On Peskin & Schroeder's QFT, page 387, the book gives a general analysis about renormalization and symmetry.

For the example of $\sigma$ mass in the linear sigma model, the classical relation is $$m-\sqrt{2\lambda}\langle\phi\rangle=0 \tag{11.104} $$

I am troubled for following sentences bellow (11.104)

Eq.(11.104) is valid at the classical level however the parameters of the Lagrangian are modified, it holds equally well when we add counterterms to the Lagrangian and then adjust these counterterms order by order. Thus, the counter term must give zero contributions to the right-hand side of Eq.(11.104). Therefore, the perturbative corrections to Eq.(11.104) must be automatically ultraviolet-finite.

My first question is, would (11.104) still holds after the quantum correction?

If we re-check the book's eq.(11.16) $$\langle\phi^N\rangle =\frac{\mu}{\sqrt{\lambda}} \tag{11.16} $$ is satisfied to all orders in perturbation theory. There is one sentence bellow (11.16) reads

the mass $m$ of the $\sigma$ field will differ from the result of the classical equations $m^2=2\mu^2=2\lambda v^2$ by terms of order ($\lambda\mu^2$).

So it appears that (11.104) no longer holds after quantum correction.

Previously, I thought (11.104) holds to all orders.

  1. The Femi constant $G_F=\frac{1}{2v^2}$ can be measure from muon decay. (PDG: $G_F=1.1663788(7)\times 10^{-5}$ $\text{GeV}^{-5}$). This imply $v=246.22 \text{GeV}$;

  2. From PDG 2022, Higgs mass $m_h=125.25 \text{GeV}$;

  3. From (11.104), $\lambda=0.129383$.

My second question is would this analysis correct?


1 Answer 1

  1. In quantum terms, the vacuum expectation value of a field operator $\langle\phi\rangle$ is defined as the minimum point of the quantum effective potential $V(\phi)$, so after calculating $V(\phi)$ with the appropriate scheme and at the specific order of perturbation, the saddle point can be examined to evaluate how much it deviates from the classical saddle point due to quantum corrections. Especially, quantum correction at one loop order is easy to evaluate because all we should do is to do the Gauss integral. (See Wikipedia and its reference).

  2. I do not know what method they used to obtain these values, but since the above calculation method can incorporate quantum effects, I believe that these results also correctly incorporate quantum effects.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.