4:00 AM
Is $\langle Tr(F_{\mu \nu}) Tr( F_{\rho \sigma}) \rangle$ gauge invariant in Non-Abelian Yang Mills theory?
4:12 AM
Hello Everyone...

3 hours later…
7:42 AM
Why is it said that a classical theory of point particles with enough conserved charges "solvable"? I can see that if there's energy conservation the order of differential equation is reduced, if there are action-angle coordinates then integrating the EOMs for angle is trivial. But how general is this? I mean what is the most generic proof of conserved charge=> solvability?
By solvability I mean that we can explicitly find out the trajectories of point particles in a classical system
4

This might have been already asked in this site but I can't find it. So here's the integral: \int_{r_\text{min}}^{r_\text{max}} \sqrt{\left(1-\frac{r_\text{min}}{r}\right)\left(\frac{r_\text{max}}{r}-1\right)}~dr=\pi \left(\frac{r_\text{min}+r_{\text{max}}}{2}-\sqrt{r_{\text{min}}r_{\text{max}}...

Hi everyone here's a definite integral with an interesting result for you all... have you seen an alternate proof of this?

1 hour later…
9:10 AM
I mean Coulomb field has conserved charges and it's not exactly solvable
Can compute $r(\theta)$ but not $r(t)$
9:33 AM
@Obliv on my way back now, it was great (if a bit rainy at the end)
9:47 AM
@Slereah I thought $r(t)$ can be found by finding out $\theta(t)$ out from specific angular momentum $l=r(\theta)^2 \dot{\theta}$ and then using $l=r(t)^2 \dot{\theta(t)}$ to find $r(t)$
I am trying to spell out what is implied here physics.stackexchange.com/questions/568109
It is a 2-body problem reduced to a one body problem and those are integrable anyway...
Hii @Slereah ... just casually pinging you in case you know the answer to this?
@Sanjana No being isometric under Poincaré selects uniquely Minkowski space
or flat spacetimes if you only require the symmetry to be local
@Slereah Do you know where I can find a proof of this?
Maybe in that Wolf book you sometimes refer to...
The Wolf book certainly does, although it's a bit of a pill to prove just that :p
Although being locally isometric under Poincaré -> flat can probably be done more simply
If it's locally isometric under Poincaré you're gonna have a zero infinitesimal holonomy under transport in any direction, so the Riemann tensor is gonna be zero
or look at the Riemann normal expression coordinate expression of quantities, you'll see that the Riemann tensor prevents symmetries by translation
Do you happen to know why the infinitesimal calculus approach was abandoned for the delta epsilon approach.
10:08 AM
Infinitesimals back then were not a properly axiomatized field, and they are very delicate to axiomatize
The delta epsilon approach was though, and that's why it stuck around
also infinitesimals were generally considered philosophically problematic
Okay, thanks.
10:46 AM
hi
@ACuriousMind glad to have u back
@Slereah by minkowski space, do u mean any spacetime with the lorentzian metric
I mean Minkowski space
@Slereah ok i got it
So Minkowski space terminology is for the spacetime in SR
And flat space terminology is for any spacetime with curvature 0
it get confusing for me because i interchange the adjectives Minkowski and lorentzian
@Slereah here, does local refer to "local, as in, finite subsets but not necessarily the whole spacetime"?
11:07 AM
Local as in true of a local flow
ie using the Killing vectors
Rather than actual isometries over the whole spacetime
okay.. what I'm thinking is that the metric can be made Minkowski at every point of spacetime, in a finite region containing that point, by a choice of co ordinates
Is this equivalent
Sure
Like what happens with Darboux's theorem in phase space stuff
@Slereah great

1 hour later…
12:28 PM
which is the the expansion parameter in an asymptotic series, such as the power series ?
12:54 PM
At Wikipedia, for Perturbation theory the following is said:
If $|n\ragle$ is the perturbated state, then the following must hold $\langle n|n\rangle=1$. From this we have:

$\langle n^{(0)}|n^{(1)}\rangle + \langle n^{(1)}|n^{(0)}\rangle=0$
And then it says:
Since the overall phase is not determined in quantum mechanics, without loss of generality, in time-independent theory it can be assumed that $\langle n^{(0)}|n^{(1)}\rangle$ is purely real.Then:

$\langle n^{(0)}|n^{(1)}\rangle = \langle n^{(1)}|n^{(0)}\rangle=-\langle n^{(1)}|n^{(0)}\rangle$
I understand $\langle n^{(0)}|n^{(1)}\rangle =-\langle n^{(1)}|n^{(0)}\rangle$
But I don't understand when they equalize it with the expression that contains no negative sign
And I don't understand what is being meant with this:
"Since the overall phase is not determined in quantum mechanics, without loss of generality, in time-independent theory it can be assumed that $\langle n^{(0)}|n^{(1)}\rangle$ is purely real"
And the consequence of that
You know, this could be answerable if you either linked the article you're talking about or defined what the $\lvert n^{(i)}\rangle$ are :P
Ah sorry
In quantum mechanics, perturbation theory is a set of approximation schemes directly related to mathematical perturbation for describing a complicated quantum system in terms of a simpler one. The idea is to start with a simple system for which a mathematical solution is known, and add an additional "perturbing" Hamiltonian representing a weak disturbance to the system. If the disturbance is not too large, the various physical quantities associated with the perturbed system (e.g. its energy levels and eigenstates) can be expressed as "corrections" to those of the simple system. These corrections...
For some reason the highlight link doesn't work but it is in the section
Time-independent perturbation theory
so what exactly is unclear about that passage?
Since the overall phase is not determined in quantum mechanics.
What phase? Why is it mentioned here? And is this the case in qm only? And how from this one can say that we have a real value and finally:
$\langle n^{(0)}|n^{(1)}\rangle = \langle n^{(1)}|n^{(0)}\rangle=-\langle n^{(1)}|n^{(0)}\rangle$
I can get this $\langle n^{(0)}|n^{(1)}\rangle =-\langle n^{(1)}|n^{(0)}\rangle$
from this $\langle n^{(0)}|n^{(1)}\rangle + \langle n^{(1)}|n^{(0)}\rangle=0$
But not the eq. which implies that the inner product between the unpertrubed eigenstate and a correction term of it is orthonormal
which would be $\langle n^{(0)}|n^{(1)}\rangle = \langle n^{(1)}|n^{(0)}\rangle=-\langle n^{(1)}|n^{(0)}\rangle$
@imbAF You should know that a quantum state vector $\lvert n^{(1)}\rangle$ is the same state as itself multiplied by a phase $\mathrm{e}^{\mathrm{i}\phi}\lvert n^{(1)}\rangle$, right? So you have a degree of freedom in defining $\lvert n^{(1)}$.
1:04 PM
Yeah I know
this complex exp. represents a degree of freedom ?
from this the text expects you to understand that, in general, $\langle n^0\lvert n^1\rangle = \langle n^1 \lvert n^0\rangle^\ast$, so you can use this to get $\langle n^0\vert n^1\rangle = \langle n^1\lvert n^0\rangle$ by choosing $\phi$ such that $\langle n^0\vert n^1\rangle$ is real
Ok
so you choose no phase at all
But still that doesn't explain the minus sign
you said you can get $\langle n^0\vert n^1\rangle = -\langle n^1\vert n^0\rangle$, so what minus is there left to explain?
you use that and $\langle n^0\vert n^1\rangle = \langle n^1\vert n^0\rangle$ and you have your result, namely that $\langle n^0 \vert n^1\rangle = -\langle n^0\vert n^1\rangle$, hence $\langle n^0\vert n^1\rangle = 0$.
The non-minus part*
Ok I see
But the assumption that we have no phase, doesn't that affect our method at all?
If you had a complex phase, you wouldn't been able to claim
@ACuriousMind this
And as a result you wouldn't been able to show the orthonormality between the unperturbed eigenstate and a correction term
Or am I wrong?
@imbAF You asking that question indicates that you did not understand the point of the phase at all
if you just say "Ok" when I say something you don't understand that is not useful
1:12 PM
I fail to see how I am not understanding
what I should be understanding actually
I know that a complex global phase
does not change the physics
such as pdf
the point was that there is a set of states parametrized by a phase $\mathrm{e}^{\mathrm{i}\phi}$ that you can choose as $\lvert n^1\rangle$, since they're all the same physical state. You choose the one from among these for which $\langle n^0\vert n^1\rangle$ is real.
What is that I am missing
There is no "assumption that we have no phase here", your question does not make sense - we choose precisely the state vector that makes the inner product real, it's not an assumption, but a choice in the formalism we can freely make
Ok
An additional thing. In the wiki article it explains the instances in which PT is used
And it essentially says that we use it to describe a complicated system from a simpler one
But I feel like this is a very vague description of when PT is relevant for use. When you make such a claim, you must somehow know or be aware that your system (the complicated one) is somehow linked with a simpler one
Can it not happen that you might consider a complicated system
Have you looked at the examples?
1:19 PM
I have not
No yet
in that case you're doing that thing again where you ask questions that should become obsolete once you see how something is used :P
Yeah
I went with the wiki definition
Since in my lecture it was only presented, but not when and why would one consider
So I wanted to analyze that part of the wiki, which looked vague to me
But yeah, I need to encounter a case where the use of PT makes sense
sometimes people write vague things
and is the right approach to a considered system
not everything is meant to be definition in the rigorous mathematical sense
1:40 PM
I am wondering what "identical particles" precisely means. ignore spatial d.o.f. then, it seems that if I have two spin-1/2 particles with one of them having $s_z = +1/2$ and the other having $-1/2$ that I should be able to distinguish between the two particles.
however, identical particles as described in sakurai is more something like if I have two spin-1/2 particles of the same species, then they are identical and we cannot distinguish between them even if they occupy distinct states
it depends on "how" you "have" them
for instance, if the two particles are trapped in different traps, of course you can distinguish them
but if they're both conduction band electrons in a metal or something like that, you probably can't
hm the setup is a weakly interacting gas of identical particles in a regime in which the gas follows Boltzmann statistics and in which collisions must be treated quantum mechanically.
in that case your setup contains the answer: You can't track any particles in the gas individually because they're identical by hypothesis
to my understanding, they claim that in this setup two gas particles are indistinguishable when they have the same $s_z$ value (or whatever direction you wish to set)
but already the abstract hints at your summary being incomplete, since they call it a "spin polarized gas"
part of that definition may well be that the particles occur in the two distinct spin states and don't switch, making a particle with spin-up distinguishable from one with spin-down
I'm not familiar with the specific setup and can't tell you whether that's really the case
1:56 PM
hm they seem to say that 1) start with weakly-interacting (in the sense of short-range scattering) gas of identical particles. 2) polarize the electronic spins using a large magnetic field. 3) now the remaining "trackable" degree of freedom is the nuclear spin. 4) the short-range scattering is assumed to not affect nuclear degrees of freedom.
so i guess this seems to fit what you are suggesting is happening
@ACuriousMind if two particles r spin up and spin down, then the state is $(Anti)Sym |up, down \rangle$. Then does this state mean the particles r still indistinguishable despite one being spin up and one being spin down?
2:36 PM
@ACuriousMind I want to ask you something about degenerate time independent PT.
In how we deal with this case.
If my understanding is correct what happens in this case, we try to diagonalize the distrubance operator $V$ (H=H_0+V) in the sub-hilbert space belonging to a degenerate eigenvalue. And we find the corresponding eigenvectors to the eigenvalues of the diagonalized disturbance operator. Then we express these eigenvectors as linear combination of the unperturbed eigenvectors of the sub-hilbert space
The eigenvectors of the disturbance operator represent the first order corrections of the degenerated eigenvalue
Is this an accurate description of what happens?

3 hours later…
5:58 PM
"Must an operator $T \in \mathcal{L}(V)$ have any invariant subspaces other than $\{0\}$ and $V$?"
I think the answer is no
We can always avoid mapping a dimension to itself, assuming we have more than 1 dimension in the space
so for dim>1, no
@ACuriousMind while you were out partying and enjoying wacken, I was at home studying the blade
What was it like? Did u go to that festival in the past
Yeah ACM... how was it? how are you??!
@Obliv The answer depends on the underlying field (reals or complex numbers?). Hint: What would a 1-dimensional invariant subspace look like? What would it mean for an operator to have none?
@Obliv I've been going there since 2016, it's my usual summer vacation by now :P
currently waiting in (digital) line to buy tickets for next year
it was pretty amazing as usual, and also physically draining as usual (the first proper hot shower back home feels like being reborn)
6:25 PM
@ACuriousMind underlying field could also be a finite field like $\mathbb{Z}_p$ for some prime $p$? A 1-dim invariant subspace for an infinite field, i'm not sure, but for finite field I think it'd just be a cyclic subspace (cyclic in the sense that the operator maps each element to an element belonging to the subspace of smaller size)
@ACuriousMind Oh nice, probably a great way to meet people and discover new bands.
@Obliv it could be $\mathbb{Z}_p$, sure, but that's not really what I was after: For a one-dimensional invariant subspace $W\subset V$, what can you say about $Tv$ for some $v\in W$? Hint 2: There's a special name for this kind of vector.
well $Tv \in W$ must be true by definition..
I wanna say eigenvector because that's the chapter this topic is in
but idk :P
@Obliv it is an eigenvector. Can you figure out why?
I shall read further because I completely forgot what eigenvectors & values are. I just know eigen means "right" or something in german
@Obliv well, you can only discover new bands if you're not already spending all your time running from one band you already know to the next :P
6:28 PM
True
ohh
So you can span $v$ and this becomes an invariant subspace of $V$, also called an eigenvector?
or no the vector spans the space but span $\neq$ the vector

2 hours later…
8:30 PM
youtube.com/watch?v=JzhlfbWBuQ8&t=14m13s I'm guessing this has to do with deBroglie wavelengths?
to his comment about it could be bowling balls??
I don't think that was in earnest, I could be wrong, but how could a solid produce a diffraction pattern when going through slits?
9:15 PM
Do local $U(1)$ gauge symmetry imply that electric charge is zero?
I know it is an old question and answered in detail by Qmechanic here. But what is the resolution of what the commentor Friedrich is asking there?
Valter's answer here also addresses Friedrich's concern but the reply seems to be in tension with what Qmechanic said in the 2nd last paragraph!
And then comes Lubos Motl's answer which says something which is completely different :) (I think that is really wrong). Valter is saying that he was doing that calculation for the first time, though so he wasn't sure about that... So I am not either...
9:55 PM
@Sanjana The resolution is that eq. (18) is not a special case of eq. (24), and you would have to explain why you think so
10:11 PM
before you claim this is obvious: the problem is probably the off-shell/on-shell nature of the equations: The equality $Q(\epsilon) = 0$ is an off-shell identity, but when you restrict it to "the case of constant $\epsilon$", it just becomes a tautological 0=0 even before using the equations of motion - just like the 2nd Noether identities in all the other posts
but you would have to use the e.o.m. to make $\mathcal{J}(\epsilon) = J(\epsilon)$ and hence derive (18) in some fashion, but you can't go anywhere from 0=0 :P
10:35 PM
@ACuriousMind But Valter's charges are evaluated on-shell, right?
And it is shown to be proportional to the usual electric charge (obtained by applying Noether's 1st theorem to global $U(1)$). Both are just zero on-shell? (I read somewhere that this is just the same as photon is neutral...)
@Sanjana yes, but Valter's charges are not the charges $Q(\epsilon)$ from Qmechanic's answer
@ACuriousMind Yeah I confused that at first...
Yeah, true that these $\mathcal{J}$ is conserved off-shell and $J$ is conserved on-shell. So let's forget about the $\mathcal{J}$ for now and go on-shell and with arbitrary parameter for now. What's bugging me right now is $Q:= \int J^0 d^3 x \propto \text{flux}$ and $Q=0$ (c.f. Valter's proof) implies $\text{flux}=0$ ?
@Sanjana note that Valter's answer is specific to the case of free Maxwell theory
Oh. So it is just saying that the photon is neutral?
so there is no confusion - free Maxwell theory, i.e. electromagnetism in vacuum of course has no charges
10:47 PM
Okayyyyieee
@ACuriousMind What about Lubos' answer which states $J^\mu =0$ altogether?
@Sanjana he is in agreement with Qmechanic, since the coefficients he's talking about are Qmechanic's $\mathcal{J}$, which do vanish on-shell.
@ACuriousMind How do you say that? I thought $\mathcal{J}$ is $J$ off-shell but as you said this can't really done.
@Sanjana sorry: they vanish on-shell and at constant $\epsilon$
11:03 PM
@ACuriousMind On second thoughts, I think on-shell is enough...
Ok, so Noether's 2nd current which obeys conservation law-like identities off the shell. But when you go on-shell it vanishes...
@ACuriousMind I can see this for the Maxwell case, but not in general. Why going on-shell should give me $\mathcal{J}=0$? With the constant $\epsilon$ if it is required. I can't see this from QMechanic's equations...
11:23 PM
And the reason why I am pressing on it for the general case is because Lubos is saying that this is valid in general?
And also, I don't think Lubos is using on-shell condition in his argument. My understanding is that he is just obeying the Noether's procedure for 1st theorem and applying it to gauge symmetry, but not integrating by parts to put the derivative on the current. Instead now since it is a true symmetry of the action and not an artefact of the first step of Noether's procedure, and $\epsilon$ is arbitrary $J=0$ offshell.
11:52 PM
Anyway... a more practical question is how to show that the charges corresponding to $\partial_mu F^{\mu \nu}$ vanish without using EOM? I am getting a charge=flux at infinity. Does this vanish due to gauge invariance? I cannot use Gauss law because that is an "EOM" and would make me go on-shell. Qmechanic finds a superpotential for the current. I can't find that too..