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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.

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  • $\begingroup$ See: arxiv:1703.10439, Ellis, 2007; pg92 and this answer on PSE $\endgroup$ Jun 10, 2018 at 13:11
  • $\begingroup$ Spontaneous symmetry breaking only occurs in the thermodynamic limit. However, there it is not clear what the density operator really should be. From a hand-wavy point of view, I would say that $\rho=|\theta\rangle\langle\theta|$ for any ground state $|\theta\rangle$ (in the tdyn. limit) is the "correct" symmetry broken state. $\endgroup$ Jun 25, 2018 at 23:24

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