Timeline for Can we do path integrals in gauge theories without fixing a gauge?
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| Jan 30, 2018 at 20:41 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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| Jul 20, 2017 at 7:54 | comment | added | tparker | @AccidentalFourierTransform It's worth noting that discretizing spacetime breaks the gauge symmetry, but not completely down to the identity. You still have a residual gauge symmetry group of arbitrarily high dimension. (Possibly even (countably) infinitely-dimensional - do you need to gauge-fix a gauge theory defined on an infinite discrete lattice?) | |
| Jul 19, 2017 at 18:38 | comment | added | AccidentalFourierTransform | @user1504 note that the discretisation of space-time breaks the gauge invariance, so by discretising the theory you are effectively fixing the gauge. In this sense, both your answer and mine are technically incorrect or at least rather unclear. I plan to edit mine in the future. | |
| Jul 19, 2017 at 18:36 | comment | added | AccidentalFourierTransform | @tparker 1) by $p^t$ I mean the transpose of $p=(p^0,...,p^3)$, that is, $(pp^t)^\mu{}_\nu=p^\mu p_\nu$. 2) yes, and you can also have that in a relativistic theory. E.g., a skew tensor field $F_{\mu\nu}$ describes massless fields with no longitudinal states, whether it comes from a vector $A_\mu$ or not. But if you want - and this seems to be what nature chose - $F$ to be the exterior derivative of $A$, then you must have a gauge symmetry regardless of whether the theory is relativistic or not. Mainly, because a massless particle has 2 d.o.f. while $A$ has 4 components. You need a redundancy. | |
| Jul 19, 2017 at 18:16 | comment | added | user1504 | What do you mean when you say that "the discretized integral diverges, by the usual arguments"? This is not true of lattice gauge theories. | |
| Jul 18, 2017 at 0:30 | comment | added | tparker | What does your notation $p^t$ mean? Also, could it possible to have a massless field with no longitudinal states in a continuum theory that is not Lorentz invariant, without needing to explicitly fix a gauge? | |
| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
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| Jan 19, 2017 at 15:46 | comment | added | AccidentalFourierTransform | @ACuriousMind I guess I don't really remember the details, but in my memory the quantisation of systems with 1st class constraints begins by turning all of them into 2nd class constraints, or by imposing several gauge-fixing conditions. In this sense, we are not really quantising a system with 1st class constraints, but one that is physically equivalent and has no 1st class constraints. But perhaps this is not true, I should read the book again... | |
| Jan 19, 2017 at 15:00 | comment | added | ACuriousMind♦ | "As of today, nobody knows how to quantise a classical theory with first class constraints." I don't know what that's supposed to mean, especially given the reference you give for it - the entire book is about how to quantize such theories! Both the Dirac-Bergmann recipe and the BRST method produce perfectly fine quantum theories. | |
| Jan 19, 2017 at 14:17 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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| Jan 19, 2017 at 14:05 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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| Jan 19, 2017 at 13:48 | history | answered | AccidentalFourierTransform | CC BY-SA 3.0 |