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Aug 31 '18 at 10:54 history bounty ended SRS
Aug 26 '18 at 20:04 comment added Norbert Schuch ... connection, which can be explained heuristically with some mean field type arguments, yet (AFAIK) has resisted rigorous assesments. (Not to mention the quantitative relation between the LRO value and the magnetization.) Note that this is not purely mathematical -- e.g., in QMC it is exxactly the LRO which is measured to compute the (alledged) spontaneous magnetization.
Aug 26 '18 at 20:03 comment added Norbert Schuch @tparker Then you should just say it's trivial and that's it. The next step would be to explain the heuristic connection between the two concepts given some mean field (=clustering) ansatz. But the point is the concepts are different: E.g. -- if you forget the quantum matter and look at Gibbs states -- LRO is also observed in Gibbs states, which DO NOT break symmetry. The mystery is: Whenever there is a symmetry and the Gibbs state has LRO, then the Gibbs state w/ a small perturbation is symmetry breaking. (Similarly for finite volume ground states.) This is a pretty non-trivial ...
Aug 26 '18 at 16:01 comment added tparker @SRS Not an exact one that holds in general. The correlation function's behavior is very system-dependent at distances less than or comparable to the correlation length $\xi$. But the asymptotic form of the correlation function for distances much longer than the correlation length is typically an exponential decay $\sim \exp(-|x-y|/\xi)$ above the critical temperature, a power-law decay $\sim 1/|x-y|^p$ at the critical temperature, and of the form $\sim C + \exp(-|x-y|/\xi)$ below the critical temperature.
Aug 26 '18 at 15:55 comment added tparker @NorbertSchuch The nontrivial fact, which requires nonlocal measurements to verify, is that physical states are always experimentally observed to respect cluster decomposition. There is still some theoretical disagreement as to why. Once you accept this, then the equivalence between SSB and LRO becomes trivial (at least in translationally invariant systems).
Aug 26 '18 at 13:48 comment added SRS @tparker Thanks for the answer. Is there a general formula when $|x-y|$ is finite (whose limit $|x-y|\to \infty$ gives your first equation)?
Aug 26 '18 at 2:08 comment added Norbert Schuch But if you only look at states with the clustering property, this is trivially the same. And can thus be determined by local measurements.
Aug 26 '18 at 1:59 comment added tparker @NorbertSchuch The existence of a symmetry-breaking local order parameter $m(x)$ such that $$\lim_{|x-y| \to \infty} \langle m(x)\ m(y) \rangle = C \neq 0.$$ Unlike SSB, this can't be determined via local measurements.
Aug 26 '18 at 1:31 comment added Norbert Schuch I'm not sure I would call this a "definition", since it very much depends on the experimental setup, but so it be -- and I think this can indeed be closely related to the formal definition of SSB. Ok, then: How do you define LRO?
Aug 26 '18 at 0:40 comment added tparker @NorbertSchuch (a) The cluster property certainly holds for the non-cat degenerate ground states, which are the ones that are physically observed. (b) I'm defining SSB as "when experimentalists measure the symmetry-breaking order parameter $m(x)$, they find a nonzero value." (c) The second equality follows from the first.
Aug 26 '18 at 0:39 history edited tparker CC BY-SA 4.0
added 16 characters in body
Aug 25 '18 at 23:11 comment added Norbert Schuch The clustering property is NOT true in this form for degenerate ground states, which is exactly what is needed in the case of SSB! Also, how do you define SSB - your argument seems rather circular! Wouldn't you define SSB as "if I put a small perturbation, I get a spontaneous magnetization", and LRO as "there are long-range correlations"? --- I mean, where does the claim of equality in the second equation even come from?! --- To me, your answer sounds a bit like "it's true because we know it's true".
Aug 25 '18 at 19:16 history answered tparker CC BY-SA 4.0