[Physics] One-loop effective potential of Standard Model

1pi-effective-actionhiggsquantum-field-theoryrenormalizationstandard-model

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

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 does the formula really break down? Or is there a way of making sense of the imaginary potential?

Best Answer

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.

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