just saw "what is diffeomorphism invariance" pop up in the feed, asked a few years ago and not closed, with only one (not very popular) answer. slightly surprising
@naturallyInconsistent seemed like something someone might have wanted to give a encompassing answer or maybe a diatribe on (given that it wasnt flagged as vague either)
@qwerty IIRC, ACM would be writing an article-sized answer somewhere. The whole thing is horrible and ACM had had enough of repeating his answers bit-by-bit.
what do you mean with exist? I guess so, it is usual mathphy topic. Kato's book is a classic, but I have problems with reading old fonts etc., and despite that, it is quite dense. Though I like what I've read so far, and it is rigorous
I'm not toooooo sure of the rigour involved when, admittedly by the authors, the convergence radius of many of the stuff we do in perturbation, is zero
Mhhh, coming from someone not too expert on this, but... Even if the perturbation series doesn't work, an asymptotic expansion is still something rigorous, isn't it?
The first thing coming to my mind is a $C^\infty$ function that is not analytic: you can't write is as a series, but a Taylor expansion is totally rigorous
If you cannot prove that an expansion converges in some sense to the result that you want it to converge towards, then it is not clear what you mean by rigorous there.
That is not to knock on the utility of asymptotic series, but that seems to be the main showstopper to proving that, say, QED is rigorously correct.
Even though we have way too many good results from there
if a civilisation has an oracle to solve the Halting problem, can they know all truths about the natural numbers? or can they know more truths than we can know, but not all?
can this civilisation breach some fundamental limitations that we have, or is the Halting problem irrelevant here? suppose the civilisation can perform countably infinitely many computations in a finite time (which solves the Halting problem)
physics.stackexchange.com/q/497666 Is this answer wrong? I mean i think $n^0$ described the interacting theory’s ground state, not the free theory’s ground state
Can I say that Noerther charge is $\Lambda=\frac{\partial \mathcal L}{\partial \dot p}\frac{\partial q}{\partial \zeta}$, where $\zeta$ is the continuous parameter with which we do the mapping from one configuration to another that leaves the Lagrangian invariant.
is the vacuum of an interacting QFT time-dependent?
for instance, we have that $U^\dagger(t)$ is the interaction picture time evolution operator, which relates the free theory in its heisenberg picture to the interacting theory in its heisenberg picture. then, $U^{\dagger}(t) a^{\dagger, \text{free}}_{k} \lvert 0 \rangle = U^\dagger (t) a^{\dagger, \text{free}}_{k} U(t) U^\dagger (t)\lvert 0 \rangle = a^\dagger_{k} (t) \lvert \Omega (t) \rangle$
where seemingly we would identify $U^\dagger(t) \lvert 0 \rangle$ as the vacuum of the interacting theory
Consider a system with Hamiltonian $\hat{H}=\hat{H_0}+\hat{V}$. We define the interaction picture kets $|\psi(t)\rangle _I$ by
$$\tag{1} |\psi(t)\rangle _I=\exp\left(\frac{i}{\hbar}\hat{H}_0(t-t_0)\right)|\psi(t)\rangle_{S}$$
and interaction picture operators $\hat{O}_I$ by
$$\tag{2} \hat{O}_I= \...
I think your computation is correct. but how would you know that $|\Omega(t)\rangle$ is the ground state of the interacting theory?
I mean it is clear that any eigenstate of the corresponding Hamiltonian has trivial time-evolution
so either your results imply that, or there is a mistake
There is, actually, not much sense to "eigenstate is time-dependent" if the Hamiltonian is time-independent, other than saying that the time evolution is trivial.
See e.g. the differential equation your $|\Omega(t)\rangle$ obeys.
the state you wrote is the interacting-picture time-evolved free ground state. Why do you expect it to be the interacting ground state, too? Or am I missing something? (sorry for my initial stupid mistake, in any case--I should've read more carefully my own equations lol)
It is, in fact, the ground state of the interaction-picture transformed free Hamiltonian, i.e. the eigenstate of $UH_0U^\dagger$, which in turn is completely fine with the fact that your state $\Omega(t)$ is annihilated by all interaction-picture transformed annihilation operators.
ah, you mean $t_0$. that's just the reference time. It should not depend on that, and I believe you could trace it down, i.e. write out everything properly...
or it must cancel out or so
but that's really not the initial question you asked
@User198 I don't know if that's the historical reason, but in the context of QFT, the Noether charge associated to the $\mathrm{U}(1)$ symmetry of the Dirac lagrangian is the electric charge
In the field theory case the Noether theorem implies a continuity equation $\partial_\mu j^\mu=0$ and so we decide to call the spacial components of $j$ "current density" and the time component "charge density"
Continuity equations are all over the place in physics, e.g. fluid dynamics, nonetheless, I think that electrodynamics is the first example that comes to mind
@SillyGoose consider e.g. the Gell-Mann & Low theorem/formula in the "adiabatic interaction framework". What you do is to construct some Hamiltonian $H_\eta(t)$, with $\eta>0$ such that physically you would expect that the ground state of $H_\eta(-\infty)=H_0$ evolves adiabatically to the ground state of $H_\eta(0)=H$. Here I already put $t_0=0$, but one could generalize (for whatever reason).
Since the (full) Hamiltonian $H$ does not depend on time, neither its energies nor its eigenvectors do so. It does not make sense that $\Omega$ depends on some time, and also not on some references time. One needs to keep track of all conventions and so on here.
Open ended question: Ik I can go and study finite dim. rep theory of graded superalgebras but I would just ask the people here: what could be a possible mathy motivation for considering chiral and real superfields while discussing SUSY and not something else? It seems kinda adhoc when people just say we are gonna impose these conditions and construct Lagrangians outta these new irreps of SUSY.
How would you motivate it without pulling it outta the hat (or saying... reality and chirality turned out to be good in modern physics and we just wanna see what happens here...) Orrr... is there a natural reason in the math lingo, for which we consider chiral/real superfields?
Of course, to formulate this rigorously is a different story, and I don't think this is true at all in the end, and there are also physical counter examples IIRC.
anyway, GL then (modulo many important caveats) that the ground state at $t=0$ $\psi_\eta$, or $\psi_\eta(0)$ for emphasize, of $H_\eta$ is related to the true ground state by $\Omega=\lim\limits_{\eta\to 0^+}\psi_\eta(0)/(\langle 0|\psi_\eta(0)\rangle)$. Any good many-body book should derive this, and discuss all caveats I skipped or only mentioned.
in our lecture today, we considered the pion decay, and in the feynman diagram, the current flow points in the positive direction, which is the opposite of the traveling pyon,opposite to each momenta direction. Are we using the conventional current notion here?
