Missing counterterms in $\phi^3$ + $\phi^4$ theory in 1PI effective action

Question

I hope I'm just overlooking something. The Lagrangian is as follows: $$\mathcal{L}=\frac{1}{2}(\partial_\mu \phi)^2-(\frac{1}{2}m^2\phi^2+\frac{1}{3!}g\phi^3+\frac{1}{4!}\lambda\phi^4)$$ and I just want to calculate the saddle point approximation (SPA) for the 1PI effective action.

First, the counterterms from the "classical part" is $$\frac{1}{2}\delta_{m^2}\varphi^2+\frac{1}{3!}\delta_g\varphi^3+\frac{1}{4!}\delta_\lambda\varphi^4$$ and these are (up to the "source counterterm" which only enforces the one point function fluctuation to vanish, see Peskin.) the only ones available to counter anything from the SPA.

The divergent part of the SPA for a homogeneous background is given by (omitting the multiplicative constants here): $$(m^2+g\varphi+\frac{\lambda}{2}\varphi^2)^2/\epsilon.$$ This can be "seen" (if one chooses not to explicitly compute this) by dimensional analysis and the fact that the quadratic part of the lagrangian after expanding around $\varphi$ gives me this "mass term" (see Peskin, chapter 11.4 for an analogous calculation.)

Expanding this product gives me in particular this term here: $2m^2g\varphi/\epsilon.$

But this is weird: Which term (by counting $\varphi$) could counter this divergence? It does not seem to be the "source counterterm", as the one loop contribution for $\langle \eta \rangle$ (the one point fluctuation) does not have the same prefactors.

(They are given by something like $(g+\lambda\varphi)(m^2+g\varphi+\frac{\lambda}{2}\varphi^2)$ instead, so I cannot counter both using one $\delta J(x)$.)

Btw: I did compute everything.

EDIT: I did not forget the field renorm. It disappears due to assumption of homogeneity of $\varphi$. But even with this, i would still only get a quadratic contribution.

EDIT2: Regarding the source counterterm: It (i.e. the combination $\delta_J\phi$) vanishes when legendre transforming. So it does not appear anymore when evaluating the saddle point approximation (SPA)

EDIT3: Not even sure if this should be an edit or just answered directly: But a comment stated that the result is wrong. Well, the SPA is given by tr log $\frac{\mathcal{L}}{\delta \phi\delta\phi}$. And then you can just do the calculation to get the effective mass: $m^2+g\phi+\lambda/2 \phi^2$. And well, if you really want to think of the tadpole instead, remember we did a shift in the fields. So we generate another vertex from the original lambda interaction.

ALSO: If people claim the source counterterm does not vanish/ can be used: State this as an answer including derivation of the fact. This would deviate from standard literature such as P&S and Schwartz and would be very enlightening to see.

Answer 1

First of all, this 1PI thing is quite comparable to perturbation series. Since you use Peskin/Schroeder, check out the reference to that Coleman Weinberg potential project: Coleman and Weinberg in their paper literally resummed all one-loop Feynman diagrams to obtain the tr log. And what I am going to do here is completely standard.

(On a different note, I don't know about Schwarz, but P&S generally doesn't do a good job on the formal aspects of QFT. You might want to try some modern reference like Srenicki or Fradkin.)

Let's agree that the bare Euclidean action (including source) has to look like $$ S_b[J] = \int d^4x \, \frac{Z}{2} (\partial \phi)^2 + \sqrt{Z}(J + \delta J) \phi + \frac{Z}{2} (m^2+\delta m) \phi^2 + \frac{Z^{3/2}}{3!} (g+\delta g) \phi^3 + \frac{Z^2}{4!} (\lambda+\delta \lambda) \phi^4. $$ We will identify $\delta J$ as the counter term that cancels UV-divergence in $\langle\phi\rangle$ correlation function when the renormalized value of $\langle\phi\rangle$ is zero.

I emphasize that, especially given that you want to stay around $\phi = 0$, you always need a non-vanishing $\delta J$ to cancel any $Z_2$-breaking effect coming from the $\phi^3$ term. You break $Z_2$ symmetry explicitly, and there is a (rather unnatural) price to pay to force a $Z_2$-symmetric vacuum. When you subsequently play around with the source to move away from $\phi = 0$, you only change $J$ but leave the "base line" $\delta J$ alone. (This is not seen in P&S because they only have $\phi^4$.)

So it should be plenty clear now: Legendre transform is between the pair $\varphi$ and $J$. The counter term $\delta J \phi \rightarrow \delta J (\varphi + \delta\phi)$ stays in the action unmolested by the procedure. You discard $\delta J \delta\phi$ in a one-loop calculation, and the remaining $\delta J \varphi$ cancels your problem.

As for the mismatch of the coefficients: they just can't be. As I mentioned above, that tr log is nothing more than the resummation of all vacuum one-loop diagrams in the presence of $\varphi$. The counter terms designed to cancel the first couple diagrams should work perfectly on tr log.

Actually, let's expand: $$ \frac12 \log(k^2 + m^2 + g\varphi + \lambda \varphi^2) = \frac12 \log(k^2 + m^2) + \frac12 \frac{g\varphi + \lambda \varphi^2}{k^2 + m^2} - \frac14 \left(\frac{g\varphi + \lambda \varphi^2}{k^2 + m^2}\right)^2 + \dots $$ First term is $\varphi$-independent, and "..." are UV-convergent. These terms are easily recognizable as vacuum Feynman diagrams. If you find a mismatch, you botch your calculation.

edit

I admit that I wrongly assumed OP wants $\varphi = 0$. But the answer just get trickier with that...

That theory is unnatural

This is a bit of a digression. The general rule of writing down renormalizable QFT is that all relevant term consistent with the symmetry must be included. Otherwise the missing term just grows back under RG flow anyway. So the theory should naturally have an intrisic $J_{i} \neq 0$.

Of course, you don't plan to flow the theory. So we may have a bare theory with a bare $J_i + \delta J$, but we are just at a special point along the flow where the renormalized $J_i$ vanishes. I am ok with that.

But that raises a more general issue: when you try to find the ground state, it should be the one in the presence of the intrinsic $J_i + \delta J$. You can impose another external $J_e$ and Legendre w.r.t. it, but you have no right to touch the intrinsic part, $\delta J$ included.

Alternatively, you can say that there is no bare $J$ whatsoever. Then all $J$ is external under your control, and this corresponds exactly to what you propose to do with your Legendre procedure. But not having a $\delta J$ means stuff blows up like nuclear warhead.

What does "no tadpole" really means?

It is really an implicitly way to implement the inverse relation $J_{e} = J_{e}[\varphi]$ I mentioned above. But it is not always the right answer.

In a $Z_2$ symmetric theory (i.e. without $\phi^3$), there is no spontaneous tadpole; any tadpole diagram is connected to $J_e$. From here it is not difficult to figure out that the subtraction of $J_e[\varphi]\varphi$ kills off all tadpoles. If there is a $J_i \neq 0$? Tough luck. I have no idea how to obtain the inverse $J_e[\varphi]$ in general, either.

However, if we fine tune the theory to have the renormalized $J_i = 0$, we go back to the scenario where only $J_e \neq 0$ can induce (renormalized) tadpoles. Then you can impose a "no renormalized tadpole" condition. So no, there is no over-subtraction.

Does $\delta J$ really cancel the UV problem?

Are you kidding me? Draw the Feynman diagram of a $\phi^3$ vertex with two legs connected into a loop, and link the external leg to $\varphi$. Write down that integral with Feynman rules. It is precisely $$ \frac{g\varphi}{2} \int \frac{1}{k^2 + m^2}, $$ right?