# One-loop effective potential of Standard Model

The one loop Coleman-Weinberg contribution of a scalar field to the effective potential (in MSbar scheme) is: $$\mathrm{const.} \times m^4(\phi_c) \left( \log \left( \frac{m^2(\phi_c)}{\mu^2}\right) -\frac{3}{2} \right)$$

Now, I have a problem with this formula. In theories with spontaneous symmetry breaking, like the standard model, the background field dependent mass will actually be negative. For the SM effective potential (usually calculated in Landau gauge) we have the Higgs field and the Goldstone fields, with:

\begin{eqnarray} m_H(\phi_c) = 3 \lambda \phi_c^2 - m^2 \\ m_G(\phi_c) = \lambda \phi_c^2 - m^2 \end{eqnarray} At the Higgs VEV $m_H(v)=2m^2$ and $m_G(v)=0$. Where $m$ and $\lambda$ are the renormalized parameters of the tree level Higgs potential.

My problem is that this implies, for example, for $\phi_c < v$ the Goldstone boson $\phi_c$ dependent mass is negative, and the logarithm is complex, also as $\phi_c \to v$ the real part of the log goes to $-\infty$.

Am I doing something silly or the formula really does break down? Or is there a way of making sense of the imaginary potential?

## 2 Answers

You are correct in observing that the effective potential is complex for certain values of the background field. This is somewhat of a thorny issue. In principle, when deriving the effective potential one writes it as a Legendre transformation of the generating functional of connected diagrams $W[J]$. This assumes that the effective potential is convex in the variable $\phi$. When there is spontaneous symmetry breaking present at the classical level this is obviously not true, as evidenced by the negative curvature at the origin of field space. The conventional wisdom is that our whole derivation does not breakdown into nonsense. Instead, when a field dependent mass turns negative then the corresponding field value does not represent a stable state. This is signaled through the effective potential by it acquiring a non-zero imaginary part that plays the role of a decay rate of this state. However, see [Precision decay rate calculations in quantum field theory] for how to calculate the actual physical decay rate properly.

Your second observation, that the logarithm diverges as $\phi \rightarrow v$ is correct but also nothing to worry about. This is because the prefactor of the logarithm, $\left(m_G^2\right)^2$, also goes to zero in this limit (faster than the logarithm diverges).

On a more general note, there is nothing inconsistent about evaluating the effective potential for general field values. However, the effective potential does not in general represent a physical quantity. Only when evaluated at extrema.

I recommend [Consistent Use of Effective Potentials] for a deep treatment of the effective potential and how to derive physical quantities from it.

• The effective potential is always convex. It is its one-loop approximation that has problems. Functional RG approaches give a convex effective potential in the broken symmetry phase (it is flat between $-\nu$ and $\nu$). – Adam Aug 19 '18 at 21:01
• Yes you are of course correct. I should have written that it is a problem due to the truncation of our perturbative calculation. Could you give me a reference to read more about the functional RG approach to the effective potential? – subtlesplendor Aug 20 '18 at 7:47
• For the specific case of the convity in the ordered phase, you can look at arXiv:1109.4024, where the flow equation of the effective potential is solved numerically. For nice introduction about FRG (but not much discussion about convexity), I really like arXiv:0702.365 – Adam Aug 20 '18 at 8:40

Let us assume you are working with the Standard Model (SM) so that the Higgs field corresponds to a $SU(2)\times U_Y(1)$ doublet. Computing the Coleman-Weinberg potential indeed leads to a contribution similar to what you wrote, let us first correct that: $$V_{eff} \sim \phi_c^4 \ln\left(\frac{\phi_c^2}{\mu}\right)$$

Additionally, you seem to be mixing the fact that to arrive to such expression one basically has to employ an expansion of the path integral around $\langle\phi_c\rangle$, which is no generic value; $\langle\phi_c\rangle$ must be a critical point of the tree-level potential for the derivation to apply. (see [Coleman-Weinberg Mechanism]) and this is the one appearing in your mass corrections; it is specified by the theory and its couplings.

In the case of the SSB in the SM case $\sqrt{2}\langle\phi_c\rangle = v \approx 200$ GeV and it is technically this expectation value that should appear in your corrections as said above. As shown in Schwartz book "Quantum Field Theory and the Standard Model", section 34.2.3, one gets: $$V_{eff}(h) = \frac{\lambda}{4}h^4 + \left(\frac{9}{64\pi^2}\lambda^2 - \frac{3}{64\pi^2}Y^4_t\right)\phi^4\ln\frac{h^2}{v^2}$$ where $Y_t$ is the top-quark Yukawa coupling for the Coleman-Weinberg potential for the Higgs (actually the most relevant contribution only). There you can see what I meant previously.

On the other hand, for values of the field that are large, you indeed have to more careful and resum these large logarithms separately. (see section 34.2.2 of Schwartz). Small numbers on the other hand pose no threat since you have a power to the fourth in front of the log.

An alternative argument for you to put your soul to rest on the matter, is to consider the unitary gauge, where you can get rid of the Goldstone bosons thus rendering your worries futile, while remembering physical quantities are supposed to be gauge independent. All in all, it is true that effective theories have a regime of validity and one should not expect them to hold for totally arbitrary values in our case it would be the electroweak scale most likely, $v$.