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I am currently working on generalized symmetries and i was reading https://arxiv.org/abs/2301.05261. In footnote 23 the authors state:

To be precise, by spontaneous symmetry breaking, we mean a phase where an order parameter constructed from operators charged under a symmetry acquires a nonzero vacuum expectation value in the thermodynamic limit.

So in general they mean something that looks like this

$\langle O \rangle \neq 0 $,

where $O$ is charged under the symmetry. So far i have no problem, but then the authors state the above definition is also valid in the case in which the operator $O$ has 0 charge under the symmetry, which from my point of view is equal to an operator that does not transform. This confuses me a lot so my question is:

Is it proper to say that we have a spontaneously broken symmetry even if the order parameter $\langle O \rangle \neq 0 $ is constructed using an operator that has 0 charge?

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  • $\begingroup$ I think you are referring to an earlier version of the paper, it would be better to mention that. I could not find those statements in the new version. $\endgroup$
    – Nandagopal Manoj
    Commented Dec 5, 2023 at 21:22
  • $\begingroup$ You say "... operator O has 0 charge under the symmetry." Are you referring to the expectation value of the form $\langle O^\dagger(x) O(y)\rangle$, where $O$ is charged under the symmetry? $\endgroup$
    – Nandagopal Manoj
    Commented Dec 5, 2023 at 21:24
  • $\begingroup$ I apologize. Yes indeed i was referring to an older version of that paper. In the new version i am not able to locate the exact statement i found previously but nonetheless the claim is the same. They used an operator with no charge to establish SSB. $\endgroup$
    – Truth and Beauty and Hatred
    Commented Dec 6, 2023 at 10:03
  • $\begingroup$ Something of the form $\langle O^{\dagger}(x) O(y) \rangle$ would be a good example since the full object in the expectation value is neutral under the symmetry. In particular in the new version of the paper this is done at the end of page 12, where they evaluate the expectation value of the operator $W^{\dagger}[C_p] which, from my understanding, is neutral under the symmetry they are considering. $\endgroup$
    – Truth and Beauty and Hatred
    Commented Dec 6, 2023 at 10:16
  • $\begingroup$ We use operators like $\langle O^\dagger(x) O(y)\rangle $ where $x$ and $y$ are well seperated to diagnose spontaneous symmetry breaking (this is commonly referred to as long range order). Are you familiar with this? I agree that the higher form case is more subtle, but this answer of mine may help. physics.stackexchange.com/questions/743794/… $\endgroup$
    – Nandagopal Manoj
    Commented Dec 6, 2023 at 18:24

1 Answer 1

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We use uncharged operators like $\langle O^\dagger(x) O(y)\rangle$ where $O$ is charged under the relevant symmetry and $x$, $y$ are well separated to diagnose spontaneous symmetry breaking (this is commonly referred to as long range order).

The higher form case is explained in this answer.

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