Followings are two independent questions as implied by the title:

(1) Considering a quantum Hamiltonian $H$ possesses some symmetries described by a symmetry group $G=\left \{ g_1,g_2,...,g_n \right \}$ (assumed to be finite for simplicity), and $\psi$ is an eigenstate of $H$. In general, one can construct a symmetric eigenstate $\phi=g_1\psi+...+g_n\psi$ preserves the symmetries $G$ that the Hamiltonian $H$ has, i.e., $g\phi=\phi (\forall g\in G)$.

And my question follows: If it is really so simple like the above argument, then the so-called spin liquid ground state (a spin state that preserves almost all the symmetries of the Hamiltonian) should generally exist, so why it is so ''hard'' to find a spin liquid ground state for the condensed matter community?

I guess that the key issue is whether the construction $\phi=g_1\psi+...+g_n\psi$ exists, i.e., whether $\phi=0$. And it seems unsure whether or not $\phi\neq 0$ from the math point of view.

(2) If a quantum Hamiltonian $H$ has an eigenstate degeneracy, then there must exist a TR breaking eigenstate. The reason is simple: Let $\psi_1,\psi_2$ be two degenerate eigenstates with TR symmetry (if they are TR breaking, then the statement already holds) and tuning the phase of the wavefunctions $\psi_1,\psi_2$ such that $T\psi_1=\psi_1, T\psi_2=-\psi_2$, then the superposition eigenstate $\psi_1+\psi_2$ breaks TR symmetry.

From the above point of view, for a TR symmetric Hamiltonian $H$, once it has ground state degeneracy, then the TR symmetry must be spontaneously broken, does my understanding correct? If it is so, for example, for a TR symmetric $Z_2$ spin liquid with a 4-fold topological degeneracy on a torus, how to understand this TR symmetry?

  • $\begingroup$ I am not so shure about this. You can turn the argument in your example around and find unbroken states from the broken ones. But the actual physical situation is decisive. In the example of a hydrogen atom, disregarding spin issues and associated level splittings, the eigenprojector upon the three $1p$-states is invariant. If we demand that the initial state is, say, the state with $m=1$ then the symmetry is broken. But we can make linear combinations which are invariant. This may be impossible if the degeneracy is infinite. Then there would be no invariant state. $\endgroup$ – Urgje Oct 25 '14 at 14:09
  • $\begingroup$ Can you calculate $\langle\phi|\phi\rangle$ and see if it is zero? You can try some simple systems to verify this first. $\endgroup$ – 喵喵是我的猫猫 Oct 26 '14 at 10:38
  • $\begingroup$ Your second question I don't understand, it seems the wave function will be zero with the condition of your TR symmetry assumption and phase tuning procedure. $\endgroup$ – 喵喵是我的猫猫 Oct 26 '14 at 10:48
  • $\begingroup$ @luming Thanks for your comment. $\psi_1\pm \psi_2\neq 0$ because $\psi_1$ and $\psi_2$ are two linear independent eigenstates. $\endgroup$ – Kai Li Oct 26 '14 at 12:01

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