Timeline for Does non-conservation of number of particles imply zero chemical potential?
Current License: CC BY-SA 3.0
21 events
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| Apr 1, 2023 at 10:21 | comment | added | Arnold Neumaier | @IvanBurbano: For neutrall massless particles, yes. But for charged particles represented by $U(1)$ invariant complex fields, there is a conservation law, which makes the charge (number of particles minus number of antiparticles) commute with the Hamiltonian. | |
| Apr 1, 2023 at 1:35 | comment | added | Ivan Burbano | Ok, so then the conclusion is that for any massless particle $\mu=0$? | |
| Mar 31, 2023 at 9:35 | comment | added | Arnold Neumaier | @IvanBurbano: Photons have zero mass, hence even small interactions produce an infinite number of soft photons. | |
| Mar 29, 2023 at 20:25 | comment | added | Ivan Burbano | Thus, what is the special thing about photons that prohibits the existence of a sector in which $[H,N]=0$? It seems like a naive application of your argument would indicate that for any type of particle $\mu=0$. | |
| Mar 29, 2023 at 20:23 | comment | added | Ivan Burbano | I just realized Haag autocorrected to Hash in my comment. Sorry. | |
| Mar 29, 2023 at 9:00 | comment | added | Arnold Neumaier | @IvanBurbano: Yes to the second question. I have no simple reference for the first, just generalities about algebraic quantum field theory. | |
| Mar 28, 2023 at 17:18 | comment | added | Ivan Burbano | Would weakly mean that $tr(\rho[H,N]A)=0$ for all operators $A$, where $\rho$ is the gran canonical ensemble? | |
| Mar 28, 2023 at 12:41 | comment | added | Ivan Burbano | This sounds really interesting. Kind of thing one would find in Hash’s book maybe? Do you have any reference suggestions for this? | |
| Mar 28, 2023 at 8:04 | comment | added | Arnold Neumaier | @IvanBurbano: The grand canonical ensemble belongs to a different representation of the field algebra. The commutator equation holds only weakly, hence depends on the sector. | |
| Mar 27, 2023 at 21:17 | comment | added | Ivan Burbano | I guess I am also confused by the claim that $[H,N]=0$ for a grand canonical ensemble in equilibrium. The statement that $[H,N]=0$ is an operator equation, independent of the state, e.g. the grand canonical ensemble. | |
| Mar 27, 2023 at 20:44 | comment | added | Ivan Burbano | Therefore, for any interacting theory we always have a zero chemical potential? | |
| Mar 27, 2023 at 19:11 | comment | added | Arnold Neumaier | @IvanBurbano: Yes, but in real life matter fields are never free. | |
| Mar 27, 2023 at 15:49 | comment | added | Ivan Burbano | Isn't $[H,N]=0$ for any free field theory? | |
| Nov 27, 2020 at 13:19 | comment | added | Arnold Neumaier | @mithusengupta123: [H,N] is always zero for a grand canonical ensemble in equilibrium. This is by definition. Look at examples to see it. | |
| Nov 26, 2020 at 5:27 | comment | added | Solidification | @ArnoldNeumaier Sorry for commenting on this old post. I found it interesting. One confusion. Isn't [H, N] always nonzero for a grand canonical ensemble because the system is connected to a reservoir? If $[H, N]\neq 0$ is always true for a GC ensemble, then $\mu=0$ always which is certainly false. What's wrong with my logic? | |
| Jan 17, 2017 at 17:01 | comment | added | Arnold Neumaier | @richard: done! | |
| Jan 17, 2017 at 17:01 | history | edited | Arnold Neumaier | CC BY-SA 3.0 |
added requested details
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| Jan 17, 2017 at 14:41 | comment | added | richard | Could you please include some formulae? | |
| Jan 16, 2017 at 19:19 | comment | added | Arnold Neumaier | Which argument is unclear? | |
| Jan 12, 2017 at 8:22 | comment | added | richard | Thank you for your answer but I didn't catch the reasoning! | |
| Jan 9, 2017 at 18:59 | history | answered | Arnold Neumaier | CC BY-SA 3.0 |