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I'm currently reading this paper and I found the following statements confusing:

On page 4:

To study symmetric SRE states, we further demand that the ensemble of states ${|\psi_{I}\rangle}$ does not break the symmetries spontaneously. For exact symmetries this simply means that each individual state ${|\psi_{I}\rangle}$ does not break the symmetries (i.e., is not a cat state), which is guaranteed by the symmetric SRE condition.

My question is, what does it mean that a state break some symmetry spontaneously? To my knowledge, a state satisfies some symmetry if $U_g|\psi\rangle = e^{i\theta}|\psi\rangle$. For cat state $|\psi\rangle = |\uparrow\uparrow\uparrow\uparrow...\rangle + |\downarrow\downarrow\downarrow\downarrow...\rangle$ and the $\mathbb{Z}_{2}$ symmetry that takes spin up to spin down, $|\psi\rangle$ seems to still be symmetric. Then why it's said that a cat state break the symmetry?

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2 Answers 2

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It is hard to be precise without further context, but on general grounds a state breaks a symmetry spontaneously if the Hamiltonian has a certain symmetry but its ground state (or some other eigenstate of interest) does not have the same symmetry. In QM symmetries are represented by unitary (or antiunitary) operators.

In formulae, say $U$ represents the symmetry and $|\psi\rangle $ the state, spontaneous symmetry breaking (SBS) means

\begin{align} [H,U ]&= 0\\ [|\psi\rangle\langle\psi|,U] &\neq 0. \end{align}

In some cases one is interested in other statistical states such as the Gibbs state $\rho_G =\exp{(-\beta H)}/ Z$.

Given the definition above one easily realizes that for $\rho_G$ SBS can happen only in the thermodynamic limit (TDL) where the dependence of $\rho_G$ on $H$ is no longer necessarily smooth.

For other states the following holds. If $|\psi\rangle $ is a non-degenerate eigenstate of $H$, it necessarily has the same symmetry as $H$. This holds at finite size and also in the thermodynamic limit although in the TDL the definition of a state is more complicated. In other words a necessary condition for $|\psi\rangle $ to break the symmetry is to correspond to a degenerate eigenvalue. Indeed if the eigenvalue is degenate, there is no well defined notion of state in the corresponding subspace. Any pick is as good as any other.

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  • $\begingroup$ Thanks for your answer. I'm still wondering why degenerate eigenvalue would lead to spontaneous symmetry breaking. Take Ising model as an example, the ground state is degenerate with states where spin all up and spin all down. Every state in the groud state subspace seems to still commute with the symmetry operator, thus not a spontaneous symmetry breaking state according to the defination. $\endgroup$
    – dhq
    Commented May 24 at 6:38
  • $\begingroup$ First, at finite size in the Ising model there's no degeneracy. The ground state is unique and in the SBS region (in the thermodynamic limit) there is an exponentially small gap. Indeed the ground state is something like $\psi_0 = |\mathrm{up}\rangle +|\mathrm{down}\rangle$ and the excited state is $\psi_1= |\mathrm{up}\rangle -|\mathrm{down}\rangle$. These are true eigenstate only deep in the ferromagnetic region but you get the idea. These states, in accordance with my argument above, are indeed symmetric (the symmetry is spin flip here). $\endgroup$
    – lcv
    Commented May 24 at 10:01
  • $\begingroup$ The symmetry broken states are instead clearly $|\mathrm{up}\rangle ,\ |\mathrm{down}\rangle$. But, since they are degenerate (in the thermodynamic limit), you need some recipe in order to single them out. $\endgroup$
    – lcv
    Commented May 24 at 10:03
  • $\begingroup$ The symmetry is the spin flip (that sends up to down and viceversa). So the external field is along $\sigma^x$. You see that the symmetry sends a symmetry broken state into another one. Again a common feature. $\endgroup$
    – lcv
    Commented May 24 at 10:07
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A more mathematical way to describe a broken symmetry is as follows. Suppose the fields of the theory transform in some non-trivial representation of a group $G$. We say that $G$ is a symmetry of the theory if the action $S$ is in the trivial representation of $G$. More precisely, if $G$ is a compact gauge group, we can view $S$ as an $L^2$ function of $G$ itself. What I mean by this is that if we imagine a fixed field configuration $\phi_e$ and its transformation $\phi_g = U(g) \phi_e$, then we can define $f(g) = S[\phi_g]$ as a function which takes a group element as an input and outputs a number. If you regulate/renormalize properly, you can integrate $\int dg |f(g)|^2 $ to get a finite answer. This implies we can decompose (essentially, Fourier transform) $f(g)$ via the Peter-Weyl theorem, and it makes sense to say the action is in a particular representation. This was a fancy way to say that $G$ is a symmetry if the action is invariant under the field redefinitions corresponding to the group action on the fields. The reason I like this way of describing it is because you can use the Peter-Weyl theorem to add a small contribution from a non-trivial, but low dimensional, representation to make precise what "softly" breaking a symmetry means.

Then, we can define a spontaneously broken symmetry as being when the ground state manifold is in a non-trivial representation of $G$, and unbroken when the ground state manifold is in the trivial representation of $G$. This also applies to subgroups of $G$. I like this way of describing it because it instantly unlocks the tools of Goldstone's theorem (the goldstone bosons are the generators of this unbroken subgroup) and explains Elitzur's theorem very quickly, as a physical state is never in a non-trivial representation of a gauge group, lest those degrees of freedom become physical.

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