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I do not participate on this site any longer, except to respond to comments regarding my own text, if that text is unavailable in another form. I do not accept the political moderation atmosphere here, it is not compatible with open science. I participate at physicsoverflow.org.


Jun
29
comment On-shell symmetry from a path integral point of view
@drake: Yes, I know about the reexponentiation, it doesn't matter for the case in the question where you are only shifting fields by other independent fields, since then the determinant is one. The problem with compensation is that the classical variation is order 1, while the determinant variation is order $\hbar$, so this only can happen if there is no classical limit. Maybe 2d Gross-Neveu type models (four Fermi), where Bosonization can reveal new symmetry, but in this case it's not going to be a determinant compensation either, but a symmetry which is only evident in one formulation.
Jun
29
awarded  Popular Question
Jun
28
comment On-shell symmetry from a path integral point of view
Yes, we agree... (I edited the comment too late, hence time travel!)
Jun
28
comment On-shell symmetry from a path integral point of view
Your Ward identity formula is not really general. what should the Ward identity be for the formal transformation on $S = \int |\partial \phi|^2 + |\partial \eta|^2 + \phi^2 \eta^2 d^4x$ under the formal non-symmetry transformation $\delta \phi = \epsilon \phi \eta$, $\delta \eta = - \epsilon \eta $? The reason I ask is because the general formula is false for this case, as the transformation is not a field shift, but a field-value dependent field shift. so there is a determinant in the measure.
Jun
28
comment On-shell symmetry from a path integral point of view
Excellent! Thanks. I got it. The formal current thing is the idea that I was missing.
Jun
28
comment On-shell symmetry from a path integral point of view
I have seen a million people ask this question about SUSY closure, why it is justified to use eqns of motion. I have seen people give glib or wrong answers, or claim it's a Ward identity. But I never saw the actual answer you have above, not this identity. The derivation of the Ward identity and the Schwinger Dyson equation with coinciding insertions are exactly the same, but the field transformation is of a different kind. That's why I am asking for a source, because I checked to see if it appears in Schwinger and Dyson's papers, no, and 6 reviews, no. There SD is just Eherenfest/Heisenberg.
Jun
28
comment On-shell symmetry from a path integral point of view
Why I am mystified: the Ward identity is a special sort of Schwinger Dyson equation, for shifting the fields by a symmetry, while the equation of motion identity is for just shifting the fields. The identity for equation of motion times a local operator is what I never saw in books, so I had to do myself. I see the transformations involved, they are similar but not the same. I am taking your word for it that this is called "Schwinger Dyson equation", but I am wondering why I never saw the identity for ${\delta S\over \delta\phi}O$ anywhere (I read the books you gave, no). Sohnius didn't too.
Jun
28
comment On-shell symmetry from a path integral point of view
The Ward identity is doing an infinitesimal symmetry transformation, that's what the $\delta \phi_i$ are in your comment, while the identity I derived is doing a shift of the field $\phi$ (not a symmetry). In papers and books, they always derive the Schwinger dyson equation as the operator equation ${\delta S\over \delta \phi} =0$, where $S = \int L(\phi) + J \phi$, including a source term J, and this is sufficient to derive perturbation theory (like Schwinger's/Dyson's papers), but it's not enough to say exactly when the equation of motion works with a coinciding insertion.
Jun
28
comment On-shell symmetry from a path integral point of view
I am not claiming originality, but I want to know who writes this, because it just doesn't commonly appear, but it appears on Wikipedia. It's not the Ward identity (although the derivation is nearly identical), it's something else, involving the question of equation of motion times local operators, and this is something that confused the heck out of me for years until I figured it out. Is it in a textbook? I skimmed a few just now and didn't find it. Is it in some review paper?
Jun
28
comment On-shell symmetry from a path integral point of view
While what you write is the Schwinger Dyson equation as it appears on Wikipedia, namely $\langle {\delta S\over\delta\phi}O\rangle = \langle {\delta O \over \delta \phi}\rangle$, trivially equivalent to my answer, I read the original articles of Schwinger and Dyson, and a lot of related literature, and please, for my sanity, could you tell me who writes this down first? I had to come up with it myself, and I didn't even know it was called Schwinger Dyson equation, SD was the simpler $\langle {\delta S\over \delta \phi} \rangle=0$ in all sources I ever read. I had to rederive it myself.
Jun
28
awarded  Necromancer
Jun
28
revised On-shell symmetry from a path integral point of view
final fixes
Jun
28
awarded  classical-mechanics
Jun
28
comment On-shell symmetry from a path integral point of view
@drake: I understand what you are saying, I agree with you on the physics, and the content you gave was fine, although not written the way or with the meaning I intended at all. I am sorry for sniping at you, I got annoyed. Thanks for making the answer nice, and good luck.
Jun
28
comment On-shell symmetry from a path integral point of view
@drake: I just noticed when I did the example that I was missing the other O's, it was dimensionally wrong too. I give up on edits, the editor is impossibly slow (every keystroke takes 10 seconds to register, and the time gets longer and longer). The point is: equation of motion for the $\phi$ field, times the other operators, is only nonzero at the collisions, where the contribution is the variation of these operators under the variation of $\phi$, I did it from memory, I made a botchy notation mess, but I really understand this, and it was annoying that the fixes were changing the meaning.
Jun
28
comment On-shell symmetry from a path integral point of view
@drake: Yes! You are right, of course, stupid me. I just noticed this too, I am sorry.
Jun
28
revised On-shell symmetry from a path integral point of view
more fixes
Jun
28
revised On-shell symmetry from a path integral point of view
more fixes
Jun
28
comment On-shell symmetry from a path integral point of view
But I understood where the confusion came from--- you thought the $\phi_k$'s were other fundamental fields in the Lagrangian--- not necessarily (and the relation would be trivial then). The $\phi_k$'s are just arbitrary composite local operators, and the right hand side has the variational derivative. Since it is not clear to you yet (and you know your stuff) I changed the notation, and fixed the "dimensional inconsistency" (which just means you didn't like it that I factored out the stupid delta function without changing the notation for $\delta \phi$, ok fixed)
Jun
28
revised On-shell symmetry from a path integral point of view
fix