Timeline for Spin-statistic for free quantum fields
Current License: CC BY-SA 4.0
14 events
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| Mar 30, 2021 at 17:27 | comment | added | Simon Henry | @gmvh for the reccord I do think you are right that the problem is with locality in this sense, but I need to think more about it. I previously interpreted locality as simply meaning "finite propgation speed" which is a condition that does make sense for the 1-particle space and do propagate through second quantization, but that condition on the Hamiltonian being locally expressed from the field operators is definitely something else and is indeed unclear from the 'fock space' point of view. | |
| Mar 30, 2021 at 13:59 | history | edited | gmvh | CC BY-SA 4.0 |
Edited in response to an extensive edit of the question
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| Mar 30, 2021 at 13:21 | comment | added | Simon Henry | @user1504 : maybe you are right, but the approach I'm talking about does know about the dimension of space time : the key input is the classification of irreducible representations of the Poincarré group and this would look quite different in lower dimension. | |
| Mar 30, 2021 at 13:21 | comment | added | Simon Henry | @gmvh I understand that, but what I'm trying to understand is how the point of view presented in some of the textbook written by mathematician (like the one linked) fits into this picture. | |
| Mar 30, 2021 at 13:13 | comment | added | gmvh | Yes, it might be tricky to reason from the Fock space side precisely because it would amount to a proof of spin-statistics. The direction usually taken in the physics literature of course doesn't prove spin-statistics, it merely shows that the specific theory under consideration can't be quantized with the opposite statistics (which doesn't exclude the existence of other examples that can be). | |
| Mar 30, 2021 at 13:08 | comment | added | user1504 | I don't think Fock space reasoning alone will give you spin-statistics. The spin-statistics theorem is false in 1, 2, and 3 dimensions. There has to be some use of a fact about dim >= 4. | |
| Mar 30, 2021 at 13:01 | comment | added | gmvh | In physics texts, one usually works the other way around and starts from the local Hamiltonian of a specific free theory (Klein-Gordon or Dirac). Then one quickly finds that imposing the wrong kind of commutation relations on the creation and annihilation operators (which directly implies the wrong kind of Fock space) leads to a badly behaved Hamiltonian in terms of Fock space states. Going this way around might be tricky. | |
| Mar 30, 2021 at 12:56 | comment | added | Simon Henry | that's interesting. I'll need to think about it, so far I still don't quite see why the type of Fock space used changes things, but I can imagine it playing a role. thanks | |
| Mar 30, 2021 at 12:45 | comment | added | gmvh | The Hamiltonian has to be local when written in terms of the field operators $\phi(x)$. Whether that's the case is not readily apparent from its form in terms of the Fock space states. | |
| Mar 30, 2021 at 12:42 | comment | added | Simon Henry | Unless you mean something different than me when you use this word, the Hamiltonian obtain in that case is locale and I have discussed that already in the question. If you mean something different then the question is "Why is the Hamiltonian is local when you take the "right" fock space and not local when you take the wrong one ?". | |
| Mar 30, 2021 at 12:39 | comment | added | gmvh | The Hamiltonian has to be local. | |
| Mar 30, 2021 at 12:38 | history | edited | gmvh | CC BY-SA 4.0 |
added 6 characters in body
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| Mar 30, 2021 at 12:37 | comment | added | Simon Henry | The first sentence do do not make sense: A representation of the Poincarré group already defines a Hamiltonian as the generator of the time translation. Now If the representation I described in the question indeed do not have positive energy Hamiltonian when the spin-statistic theorem is violated then that would be an answer to the question. But so far I don't see that clearly. | |
| Mar 30, 2021 at 12:20 | history | answered | gmvh | CC BY-SA 4.0 |