In the quantum$^1$ system with a continuous symmetry (in the Thermodynamic limit) relating the ground states $\newcommand{\ket}[1]{\left|#1\right>}\{\ket{\theta}\} \newcommand{\bra}[1]{\left<#1\right|}$ . Before spontaneous symmetry breaking the density operator of is: $$\rho=e^{-\beta \hat H}$$ After spontaneous symmetry only a subset of states in this operator. In the $T\rightarrow 0$ limit only those states with the ground state energy survive.
My question is the following:
At $T=0$ do we always have: $$\rho=\ket{\theta_1}\bra{\theta_1}\tag{1}$$
Could we have e.g. $$\rho=\frac{1}{2}\ket{\theta_1}\bra{\theta_1}+\frac{1}{2} \ket{\theta_2}\bra{\theta_2}\tag{2}$$ or more realistically: $$\rho=\mathcal{N}\int^{\theta_1+\Delta \theta}_{\theta_1-\Delta \theta}d\theta \ket{\theta} \bra{\theta}\tag{3}$$ please can someone explain why which are the possible density operators after SSB in the $T\rightarrow 0$ limit and why.
(I feel we get something like (3) with the Heisenberg model)
$^1$ although an answer explaining what is different in the classical system would also be appreciated.
