Timeline for Why does a fermionic Hamiltonian always obey fermionic parity symmetry?
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Sep 6, 2019 at 2:22 | comment | added | Rococo | There is a nice argument, similar to those in the given answers, in these lecture notes: lassp.cornell.edu/clh/Book-sample/1.1.pdf | |
| Aug 8, 2017 at 15:40 | answer | added | user51055 | timeline score: 3 | |
| S Jun 1, 2017 at 20:12 | history | bounty ended | tparker | ||
| S Jun 1, 2017 at 20:12 | history | notice removed | tparker | ||
| Jun 1, 2017 at 6:14 | answer | added | tparker | timeline score: 11 | |
| May 31, 2017 at 15:06 | answer | added | Ruben Verresen | timeline score: 20 | |
| May 25, 2017 at 23:53 | comment | added | Ruben Verresen | @Rococo Hm, are you sure? Can you write down an eigenstate which has an arbitrarily low energy? | |
| May 25, 2017 at 23:52 | comment | added | Ruben Verresen | @tparker This would still allow for spontaneous symmetry breaking. I mean, there are many effective models we know and love which have less symmetries than our microscopic standard model. | |
| May 25, 2017 at 5:29 | history | tweeted | twitter.com/StackPhysics/status/867613489094897664 | ||
| May 25, 2017 at 4:01 | comment | added | TLDR | Here's a simple physical example: a quantum field theory that describes an adiabatically charging conducting shell. | |
| May 25, 2017 at 3:48 | answer | added | user2820579 | timeline score: 0 | |
| S May 25, 2017 at 3:08 | history | bounty started | tparker | ||
| S May 25, 2017 at 3:08 | history | notice added | tparker | Draw attention | |
| May 25, 2017 at 3:07 | comment | added | tparker | Great question, which I've wondered myself and never gotten a good answer two. Someone once argued to me that condensed-matter effective Hamiltonian degrees of freedom are collective excitations of actual Standard Model electrons, and the Standard Model Hamiltonian doesn't contain any terms that violate fermion parity so the effective Hamiltonians can't either. I'm not sure I buy that argument though. | |
| Mar 20, 2017 at 8:48 | comment | added | FraSchelle | Just remarked that my previous comment is the same as @RGWinston. Clearly open systems are described with such operators $c+c^{\dagger}$. About superconductivity : it is not an open system per se because the total number of fermion modes is conserved, i.e. the Hamiltonian you start with conserve the number of fermion modes $N\sim c^{\dagger}c$. Some just transmute to quasi-bosonic degrees of freedom (the Cooper pairs), but nothing forbids you to count a Cooper pair as two fermionic degrees of freedom (i.e. 2 electrons still, in a picturesque language). | |
| Mar 20, 2017 at 8:42 | comment | added | FraSchelle | I guess there is no fundamental reason for this, especially in condensed matter systems. But definitely a term like $c+c^{\dagger}$ would not conserve the number of particle, say differently the system would be intrinsically open, which renders the mathematical description of such a system a hard task. So it might well be a pragmatic point of view indeed : how to deal with non-parity conserving systems ? Somehow related to your question is the paper arxiv.org/abs/physics/9808029 on coherent states of fermions, generated by the fermionic displacement operators $\sim\exp{c+c^{\dagger}}$ | |
| Mar 16, 2017 at 21:58 | comment | added | Ruben Verresen | @RGWinston That is exactly the point: you are happy to accept that closed systems need not have conserved particle number (at least when talking about effective particles). The question is why we are not allowed to imagine a closed system without conservation of fermionic parity. | |
| Mar 16, 2017 at 21:53 | answer | added | Mikewins | timeline score: 2 | |
| Mar 16, 2017 at 21:49 | comment | added | RGWinston | Hmm... I think we do have to think of superconductors in the effective picture as an open systems, because they don't have a conserved particle number. The BCS wavefunction is a superposition of all possible different even particle numbers. | |
| Mar 16, 2017 at 21:37 | comment | added | Ruben Verresen | @RGWinston With that same reasoning you would also call the effective Hamiltonian for a superconductor ($H = \cdots + c^\dagger c^\dagger + \cdots$) an open system. So my question is, why do even effective fermionic Hamiltonians for closed systems have $P$ symmetry? | |
| Mar 16, 2017 at 21:34 | comment | added | RGWinston | I don't have a general mathematical answer, but I think a Hamiltonian like (c_i^\dagger + c_i) -with the second term added to make it Hermitian- violates it being a closed system as they would inject particles into, or remove them from, the system | |
| Mar 16, 2017 at 21:32 | history | edited | Ruben Verresen | CC BY-SA 3.0 |
example hamiltonian was not hermitian
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| Mar 16, 2017 at 20:25 | history | asked | Ruben Verresen | CC BY-SA 3.0 |