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Nairit Sahoo

4:00 AM

Is $\langle Tr(F_{\mu \nu}) Tr( F_{\rho \sigma}) \rangle$ gauge invariant in Non-Abelian Yang Mills theory?

123

4:12 AM

Hello Everyone...

Nairit Sahoo

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

Sanjana

4

Q: Proof of a definite integral whose result is the difference of arithmetic and geometric means

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?

Slereah

9:10 AM

I mean Coulomb field has conserved charges and it's not exactly solvable

Can compute $r(\theta)$ but not $r(t)$

ACuriousMind

9:33 AM

@Obliv on my way back now, it was great (if a bit rainy at the end)

Nairit Sahoo

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...

Sanjana

Hii @Slereah ... just casually pinging you in case you know the answer to this?

Slereah

@Sanjana No being isometric under Poincaré selects uniquely Minkowski space

or flat spacetimes if you only require the symmetry to be local

Sanjana

@Slereah Do you know where I can find a proof of this?

Maybe in that Wolf book you sometimes refer to...

Slereah

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

user 20458579510081670432

Do you happen to know why the infinitesimal calculus approach was abandoned for the delta epsilon approach.

Slereah

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

user 20458579510081670432

Okay, thanks.

Ryder Rude

10:46 AM

hi

@ACuriousMind glad to have u back

@Slereah by minkowski space, do u mean any spacetime with the lorentzian metric

Slereah

I mean Minkowski space

Ryder Rude

@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"?

Slereah

11:07 AM

Local as in true of a local flow

ie using the Killing vectors

Rather than actual isometries over the whole spacetime

Ryder Rude

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

Slereah

Sure

Ryder Rude

Like what happens with Darboux's theorem in phase space stuff

@Slereah great

imbAF

12:28 PM

which is the the expansion parameter in an asymptotic series, such as the power series ?

imbAF

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

ACuriousMind

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

imbAF

Ah sorry

Perturbation theory (quantum mechanics)

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

and it addresses normalization

ACuriousMind

so what exactly is unclear about that passage?

imbAF

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 $\langle n^{(0)}|n^{(1)}\rangle = \langle n^{(1)}|n^{(0)}\rangle=-\langle n^{(1)}|n^{(0)}\rangle$

ACuriousMind

@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 $\lvert n^{(1)}$.

imbAF

1:04 PM

Yeah I know

this complex exp. represents a degree of freedom ?

ACuriousMind

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

imbAF

Ok

so you choose no phase at all

But still that doesn't explain the minus sign

ACuriousMind

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 $\langle n^0\vert n^1\rangle = 0$.

imbAF

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?

ACuriousMind

@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

imbAF

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

ACuriousMind

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.

imbAF

What is that I am missing

ACuriousMind

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

imbAF

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

ACuriousMind

Have you looked at the examples?

imbAF

1:19 PM

I have not

No yet

ACuriousMind

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

imbAF

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

ACuriousMind

sometimes people write vague things

imbAF

and is the right approach to a considered system

ACuriousMind

not everything is meant to be definition in the rigorous mathematical sense

Silly Goose

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

ACuriousMind

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

Silly Goose

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.

trying to follow this paper

ACuriousMind

in that case your setup contains the answer: You can't track any particles in the gas individually because they're identical by hypothesis

Silly Goose

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)

ACuriousMind

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

Silly Goose

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

Ryder Rude

@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?

imbAF

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?

Obliv

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

Sanjana

Yeah ACM... how was it? how are you??!

ACuriousMind

@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)

Obliv

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.

ACuriousMind

@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.

Obliv

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

ACuriousMind

@Obliv it is an eigenvector. Can you figure out why?

Obliv

I shall read further because I completely forgot what eigenvectors & values are. I just know eigen means "right" or something in german

ACuriousMind

@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

Obliv

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

Obliv

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?

Sanjana

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...

ACuriousMind

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

ACuriousMind

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

Sanjana

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...)

ACuriousMind

@Sanjana yes, but Valter's charges are not the charges $Q(\epsilon)$ from Qmechanic's answer

Sanjana

@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 $\text{flux}=0$ ?

ACuriousMind

@Sanjana note that Valter's answer is specific to the case of free Maxwell theory

Sanjana

Oh. So it is just saying that the photon is neutral?

ACuriousMind

so there is no confusion - free Maxwell theory, i.e. electromagnetism in vacuum of course has no charges

Sanjana

10:47 PM

Okayyyyieee

@ACuriousMind What about Lubos' answer which states $J^\mu =0$ altogether?

ACuriousMind

@Sanjana he is in agreement with Qmechanic, since the coefficients he's talking about are Qmechanic's $\mathcal{J}$, which do vanish on-shell.

Sanjana

@ACuriousMind How do you say that? I thought $\mathcal{J}$ is $J$ off-shell but as you said this can't really done.

ACuriousMind

@Sanjana sorry: they vanish on-shell and at constant $\epsilon$

Sanjana

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...

Sanjana

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.

Sanjana

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..

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