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It is said that domain wall formation is the signature of in spontaneous symmetry breaking but not explicit symmetry breaking. Why is this so?

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  • $\begingroup$ Do you have a citation? I've never heard anyone say that. You can have domain walls without spontaneous symmetry breaking. $\endgroup$
    – tparker
    Jun 16, 2018 at 0:11
  • $\begingroup$ @tparker "...the special phenomenon associated with spontaneous symmetry breaking is the existence of the domain wall." Leonard Susskind. See around 7.12 minutes here youtube.com/watch?v=7cKvhkFhXDk $\endgroup$
    – SRS
    Jun 16, 2018 at 17:35
  • $\begingroup$ I see what's Susskind's getting at, but I wouldn't have phrased it quite that way. It's true that you don't get (large) domain walls with explicit symmetry breaking, but you do get lots of domain walls in the high-temperature phase where the symmetry is respected. $\endgroup$
    – tparker
    Jun 16, 2018 at 17:43

1 Answer 1

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Explicit symmetry breaking means that you have added a term in the Hamiltonian that explicitly breaks a symmetry of the initial Hamiltonian.

Now imagine a classical Ising model in the ferromagnetic phase, meaning that on average all spins are pointing in one direction (say, up). This spontaneously breaks the symmetry of the Ising Hamiltonian (that would naively predict that only zero magnetization is possible). Now flip the spins of a connected region of the system in the opposite direction (they will be mainly down). You have now a domain wall (the border between a mainly up region and a mainly down region). This is possible only if the system is in the spontaneously broken phase. Indeed, in the paramagnetic phase, all the spins are random, so if you flip the spin of a region in space, that can't create a domain wall.

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