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Let the Hamiltonian $H$ be invariant under the symmetry $U$ such that: $$U H U^\dagger = H \, .$$ Consider a bias of ground states $\{\left|GS,n\right>\}$. It is often said that if $\left|GS,n\right>$ are not invariant under the symmetry $U$ then the system chooses one of these and hence spontaneous breaks the symmetry (e.g. ferromagnets and the Higgs).

But why do we emphasis a specific bias and which basis do we emphasis in general? Would it not be possible to write a basis consisting of a superposition of the states $\{\left|GS,n\right>\}$ which is invariant under the symmetry - would this not be a natural choice?

DanielSank
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  • Also should that be $H$ rather than $U$ on the right hand side of your equation? – By Symmetry Jun 06 '18 at 19:51
  • @BySymmetry Yep sorry that was a typo – Quantum spaghettification Jun 06 '18 at 20:07
  • For a ferromagnet, such a macroscopic superposition state would collapse into one of its two configurations. On the other hand, from an all-up state to tunnel into its all-down partner state, the system goes through an infinite energy barrier. But for a " gauge symmetry" with infinite symmetry generators, the ground state is always a superposition. This is known as Elitzur's theorem, i.e., gauge symmetry cannot be spontaneously broken. This would be your statement. – pathintegral Jun 06 '18 at 21:07
  • R. Shankar's new book has an excellent explanation on why and how symmetry breaking phase transitions can only happen for infinite systems. Should be relevant to your question. – pathintegral Jun 06 '18 at 21:13
  • @BySymmetry I believe this not to be a duplicate of you linked question but is off https://physics.stackexchange.com/questions/287760/qft-superposition-and-the-higgs-ground-state?rq=1 and https://physics.stackexchange.com/questions/243291/superposition-of-distinct-vevs-in-spontaneous-symmetry-breaking . The reference mentioned in one of the comments: Weinberg's QFT Vol. 2, section 19.1, is also good. – Quantum spaghettification Jun 07 '18 at 04:09

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