Consider the familiar $\phi^4$ theory Lagrangian for a complex scalar field

$$\mathcal{L} = \frac{1}{2}|\partial_{\mu}\phi |^2+\frac{1}{2}m_0^2| \phi | ^2 - \frac{\lambda}{4!}| \phi | ^4$$

with $m_0^2 > 0$.

In the context of phase transitions, for high temperatures, $T>T_c$, the comlpex field $\phi$ has an expectation absolute value $0$, $\langle |\phi| \rangle = 0$, while for $T<T_c \implies \langle |\phi| \rangle \neq 0$. Hence, the symmetry is sponaneously broken. A typical example of this behaviour is the Higgs field.

According to Wikipedia, in QFT a true vacuum is globally stable, while a false vacuum is only locally stable.

Let's then consider the Higgs field. To my understanding, in the early days of the Universe $\langle |\phi| \rangle = 0$ would be a true vacuum (while in the highly symmetric state). However, as $T$ diminished, the symmetry was broken and the 'previous vacuum' became unstable: a new true vacuum emerged with non-zero expectation value.

However, the Higgs field is said to be in a metastable state (false vacuum). Ho can I conciliate all of this information?


1 Answer 1


Short answer: Even if the Higgs potential has another minima at zero temperature (doubtful), the probability to tunnel to said minima is ridiculously small. Yet during the phase transition the probability to tunnel (or thermally jump) is sizable.

Long answer:

With your zero-temperature potential $$V(\phi) =-m_0^2 \phi^2/2+\lambda/4! \phi^4, $$ there are additional terms at high temperatures:

$$V(\phi) =(a g^2 T^2-m_0^2) \phi^2/2 +\lambda/4! \phi^4$$, where g is the weak gauge coupling and I neglected similar terms from fermions etc. Point is, for large enough temperatures the quadratic pre-factor, $a g^2 T^2-m_0^2$, is positive. So the symmetric, $\phi=0$, minima has the lowest energy. While at the critical temperature the broken minima, $\phi\neq 0$, has the same energy as the symmetric one.

The phase transition can, depending on the details, be either first-order or second-order. In either case, the transition from the symmetric to the broken minima happens (in most cases) quite rapidly as the temperature is lowered.

There's then the issue of metastability at zero temperature. Here quantum effects generate a new minimum at large $\phi\sim 10^{17} GeV$ values. Though, there are some caveats. First, the generation of this new minima assumes only the Standard Model. And since there is almost certainly some new physics lurking around, the presumed metastability should be taken with a grain of salt. Second, even barring new physics, the lifetime of the Standard Model is around $10^{100} $ years~arXiv:1707.08124 .


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