The h Bar

01:26

@Relativisticcucumber then, is it ok if miao miao suggests that you find a more sensible route? Like, QFT is not meant to be learnt without first learning the stuff, like SR, stat therm, etc, before coming to QFT

like, there are necessary suffering if you wanna learn, but there is no point of unnecessary suffering

Relativisticcucumber

03:50

@naturallyInconsistent what do you propose is a more sensible route?

naturallyInconsistent

@Relativisticcucumber What have you learnt before, and where do you want to end up understanding?

jigglypuff~

Relativisticcucumber

@naturallyInconsistent i have taken mech, e&m, stat mech, quantum, electromagnetism and special relativity, general relativity, modern physics, experimental physics, ode, linear, analysis, multivariable calc, and probability. i would say out of these subjects i think i have at least a decent undergrad foundation in quantum and a good working knowledge of gr for my level.

i also think i know undergrad linear, probability, and ode at a good level. the other stuff, i know some of, but i wouldnt say i have mastered it. my goal is to learn qft.

but at this point you probably know more about my level than i do XD

my goal in learning qft is to understand the standard model and to eventually learn AMO stuff, i dont have an interest in the math phys side of things really

naturallyInconsistent

Oh, that is then confusing. You seem to have done the important topics prior to QFT, yet you seem to keep having difficulties with them. For example, the SR arguments that we had had with choosing the action to have SR symmetries.

But yes, you should do Atomic and Molecular physics from the QM side first, so that the mathematical tools of QM is not all just crazy shit.

Relativisticcucumber

@naturallyInconsistent in that SR bit i think what i didnt get is the starting point is our lagrangian is already relativistic. i still dont see why a lagrangian that obeys certain symmetries necessarily has EOM that obey the same symmetries but i can accept that as fact and then be fine. but i think this confusion was actually not at all about SR

naturallyInconsistent

You might also want to work out the S matrix (and transfer matrix) for the 1D square well and barrier for Schrödinger equation; the textbooks usually solves the problem, but stop short of expressing the solution as an S matrix. It baffles the mind, why they are already at the goal and won't just do it.

Relativisticcucumber

04:03

but still there is some problem bc that should not have taken days to understand i think

so i would like to find out the root of the issue

@naturallyInconsistent hm interesting. that's a good idea

i think i learn best with extreme clarity. like i often struggled in gen phys when my prof would not explicitly state the limitations of the model or theory at hand

e.g. constant acceleration cases or smth

bc i think i have a large imagination XD so i would automatically think of problematic cases and be confused

idk

naturallyInconsistent

A&M from QM side, as is normal curriculum route, and basic solid state physics, is helpful. Because then you can smoothly connect to the condensed matter introduction to QFT, where they have a simple conversion of the QM stuff to QFT notation. That familiarity helps.

@Relativisticcucumber Ahhhhh. I have friends like you. When I teach, I make sure to avoid this problem. I always have to remind profs that no prof listening to a lecture ever requires the pedantic rigour---it is purely for the sake of helping the student understand, that profs should state things clearly.

I'm not sure what it is that NR QFT helps with, but at least it doesn't seem as alien as the sudden jump to SR QFT is.

Relativisticcucumber

04:40

yeah idk what the problem with me is lol im just trying to improve everyday @naturallyInconsistent

@naturallyInconsistent this is nice. but i also recognize that i need to adapt so BLAH

im sure ill find a way.

thank you for the advice

naturallyInconsistent

meowth

Silly Goose

05:52

Any subject looks alien when u formalize it enough;)

@Relativisticcucumber steven weinberg enters the chat

naturallyInconsistent

No, the maths texts introducing rigorous uni maths to beginners is now somewhat static and familiar, and they are very formalised.

Lalit Tolani

06:21

@ACuriousMind Can you please unfreeze this room chat.stackexchange.com/rooms/130073/…

Sanjana

07:00

In the superconformal algebra in 4 dimensions, $[D,Q]=1/2 Q$(indices are implicit) implies that the scaling dimension of $Q$ is $\frac{3}{2}$. But I expected it should be $\frac{3}{2}$ in $4$ Dimensions because $Q$ is a spinor and Weyl/Dirac spinor has a scaling dimension of $\frac{d-1}{2}$ classically as can be easily seen by computing the mass dimension of the Weyl/Dirac Lagrangian. I know mass dimension gets renormalized. But I guess something else is happening here...

Why are the calculations not matching/expected to match?

Qmechanic

07:33

@LalitTolani : Done.

ACuriousMind

08:03

@Sanjana You talk about deriving $\frac{d-1}{2}$ from the Weyl or Dirac Lagrangian. But $Q$ is not a field and it does not obey such a Lagrangian, so why should it have that scaling dimension?

Sanjana

@ACuriousMind So scaling dimension is not expected to be the same for "objects" which transform in the same way under Lorentz group?

ACuriousMind

@Sanjana No - if the representation of the Lorentz group determined the scaling dimension uniquely, then the scaling dimension would not contain any additional information, righti?

Also, if it did, why would you derive the scaling dimension from the Lagrangian instead of a general argument about spinor transformations? :P

Sanjana

@ACuriousMind Righti.

@ACuriousMind So what's the general method to determine scaling dimension for an object, be it a field or some generator?

Slereah

Apply some general coordinate transform to it and look at how it scales with it?

Sanjana

Evaluate $[D, \cdot]= \Delta \cdot$?

Slereah

08:08

also that yes

They're basically irreps of the dilation group

ACuriousMind

@Sanjana I mean, you already did it above, no? You looked at $[D,Q] = 1/2Q$ and said "Hey, this means it has scaling dimension 1/2".

(although you made a typo and wrote 3/2 both times :P)

Sanjana

@ACuriousMind oh, yes!

@Slereah Going by the definition, I see. But I am confused...where to even start from?

Slereah

According to some paper I once read, "The orders $k$ are "spins" with respect to the dilation operator $$[X_k, D] = k X_k, D = x^a \partial_a$$"

lol

ACuriousMind

@Sanjana How would you end up in the situation that you know you have a bunch of operators but not know how the $D$ acts on them?

Why do you need a general algorithm to determine information that is usually either obvious or explicitly assumed/defined to be given

Ryder Rude

what is the dilation operaror

i think it's $D[\phi(x)]= \phi(diag (a,b,c,d) x)$

Sanjana

08:20

@ACuriousMind Okay, I see.

Tbh, I know how to "derive" conformal algebra and Poincare-SUSY algebra, but nothing like that for superconformal algebras.

E.g. I would define conformal transformations to be angle preserving transformations, look at the most generic transformation and eventually find out the generators of these transformations and the algebra. For SUSY, I would use Coleman-Mandula and facts such as $Q$ transforms as a spinor under Lorentz transformations etc.
For superconformal algebra I see the algebra just "stated"...

So I guess I am asking where does the $[D,Q]=\frac{1}{2} Q$ comes from?

This question logically precedes the first

@RyderRude Generator of scaling transformations $z \to \lambda z$

One particular representation is $x^\mu \partial_\mu$

Mattia

08:41

0

Q: Conservation of angular momentum in particles reactions

In particles or nuclear reactions the square modulus of total angular momentum (the sum of spin and orbital angular momentum) is conserved. When calculating the density of states to compute cross section for reactions, the spatial density of states is multiplied by the spins multiplicity $(2s_a+1...

Any suggestion for this question?

Silly Goose

this collaboration seems quite interesting: projects.iq.harvard.edu/ultra-qm/overview. "ultra-quantum matter" supposedly finds gauge theory as necessary to describe macroscopic quantum matter

naturallyInconsistent

@Mattia oh hi

Mattia

Hi

naturallyInconsistent

08:57

What is it in particular that you are asking about that?

Mattia

@naturallyInconsistent so you are saying that the cross section is calculated in momentum eigenstates so we don't see angular momentum conservation? But then why dowe multiply the cross section by the the spin multeplicity factor?

naturallyInconsistent

Well, if you really wanted to, you can compute everything in spin spherical harmonics

Ryder Rude

@Sanjana oh. makes sense

in 2D conformal field theory, can spacetime be modelled using complex numbers

since holomorphic functions preserve angles

naturallyInconsistent

The thing is that, in linear momentum parameterisation, all the energy-momentum parts are parametrised properly, but we would still have the spin degrees of freedom. However, because simply changing the origin from which a beam is computed, the orbital angular momentum would have the illusion of changing, the angular momentum of such beams would appear to not be conserved. That is sad, but it is life. The angular momentum conservation appears as some behaviour in the actual S-matrix calculations

Sanjana

@RyderRude Yes, holomorphic transformations give rise to 2D conformal transformations...but what did you mean by metric modelling?

Ryder Rude

09:04

@Sanjana i just mean that we represent spacetime points by complex numbers

Slereah

You can in 2D yes

Ryder Rude

oh

Slereah

ie : en.wikipedia.org/wiki/Uniformization_theorem

Ryder Rude

does this generalise to quaternions in 4D? are they good for conformal theories

Slereah

Not particularly no

Sanjana

09:05

@RyderRude Whether an even dimensional real manifold in general can be described using complex geometry depends on triviality of Nijenhuis tensor, btw...

Ryder Rude

oh

Relativisticcucumber

@Mr.Feynman :

6

naturallyInconsistent

jigglypuff~~

Relativisticcucumber

HONK

Mattia

@naturallyInconsistent okay but if we use linear momentum and intrinsic spin eigenstates why do we multiply the density of states by the spin multeplicity factor? Shouldn't we consider if ther is the possibility of spin change before doing it?

Relativisticcucumber

09:11

@ACuriousMind do you like it

naturallyInconsistent

@Mattia I'm not sure what you mean by that? After completely specifying the possible input and output states (with spin), there should not be need for more multiplicity factors.

H O N K

ACuriousMind

@Relativisticcucumber lol

@Sanjana It is extremely annoying that everyone just states it, but the computation is in the original HLS paper

naturallyInconsistent

very lol

Sanjana

@Relativisticcucumber This is super realistic. I saw one picture of ACM: He smiles exactly like that.

@ACuriousMind Thanks...i didn't know this.

ACuriousMind

(specifically chapter 5, "complete algebraic structure")

Silly Goose

09:17

now where is the proof of $1+1 =2$ stated...

ACuriousMind

@SillyGoose I believe the canonical references is the 360 page proof by Russell and Whitehead

Slereah

@ACuriousMind You can find a shorter proof with Peano

ACuriousMind

you don't say! :P

Slereah

\begin{eqnarray}
1 + 1 &=& 1 + s(0) = s(0 + 1) = s(1) = 2
\end{eqnarray}

Silly Goose

@ACuriousMind hate when they leave that out ! ;)

Slereah

09:23

I wonder what the original Peano paper looks like

It is old enough to be in Latin apparently

Unless that's just Italian

naturallyInconsistent

@ACuriousMind >✱54.43: "From this proposition it will follow, when arithmetical addition has been defined, that 1 + 1 = 2." – Volume I, 1st edition, p. 379 (p. 362 in 2nd edition; p. 360 in abridged version). (The proof is actually completed in Volume II, 1st edition, page 86, accompanied by the comment, "The above proposition is occasionally useful." They go on to say "It is used at least three times, in ✱113.66 and ✱120.123.472.")

Slereah

"Ioseph Peano"

Nope that's Latin

naturallyInconsistent

How do we make quotations here?

Slereah

Looks like the original Peano book doesn't have a successor operator

He just uses +1

So 1+1 = 2 is a definition

Mr. Feynman

@naturallyInconsistent Do you mean like this?

> Mr. Feynman

Mattia

09:26

Slereah

naturallyInconsistent

@Mr.Feynman yes

Mattia

@naturallyInconsistent I I mean something like that g_f in the image

naturallyInconsistent

@Mattia Looks like they are only interested in the energy-linear-momentum conservation and performed a spin sum. Like, evaluated the expectation value ignoring the spin degree of freedom by summing all possible spin outcomes.

>test

> test

arrghh space was the problem

asdfasdf > asdf

asdfe
> asdf

Sanjana

@ACuriousMind It's rare to see an original paper so much detailed.

Mattia

09:34

@naturallyInconsistent I dont see how that is allowed

In this way the cross section is bigger but I dont see the physical reason

ACuriousMind

@Sanjana I've found most of Haag's papers to be very detailed and well written; it is perhaps unsurprising that the guy who found Haag's theorem while most other people were busy mucking around with the interaction picture is a careful person :P

naturallyInconsistent

@Mattia It is not that crazy; in practice, we start with beams or whatnot, and they come in with all sorts of angular momentum. Results are only visible with statistics, and so it is really a spin-average of all possible inputs, which also means all possible outputs. This spin-multiplicity is just going to appear in the final statistical plot.

@ACuriousMind oh, that is very nice.

Mattia

@naturallyInconsistent But if they come in with random spins or with aligned spin the cross section should be the same I think

naturallyInconsistent

09:51

@Mattia If the input has aligned spins, then the spin averaged output is not correct. Random input spins, however, make the averaging of the spin averaged output correct. In any case, it is important to take the output spins into account.

Mr. Feynman

@Relativisticcucumber lmao great job

@naturallyInconsistent start the message with > and then space

Oh you'd figured it out

Mattia

@naturallyInconsistent but that seems more like a sum than an average .

Mr. Feynman

@Sanjana I'm starting to consider the option that someone engaged you to get rid of ACM

Mattia

It's like if the spins are not aligned then the cross section is bigget

And that is strange

bolbteppa

@Jagerber48 my answers have obviously freaked you out and are too much for you so I have refocused it on the bare minimum, just forget about the deeper representation theory underlying it and first get to grips with even needing rep theory in the first place which your question is hoping to avoid

Mr. Feynman

09:56

And now that I blew up your cover you'll kill us all

Alright people, listen here

I've just been to a geology department and they use rocks as doorstops

I want a collimated neutron beam to hold doors in my department now

Ryder Rude

10:25

@bolbteppa u can also add the previous answer in a separate question

ACuriousMind

> He had been fascinated by the articles about renormalization in the Physical Review, feeling like an archaeologist coming across the hieroglyphs. “Obviously these were messages from highly intelligent people, but what did they mean?”

@Mr.Feynman you might appreciate this quote from Haag about K. O. Friedrichs' reaction to renormalization

Ryder Rude

renormalization is just limits. idk y it was ever controversial

u can find physicists saying that renormalsation is making the theory agree with predictions in an ad-hoc way

Dirac said this. Feynman did too maybe

naturallyInconsistent

@Mr.Feynman our whole company just finally got extremely spooked out by neutrons. I am extremely sure you will want to have nothing to do with neutron beams if you know what it does.

Ryder Rude

even today, renormalisation is more infamous than Haag's theorem

Mr. Feynman

11:02

@ACuriousMind I need to share this everywhere :D

@naturallyInconsistent I mean, I think a sensible (for this purpose) neutron beam would pierce the door (and whoever crosses it)

naturallyInconsistent

@Mr.Feynman other than by having the distance be really long, you have no way to collimate the beam to only go one direction. i.e. if you have it in any building, the whole building will be peppered with it. which includes yourself.

fqq

@RyderRude because if you're doing perturbative qft you can ignore completely Haag's theorem and nothing goes wrong, while you have to deal with renormalization explicitly

Ryder Rude

11:31

@fqq oh. makes sense

@fqq "The hilbert space of the interacting qft is the renormalised hilbert space" what does this mean

what is a "renormalised hilbert space"

Ryder Rude

11:56

hello. boost takes $|p\rangle$ to $|\Lambda p\rangle$, but the norm of these two states is different, becuz $|p|=2\omega _{p) \delta (0)$. how is boost still unitary despite this

ACuriousMind

12:07

@RyderRude Their norm is not different, it is undefined due to the $\delta(0)$ in both cases.

Never try to make up clever contradictions with the non-normalizable states - you will always find some, but all you're showing is that they're not actually states/vectors inside the Hilbert space

Mr. Feynman

12:23

Is CPT a theorem or just good advice?

Ryder Rude

thanks.

so it is unitary for actual states

@Mr.Feynman it is a theorem. it can be derived from Wightman axioms I think

Mr. Feynman

I'm half-quoting Witten :P

Ryder Rude

@Mr.Feynman pls see the consequences section en.m.wikipedia.org/wiki/Wightman_axioms

Mr. Feynman

There was a clip of a lecture of his own in which somebody asked a question and he replied "Well, the CPT theorem is a theorem, not just good advice"

Can't find that clip atm

Ryder Rude

lol

Haag's theorem is just good advice

surprisingly, for a unitary transform, the basis vectors need not inner product to dirac delta

this is different from finite dimension linear algebra

this just happened to hold for the Fourier transform

Slereah

12:44

I'm glad that we decided to drop the habit of writing philosophical writing in poem form early on

naturallyInconsistent

What about in song form? Without the ability to write down the musical tones

Slereah

I know a few physics song although they're not that informative usually

Ryder Rude

$\psi (x) \rightarrow k\psi(kx)$ is also unitary, right?

Slereah

Except this one I guess : en.wikisource.org/wiki/There%27s_a_Delta_for_Every_Epsilon

Ryder Rude

the basis vectors of this transform are $k\delta (x-a)$

but it's unitary anyway

why does this happen?

Fourier transform's basis vector do normalise to delta

sorry this transform is : $\phi (x')= \int k \delta (kx'-x) \psi(x)$

so the basis vectors are $k \delta (kx'-x)$

ACuriousMind

12:56

@RyderRude again, the problem is that you're not looking at vectors inside the Hilbert space

Ryder Rude

i thought it was a big fact that this held for the Fourier transform basis

is this irrelevant

@ACuriousMind yes, for the actual vectors, norm is preserved

ACuriousMind

what needs to be invariant under unitary transform is not their ill-defined "product" with each other, but their resolutions of the identity. In the relativistic case, you have $1 = \int \mathrm{d}\Lambda_p \lvert 0\rangle\langle 0\rvert$, where by $\mathrm{d}\Lambda_p$ I mean the usual Lorentz-invariant measure, and this whole thing is invariant under Lorentz transformations

Ryder Rude

yes. so we recover $\delta (x-x')$ on the left anyway (what u write as $1$

but it's just that $|p\rangle \langle p|$ need not be delta for a unitary transform

the measure is what doesnt show up finite dimensional unitary transforms. so this is y the basis vectors always inner product to kronecker delta there

@ACuriousMind but this has an integration measure too

Ryder Rude

13:28

if $|p\rangle$ is non relativistically normalised, then boost sends this to $|p\rangle \rightarrow \frac{\sqrt {\omega_{p'}}}{\sqrt{\omega _p}}|p'\rangle$, right?

and this is unitary, which means : $\int |f(p)|^2 dp = \int |f(p')|^2 dp'$

we r using the non relativistic normalised states now

Ryder Rude

13:44

but now we just have the usual measure, so why is it justified that norm of $\frac{\sqrt {\omega_{p'}}}{\sqrt{\omega _p}}|p'\rangle$ is not delta?

becuz we r not using Lorentz invariant normalisation now, so we have the usual measure

naturallyInconsistent

Why don't you just look up any one of the so many good presentations on the subject?

Ryder Rude

they dont explain this

naturallyInconsistent

They do. You just don't realise it

Ryder Rude

they just say that we will work with relativistically normalised states becuz they transform nicely

naturallyInconsistent

This is an important ingredient, no matter if a textbook chooses to cover it using relativistic normalisation, or chooses non-relativistic normalisation, they would have to cover this topic.

ACuriousMind

13:54

@naturallyInconsistent I mean, Ryder is correct that most sources do not discuss this because it would draw attention to the strange nature of the non-normalizable $\lvert p\rangle$ they'd rather avoid (otherwise they might end up having to do actual operator theory :P).

naturallyInconsistent

@ACuriousMind but with a bit of thought, the relativistic normalisation is a trivial solution to this problem, and he even wrote down the correct equations showing that it would work out.

Ryder Rude

sorry im not sure what im confused about

Ryder Rude

14:09

ok so $f(p)---> f(\Lambda p)$ is at least a rep of boost becuz it obeys the algebra

but it's not a unitary rep yet becuz $\int |f(p)|^2dp \neq \int |f(p')|^2 dp'$ where $p'=Lambda p$

but we can change the definition of the norm by defining $\langle p|p'\rangle =2\omega _p \delta (p-p')$. now it's a unitary rep of boost

so far so good

im just not sure what im confused about

it's to do with this chat.stackexchange.com/transcript/message/64858607#64858607

naturallyInconsistent

If you take this relativisitic normalisation as the definition, you see that it is a unitary rep. The issue is that, even without the relativisitic normalisation, it already is a unitary rep. The argument is a little more delicate, but it is a unitary rep nonetheless.

Ryder Rude

yes. but why r the basis vectors not required to inner product to delta when we r using non relativistic normalisation

i get that unitarity is basis independent

naturallyInconsistent

Because that is not the arbiter of what constitutes unitary representations

Ryder Rude

14:42

ok so my premise is that a unitary transform maps an orthornormal basis to an orthornormal basis

but a boost doesnt do that. the non rel normalised $|p\rangle$ are orthornormal, but $\Lambda |p\rangle$ r non orthonormal

this means that my premise doesnt capture a general unitary transform

Jagerber48

@bolbteppa Your answers weren't "freaking me out". They just assumed tons of background knowledge I didn't have. I also didn't want a deep treatise on representation theory, at most, if the answer to my question is "no" (which should be clear in the answer body if so), I want a justification for why representation theory can't be avoided. Your revised/focused answer is much improved and much easier to understand, but I'll still need to spend time processing it.

Ryder Rude

so in general, a unitary transform only preserves : $\langle \psi|\phi \rangle$ where $\psi$ and $\phi$ are actual vectors in the Hilbert space

the premise is correct as long as the basis lives inside the Hilbert space

naturallyInconsistent

> non rel normalised |p⟩ are orthornormal

They are manifestly not

Ryder Rude

so if i checked the boost rep in the angjlar momentum basis, the boost wud map orthonormal to orthonormal

becuz these lige inside the hilbert space

@naturallyInconsistent i meant delta orthonormal

naturallyInconsistent

ACM kept pointing you to the fact: Dirac delta normalisation is not a proper normalisation, and that immediately screws with your premise.

Ryder Rude

14:57

becuz the angular momentum basis is actual functions like $f(p)$ whose norm is manifestly preserved using the normal formula

@naturallyInconsistent yes

naturallyInconsistent

If you want to pretend that

> a unitary transform maps an orthornormal basis to an orthornormal basis

then you have no choice but to accept SR normalisation as your lord and saviour

Ryder Rude

this premise holds for basis which actually live in the hilbert space

@naturallyInconsistent SR normalisation does not fulfill this premise either becuz they too r outside the hilbert space

and they r not even orthonormal even in the delta sense. they r only ortho

the non rel normalisation is orthornormal..but boosts just dont map delta normalised states to delta normalised states

naturallyInconsistent

Nobody is denying that eigenfunctions within the continuous spectrum live in Rigged Hilbert space and not Hilbert space. But you can pretend the premise holds by using SR normalisation

How many times must people tell you that you are wrong?

Ryder Rude

how on earth does the premise hold in SR normalisation

naturallyInconsistent

NR normalisation is not a satisfactory "normal" for you to claim that it is "orthonormal"

SR normalisation happens to be satisfactory enough to make the pretense ok

Ryder Rude

15:02

not really. it maps $2\omega _p \delta (0)$ to $2\omega _{\Lambda p}\delta (0)$

so the normalisation of SR normalised basis changes under boosts

the premise only holds for vectors inside the hilbert space

in principle, we can just avoid talking about these rigged vectors and their norm

naturallyInconsistent

If you are already taking SR normalisation, then $\omega_{\Lambda p}\delta()$ is the definition of what is correctly normalised. That is the price to pay for keeping up the pretense

Ryder Rude

oh. the formula of the normalisation remains similar, yes

but it's a different norm

naturallyInconsistent

So what? At least the scheme would work.

Ryder Rude

instead of saying that this rigged basis forms a rep of boosts , we shud just say that the Hilbert space of functions forms a rep of boosts, with the norm formula being the invariant measure one

naturallyInconsistent

Of course, it is preferable to simply use a better definition of unitary, but if you want to insist on that premise, then with these changes, you can keep it up

Ryder Rude

15:07

in principle, we can avoid these rigged vectors

@naturallyInconsistent i will keep it but i will pretend that the Hilbert space is only of functions, and not of these rigged vectors

i cant believe im speaking against rigged vectors

@naturallyInconsistent yes. the definition of unitarity is on functions

i shud kick these vectors out

thanks everyone

SR normalisation is great tho

Ryder Rude

15:32

i just realised everything is so simple if u dont talk about rigged vectors. instead of saying that $\int dp \frac{1}{2\omega _p} f(p) g*(p)$ comes from $\langle p|p'\rangle=2\omega p \delta (p-p')$, u just say that u start with a Hilbert space where the norm is defined to be this

and then unitary transforms just preserve this

what's important is that this norm is part of trying to find a unitary rep of Lorentz algebra. books dont emphasize this. a unitary rep is not just $f(p)\rightarrow f(\Lambda p)$

Mr. Feynman

15:53

@Slereah are you also triggered by "Lagrangean" or "Hermitean" or whatever?

Slereah

I find it a bit strange

Jan 20, 2018 at 23:35, by Slereah

Who writes "Lagrangean"

naturallyInconsistent

im actually more confused by the opposite. Their names are Lagrange and Hermite, so why not Lagrangean and Hermitean? Why Lagrangian and Hermitian?

Slereah

-ian

== English == === Alternative forms === -an, -n === Etymology === From Latin -iānus, which forms adjectives of belonging or origin from a noun. === Pronunciation === IPA(key): /iːən/ === Suffix === -ian (as an adjective) From, related to, or like. (as a noun) One from, belonging to, relating to, or like. (as a noun) Having a certain profession. ==== Usage notes ==== When males with a profession are distinguished from females, males are -ian, females -ienne. The plural is -ians (one magician, two magicians). When added to a word ending in a vowel, the infix -v- is inserted (Peruv...

naturallyInconsistent

@Slereah this says that an alternative form is -an

the amusing part is that the -an page gives an example: Rome + an -> Roman; the e is eaten. So then Lagrangan and Hermitan, omg worse

ACuriousMind

@naturallyInconsistent I think the point is that the e is silent in their native languages, and what we say is Lagrang-ian and Hermit-ian

What we say really is Lagrange + ian and Hermite + ian, not Lagrange + an, Hermite + an

naturallyInconsistent

16:10

now that makes so much more sense

Ryder Rude

i think the basis vectors of this $\psi (x)\rightarrow k\psi (kx)$ transform r delta orthonormal

Ryder Rude

17:46

lets say we choose a delta normalsied basis. then $\psi _x U_{xy} U*_{ya} \phi * _a = \psi _x \phi * _x$. repeated indices r integrated over using Lebesgue measure. then we must have $U_{xy} U*_{ya}= \delta _{xa}$

so this proves that the column vectors of any unitary transform in a delta orthonormal basis must be delta orthonormal?

so if a boost in non rel normalised basis is $|p\rangle ---> \sqrt{\frac{\omega _{p'}}{\omega _p} }|\Lambda p\rangle$, then y isnt the rhs delta orthonormal?

pls help

ACuriousMind

6 hours ago, by ACuriousMind

Never try to make up clever contradictions with the non-normalizable states - you will always find some, but all you're showing is that they're not actually states/vectors inside the Hilbert space

Ryder Rude

but the proof i gave is a property of $U_{xy} U*_{ya}$ in a delta normalised basis where U is unitary

ACuriousMind

there are no "proofs" that involve talking about $\langle x\vert y\rangle = \delta(x-y)$

mathematically that is just nonsense

Ryder Rude

oh

thanks

Mr. Feynman

18:04

@ACuriousMind controversial and non-rigorous as it is, isn't "nonsense" too much? :P

ACuriousMind

@Mr.Feynman if we're talking about proofs, I don't think so

it does its job where physicists want it to, the problem is there's no way to tell which manipulations on these objects lead to correct results and which don't if you don't already know the correct result

the rule of thumb is: This expression is fine if you only use it inside of resolutions of the identity, but you shouldn't obsess over its significance outside of that

Ryder Rude

18:32

whats crazy is that this $|p\rangle ---> \sqrt{\frac{\omega _{p'}}{\omega _p} }| p'\rangle$
also happens to be delta normalised becuz $\int \frac{\omega _{p'}}{\omega _p} \delta (p-p') \delta (p-p') dp=\delta (0)$

i didnt think that square root was gonna go away

but it just gets cancelled to 1 becuz p=p' in the integral

so the non rel normalisation works out idk

@naturallyInconsistent

if this is delta 0 before and after boost, how on earth is the rel normalisation changing the norm

Ryder Rude

19:03

@naturallyInconsistent sorry this calculation is wrong as i used the same variable for different things

Obliv

19:26

does it make sense to say light carries momentum?

Mr. Feynman

@Obliv yes

The classical electromagnetic field carries energy and momentum

Obliv

maybe this answer key is wrong, I wouldn't be surprised

Mr. Feynman

What?

Obliv

OH wait nvm

I read it wrong

Mr. Feynman

It is one of the cases where the quantum theory makes it easier. Think of it in terms of the single photons. A photon has momentum in compliance with $p=\hbar\omega/c$

Obliv

19:29

yes, I agree I just thought for some reason this question answer contradicted that

but it doesn't so we're good

Mr. Feynman

It is folklore that Feynman was very skilled at math. Why is it so? I'm not saying it is not the case but Feynman is someone that would always put physics first. What is an example of Feynman's mathematical prowess?

I don't think people refer to path integrals, do they?

Obliv

idk but i remember in an interview he said he'd go to sleep dreaming about integrals lol

Mr. Feynman

Yeah I know he liked calculus very much but... that's just calculus

No matter how good you are at it, I can't see anyone creating such folklore with calculus

Obliv

where do we get $E = \frac{p^2}{2m_e}$ from

Mr. Feynman

@Obliv that's the dispersion relation of a massive non relativistic particle

Obliv

19:34

i use this for determining momentum of E energy electron, but i'm not sure where it's derived

ACuriousMind

@Mr.Feynman is it?

Feynman is not who comes to mind when I think about "very skilled at math"

Ryder Rude

@Mr.Feynman he quickly solved some problem that mathematicians wer trying go solve

Mr. Feynman

@ACuriousMind well at least in a popsci sense I guess but I heard a couple of physicist say that

Obliv

$p = \frac{\sqrt{E^2-m^2c^4}}{c}$ does this reduce to that for an electron

Mr. Feynman

@ACuriousMind that's why I asked

@Obliv read my previous message

non relativistic

ACuriousMind

19:36

I think it's just that Feynman is very popular and so people will go "he was great at math" because they just don't know how good at calculations most physicists actually are :P

Ryder Rude

i forgot what the problem was

Obliv

@Mr.Feynman i do not understand your answer. Is there a way for me to understand this in layman terms?

Ryder Rude

@Obliv u can derive this from E=1/2mv2 and p=mv

ACuriousMind

@Obliv It's just normal kinetic energy

Obliv

can i use these equations to get to it?

Mr. Feynman

19:38

@Obliv the last formula you have written is valid in special relativity, the very first is non relativistic and you get it from the last Taylor expanding with $p/mc\ll1$

Ryder Rude

i think Feynman created that backwrd in time theory and also invented renormalisation. he was probably great at math

Obliv

i just don't get where this eq. comes from it's not in my book

Ryder Rude

he got a nobel prize for renormalsation, i think

ACuriousMind

@Obliv what do you mean, "it's not in my book"

Obliv

i used the "relativistic" expression and I don't think I got the correct answer.. i'm not sure

ACuriousMind

19:40

it's just ordinary non-relativistic kinetic energy

Mr. Feynman

@RyderRude I would call that great at physics though

Obliv

so if it doesn't specify relativistic, we assume non-relativistic?

Mr. Feynman

What about Dirac?

Ryder Rude

yeah. he was great at physics math, u can say. the math that invovles lots of physical intuition

ACuriousMind

@Obliv depends on the situation, but using the classical expression, I end up at this electron moving with 0.6% of the speed of light, so why would you use relativistic expressions :P

Ryder Rude

19:45

@Mr.Feynman csnt say for sure. again he seems to be great with physics math

Von Neumann was great at mathematician math

Mr. Feynman

I would like to be more dismissive of math and just care about the physics :P

I feel like that would be the right thing to do

But I keep feeling uneasy if I don't learn math properly and abstractly

Ryder Rude

but ive no idea how to separate these two skills. they overlap too much

@Mr.Feynman this, however, can be separated frm physics

many good physicists cant do rigorous math

no doubt Feynman didnt care about learning "proper" math

ACuriousMind

you really need to stop guessing things that can easily be looked up

Feynman started as a math major

Ryder Rude

oh

it says he experimented a lot with math before college

inventing half derivatives and stuff

Mr. Feynman

20:01

@ACuriousMind true and then he switched to engineering :P

And then physics of course

user587860

20:31

@ACuriousMind This might be too broad to ask, but what kind of research qualifies to be published in Physical Review Letters?

ACuriousMind

@Supersymmetry why are you asking me? I'm not working as a physicist and I've never published a paper

user587860

Hmm, I thought you could have possibly published a paper in such journals. That's why I tried to ask

Mr. Feynman

@ACuriousMind you have no proof that is true

user587860

Haha, indeed

Slereah

For the GR book I have to prove that the Earth is round

Mr. Feynman

20:45

Use group contractions

Ryder Rude

one guy used shadows to prove it

Slereah

The shadow only prove that it is a section of a sphere at best

Ryder Rude

apparently, a flat model can also satisfy those shadows

it is a model with a tiny sun close to the earth which goes over the earth like a flashlight

this video has that model youtu.be/VNqNnUJVcVs?si=gt9uJ7v134yaXwif

Silly Goose

@Supersymmetry you could take a look at PRL and judge yourself ;)

Ryder Rude

21:02

rigged vectors behave so terribly with boosts

Charlie

My god I didn't realise that was a vsauce video and the start made me jump

Ryder Rude

lol

Obliv

@ACuriousMind so in general the relativistic momentum is larger?

user587860

22:06

@SillyGoose Yes, I am aware how high quality such research papers are of.

user587860

@SillyGoose But could you provide piece of advice for someone who's aiming to get published in such a journal? I know that as an undergraduate student, my priority should be to absorb the material, but it's my dream to publish in a prestigious journal.

Sanjana

@Supersymmetry I have a friend who has publications in PRD...Contrary to popular belief you can actually suggest the peer reviewers and hopefully the editor will listen to your suggestion; otherwise the editor will carefully choose a "trustable" anonymous referee...either way it depends on the referee. The friend's supervisor and the referee exchanged hundreds of mails over a matter of months to convince each other of the arguments presented in the paper, and finally it got published!

But imagine, if the referee wasn't convinced at the end after all that hard work...

user587860

22:36

@Sanjana Thank you very much. However, what do you suggest in particular for coming up with work that is of sufficiently high quality to get published in PRD? Of course, there's no magic formula, but I am curious how one can, for instance, self-assess which questions the community is interested in, and what possible attempts currently exist to answer those questions. Which is why you need an advisor working with you, and which I lack, to be honest.

ekardnam_

@Mr.Feynman that is a terrible idea imho

The learning just the physics thing I mean

Sanjana

@Supersymmetry :64860624 High quality? idk...I can say what Edward Witten told my friend over an email...My friend asked Witten how he gave so many groundbreaking and new ideas...He told that most of his ideas were not "new" but buried in other's ideas, he just thought about them deeply---so he advised to think deeply about what you love.

@Supersymmetry As for the other questions: Review articles often summarize where the community stands, and what needs to be done. Most papers also contain conclusion/outlook section...there are people here also who can tell what are the various problems.

ekardnam_

22:53

@Sanjana that's such a great piece of advice

From Witten

Sanjana

Also, apparently Witten likes Polchinski's book on string theory more than his one or the modern books like that of BBS :p

ekardnam_

I don't think it makes sense to try to publish a paper without an advisor and as an undergrad, you should focus on studying and doing grad school, then you will publish

I never read Witten's book, however I read somewhere they basically never even mention conformal field theory in there (maybe at the time of writing it wasn't standard to do so idk) so Polchinski maybe is more modern?

Sanjana

I agree with ekardnam_ but the prob. is sometimes grad. schools prefer students with a paper :( Not in my country though---they would conduct entrance exam, take one or two interview where they will ask hard questions and then select you...

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