Mass correction in ##\phi^4##-theory

[Markus Kahn]

Homework Statement:

Consider $\phi^4$ theory described by
$$\mathcal{L} = \frac{1}{2}(\partial_\mu \phi)^2 +\frac{1}{2}m^2\phi^2 - \frac{\lambda}{4!}\phi^4.$$
We want to renormalize the theory. Assume that the renormalization for the 1PI 4-point function has been done, resulting in a shift of $\lambda$, i.e.
$$\lambda_{r e n}=\lambda+\frac{\lambda^{2}}{32 \pi^{2}} \int_{0}^{1} d x\left[3\left(\frac{2}{\varepsilon}-\gamma_{E}+\ln (4 \pi)\right)-\ln \left(m^{2}-4 m^{2} x(1-x)\right)-2 \ln \left(m^{2}\right)\right].$$
We are left with renormalizing the 2-point function. Show that the renormalized mass $m_{ren}$ is given by
$$m_{r e n}^{2} = m^2 + -\frac{\lambda}{2(4 \pi)^{\frac{d}{2}}} \frac{\Gamma\left(1-\frac{d}{2}\right)}{\left(m^{2}\right)^{1-d / 2}}.$$

Relevant Equations:

All given above.

Before I start, let me say that I have looked into textbooks and I know this is a standard problem, but I just can't get the result right...

My attempt goes as follows:

1. We notice that the amplitude of this diagram is given by $$\begin{align*}K_2(p) &= \frac{i(-i m)^{2}(-i)^{4}}{\left(p^{2}-m^{2}+i \epsilon\right)^{2}} \frac{1}{2} \int \frac{-i \mathrm{~d} \ell^{4}}{(2 \pi)^{4}} \frac{1}{\ell^{2}-m^{2}+i \epsilon} \frac{1}{(p-\ell)^{2}-m^{2}+i \epsilon}\\&=: \frac{i(-i m)^{2}(-i)^{4}}{\left(p^{2}-m^{2}+i \epsilon\right)^{2}}i I\left(p^{2}\right).\end{align*}$$
2. We then remember that the 2-point function can be expanded as shown in this diagram. The amplitude of this diagram is then given by $$ \begin{align*}K(p) &= \sum_i K_i(p)= \frac{1}{p^{2}-m^{2}+i \epsilon} \sum_{j=0}^{\infty}\left(\frac{m^{2} I\left(p^{2}\right)}{p^{2}-m^{2}+i \epsilon}\right)^{j}+\ldots \\&= \frac{1}{p^{2}-m^{2}-m^2I(p^2) +i \epsilon}+ \ldots.\end{align*}$$ If I'm not mistaken, this is essentially what one finds in Peskins book in eq. (10.27).
3. What we now need is a renormalization condition. This was provided as a hint to the exercise and is given by $$K(p) = \frac{i Z }{p^2-m_{ren}^2}+(\text{terms regular at }p^2=m^2).$$ Note that $Z$ is defined by $\sqrt{Z}\phi_{ren}=\phi$.

And this is essentially where I'm stuck. First of all, I don't really understand what "terms regular at $p^2=m^2$" is supposed to mean. One can of course now write down $$\frac{i Z }{p^2-m_{ren}^2}+\ldots = \frac{1}{p^{2}-m^{2}-m^2I(p^2)}+ \ldots,$$ but how am I supposed to extract $m_{ren}$ and $Z$ out of this? In Peskin's book they now explain that this renormalization condition can be restated as
$$I(p^2)|_{p^2=m^2}=0 \quad\text{and}\quad \frac{d}{dp^2}I(p^2)|_{p^2-m^2}=0.$$ Assuming this is true (I need to think about this a bit more...), how is one supposed to find $m_{ren}$ and $Z$?

 

[vanhees71]

Something went wrong with the dimensions. You can never have a dimensionful quantity as the argument of a logarithm. This is particularly important in the context here (though also Peskin&Schroeder is sloppy on this point).

Concerning $Z$ you can argue without doing any calculation. The point is that at one loop in $\phi^4$ theory you have only a tadpole contribution, which is an effective one-point function (there's only one spacetime point involved for the truncated diagram). What does this imply concerning the dependence of this self-energy contribution on the external four-momentum $p$?

 

[Markus Kahn]

Thank you very much for the response!

Something went wrong with the dimensions. You can never have a dimensionful quantity as the argument of a logarithm. This is particularly important in the context here (though also Peskin&Schroeder is sloppy on this point).

I hope you mean the $\log$ that will eventually show up in $I(p^2)$, if not, I'm not really sure what you mean. I just went back to my QFT1 lecture notes (Chp. 11.2) one more time to check, and my Prof. got for this integral two different expression, depending on the way he calculated it:
Feynman parameter:
$$I\left(p^{2}\right)=-\frac{1}{32 \pi^{2}} \int_{0}^{1} \mathrm{~d} z \log \frac{z (1-z) p^{2}- m^{2}-i \epsilon}{\Lambda_{\mathrm{cut}}^{2}}.$$
Wick rotation:
$$
\begin{aligned}
I\left(p^{2}\right)=&-\frac{1}{16 \pi^{2}} \sqrt{\frac{p^{2}-4 m^{2}-i \epsilon}{-p^{2}}} \arctan \sqrt{\frac{-p^{2}}{p^{2}-4 m^{2}-i \epsilon}} \\
&-\frac{1}{32 \pi^{2}} \log \frac{m^{2}}{\Lambda_{\mathrm{cut}}^{2} \mathrm{e}^{2}}.
\end{aligned}
$$
I'm not sure, but these seem to me like unitless quantities inside the logarithms.

Concerning you can argue without doing any calculation. The point is that at one loop in theory you have only a tadpole contribution, which is an effective one-point function (there's only one spacetime point involved for the truncated diagram). What does this imply concerning the dependence of this self-energy contribution on the external four-momentum ?

I'm sorry, but you are too quick for me here. We never really worried about the tadpoles so far, so I'm not really sure what the answer to your question is (although I see that the truncated tadpole would only involve one spacetime point...)

 

[vanhees71]

These expressions make sense, because now you have introduced a renormalization scale, and this you must do in some way. Here you have regularized by a cutoff, and the cutoff brings in the scale.

Now you can calculate the physical mass $m_{\text{phys}}$ in terms of the bare mass $m$ and the cutoff as well as the residuum of the Dyson-resummed propagator, which gives you the wave-function renormalization $Z$ to this one-loop order.

 

[Markus Kahn]

Now you can calculate the physical mass in terms of the bare mass and the cutoff as well as the residuum of the Dyson-resummed propagator, which gives you the wave-function renormalization Z to this one-loop order.

I'm sorry, but I don't understand how to do that...

What I have tried (thought about) so far:
$$ \frac{1}{p^{2}-m^{2}-m^2I(p^2)} \approx \frac{1}{p^2-m^2} + \frac{1}{p^2-m^2}m^2I(p^2)\frac{1}{p^2-m^2}.$$
Can we use this maybe like this:
$$\frac{i Z }{p^2-m_{ren}^2}\overset{!}{=}\frac{1}{p^{2}-m^{2}-m^2I(p^2)} \approx \frac{1}{p^2-m^2} + \frac{1}{p^2-m^2}m^2I(p^2)\frac{1}{p^2-m^2}?$$
The issue with this is that we now would have to (I think) assume that $Z\approx 1+\delta Z$ or something like that... But in this case we could just deduce that $m^2\approx m_{ren}^2$, which doesn't seem very useful..

 

[vanhees71]

The physical mass $m_{\text{ren}}$ is the value of $p^2$, where the propagator has a pole, i.e., you have to solve for
$$m_{\text{ren}}^2=m^2[1+I(m_{\text{ren}}^2)].$$
Note that this is not $\phi^4$ theory but the one-loop self-energy of $\phi^3$ theory. Here you'll get a non-trivial $Z$ because of the opening of the two-particle contribution to the spectral function. It's nicely explained in the manuscript how to calculate it in the given order of the coupling, $\kappa^2$ or $\hbar^1$ (one loop) on p. 11.1ff. Also note that the author obviously uses the east-coast convention for the Minkowski pseudo-metric, i.e., $\eta_{\mu \nu}=\mathrm{diag}(-1,1,1,1)$ rather than Peskin Schroeder who uses the west-coast convention.

 

[Markus Kahn]

The physical mass $m_{ren}^2$ is the value of $p^2$, where the propagator has a pole

Alright, this makes sense. Then we have
$$m_{\text{ren}}^2=m^2[1+I(m_{\text{ren}}^2)] \approx m^2[1+I(m^2)].$$

Note that this is not $\phi^4$-theory but the one-loop self-energy of $\phi^3$-theory.

When exactly did that happen? Where in post #1 did I make a mistake so that I ended up in $\phi^3$ theory?

 

[vanhees71]

I think, you just used two sources, Peskin&Schroeder and the ETH manuscript. In the manuscript a theory with cubic an quartic terms is discussed, and at the one-loop order you have two self-energy diagrams. The tadpole diagrams can be neglected in such theories (in the operator formalism you can normal-order the interaction pieces of the Lagrangian in the interaction picture; this is fine as long as you don't deal with gauge theories, where normal-ordering can destroy gauge invariance for proper vertex functions, e.g., in scalar QED).

I didn't check the integrals in the manuscripts, but they look correct.

For pure $\phi^4$ theory, you may also have a look at my lecture notes on QFT:

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[Markus Kahn]

@vanhees71 Thanks a lot for the explanations and I will be sure to check out your lecture notes!

Just as a quick check, the issue is that I basically conflated the following, right?

i.e. I assumed that $\phi^4$ has this one extra loop diagram that appears due to a $\phi^3$ interaction term, when $\phi^4$-theory doesn't have an $\phi^3$ interaction term.

 

[vanhees71]

Exactly.