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I am just reading some material about symmetry breaking and so-called effective action/potential

Consider a lagrangian \begin{equation*} \mathcal{L}=\frac{1}{2}(\partial \phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{4!}\phi^4=\frac{1}{2}(\partial \phi)^2-V(\phi) \end{equation*} Now it is known for $m^2>0$, symmetry is unbroken, $m^2<0$, symmetry is broken. But people usually spend pages on discussing the case $m^2=0$.

My first confusion is: for $m^2=0$, $V(\phi)=\frac{\lambda}{4!}\phi^4$ attains its minimum at just one point $\phi=0$, and the symmetry seems unbroken. Why is it necessary for us to consider $m^2=0$?

My second confusion is:

In the P&S's book "An Intro to QFT", 11.1, it is shown $V(\phi)$ for $m^2<0$ has two minima $\pm \phi_0$, called expectation value of $\phi$. Why is it called "expectation value"? Is it an average of something? In field theory, what is the physical motivation of minimizing a potential $V(\phi)$?

Sven2009
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  • Just a heads up: as the rule of the site says, you should make only one question per post. In any case, could you bring up some references on the $m^2=0$ case? – Davide Morgante May 31 '20 at 14:54
  • The vacuum expectation value (vev) is the expected value of the field in the vacuum $\langle0|\phi|0\rangle$. In the unbroken phase, the vev will be zero. The fact that it is non-zero is exactly what is meant by a broken symmetry. – GrassyNol May 31 '20 at 15:06
  • Near duplicate, as well as. – Cosmas Zachos May 31 '20 at 15:18
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    If $m^2=0$, you're right that there is no spontaneous symmetry breaking at the classical level. But when you take quantum corrections into account and construct the effective action, you'll see that symmetry can be broken. You can see the Chapter "Effective Potential" in A. Zee's QFT book. – SRS May 31 '20 at 15:48
  • @GrassyNol But I have seen some definition saying that in the quantum case, the symmetry is broken if the vacuum state is not unique. What is the relation between this definition and a non-vannishing VEV you mentioned? Are they equivalent? – Sven2009 Jun 01 '20 at 12:31
  • @SRS A.Zee's book says quantum fluctuation might bring symmetry breaking in Chapter "Effective Potential" . But is this "quantum fluctuation" the same as the "quantum corrections" you mentioned? Also what quantity does "quantum corrections" contribute to? is <0|$\hat \phi$|0>? – Sven2009 Jun 01 '20 at 12:40
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    Yes. You're right. In general, $\langle 0|\hat{\phi}|0\rangle=v+(\hbar\text{dependent quantum corrections})$. $v$ can be obtained classically by minimizing the classical potential. – SRS Jun 01 '20 at 14:24

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