In section 11.4 of Peskin & Schroeder's "Introduction to Quantum Field Theory", the authors calculate the effective potential of the linear sigma model to one-loop order:

$$\begin{align*} V_{\text{eff}}(\phi_0)&=-\frac 1 2 \mu^2\phi_0^2 + \frac \lambda 6 \phi_0^4 \\ &\hspace{1cm}-\underset{\text{divergent term}}{\underbrace{\frac{\Gamma(-d/2)}{2(4\pi)^{d/2}}\left[(N-1)(\lambda\phi_o^2-\mu^2)^{d/2}+(3\lambda\phi_0^2-\mu^2)^{d/2}\right]}}\\ &\hspace{1cm}+\frac 1 2 \delta_{\mu}\phi_0^2 + \frac 1 6 \delta_{\lambda}\phi_0^4 \tag{11.84} \end{align*}$$

where $d$ is the spacetime dimension and $\delta_{\mu,\lambda}$ are counterterms, and $\phi_0^2\equiv \sum_i \phi_{i,0}^2$ is the minimum energy field configuration (constant in spacetime). This expression is divergent for $d=0,2,4$, since the gamma-function has a pole at those values. The authors say the following:

Thus, Eq. (11.84) becomes a finite expression in the limit $d\rightarrow 2$ if we set $$\delta_{\mu}=-\lambda (N+2) \frac{\Gamma \left(1-\frac d 2 \right)} {(4\pi)} + \text{finite.}$$

But I do not think this is true. When I evaluate the divergent term in Eq. (11.84) above in the limit $d\rightarrow 2$ and keep the divergent terms unevaluated, I get:

$$\begin{align} \frac{\Gamma(-d/2)}{2(4\pi)}\left[(N-1)(\lambda\phi_o^2-\mu^2)+(3\lambda\phi_0^2-\mu^2)\right]&=\frac{\Gamma(1-d/2)}{(-\frac d 2 )2(4\pi)}\left[(N+2)\lambda\phi_o^2- N\mu^2\right]\\ &=-\frac{\Gamma(1-d/2)}{(4\pi)}[\underset{\color{green}{\text{[A]}}}{\underbrace{(N+2)\lambda\phi_o^2}}- \underset{\color{red}{\text{[B]}}}{\underbrace{N\mu^2}}] \end{align}$$

[Question] You can see that $\color{green}{\text{[A]}}$ is the term the author cites/cancels, but what about term $\color{red}{\text{[B]}}$? That still contributes to the overall divergence.

[2$^{\text{nd}}$ Question] Note also that if we simply cancel that "additional" divergence with, say, $\delta_{\mu}$, then the counterterm will actually depend on $\phi_0^2$. Is this an issue?

And by the way, nowhere in that section does the author take the limit/case $\mu^2\rightarrow 0$.


1 Answer 1


Since the term proportional to $\mu^2$ is constant in the fields, it is irrelevant in the effective potential because it is a rigid constant shift of the potential itself.

The fact that this term actually diverges is no issue - it can be reabsorbed in the normalization of the path integral, for instance. It would be relevant only in coupling the theory to gravity, but here that is not the case.


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