On Peskin & Schroeder's QFT pages 353-355, the book uses the Linear sigma model to illustrate the renormalization and symmetry.

We can write the Lagrangian of Linear sigma model with

$$ \begin{aligned} & \mathcal{L}=\frac{1}{2}\left(\partial_\mu \phi^i\right)^2+\frac{1}{2} \mu^2\left(\phi^i\right)^2-\frac{\lambda}{4}\left[\left(\phi^i\right)^2\right]^2 \\ &+\frac{1}{2} \delta_Z\left(\partial_\mu \phi^i\right)^2-\frac{1}{2} \delta_\mu\left(\phi^i\right)^2-\frac{\delta_\lambda}{4}\left[\left(\phi^i\right)^2\right]^2 . \end{aligned} \tag{11.14} $$ Written in term of $\pi$ and $\sigma$ fields, the second line take the form $$ \begin{aligned} \frac{\delta_Z}{2}\left(\partial_\mu \pi^k\right)^2 & -\frac{1}{2}\left(\delta_\mu+\delta_\lambda v^2\right)\left(\pi^k\right)^2+\frac{\delta_Z}{2}\left(\partial_\mu \sigma\right)^2-\frac{1}{2}\left(\delta_\mu+3 \delta_\lambda v^2\right) \sigma^2 \\ & -\left(\delta_\mu v+\delta_\lambda v^3\right) \sigma-\delta_\lambda v \sigma\left(\pi^k\right)^2-\delta_\lambda v \sigma^3 \\ & -\frac{\delta_\lambda}{4}\left[\left(\pi^k\right)^2\right]^2-\frac{\delta_\lambda}{2} \sigma^2\left(\pi^k\right)^2-\frac{\delta_\lambda}{4} \sigma^4. \end{aligned} \tag{11.15} $$

Now, let's consider the renormalization conditions. One of the conditions is to set the tadpole diagram equal to zero.

I think this condition leads to (11.16)

$$ \left\langle\phi^N\right\rangle=\frac{\mu}{\sqrt{\lambda}} \tag{11.16} $$ which is satisfied to all orders in pertubation theory.

This is where I am troubled. Why setting all the tadpole terms equal to zero can guarantee the vacuum expectation value (vev, $v$) $\frac{\mu}{\sqrt{\lambda}}$ in all orders?

I think $v=\frac{\mu}{\sqrt{\lambda}}$ only holds for the first line of (11.14). In the counter terms, there are two-points, three-points and four points interactions of the $\sigma$ field. Shouldn't all these counterterms will affect the potential shape and then affect the vev?


1 Answer 1


That's a good question.

  1. The linear sigma model (11.14) has 3 terms, and hence 3 counterterms, and hence needs 3 renormalization conditions (11.17a+b+c).

  2. Yes, since the $N$th component $$\phi^N(x)~=~ v +\sigma(x) \tag{11.8}$$ and the definition of the VEV$^1$ $$ v~:=~\frac{\mu}{\sqrt{\lambda}},\tag{11.7}$$ then condition (11.16) for a connected 1-point function is equal to $$\langle \sigma \rangle^c_{J=0}~=~0,$$ i.e. that the tadpoles for the $\sigma$ field vanish$^2$ $$ \left(\fbox{1PI}==\stackrel{\sigma}{=}==\right)_{\text{amputated}}~=~0,\tag{11.17a}$$ cf. e.g. my Phys.SE answer here.

  3. Yes, diagrams with more and more interaction terms do enter on the LHS of eq. (11.17a). For explicit calculations at one-loop, see eqs. (11.31) & (11.32).

  4. It is important to realize that the field $\phi~=~\phi_0/\sqrt{Z_{\phi}}$ and the VEV (11.7) scale differently under the RG flow. In particular, since both sides of eq. (11.7) should remain$^3$ finite/physical, one should not try to e.g. introduce infinite $Z$-factors and/or counterterms into eq. (11.7), cf. Refs. 1 & 2.


  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; p. 353-355.

  2. M. Srednicki, QFT, 2007; chapter 31. A prepublication draft PDF file is available here.


$^1$ Here the mass parameter $\mu$ in the Lagrangian (11.14) is the renormalized mass. It should not to be conflated with the physical mass $m$, which satisfies
$$ \left(==\stackrel{\sigma}{=}==\fbox{1PI}==\stackrel{\sigma}{=}==\right)_{\text{amputated}}~=~m^2-\mu^2\quad \text{at}\quad p^2=m^2.$$

$^2$ The tadpoles for the $\pi^k$ fields vanish automatically, due to a $\mathbb{Z}_2$ symmetry.

$^3$ Later in dimensional regularization one typically makes the coupling constant $\lambda\to\lambda\tilde{\mu}^{\epsilon}$ dimensionless, cf. Ref. 2.

  • $\begingroup$ Hello, may I ask you a related question about this post? I summarize my new thoughts in this post: physics.stackexchange.com/q/761623. But I don't think the question is solved! I would be appreciate it if you can look that! $\endgroup$
    – Daren
    Commented May 5, 2023 at 8:43

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