allowed to use square roots?

6. Dec 29, 2014

### vanhees71

Sure, in principle it's allowed, but Weinberg's theorem tells you how the asymptotic behavior for large momenta look, and according to this theorem, there is no term $\propto p=\sqrt{p^2}$.

For a treatment of Weinberg's theorem (reproducing the original proof by Weinberg), see

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

p. 166ff.