# Renormalization when there is spontaneous symmetry breaking

Standard quantum field theory textbooks discuss spontaneous symmetry breaking with the following Lagrangian:

$$L=\frac{1}{2}\partial_{\mu}\vec{\phi} \cdot \partial^\mu \vec{\phi}+m^2\vec{\phi}\cdot \vec{\phi}-\frac{\lambda}{4!} (\vec{\phi}\cdot\vec{ \phi})^2 \tag{1}$$

However, several questions are not clear to me

1. The vacuum expectation value of the field operator is not zero, i.e., $$\langle 0|\vec{\phi}(x)|0\rangle \neq 0$$, which is the starting point for the mechanism of the spontanous symmetry breaking. However, I'm wondering whether the following definition is still true: $$$$\langle 0| \phi_n (x) |0 \rangle=\frac{\delta W[\vec{j}]}{\delta j_n(x)}\big|_{\vec{j}=0} \tag{2}$$$$ If Eq. (2) is true, then we will end up the following path integral representation $$\langle 0| \phi_n (x) |0 \rangle=\int D\vec{\phi}\phi_n \exp{(iS[\vec{\phi}]/\hbar)} \tag{3}$$ Now since $$S[\vec{\phi}]$$ is even in $$\vec{\phi}$$, Eq. (3) actually is zero. We then conclude $$\langle 0| \phi_n (x) |0 \rangle=0$$, which contradicts with the original assertion at the beginning.

2. The fact that $$\langle 0|\vec{\phi}(x)|0\rangle \neq 0$$ indicates that Eq. (2) may no longer hold, which then indicates that Eq. (3) may no longer hold. But Eq. (2) is not true, we may have troubles to renormalize the theory as Eq. (2) is the starting point to define the generating functional of 1PI vertex function. At least, the effective action technique needs to be modified?

3. The straightforward way to renormalize the theory is to expand the Lagrangian around some randomly chosen minimum the potential, say $$\vec{v_c}$$. This is equivalent to replace $$\vec{\phi}=\vec{\varphi}+\vec{v_c}$$. Then we have $$\langle 0| \varphi_n (x) |0 \rangle=0 \tag{4}$$ and can apply the standard technique of the effective action. The problem with this approach is that the new Lagrangian not only contains $$\varphi^4$$ term but also $$\varphi^3$$ term, which will make the calculations more tedious.

Does anyone know the answers or any relevant references that discuss these issues?

As it stands, the right-hand side of (3) is indeed zero, because it corresponds to summing over all of the degenerate vacua. In other words, the $$|0\rangle$$ on the left-hand side of (3) is actually $$|0\rangle=\int dc\ |0(c)\rangle$$, where $$|0(c)\rangle$$ is the vacuum state in which the expectation value of $$\phi_n$$ is $$c_n$$. These different vacuum states belong to different superselection sectors, which means $$\langle 0(c)|A|0(c')\rangle = 0 \hskip1cm \text{if }c\neq c'$$ for all observables $$A$$. Therefore, the left-hand side of (3) is $$\int dc\ \langle 0(c)|\phi_n(x)|0(c)\rangle = 0,$$ which is consistent with the fact that the right-hand side is zero.

That isn't what we want, of course. We want to consider just one of the degenerate vacuum states $$|0(c)\rangle$$, which is the intended meaning of the left-hand side of (3). To enforce this on the right-hand side, we can use this recipe:

• First define the right-hand side of (3) in finite volume, with the help of Wick rotation as usual.

• Include an explicit symmetry-breaking term in the action $$S$$. Thanks to the Wick rotation, this ensures that only one of the degenerate vacuum states $$|0(c)\rangle$$ contributes to the right-hand side.

• With the explicit symmetry-breaking term in place, take the infinite-volume limit.

• After taking the infinite-volume limit, take the limit as the explicit symmetry-breaking term goes to zero.

The point of this recipe is that the two limits don't commute: if we had set the symmetry-breaking terms to zero before taking the infinite-volume limit, we'd be in the undesirable situation described in the first paragraph. To get what we want, we need to take the limits in the right order.

That addresses part 1 of the question. Part 2 is addressed similarly. The answer to part 3 is yes, the $$\phi^3$$ term does make the calculations more tedious. I don't know of any clever ways to avoid that.