The h Bar

Daniel Underwood

02:06

You can walk through Maxwell's house on a virtual tour: clerkmaxwellfoundation.org/india-street/housetour.html

PM 2Ring

02:50

> An astronomical system positing that the Earth, Moon, Sun, and planets revolve around an unseen "Central Fire" was developed in the fifth century BC and has been attributed to the Pythagorean philosopher Philolaus. en.wikipedia.org/wiki/Pythagorean_astronomical_system

naturallyInconsistent

04:32

@Obliv It is the standard $\sin\vartheta\approx\tan\vartheta=\frac xf$ at work here.

@Supersymmetry why not? If you want people to entertain your ideas, bring evidence, and then we can have a conversation over them.

Ryder Rude

09:52

hello

the last expression should hav $\frac{1}{2\omega _p}$ instead

where is this wrong

Ryder Rude

11:46

@Mr.Feynman hi. do u know the answer to above

Mr. Feynman

12:16

@RyderRude why?

Ryder Rude

@Mr.Feynman becuz i thought the purpose of this normalisation was to have a manifestly lorentz invariant inner product which shud hav a $\frac{1}{2\omega _p} dp$ which is a Lorentz scalar

Mr. Feynman

However, there is also something convention dependent here. Typically one writes $$\lvert\psi\rangle=\int\frac{d^3p}{(2\pi)^3\sqrt{2E_p}}\tilde{\psi}(p)\lvert p\rangle$$

user587860

@naturallyInconsistent No, certain things are not up to debate and there's no entertainment of ideas. This is because we're doing science. No theory gets credit until eventually we perform uncountably many experiments, and so it is proven to be compatible with what we observe in reality.

Ryder Rude

@Mr.Feynman is the inner product with this normalisation not suposd to be $\int dp \frac{1}{2\omega _p) |\psi (p)|^2$

user587860

@naturallyInconsistent Some facts that we already know are so successful, and has a so rich history that there's no need for foolish attempts to re-question their validity in any regime.

Mr. Feynman

12:24

I'm not saying this solves your problem. I'm saying that your problem should be about those factors absorbed inside coefficients

I mean, because of this it is not manifest

Ryder Rude

oh. so the inner product with this normalisation is not manifestly Lorentz invariant if we dont write these absorbed coefficients?

here's my entire thought: if we want the inner product in some basis to be :$\int dp \frac{1}{2\omega _p} |\psi (p)|^2$, where $\psi (p)$ is the full component in this basis with everything absorbed, then we require $\langle p|p'\rangle =\frac{1}{2\omega _p} \delta (p-p')$

this shud b correct, right?

Mr. Feynman

@RyderRude that's what I think the issue is, yes

@RyderRude your last equation is not LI

Ryder Rude

oh. so that normalisation is bad

but the inner product i get in this basis has a good form

so books r not using this basis. they r using Lorentz invariant normalisation,right? @Mr.Feynman

so their inner product is not the one I wrote. it is the one with $2\omega _p dp$

bolbteppa

From $f(0) = \int f(x) \delta(x-0) dx = \int <f|x><x|0>$, if we set
$$f(0) = \int \frac{1}{N} f(x) N \delta(x) dx = \int \frac{1}{N} f(x)<x|0> dx = \int <f|x><x|0> \frac{dx}{N} $$
with $<x|0>=N\delta(x-0)$ then we have to choose $\frac{dx}{N}$ as our integration volume, thus e.g. $I|y> = \int \frac{dx}{N} |x><x|y> = \int \frac{dx}{N} |x> N \delta(x-y) = |y>$ is consistent. Now $<\psi|\psi> = \int \frac{dx}{N} <f|x><x|f> = \int \frac{dx}{N} \psi^*(x)\psi(x)$

Mr. Feynman

12:39

Also, if you look look at the definition I used for $\psi$ or the one you used, the measure is not invariant in either case, so the invariance of the coefficients is a problem too, I would say

bolbteppa

You are technically doing this all the time in QFT when you change the integration volume to be '1 particle per unit phase volume' which is $\frac{V d^3 p}{(2 \pi)^3}$ (with $h^3 = (2 \pi)^3$)

Ryder Rude

@Mr.Feynman yes, but if psi transforms, then it kills the whole point of manifest Lorentz invariance. with the choice of the alternate normalisation i gave, im getting manifest lorentz invariance where $\psi(p)$ transforms like a scalar

Mr. Feynman

Your original choice had measure $d^3p$, which is not invariant

Ryder Rude

im getting $d^3p \frac{1}{2\omega _p}$ tho

in the other normalisation

the book normalisation is giving $dp 2\omega _p$. this wud mean psi wud hav to transform too

Mr. Feynman

Ok you have it in your final formula but not in the initial one (and it is fine, just pointing out there is no difference in that)

Ryder Rude

12:49

yes

one thing is that book nornalisation also gives $ I =\int \frac{1}{2\omega _p} |p\rangle \langle p|

this has a lorentz invariant measure

but the inner product is screwed up

@bolbteppa is this final inner product assuming $\langle x|x\rangle =N\delta (x-x')$

@bolbteppa yes. u have written $<x|0>=N\delta$

@bolbteppa the derivation i wrote disagrees with this tho. i assumed $<x|x'>=N\delta$ and got a $Ndp$ in the inner product

@bolbteppa can u see whats wrong

bolbteppa

@RyderRude Where is the $1/N^2$, i.e. $\frac{dp dp'}{N^2}$

Ryder Rude

why wud that show up

bolbteppa

$<f|f> = <f|I I|f> = <f| \int \frac{dp}{N}|p><p| \int \frac{dp'}{N} |p'><p'|f>$

(I wrote $f(0) = \int f(x)\delta(x-0) dx$ above, but you can choose do this in momentum space instead, $\hat{f}(0) = \int \hat{f}(p)\delta(p-0)dp =...$ etc...)

Ryder Rude

oh

so we have to write $\psi = \int \frac{dp}{N} \psi (p) |p\rangle$

yes. this is the identity relation

thankss

wow this was unexpectes. im not uses to non-orthonormal bases

Ryder Rude

13:20

i thought we were using the actual basis components but we r using $\psi (p):=\langle p|\psi \rangle$

this is equal to the basis component only in delta normalised bases (or unity normalsied in finite dimensions). so it is a confusing definition of $\psi (p)$

Ryder Rude

13:36

@Mr.Feynman u were correct. the basis component transforms too. only after we take 1/N out of it does it take a manifestly lorentz invariant form

Ryder Rude

15:18

i think $\psi (p):= \langle p|\psi\rangle is the better object to work with than the basis component

nickbros123

I was reading zangwill, modern electrodynamics, 1st chapter, where he introduces the "time derivative of a flux integral over a surface", and the surface is assumed to be time dependent, $S(t)$. My question is, if the surface in question itself changes over a time period, how does the formula $\delta [\iint_{S(t)} \vec B \cdot \hat n dS]=\iint_{S(t)}\delta \vec B\cdot\hat ndS+\iint_{S(t)}\vec B\cdot\delta(\hat ndS) $ represent the "change in" the flux integral between $t \to t+\delta t$

Ryder Rude

@nickbros123 whats the problem with the formula

bolbteppa

15:38

The variation of the surface integral of $B$ over some surface is (via the product rule) the integral of the varied field over the original surface and the original field over the varied surface. The integral of $B$ over the varied surface is the integral of $B$ over two nearby surfaces at $t$ and $t+\delta t$, which can be re-written as a volume integral over a volume with these as two of the sides of a 'cylinder' like object, minus the surface integral over the sides joining those two surfaces

nickbros123

@bolbteppa , sorry, is there a (doesnt even hav to be rigorous) proof of the product rule here, because im not able to see how the product rule manifests like this, as in, there is a sort of complicated dependence of the surface integral on the surface $S(t)$. when one speaks of variation

it could be my misunderstanding of notation, although

bolbteppa

$\delta [B(t) dS(t)] = B(t+\delta t) dS(t+\delta t) - B(t) dS(t) = [B(t+\delta t) - B(t)] dS(t + \delta t) + B(t)[dS(t + \delta t) - dS(t)] = \delta B dS + B \delta dS$

nickbros123

oh, right

@bolbteppa, one more clarification: when u say $dS(t+\delta t)$ you are accounting for here, the surface itself at $t+\delta t$ ?

bolbteppa

15:55

Yes

nickbros123

@bolbteppa ah, makes sense now- thanks

its just i never really "rigorously" understood the definition of a surface integral beyond the superficial "limit of the sum the vector field, dotted with the normal, summed over the surface approximated by polyhedra"

bolbteppa

That's all there is to it, only now you're considering the surface element to vary with time

eigenvalue

16:24

Hey there!
we know per definition: M is globally hyperbolic if and only if for every $p,q\in M$ the 'causal diamonds' $J^+(p)\cap J^-(q)$ are compact.

Are the causal diamonds nonempty? or which condition/constraint guarantees that the causal diamonds are nonempty?

Slereah

17:43

The empty set is compact

Just pick for instance $J^+(p) \cap J^-(q)$ with $p$ to the future of $q$

The two light cones will be disjoint and therefore the intersection is empty

[in Minkowski space]

Minkowski space being very mild

Obliv

18:22

Can anyone shed their thoughts on this question before it inevitably gets marked for close?

nvm disregard I figured it out

i forgot about the equivalence of OPL through a lens

user587860

How does shifting the boundary conditions of a string affect the mass spectrum?

Obliv

19:29

is a fourier transform of a function just a function * a function

user587860

By shifting the boundary conditions, I do not refer to the Lorentz transformation $X^\mu (\sigma,\tau) \to \Lambda^{\mu}_{\nu}X^\nu (\sigma,\tau)$, as Lorentzian symmetry is a symmetry of the theory $S[\gamma, X] = \frac{T}{2}\int \mathrm{d}^{2}\sigma{\sqrt{-\gamma}}\gamma^{ab}\partial _{a}X^{\mu}(\sigma)\partial_{b}X^{\nu }(\sigma)$. On the other hand, the mass spectrum will depend on the vibrational modes $\tilde{a}_{\mu}, \tilde{a}_{\mu}$. Which are determined by the boundary conditions

user587860

But there are probably local actions on the worldsheet ${X^{\mu](\sigma, \tau)}^{D-1}_{\mu = 1}$ that won't leave the boundary conditions invariant (like BCC operators), and so mass spectrum changes.

Obliv

if ACM were here, he'd be able to answer your questions. Looks like he isn't :P

your ] should be a } after \mu

@Supersymmetry

user587860

Yeah, that's a typo :)

Obliv

oh it's already too late to change it, oh well.

user587860

19:35

But it seems too late to correct it

user587860

@Obliv But my idea to approach to your question would be that the point of Fourier tranformation of a function is to take your function, and return another complex-valued function that depends on frequency

Obliv

on frequency? are you sure it's not on position within the slit?

user587860

@Obliv Oh, I am answering your question in general :D

user587860

I've not read the text you've sent

user587860

Fourier transform

In physics, engineering and mathematics, the Fourier transform (FT) is an integral transform that converts a function into a form that describes the frequencies present in the original function. The output of the transform is a complex-valued function of frequency. The term Fourier transform refers to both this complex-valued function and the mathematical operation. When a distinction needs to be made the Fourier transform is sometimes called the frequency domain representation of the original function. The Fourier transform is analogous to decomposing the sound of a musical chord into th...

Obliv

19:38

oh ok, yeah this is for single slit far field diffraction. it just mentioned in case the slit had some non uniform field strength

hmm.. i'm not sure i understand what they mean by "describes the frequencies present in the original function"

like for example $Ae^{i(kx - \omega t)}$ in this situation of slit diffraction I assumed maybe we used another function for phase difference like $Ae^{i(kx-\omega t)}e^{ik\Delta}$ where $\Delta$ could be a function ?

or maybe $A$ itself depends on some parameters

idk I shouldn't get ahead of myself anyway

ill read the section on FT

user587860

@Obliv I think you can think of Fourier transform just as a change of basis, change of basis to frequency-domain. Whenever you have a function f that has a periodicity $2\pi$ for instance, this means that you can express it as $f(x) = \sum_{n=-\infty}^\infty c_ne^{inx}$

Obliv

ooh ok

user587860

19:57

(We obviously want f to be integrable over [-\pi, \pi] for c_n to make sense, to be more precise)

user587860

Then you can view Fourier transform as a more general (continuous) representation as an integral of exponentials (rather than discrete series). Which you get by taking the period $T$ to $\infty$ (hence its limits do change)

Jagerber48

Next vector/spinor question. My picture for orbital angular momentum L in an atom is this. The atom has motion around the nucleus which has some angular momentum L. Classically, the angular momentum could be represented by a vector. But when we "quantize" the system we find the state of the system is described as being in a state space which breaks down into irreducible representation of SO(3) labeled by quantum number \ell.

But note, it is possible for the system to be in a superposition of states of different \ell.

Since, classically, L is a vector, there is no problem to rehash this whole story for two electrons. There are angular momenta L_1 and L_2, we add them up to get total L. the available states of the system are actually the same, the Hilbert space breaks into a countably infinite set of irreducible representations labeled by \ell

Only in the L_1 + L_2 case we have the option to express the state using labels from the individual L1/L2 states, or the total L states (using Clebsch-Gordon coefficients).

By analogy, I want to tell a similar story for spinors (This is the story I want to tell but I don't know if it's viable). Something like, we have a spinor object that can be understood as classical mathematical (if not physical) object. When we look at the quantum version of this object, we see that the states are broken up into a countably infinite set of irreducible representations of the Spin(3) group labeled by quantum number s

Similarly we can add up two of these spinors to get S_1 + S_2 = S

using Clebsch-Gordon coefficients again

But I think this story breaks down. The difference that I see is that, in the angular momentum case, the available Hilbert space includes superpositions of states of different \ell, whereas in the spinor case, particles DO NOT have access to a countably infinite set of irreducible representations. They instead only have access to, I guess what we call, the fundamental representation.

What can be said about this difference? Why does a particle have the ability to occupy a state with any quantum number \ell, but s must be fixed to small values (possibly always just 1/2?)

Ryder Rude

22:06

I think this is just decided by experiments

Jagerber48

22:32

I guess the analogy then is not with a general vector, but with a vector whose length is fixed for some reason.

Jagerber48

22:48

I need to learn what the fundamental representation is. But this story I'm describing is annoying. If I'm forced into representation I'd like for there to be parallel narratives between vectors and spinors. Something like: When we look at a quantum mechanical vector, the state space is the space of countably many irreducible representations of SO(3). When we look at a quantum mechanical spin the state space is the space of countably many irreducible representations of Spin(3).

But the latest sentence seems wrong. It seems like in the later case the answer is that the state space is the SINGLE space that is THE UNIQUE fundamental representation of Spin(3). << Not sure if I'm getting this sentence correct or not.

ekardnam_

23:11

@Jagerber48 think about this: spherical harmonics are representations of $SO(3)$ on $\mathrm{L}^2(S^2)$, but not of $SU(2)$

the action on a function $f$ of $\mathrm{L}^2(S^2)$ of an $\mathrm{SO}(3)$ element $R$ being $R f(x) = f(R^{-1}x)$ and the action on $S^2$ the obvious one

Obliv

so a traveling wave of the form $y=f(x\pm vt)$ must obey $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2 y}{\partial t^2}$

to determine if a function of x,t represents a traveling wave it should be in that form or obey the d.e

ekardnam_

i'll leave this to you just to avoid saying wrong stuff because i dont remember precisely. but there is a difference between integer and half integer spin when you consider infinite dimensional representations basically

Obliv

but some waves are approximate solutions to $\nabla^2 \psi = \frac{1}{v^2}\frac{\partial^2 \psi}{\partial t^2}$

ekardnam_

> Why does a particle have the ability to occupy a state with any quantum number \ell, but s must be fixed to small values (possibly always just 1/2?)

this is another nice question

the thing is two states with different $s$ are different particles

but for this you would actually need to study the correct definition of particle first, a la Wigner

Obliv

why are spherical waves given as $\psi = (\frac{A}{r})e^{i(kr-\omega t)}$, like why isn't $A$ constant?

is this a physically motivated definition

ekardnam_

23:21

yeah

they spread over a surface from a point

the intensity has to go like $1/r^2$

Obliv

even for mechanical waves, not considering EM?

ekardnam_

that dependence is more like conservation of energy really

so yes also for mechanical waves

> they spread over a surface from a point

they spread over a spherical surface of increasing radius

Obliv

so the general harmonic complex waveform is just without the inverse $r$ dependence then?

ekardnam_

if its a spherical wave you usually would have the inverse r

Obliv

is there some similar $\frac{1}{r}$ dependence for plane waves?

ekardnam_

23:25

no for plane waves there is not

Obliv

hmm ok i will keep reading

ekardnam_

because "they are sourced from a plane and spread over planes of constant area"

Jagerber48

23:47

@ekardnam_ are there two fundamental “particles” that are the same in every way except s? If so why not consider them different states of the same spinor field? If the E field has 1 or 2 units of orbits angular momentum we don’t say it’s a different particle. It’s just a different state of the EM (photon) field

The EM/photon field is a vector field.

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