# Vacuum expectation value (VEV) of a Gauge theory - Spontaneous Symmetry Breaking (SSB) - Higgs Mechanism

I am dealing with a sort of scalar QED with a term of SSB

$$\begin{equation} \mathcal{L}=\left|D_{\mu} \phi\right|^{2}-\frac{1}{4}\left(F_{\mu \nu}\right)^{2}-V\left(\phi^{*} \phi\right) \end{equation}$$

with potential

$$\begin{equation} V(\phi)=-\mu^{2} \phi^{*} \phi+\frac{\lambda}{2}\left(\phi^{*} \phi\right)^{2} \end{equation}$$

So I have learned here that the classical potential is at tree-level a good approximation for the effective potential $$V_{eff}$$. Therefore we use it to calculate the vacuum expectation value of the field. The minimum of $$V(\phi$$) is

$$\begin{equation} \phi_{0}=\left(\frac{\mu^{2}}{\lambda}\right)^{1 / 2} \end{equation}$$

and this is a good approximation at TL for the vacuum expectation value $$\begin{equation} <\phi> \sim \phi_{0} \end{equation}$$

But now it is not clear to me why we are considering only the potential $$V(\phi)$$ when calculating the minimum. In fact we have terms with $$\phi$$ inside the covariant derivative $$\left|D_{\mu} \phi\right|^{2}$$ as well, meaning for example

$$\begin{equation} \phi\phi^*A_{\mu}A^{\mu} \end{equation}$$

Why these terms do not have a role in calculating the minimum of the potential? Meaning why only the self-interaction terms determine the vev?

• Great question! I assume we consider the case where there are no kinetic term contributions when it comes to the vacuum expectation value, since we are considering the vacuum. I'd like to know too if there is a more elaborate explanation. Jan 17, 2020 at 23:00
• 'cause $<A_{\mu}>= 0$ in vacuum. A trick question for ya: if the there is an external nonzero EM field $<A_{\mu}A^{\mu}> \neq 0$, would it change the Higgs VEV $\phi_0$ and consequently changing the fermion mass? Jan 17, 2020 at 23:09
• In the previous case (the scalar QED) I would say that the value of $\phi_0$ changes because of the external field and therefore the mass $m_A$ of the gauge field changes. In SM I would say that $\phi_0$ changes as well, but somehow it does not seem correct. Jan 17, 2020 at 23:31
• Possible duplicate: physics.stackexchange.com/q/75845/50583 (the answers there prove that the minimum of the effective potential is the VEV, and since you already accept that the classical potential is a good approximation to the effective potential, this answers why the classical minimum is a good approximation of the VEV) Jan 20, 2020 at 17:14