# Ambiguous points in spontaneous symmetry breaking of discrete symmetry

For a discrete symmetry: At the minimum value of the potential, $V$, in the Lagrangian density, why do we take $\phi= \langle v\rangle + \eta$? Aren't we deliberately breaking the symmetry? If we don't do this, the symmetry is intact. On the other hand, if we replace $\phi$ by $\phi= \langle v\rangle + \eta$, even when not in a ground state, then also the symmetry will be broken in $\eta$ field. Please explain.

• How is this question specifically related to discrete symmetries? Sep 11, 2014 at 7:44
• no it's not. i chose this because it is simplest example of spontaneous symmetry breaking Sep 11, 2014 at 13:01

In canonical quantization, a quantum field is a linear combination of so-called "creation and annihilation operators". That means that the field $\phi$ creates and destroys particles of type $\phi$.
The state $|0\rangle$ is the vacuum: the state with no particles. If $\phi$ is a quantum field that creates and destroys particles, it must be that $\langle 0 | \phi | 0 \rangle=0$, because the state with particles created/destroyed, $\phi|0\rangle$ must be orthogonal to the empty state $|0\rangle$.
We have no choice, then, but to write $\phi=v+\eta$, such that $\langle \eta \rangle=0$. Th new field $\eta$ now has a good particle interpretation, whereas the original field $\phi$ did not, because $\langle\phi\rangle = v$.
• @innisfree I want to make a comment. I don't think $\langle 0|\phi|0\rangle=0$ only when $\phi$ can be written as a linear combination of $a$ and $a^\dagger$. For example, in self-interacting $\lambda\phi^4$-theory, one has $\langle 0|\phi|0\rangle=0$. The actual criterion is, if $\mathcal{L}(-\phi)=\mathcal{L}(\phi)$, all $n$-point functions with odd $n$ will vanish even if $\phi$ is non-exapandable as a linear combination of $a$ and $a^\dagger$.