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.
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 $\langle\phi\rangle = v$.