One-loop Effective potential, Higgs VEV and renormalization condition

On Peskin & Schroeder's QFT, chapter 11.4, the book discusses the computation of the effective action of linear sigma model.

I am troubled for the relation between renormalization condition and Higgs VEV (vacuum expectation value) appearing in page 376-377.

First, on the bottom of page 376, the book says we can apply the usual renormalization condition (11.16) as in section 11.2. In this case, the tadpole renormalization condition reads

$$\frac{\partial V_{\mathrm{eff}}}{\partial \phi_{\mathrm{cl}}}\left(\phi_{\mathrm{cl}}=\mu / \sqrt{\lambda}\right)=0.\tag{p.376}$$

I understand this as Higgs VEV are invariant, $$v=\mu/\sqrt{\lambda}$$.

Second, the book says it's useful to apply the $$\overline{MS}$$ prescription to visualize the modification of the lowest-order result. Under the $$\overline{MS}$$, the effective potential reads \begin{aligned} & V_{\mathrm{eff}}=-\frac{1}{2} \mu^2 \phi_{\mathrm{cl}}^2+\frac{\lambda}{4} \phi_{\mathrm{cl}}^4 \\ & \quad +\frac{1}{4} \frac{1}{(4 \pi)^2}\left((N-1)\left(\lambda \phi_{\mathrm{cl}}^2-\mu^2\right)^2\left(\log \left[\left(\lambda \phi_{\mathrm{cl}}^2-\mu^2\right) / M^2\right]-\frac{3}{2}\right)\right. \\ &\quad \left.+\left(3 \lambda \phi_{\mathrm{cl}}^2-\mu^2\right)^2\left(\log \left[\left(3 \lambda \phi_{\mathrm{cl}}^2-\mu^2\right) / M^2\right]-\frac{3}{2}\right)\right) . \\ & \end{aligned}\tag{11.79}

Now, since we have an arbitrary dimensional parameter $$M$$, the Higgs VEV should depend on the value of $$M$$? In general, it's not equal to $$\mu/\sqrt{\lambda}$$, right?

I am also troubled for following description below (11.79)

The connection to $$V_{\text{eff}}$$ is undefined when the arguments of the logarithms become negative, but fortunately the minimum of $$V_{\text{eff}}$$ occur outside of this region, as is illustrated in Fig. 11.9.

Does this means that $$\phi_{\text{cl}}$$ always larger than $$\mu/\sqrt{\lambda}$$? How do we insure that?

The problem here is a problem about renormalization.

First, if we have a Lagrangian $$\mathcal{L}$$, with only so-called bare parameters, such as $$\mathcal{L} = \frac{1}{2}(\partial\phi)^{2}-\frac{1}{2}m^{2}_{0}\phi^{2}-\frac{\lambda_{0}}{4!}\phi^{4}$$. We can use this model to calculate observables and compare them with the experiments to get the parameters $$(m_{0},\lambda_{0})$$ in my model if this model is sufficient. This is the basic logic, parameters in the Lagrangian depend on experiments.

Second, to isolate the UV part physics (or you can say to perturb using the physical parameters), we isolate the Lagrangian as $$\mathcal{L} = \mathcal{L}_{1}+\mathcal{L}_{ct}$$ where $$\mathcal{L}_{ct}$$ contains so-called conterterms. As you know, for renormalizable Lagrangian, terms in $$\mathcal{L}_{ct}$$ are got by isolating parameters in the bare Lagranian, such as $$m_{0} = m + \delta_{m}$$. If we focus on no-counterterms part $$\mathcal{L}_{1}$$ only, the parameters in it depend on both experiments and counterterms, equivalently, both experiments and renormalization schemes.

Then, back to your problem. $$\phi_{cl}$$ can be view as an observable (not really in fact), and it depends on experiments. If we are using the full Lagrangian $$\mathcal{L}$$, not isolating counterterms, using bare parameters ($$(\lambda_{0},\mu_{0})$$ here), $$\phi_{cl}$$ has a specific dependence on (bare) parameters. We can use this dependence and the experiments to get an equation of (bare) parameters. However, if we isolate the counterterms and focus on the $$\phi_{cl}$$ dependence on the parameters in $$\mathcal{L}_{1}$$ only ($$(\lambda,\mu)$$ here). We can have any $$\phi_{cl}$$ dependence by having different counterterm parameters, equivalently by using different renormalization schemes. The logic is we don't care if $$\phi_{cl} = \mu/\sqrt\lambda$$, $$\phi_{cl}$$ depends on experiments, and we use the relation between it and parameters to determine the parameters in Lagrangian.

The first renormalization scheme you present is similar to the on-shell scheme. Namely, the parameters $$(\mu,\lambda)$$ in $$\mathcal{L}_{1}$$ (not bare parameters in full $$\mathcal{L}$$) can be view as "physical parameters". The relation between $$\phi_{cl}$$ and $$(\mu,\lambda)$$ is the same as that in classical theory.

The $$\overline{MS}$$ scheme have a renormalization scale $$M$$ (use $$mu$$ in most contexts, but $$mu$$ has been a parameter in Lagrangian here), the existence of $$M$$ is part of this renormalization scheme. Following the logic above, if we use $$\overline{MS}$$ scheme and derive a relation between $$\phi_{cl}$$ and $$(\mu,\lambda,M)$$ and the experiment value of $$\phi_{cl}$$, we can assign a value of $$M$$ as we like, and get a relation between $$\mu$$ and $$\lambda$$ which can help us determine the value of them.

• Thank you so much! That's incredibly helpful to me. So, to sum up, what we truly value are the quantities measured in the experiment and how they relate to one another, which should be dependent on the schemes. So, it appears that $\phi_{\text{cl}}=\mu/\sqrt{\lambda}$ only makes sense in the On-shell scheme, correct? If we examine Peskin's equation (10.19), the measured quantities $m$ and $\lambda$ have a relationship of $m=\sqrt{2\lambda}v$, which allows us to obtain $v$. Then, we can calculate $\mu$ using $\mu=v\sqrt{\lambda}$, right? Thank you! Commented Apr 21, 2023 at 10:27
• Actually, there is another point that I don't fully understand, which is the sentence I quoted at the end of my post. Could you please elaborate on why they say the argument of the logarithm is non-negative? Commented Apr 21, 2023 at 10:46
• @Daren 1. $\phi_{cl} = \mu/\lambda$ only makes sense in a specific renormalization scheme. If you make it a renormalization condition like Peskin says on Page376, $\phi_{cl} = \mu/\lambda$ makes sense by definition. Here, $\mu,\lambda$ are parameters in $\mathcal{L}_{1}$, not bare parameters in full $\mathcal{L}$. 2. Sorry, but I didn't find a $v$ in eq.(10.19) in Peskin. 3. You could find the argument of the log is non-negative by looking at the Figure11.9. The absolute value of the minima of $V_{eff}$ are larger than those of $V$ without quantum fluctuation included.
– TOAA
Commented Apr 22, 2023 at 5:38
• I think there are nice physical interpretations explaining why quantum fluctuation enlarges $\phi_{cl}$. But I don't know any.
– TOAA
Commented Apr 22, 2023 at 5:40
• Thank you so much! Commented Apr 23, 2023 at 10:08