3:17 AM
oh my god ._. i thought that $k_\mu x^\mu = k_0x^0 - k_1x^1 - ...$ ._.
index notation moment

3:39 AM
@SillyGoose omg no
I think part of the problem is that you osmosised a lot of weird misconceptions of index notation rather than having a comprehensive first introduction that tells you how to work with them in general...

@naturallyInconsistent i know right

H O N K

yes tonight i am just going to watch a lecture or read an intro section of a GR book or something
and then write it in my own notes for future me
H O N K

I usually give students the exact same advice: Start with RHB and Cartesian tensors. It gives you a very simplest best case scenario so that you can use tensors to prove vector identities. Then move on to Schutz GR and try to learn what it means to have contravariance and covariance. The point of the index notation is to keep track of these, not keep track of the metric signature. If you are having metric signature issues, you are beyond confused.
And because it is GR, you should be doing -+++ as we all should miehehehe
And then you should link the covariant basis to condensed matter reciprocal lattice vectors

1 hour later…
4:50 AM
is there anything stopping me from making the replacement $\eta^{\nu\sigma} \mapsto \eta^{\sigma\nu}$ in the second to last line?
I feel like there must be
the only thing that i started to think of is maybe contracted indices need to change together if you change them

@SillyGoose The flat Minkowski metric is famously symmetric and doesn't matter if it is both upper indices or both lower indices so this replacement is definitely allowed. It is rare that such replacements work, but this is one of the special cases that it does.

@naturallyInconsistent but then wouldn't i be able to conclude from the second to last line that $[\eta][\Lambda][\eta] = [\Lambda^{-1}]$ where the bracket notation means matrix representation of the bracketed object

In your particular case, you only needed the thing to be symmetric, not symmetric between upper and lower indices, so there is a much wider class of tensors in which this replacement works
@SillyGoose $[\Lambda^T][\eta][\Lambda]=[\eta]$ already immediately implies $[\eta^{-1}][\Lambda^T][\eta]=[\Lambda^{-1}]$ and since $[\eta]=[\eta^{-1}]$ so I am not sure what it is you think you are confused about.

2 hours later…
6:53 AM
what is a good reference for going through the classification of the finite dimensional irreducible representations of the universal cover of the restricted lorentz group?
i am up to identifying the isomorphism between the complexified lorentz algebra and a direct sum of two copies of $sl(2, \mathbb{C}$, but I haven't found a resource which runs through thoroughly the remaining steps

7:14 AM
@SillyGoose Are you just looking for Weyl and Wigner's treatment of spins?

@naturallyInconsistent i would like a general classification

7:45 AM
@SillyGoose Weinberg QFT Chapter 2

not of the infinite dimension unitary representations @bolbteppa of the finite dimensional non-unitary representations

@Sanjana what an insane set of notes
@SillyGoose chapter 5 on fields

oh i see
okay so for each pair $(n, m) \in \frac{1}{2}\mathbb{N}_0 \times \frac{1}{2}\mathbb{N}_0$ we obtain a representation of $sl(2, C) \oplus sl(2, C)$ by sending $A_i \mapsto$ usual spin-n operators and $B_i \mapsto$ usual spin-m operators. but actually we want representations of $\mathfrak{so}(1,3)$, not its complexification, so we solve for $J_i$ and $K_i$, resulting in an irrep for $\mathfrak{so}(1,3)$. And then by usual machinery, this is equivalent to projective lie group irreps of $SO^+(1,3)$?

8:12 AM
what is the intuition for irreps being tensor products of two individual irreps of $sl(2,C)$?

ACuriousMind mentioned this. lemme find
in general, the result is that $T(g_1)\otimes T(g_2)$ is an irreducible rep of $G_1\times G_2$

@SillyGoose ultimately this is due to the Peter-Weyl theorem: Equivalently we're looking at representations of $\mathrm{SU}(2)\times \mathrm{SU}(2)$, and more generally at reps of $G\times H$ for compact Lie groups $G,H$. $L^2(G)$ contains all the irreducible representations of $G$, and $L^2(G\times H) = L^2(G)\otimes L^2(H)$ so the irreps of $G\times H$ have a basis in terms of tensor products of irreps of $G$ and $H$
this is also the reason the infinite-dimensional reps of $\mathrm{SO}(1,3)$ are not of this form, because the equivalence to reps of $\mathrm{SU}(2)\times \mathrm{SU}(2)$ only works in the finite-dimensional case

8:30 AM
@ACuriousMind is the representation $\phi(x^{\mu})\rightarrow \phi(\Lambda x^{\mu})$ for scalar $\phi$ irreducible?
$\psi(x)\rightarrow \psi(R_{\theta} x)$ is reducible for rotations. does this not hold for the analogous thing for lorentz group?

8:52 AM
@SillyGoose Note that I did not understand Weinberg's version of this, but I did understand Anthony Duncan's.

9:03 AM
@ACuriousMind do you think i could understand Peter-Weyl theorem in a reasonable amount of time :P
@naturallyInconsistent hm i'll check it out

9:15 AM
@PM2Ring this is the message I would never have expected to read as soon as I woke up
It thrills me

is this the appropriate explicit construction of irreps for $\mathfrak{so}(1,3)$?
(i have set $\hbar = 1$)

9:47 AM
I think this helps
There was an excellent answer by Qmechanic, which I don't have time to search rn

i think i might have read through it
at least i read a few of qmech answers related :P maybe one of them was it
hm this stipulation seems potentially problematic: "The passage from real to complex is harmless, as the complex linear representations of the complexified Lie algebra are in 1 to 1 correspondence with the real linear representations of the real form. (Given, that we are working with representations on a
C
vector space.)"
might we have a real field?
then we could not do our complexification business by the quote?

@SillyGoose you can just determine all the complex representations and then figure out which ones restrict to real ones; it's easier than to restrict to the reals from the start

10:03 AM
oh no wait i think something must be wrong in my definition bc $K_i$ is not hermitian

that's normal :P
the boost generators are not unitary in the finite-dimensional representations
remember that you're not looking at unitary representations of SO(1,3) here

oh right

10:27 AM
$\psi (x)\rightarrow \psi(R_{\theta} x)$ is reducible for rotations, but $\phi(x)\rightarrow \phi(\Lambda x)$ is not reducible for Lorentz transforms of Klein gordon solutions, right?
sorry take the 1-particle space of Klein Gordon $\psi (p^{\mu})\rightarrow \psi(\Lambda p^{\mu})$ with on-shell p ^(mu)
if this was reducible into a direct sum of (j,m) reps, then it wud mean boosts are non-unitary here. but boosts are unitary here, so it should be irreducible
or maybe it's reducible but not reducible into a direct sum of (j,m) reps

3 hours later…
1:09 PM
latex question
how do i add some comment inside the square brackets in a citation
like [5, pag. 30]

@ekardnam_ why don't you just ref instead of quoting the page, which may change?
Oh the page is of the source

@SirCrackpot its the page of the book

Sry

anyhow i find out its \cite[pag. 30]{nameofyourbibtexentry}

1 hour later…
2:46 PM
This Arnold Neumaier guy seems to know everything

3:01 PM
How can i physically interpret that near $p^2=m^2$, the dressed propagator looks like the free propagator?

@DIRAC1930 He is rather extremely wrong about Lattice QCD. He thinks that LQCD would not need to consider loops.
@ekardnam_ You can also define the bibtex entry in the references file with the page numbers already there, so then you don't have to write which page it is, if you reuse that particular citation-to-page often.

@ACuriousMind Is there no physical interpretation to my question?

3:23 PM
Won't this mean that the field operators near $p^2=m^2$ look like free field operators?

Why don't you read a standard textbook covering this topic?

$p^2 = m^2$ is gonna be precisely the case closest to a free particle
Since that's just the case where the field contains exactly the energy of one particle
It is not lower (due to being affected by some potential energy) or higher from being a system of bound particles

@Slereah Why do you not use the sensible convention?
Oh wait, you mean that system of bound particles is higher in energy because more particles means more rest energy?

So if I operate on the vacuum with $\hat{\Phi}(p^2=m^2)$, I will create what looks like a free particle?

@Slereah that mass hyperboloid being isolated is protected, but its location is moved around by renormalisation.

3:32 PM
yeah there's plenty of nuance to what I said
But that's the rough idea

So how do I interpret this physically? If I take the non rel limit, all I essentially have is the free field operator since $p^2 \approx m^2$

Like this is the typical figure of a spectral function

@DIRAC1930 the non rel limit is the one in which $E\gg |\vec{p}|$, so you want the energy to be basically just mass energy
If you take $p^2=m^2$ you just have a relativistic free particle

Ah yes sorry
I meant $E \approx m$

@Slereah Actually, it seems to not be about typicality. For the QFT that we can even do Feynman diagrammatics on, Källénâ€“Lehmann is a necessity, isn't it? That's why it is in all textbooks on the subject

3:39 PM
I guess?

More than a necessity I would call it a theorem
I mean, it's a fact about how a field is expanded, it's not directly related to perturbation theory as far as I remember

I'm trying to understand this in terms of non-rel QFT i.e. the poles of the GF become the excitation energies of the system

You may be interested in some many body theory book
Condensed matter people deal with this all the time, they're called quasiparticles

@SirCrackpot Remember, in realistic models we have, say, bound states of protons and electrons to make hydrogen atoms at least, and that is some bound state that has the field operators giving a state at energies strictly smaller than the $m^2$ of each of the "free dressed particle" precisely because it is a bound state.

Page 72 in the pdf

3:43 PM
anything more than one particle and you will never really get a "free" state, except asymptotically

I'm trying to understand the 3rd paragraph
Starting at 'The Green's function of a liquid, near it's pole...'

@DIRAC1930 What else would you expect it to look like? A single on-shell particle usually cannot decay due to kinematic constraints.
all it can do is...propagate

If I have a particle in front of me, what do I have? How does this fit into that picture? All I currently know is that I have a dressed particle.

Are bound states the only hindrance to perturbation theory?

@DIRAC1930 What is "a particle" in this context

3:46 PM
@DIRAC1930 what "picture" do you mean?

A particle can mean quite a lot of things in QFT

is this going to be another round of you trying to bypass the rule that we don't really know the interacting/dressed space :P

Lets just say if I had an electron in front of me

In which state :p

how do you know it's there
you can't see electrons

3:48 PM

Like what people call a particle in QFT is a momentum eigenvector
It's not really what we think of as an electron in experimental physics
It's not a mostly localized packet

How do I go from QFT to the real world

@Slereah and Pauli-Lubanski (?)

@DIRAC1930 with precaution
Haag's book has a chapter on what this all means on an experimental level, it's not super clear

Isn't the answer to Dirac's questions just rep theory of the Poincaré group?

3:50 PM
Feb 5 at 15:26, by ACuriousMind
@DIRAC1930 Then perhaps you should start learning a lot more technically focused QFT (like how to compute jets in colliders, or how to do lattice QCD) instead of worrying about all these abstract issues in QFT in general :P

Does anything I was talking about before have anything to do with the fact that we can treat particles as being free when doing non rel QM using the Schrodinger equation?

@ACuriousMind Do we really not? I thought with infra-particles we treat IR divergences properly, at the cost of the propagator no longer looking like what we always assume it to be? Whereas we can use CPT to deal with UV divergences (I do not know if the combination works, but I surely do hope so)

why do you keep asking the same questions

You're one to talk, giving the same answer :P

I swear I've only asked this once
It's not like that time when I kept asking about asymptotic states lol

3:52 PM
I'm not even sure what a typical dumb electron shot from a cathode ray tube is supposed to look like as a wavefunction

@DIRAC1930 but that's the same issue!

@SirCrackpot Poor Sisyphus

lol

Probably some coherent state but idk

because the "particles" we usually talk about are asymptotic states

3:52 PM
@naturallyInconsistent I'm in a mood for banter

@SirCrackpot arrives with batter
oops

But if I have two particles near each other, I can still use the Schrodinger equation

@DIRAC1930 are you sure you can?

Does anyone have anything to say about the hideous passive-active question I asked yesterday? :P

The real truth is that I don't think we really have a good QFT theory that is usable at all levels of physics

3:54 PM
@naturallyInconsistent Yes

I mean, all I need is "it's just a name"

We have a patchwork of various elements that sort of overlap

@Slereah Isn't that the point of EFT

@SirCrackpot the only thing that I have to say is that I don't want to talk about active and passive transformations :P

@SirCrackpot Might I suggest "It is a name that ACM is particularly allergic to"?

3:54 PM
We have the formal theory, we have the more approximative theories, we have the way this relates to experiment
And all those things have some links between them

@ACuriousMind don't be so passive, come on :P

But it is not entirely clear how everything coheres
Doing an entire theory like that is extremely hard, even for classical mechanics

really, we don't even have that for QM without QFT, at least not one everyone agrees on, cf. the measurement problem

Typically you have to do some handwaving to avoid complications

if you want to actually model what the measurement apparatus does it all gets a bit hazy, too

3:57 PM
You can kind of do it but yeah, in experiment nobody actually does :p
You're not gonna do the wavefunction of your bubble chamber

in a sense I consider treating the measurement apparatus as a black box as about as bad as just assuming all the stuff with asymptotic states and particles works - it's a crucial assumption you need to apply the theory to anything but if you dig into it too deep there's no foundational bottom in that rabbit hole

You also have to solve problems like "how does state preparation even work"

QFT sucks
GR is sexy physics

I keep telling people
I mean GR has its host of issues as well certainly

Look what you've done to me you monster

3:59 PM
@Slereah easy, I just use my machine that spits out identically prepared states :)

@ACuriousMind But then which one is it!

don't ask me how that works, that's the experimentalists' job

If I take the non rel limit and ignore transitions involving antiparticles I can always write, $\hat{\Phi}^\dagger=\sum_p \psi_p \hat{a}^\dagger$. So is the issue of interpreting rel-QFT in that a) I cant write a wavefunction, and b) I need to include transitions involving antiparticles because now there is sufficient energy, hence motivating the field operator expansion in terms of 2 types of creation operators which muddy's physical interpretation

@Slereah at least not like "the metric tensor is just a formal object and doesn't really exist"

A lot of experimental QM sort of relies on analogy with classical mechanics I think?

4:00 PM
@DIRAC1930 those are both issues but I wouldn't commit myself to claiming they're the only ones

Like we assume that in a bubble chamber, the measurement is sort of gonna be like position since you can see its trace

@Slereah this is a really confusing terminology. more standard is to say that any state of the 1-particle space is a particle

@Slereah there's actually a pretty good explanation originally by Mott on how the bubble chamber trajectories emerge

I'm not sure if anyone has ever formally proven that a bubble chamber's measurement is close to a position measurement

Dang they just told me I looked like a law students

4:01 PM
@ACuriousMind Yeah I saw it once, but can you check it

but some treatments popularise the idea that a particle is a discrete chunk of energy @Slereah

it's not that the chamber is a "position measurement"

How does one unlock the physics person look?

Like it relies on previous assumptions about how atoms work

everything relies on everything else

4:02 PM
Alas

physics is a house of cards and at the very bottom is a bunch of "obviously"s

In the philosophy of science, observations are said to be "theory-laden" when they are affected by the theoretical presuppositions held by the investigator. The thesis of theory-ladenness is most strongly associated with the late 1950s and early 1960s work of Norwood Russell Hanson, Thomas Kuhn, and Paul Feyerabend, and was probably first put forth (at least implicitly) by Pierre Duhem about 50 years earlier.Semantic theory-ladenness refers to the impact of theoretical assumptions on the meaning of observational terms while perceptual theory-ladenness refers to their impact on the perceptua...
They were right ðŸ˜”

@SirCrackpot if they think you're a law student then your hair is too neat :P

In the non rel real world, if I have two electrons together, what will they do?

Shake hands?

4:03 PM
how did you get them there?
they're like charges, they'll repel each other
this isn't even a QM question :P

So something weird must be going on between the rel and non-rel limit

Plenty of weird things happen certainly

yes, it's not even clear to me that limit exists in a proper sense

usually people just match a few interesting quantities, but I think that - much like the "classical limit" - there's not actually a well-defined procedure that turns a consistent relativistic theory into a consistent non-relativistic theory

4:06 PM
@DIRAC1930 they've got a lot to teach and not that much time :p

@DIRAC1930 Because of time constraints, the correct avenue to cover such topics is in thick books, where they have the space to properly introduce the topic.

Plus really what are most physics students gonna say if they try to teach you all the nuances

@ACuriousMind does Haag Ruelle not completely solve the asymptotic state problem?

@DIRAC1930 the converse question would be - why do you care? These limits etc. are theoretically interesting but practically useless: Just use the appropriate theory for the appropriate regime

Ideal QFT course would be 1) Calculate Lamb shift (optional) 2) Calculate lifetime of positronium (optional) 3) Show how all of QM comes out of QFT

4:08 PM
the purpose of QFT is not to give you back QM, it's to build QED and QCD, to explain the particle zoo and other collider observations

@ACuriousMind Actually, I think such a scheme already exists. The rel -> non-rel should exist as a linearisation starting from rapidity variables. I think it is the quantum -> classical limit that isn't well-established.

@ACuriousMind I guess it depends on the person

I am reading about defining the Kelvin Scale using carnot maschines, i reached this part where we have $1- \eta = \frac{T_2}{T_1}$ where $T_1 > T_2$ the text just mentions from this the conclusion, how exactly is this now deduced from this statement? i understand that $\eta$ is universal, so the left side is equal to a constant value. then you have T_1 = Constant T_2 .. ?

@DIRAC1930 How can starting with two optional things be a tolerable start? If anything, my stillborn textbook is about simply skipping QM and introducing students directly into QFT land, so that the usual QM stuff are done in QFT language.
@Mad It is a standard observation that efficiency $\eta=1-\frac{T_C}{T_H}$ and so rearranging this expression, you can work backwards and define temperatures using efficiencies.

@naturallyInconsistent I know that that's what people do - to the "few interesting quantities" I mentioned
I'm not convinced this really works universally to produce a consistent non-relativistic theory

4:13 PM
@naturallyInconsistent this is what i wrote, if i substitute this, i will get naturally zero.

@naturallyInconsistent i dont understand why you rewrote the equation that i wrote.

but again, that's not really what we need - just the predictions of the non-relativistic and relativistic theories need to match in the regime where they overlap, there is nothing physical/experimental telling us that such a limit on the level of the theories needs to even exist

Things get a little weird when you try to apply kinematic limits to systems rather than to theories I think

yes... all we know is that the relativistic theory applies to a larger domain, the non rel theory is defined to be some approximation of it that u can afford to do in some situations @ACuriousMind

4:15 PM
For instance go from special relativity to classical mechanics

but a limit on the level of the theories would need to exist to produce the kind of universal reductionist explanation of QM from QFT that @DIRAC1930 wants

@Mad That is not what I did. I wrote down what is the quintessential and original discovery of Carnot and Clausius and then explained to you how, by carefully arguing things so that temperatures were not defined prior to that point, then the efficiencies can be used to define the temperatures.

so the non rel theory does not have a concrete definition

Take a point, look at the light cone it describes, then contract it to the Galilean group

@naturallyInconsistent I just derived the expression you wrote, its the same one that i wrote above! you literally just moved it on the other side of the equation sign.

4:16 PM
Points on the past and future light cone fuse together, since the light cone is flat classically
Their time interval is zero

@ACuriousMind I feel like there must be a simple answer to this

@Mad Correct, but I did not start from yours. I am trying to tell you that people originally started from what I wrote down, and then discovered that the logical flow could be reversed.

but if those two points are timelike separated, then from another perspective, their interval shouldn't be zero after the contraction

@DIRAC1930 if so, it has eluded us for about a century now :P

I think pretty much all "limits" in physics are super tricky and probably not formally that possible to do for the systems themselves

4:20 PM
i am satisfied with a vague idea of that limit works personally

Too brutal of a limit in most cases :p

All I want to do is take QED and then take some limit and reproduce the Schrodinger equation

You're gonna have to take a few different limits I fear :p

there are known ways

@DIRAC1930 what do you mean "the Schrödinger equation"
the Schrödinger equation of what system

4:23 PM
she means either the Pauli or the Dirac eqn
but the Dirac eqn interpreted as a wave eqn

Lets say something like $\sum_n \hat{H}^{(1)}_n+ \hat{H}^{int}$ where $\hat{H}^{(1)}_n$ is a single particle Hamiltonian and $\hat{H}^{int}$ is the interacting part

@naturallyInconsistent I think i undesrstand the statment, do they pick the cold bath to have the temperature of the triple point or the hot bath?Because its a quotient so you need comparrision

@DIRAC1930 well, the free/single particle part you get from one-particle states obeying the same equation of motion as the field, see the answers to this question of mine and the interaction potential you get from the tree-level 2 -> 2 diagram(s), see this answer of mine
you could probably also produce "multipotentials" by looking at 3->3 and higher diagrams, but I can't quite remember if I've ever seen that used for anything

@Mad There will be a cold bath and a hot bath. To define temperatures above the 3ple point, the cold bath is the 3ple point, and to define temperatures below the 3ple point, the hot bath is the 3ple point.

@naturallyInconsistent Ok so its relative i see. thanks!

4:27 PM
the threeple point

@ACuriousMind always fun!
I mean, if minds are out of the gutter, it can always be spidey~

reminds me (although it is not really related) that I just recently had to deal with some old code that called a secondary connection the "2th connection" and it immediately bothered me subconsciously but it took me a while to figure out why

@ACuriousMind :D

the tooth?

I swear to tell the tooth, the full tooth, and nothing but the tooth

4:35 PM
what r some interesting things in philosophy of cloning?

@ACuriousMind my hair is a MESS
I mean, a combed mess

@ACuriousMind reminds me of plusve and minusve

@SirCrackpot you mean, like a barrister's wig?

Lol

@Loong hey, that's some of my most beloved shortforms!

4:37 PM
@Loong I actually haven't seen that in a while - "thanks" for reminding me!

I mean I have some tufts that even spray, wax or gel can't tame

@SirCrackpot all that is left is fire

@SirCrackpot it's that fact that you seem to use spray, wax or gel that makes you not a stereotypical physics student :P

And those are in the back of my head, so whatever I do I always have this neat "fountain" of hair at the top of my head

Also, are you an orc? Because nobody should have hair that wild

4:38 PM
@ACuriousMind I don't! I just tried to tame my hair :P

@SirCrackpot have you consulted "How to Train a Dragon"?

Well, I also had a long coat and a suitcase so maybe that's my fault but really that would be ok in the 50s...

lol
guy over here carrying around a suitcase wondering why people tell him he looks like a lawyer

In the 50s I could have been everything!

I'm sorry to tell you it's 100% the suitcase
2

4:42 PM
It's comfier to carry around though :P

lawyers have to memorise all the laws

I guess my mental image of a physicist is just plainly wrong

or maybe they can google in courtroom

@SirCrackpot does your pants look like this

@naturallyInconsistent oh yeah i have seen this before, but i only need it once in this case

4:54 PM
@ekardnam_ sure
@SirCrackpot omg, please don't look like the ppl in Young Sheldon

@ACuriousMind does the probability of having long/disheveled hair coincide with how far along you are in your physics degree?
I don't help the statistics in any way :P