The h Bar

imbAF

00:32

Can anyone give me good argument as to why the point x, is in position (0,1,0) in k-space ?
https://en.wikipedia.org/wiki/Brillouin_zone#/media/File:Brillouin_Zone_(1st,_FCC).svg

qwerty

01:31

@ACuriousMind Actually all my undergrad textbooks are kinda in agreement on this point. I think I must have picked up the obsolete terminology off lecture notes/ the lecturers even though the books all made this point, because I seem to recall it (and transformation rules in general) being discussed quite emphatically

PM 2Ring

@qwerty If someone claims to be an experimental cosmologist, be suspicious. Or very afraid.

qwerty

@PM2Ring lol. yes. I think the counterpart is "observational" in astro. but people who just run a lot of simulations seem to have things in common with people who run experiments too.

PM 2Ring

Right. Big simulations are very similar to real-world experiments, but less messy. But of course, they can only "reflect" your model back to you, they can't give you novel real-world data.

qwerty

@PM2Ring yeap

PM 2Ring

I have lots of fun exploring the celestial mechanics of the Solar System via JPL Horizons. ssd.jpl.nasa.gov/horizons It spans 20,000 years. And it's fitted to a huge amount of ground- and space-based observations, so it's almost like observing the Real World.

But you'll never discover a new comet looking through Horizons ;)

OTOH, you may discover neat new patterns by exploring the 1.4 million or so asteroids in their database.

Relativisticcucumber

01:47

@qwerty hopefully next year i will be a baby experimentalist

PM 2Ring

A few former mods were experimentalists. At least, they did stuff with particle beams, etc, while earning their degrees.

Relativisticcucumber

i saw a poster at my uni and the first citation was emilio :D

but i think emilio is not a mod right? just mega user

PM 2Ring

And of course Anna V, one of our most prolific members, was an experimentalist before she retired. But she never visits chat.

qwerty

@ACuriousMind I've just typed up my understanding of what you wrote here and I think I'm getting your point at last :)

Relativisticcucumber

@PM2Ring oh also john rennie right?

but chemistry iirc

PM 2Ring

01:51

@Relativisticcucumber Not a mod. Definitely an experimentalist.

qwerty

@Relativisticcucumber eee exciting :)

PM 2Ring

@Relativisticcucumber Yes, John was a chemist. But certainly working with stuff like colloids, where physical properties are very important.

Anna V is one of the most senior active members on the whole network. She's in her mid 80s, and still pretty sharp.

63

A: Congratulations to anna v, our second 200k user

Thank you, Rob, it has been a nice trip these ten years. I was forced to retire from research in 2000, because of civil service rules, and did not want to stay on (with various grants) because money for research in Greece was, and still is, not enough for high energy experiments. That require a n...

qwerty

@PM2Ring googling her comment on white cells just has me even more confused lol

PM 2Ring

02:08

@qwerty Hey, she's a physicist, not a biologist. ;) But yeah, brains have both grey and white neurons.

qwerty

think it could be this but no idea if the source is in any way reliable innovations-report.com/studies-and-analyses/report-39198

PM 2Ring

Could be...

Growing Mind

03:54

Hello Everyone...

Michael

04:15

Hi @GrowingMind

Growing Mind

Hi @Michael

Michael

What do I do when 2 parts of the same equation have differing unit superscripts? Like a/b, a is 3 cm s^-2 and b I'd 5 cm ^2 s -2

Is* not I'd

Please email [email protected]

I'm going to bed

qwerty

04:31

@Michael en.m.wikipedia.org/wiki/Dimensional_analysis#Simple_cases

no one is going to email...

naturallyInconsistent

05:01

@PM2Ring holy F mad respect. She's incredibly lucid!

@qwerty note that ACM never claimed that his way of looking at things is the conventional terminology for it. Those of us who agree with ACM clearly see that the terminology is obsolete, but that is still the norm, and basically the only way to discuss these things, bad as it is, on both the physics side and the maths side.

qwerty

@naturallyInconsistent right. it's just that Schutz explicitly said it was obsolete, and Hartle also didn't really use contra /co very much, so I think I must have gotten it from lecturers or other notes...

naturallyInconsistent

@SillyGoose Think of it as a Venn diagram. Just like how the Riemann integral can integrate some stuff that Lebesgue integral cannot and vice versa, due to the possibility of the Legendre transform not working over the entire phase space, there are Lagrangians that dont give rise to a good Hamiltonian and Hamiltonians that dont give rise to a good Lagranigan. You can always still find another way to describe them, though, hence all that constrained Hamiltonian talk. @Arjun

qwerty

ooo maybe it was an older book like d'inverno.

naturallyInconsistent

@qwerty but even ACM's stuff is somewhat confusing. For example, he would say raising v.s. lowering indices, but if you push for what the raised indices v.s. lowered indices mean, it would be contravariant v.s. covariant indices. Same thing with vector v.s. covector bases; there simply is a physical thing we are trying to talk about, and so it is necessary to think about them and have a good way to express things in terms of them.

Lucky Chouhan

@qwerty I don't know what to say but alphabets make sense only when we use them rather telling the story of their origin. People read and talk using alphabets without actually realizing how much we rely on them for every letter but still we enjoy. It is not like math or physics where we first should have solid foundation, here none cares how much you polish the banister.

@naturallyInconsistent have you celebrated your birthday?

naturallyInconsistent

05:16

@Arjun ACM had quite a few posts on the main site about this and you should check the hbar transcript for the link to them. position-velocity space is not identify-able with position-momentum space, until you impose the Legendre transform's conditions on them, which might fail. You are correct, however, to note that the usual L=T-V prescription is lost when we transition to Hamiltonian-first, but as ACM noted, that not actually physics in itself. A prescription is just a way for you to guess

Under the Hamiltonian-first method, you would just have another guessing prescription to make

@LuckyChouhan yes

Lucky Chouhan

@naturallyInconsistent you turned __ ? :P

naturallyInconsistent

@LuckyChouhan for appropriate substitution into that variable, yes :P

qwerty

@LuckyChouhan do runes not make sense if no one uses them anymore? what about bone oracle script? it's just a different topic that's all. alphabets aren't the content of what you're trying to convey with them but they are interesting as a separate topic

naturallyInconsistent

RobWords presented a good case that English really made more sense if we go back to using runes for it.

Lucky Chouhan

@naturallyInconsistent yeah, we want to find the value (or approximate value) of that variable. :')

naturallyInconsistent

05:19

@LuckyChouhan That value is known and it needs to be smaller

Lucky Chouhan

@qwerty yeah but here Rob wanted to say something totally different rather than saying anything about runes and linguists. Anyway,

@naturallyInconsistent It is hard to beat Miao Miao~~

qwerty

@naturallyInconsistent yes I haven't gotten to the bottom of it all. it's just one of those things where I've gotten comfortable-ish with the abstract mathematical objects and separately with index gymnastics over time, but pushing back to the fundamentals I will need to brush up/re-learn it to see if I have a better perspective

naturallyInconsistent

@imbAF A serious problem with maths-first approaches to pedagogy is that the introduction of topics can be jarring as you felt. Things just dropped out of the sky out of nowhere. Instead, what really happened was that Euler-Lagrange found calculus of variations to be a great mathematical tool, and then discovered that if we wrote L=T-V, then N2L (for constant mass) can be recast as EL equations. That made solving a lot of irritating mechanics problems way too easy.

Lucky Chouhan

@qwerty as you yesterday said something like this "I've got some energy to do physics today.." here so don't you study physics daily?

naturallyInconsistent

@LuckyChouhan some days are for fun

qwerty

05:24

@LuckyChouhan I've finished my studies :p

naturallyInconsistent

@qwerty there is no need to "see if". Definitely will have a better perspective. Miao miao was lucky to have learnt it correctly the first time around.

No thanks to the standard curriculum, though. Soooo bad

Lucky Chouhan

@naturallyInconsistent once Niels Abel said about Gauss writing style "“He is like the fox, who effaces his tracks in the sand with his tail."

@qwerty I see, may I ask so do you teach at university?

naturallyInconsistent

@Mr.Feynman This is sooooooo important. I really don't understand why some people would elect to hide indices when the point of confusion is literally coming from their omission. Also, I also want to learn how van der Waerden notation spinor indices would recast these operations as. Anyway, stars

@qwerty A while back, ACM pointed out that the only way to properly prove these things is via the Coleman-Mandula theorem. Otherwise, I really think that there is no possibility to prove such things: it is more about having tried the basic assumption out in a mathematical way and showing agreement between experiment and the theoretical assumption. See ACM's answer to that linked question.

@LuckyChouhan Which I think is very bad. It gives all the historians a headache, at least. Of course, present findings in a logical manner, but leave crumbs as a separate log if the presentation ends up being wildly divergent from chronological order.

Lucky Chouhan

@naturallyInconsistent do you think concepts make sense when they are introduced after stating the origin of them here origin means not to preach the whole history but just a problem which motivated people to make the concept to use for future reference and solve the problem in systematic way?

naturallyInconsistent

@LuckyChouhan Most of the time, yes. But there are cases where it does not, e.g. quantum, and symplectic

Lucky Chouhan

05:36

@naturallyInconsistent I don't know anything about quantum and symplectic :(

naturallyInconsistent

@imbAF this is very widely covered. It is correct and you should just derive it on your own. The Lagrangian is a sum of quadratics; you have one term coming from left neighbour and the other term coming from right neighbour.

@LuckyChouhan if you want to stay in physics chat, it would help to learn some quantum.

@qwerty lol

@imbAF arbitrary definitions do not require explanation.

Lucky Chouhan

@naturallyInconsistent sure, I need to learn. Have you ever taken Olympiad exams?

naturallyInconsistent

@LuckyChouhan once or twice, just to produce answer key for students. Too difficult and not very illuminating.

qwerty

@naturallyInconsistent I will never forget this one question where you had to know about the *bulk modulus *

first and last time I had to learn that in physics

naturallyInconsistent

@qwerty how can that be? When in uni physics you had to learn about waves and acoustics, they would point out that in sound waves the relevant quantities included bulk modulus

Lucky Chouhan

05:42

@naturallyInconsistent I find problems of IMO interesting, but IOI problems doesn't appeal to me, I like programming but not the type of problem which are asked in IOI.

qwerty

@naturallyInconsistent it honestly never came up! it seemed a bit niche

maybe if you were into modelling that seriously

Lucky Chouhan

What motivated you to take "naturally Inconsistent" as your username? There is a user on MSE and his username is "MathStackExchange is very bad" lol, there are many people with interesting usernames :)

naturallyInconsistent

@qwerty It is just a addendum kind of knowledge that is just suitable for exam MCQ. Like, you learnt in deriving the wave equation in the first go that you have tension in the string and mass density along the string, and then you have springs with stress and strain going with Young's modulus. Once you are happy with Young's modulus, bulk modulus is just plug and chug symbol swapping.

@LuckyChouhan I explained this before. To be only as inconsistent as Nature forces us to, but not much more inconsistent.

Lucky Chouhan

"To be only as inconsistent as Nature forces us to, but not much more inconsistent." ~ Miao Miao 2024. Thank you,

qwerty

@naturallyInconsistent oh it was a long form big question in the Olympiad. I was a cheesed off teenager as I hadn't heard of it nor did I think of it :p

naturallyInconsistent

05:54

@qwerty thats also why I think it is not always productive to do Olympiad. They often tend to just be exclusionary.

The time can be much better spent upon the standard curriculum

But of course, there is always people who get fun out of competitions, and they can just continue on their own pace

Arjun

06:33

@naturallyInconsistent When you say good Lagrangians/Hamiltonians are you hinting at them not being single-valued?(and or not being differentiable/continuous etc)

naturallyInconsistent

@Arjun Well, take the trivial example. If you try the naïve covariant Lagrangian where time is also a variable with a velocity and your integration parameter is closer to proper time, then the resultant covariant Hamiltonian is identically zero. Clearly not good.

Arjun

@naturallyInconsistent Hmm,Are there cases where good Lagrangians give bad Hamiltonians when taken a Legendre transform of?

naturallyInconsistent

But that covariant Lagrangian is a good Lagrangian

Arjun

@naturallyInconsistent Oh,sorry I haven't studied str :(

naturallyInconsistent

I'm not sure where it would fit into your standard curriculum. Usually analytical mechanics

Arjun

06:43

I have it next sem

@naturallyInconsistent In such cases,do we have no other way of writing the Hamiltonian,i.e not simply taking the Legendre transform of the Lagrangian,to get the correct eom's ?

naturallyInconsistent

Part of the issue is that we want the Lagrangian to manifestly have the Lorentz invariance, because then we can omit a lot of cross-checks. But a Hamiltonian never will have the Lorentz invariance, so there is no good reason to pick a covariant Hamiltonian in the first place. We can just take the covariant Lagrangian, check out what EoM and other entities it gives us, and pick a Hamiltonian that gives us the correct dynamics and work with that.

Otherwise, we can also pick to do the constrained Hamiltonian route, but that is arcane magic that miao miao don't understand. Luckily, most of physics does not need it.

Ryder Rude

@MoreAnonymous i meant that non local hidden variable theories r non sense cuz one would have to write action at a distance lagrangians that r also consistent with relativity

QM non locality is fine cuz the theory has lorentz invariant and local actions

it is all consistent with relativity

Growing Mind

07:02

Hi @RyderRude

Ryder Rude

@GrowingMind hi

Mr. Feynman

08:16

@ACuriousMind so you're an ACM now

ACuriousMind

@qwerty Oh, I'm not (sadly) not saying the terminology is obsolete, many people still use it and seem to think it's important. I just don't agree :P

Mr. Feynman

@naturallyInconsistent I think there was some SE answer talking about physicists having double standards and hiding only the spinor indices :P

qwerty

@ACuriousMind I think I vaguely recall mathematicians also seem to talk about "the connection" more often than the "covariant derivative" as well...

also, eh, we're all still learning from books written in the 70s mostly, so it makes sense.

twas indeed d'inverno's terminology I followed

ACuriousMind

08:34

@qwerty Depends on the context - a covariant derivative and a connection are two sides of the same coin (depending on the definition of connection you start with the equivalence is more or less easy to arrive at)

naturallyInconsistent

@ACuriousMind well, I think you should. Once we find a better way to do things, we should push the improvement out

ACuriousMind

@naturallyInconsistent Uh, I think you misunderstood me - I would very much like the terminologyto change. But it's not yet obsolete since the majority of people still use it and there is no general consensus that that should change.

naturallyInconsistent

lol we are arguing due down to definition again

qwerty

@ACuriousMind I think someone once told me they're synonymous but I never looked into it. From wiki "This new derivative – the Levi-Civita connection – was covariant in the sense that it satisfied Riemann's requirement that objects in geometry should be independent of their description in a particular coordinate system. " If I had to guess, you'd disagree with the historical reasons for the name, but because it's a well-defined mathematical operator you're fine with the name?

Mr. Feynman

I don't like thinking of them as the same thing, although in practice they're all equivalent :P

I like to think of them in a hierarchical way: connection->covariant derivative-->parallel transport

Slereah

08:53

also "connection" and "covariant derivative" isn't necessarily the same thing

Connection is often meant as the field of distributions over the bundle, not the derivative

qwerty

ok. I will complain to the resident mathematician who provided this claim eyes

Slereah

the two being related by $$\nabla_X s = \pi_V (ds(X))$$

Mr. Feynman

@Slereah That's not the definition I was using, but it doesn't change much. Even in the with a less advanced definition, a connection is a map $\nabla: \mathcal{E}(M)\times\mathcal{T}(M)\to\mathcal{E}(M)$ that maps $(s, X)\to \nabla_X s$

Slereah

Where $\pi_V$ is the vertical projection

Mr. Feynman

The connection is the map; the covariant derivative is the image of a section through this map. Saying that the connection is the covariant derivative is no different from saying that $f(x)$ is a function

Slereah

08:59

I mean they're "the same thing" in the same sense that vector fields and derivatives are the same thing :p

Mr. Feynman

Yeah

I had seen a nice diagram of it, let me see if I can find it

Can't find it, but I could find this. It's from a set of notes by Schuller

(parallel transport and covariant derivative are reversed with respect to mine, but the context is a bit different since he's concerned with principal bundles and I was concerned with vector bundles)

qwerty

iiiinteresting

Mr. Feynman

Incidentally, the notes are from a course named The geometric anatomy of theoretical physics

Claudio

DANG, that's a good name ngl

Mr. Feynman

I'm wondering just now why I never watched the course. I've been too much of a purist, wanting to read books on my own and look where it got me :P

qwerty

09:08

that's a very nice name

@Mr.Feynman what are you doing your MSc on btw?

Mr. Feynman

@qwerty As you can easily guess I'm working on detectors and extremely practical stuff I will not mention (of course not because this is a blatant lie)

qwerty

ahhh the air of secrecy implies the Manhattan project

of course mr feynman

Mr. Feynman

Theoretical physics

My master's is not focused but I am (was) into HEP

qwerty

nice :)

Mr. Feynman

What was you MSc on?

qwerty

09:14

MSc's aren't popular in australia. we mostly follow the british model (3 +1 years undergrad including honours research), then phd.

Mr. Feynman

Oh, I see. So are you PhDing now?

qwerty

I finished a short while ago, cosmology

Mr. Feynman

Ah, I see. Good stuff :D

Damn, I sounded like Jesse Pinkman on that one

ACuriousMind

09:47

@qwerty Yes. "Covariant derivative" as a technical term is fine, acting as if the "covariant" alone means anything important is not :P

qwerty

@ACuriousMind if you were to go back in history a la xkcd.com/567 , would you rename it? and don't say the "ACM derivative"... :P

ACuriousMind

10:05

@qwerty just to have consistency I think "connective derivative" would be fine?

Slereah

10:38

@ACuriousMind Isn't "covariant" just physicist for "equivariant" :p

Ryder Rude

@Mr.Feynman one can also develop another hierarchy starting from the metric

these ideas are all inter-related

in LQG, they get rid of the metric and quantise something like the "frame field", which is supposed to be a set of basis vectors at different points of spacetime

this idea is also related to having a connection

as the Christoffel symbols describe how the basis vectors change

what LQG quantises is something dual to the frame field. i think it should've been an element of SO(3) but they make it SU(2) without no justification, which is extremely ad-hoc

it's their classical theory that has this SU(2) valued field, even tho GR has no complex numbers

ACuriousMind

10:56

@Slereah "Equivariant derivative" is also fine with me; you're right that in this case that's what's meant, but my problem with "covariant" is that it also has other meanings (the "equivariant" meaning is not same as when it's used in contrast to "contravariant")

Slereah

you could probably write a whole book on what is meant by "covariant" really

ACuriousMind

I have no desire for that kind of suffering :P

qwerty

@Slereah I'm sure such a book would somehow annoy and bore everyone to death in equal measure

Slereah

That was Polanyi's whole thing about personal knowledge rly

That a lot of science isn't some written down rigorous knowledge but some vague standards that good scientists know but do not typically write down

qwerty

ah yes the good ol "I know it when I see it" definitions

oh I had no idea of the history of that phrase en.m.wikipedia.org/wiki/I_know_it_when_I_see_it

Slereah

11:09

Typically the actual rigorous thing tends to be written somewhere but it is usually in some boring paper that nobody has ever read :p

qwerty

@Slereah :( such papers need more citations

Slereah

11:53

One such thing being the notion of a Nice Function

Physics mostly runs on Nice Functions and Nice Spaces but what those are is a bit vague typically

Nice enough so that all the math mistakes are correct

Mr. Feynman

@Slereah please write it

I have a thing for wasting time on completely inconsequential stuff :P

Silly Goose

12:11

i am confused by the continuity equation in tensor notation. we have that $\partial_\mu j^\mu = 0$. Expanded, this is $-\partial_0 j^0 + \sum_i \partial_i j^i$. Seemingly this says $-\partial_t \rho + \nabla \cdot \vec{j} = 0$, which has an incorrect relative minus sign. I am wondering where I am going wrong.

ACuriousMind

12:22

@SillyGoose Your expansion is wrong: where is the - in front of $\partial_0 j^0$ supposed to come from? The summation convention just means you sum, i.e. it's $\partial_\mu j^\mu = \partial_0 j^0 + \sum_i \partial_i j^i$

Silly Goose

bleb i think i made this mistake before ._.

Slereah

@Mr.Feynman sites.pitt.edu/~jdnorton/papers/decades.pdf

Mr. Feynman

12:46

@SillyGoose if it makes you feel any better, I do that once a month

Ryder Rude

12:56

Child's Play or Hellraiser?

don't spoil any

i will pick one

2

original movies

Michael

13:44

What do I do when 2 parts of the same equation have differing unit superscripts? Like a/b, a is 3 cm s^-2 and b is 5 cm ^2 s -2

ACuriousMind

@Michael "a/b" is not an equation, just an expression, and what is supposed to be the problem here?

naturallyInconsistent

14:35

I already picked a lull in the rain to go out, and it was correctly a lull, but then suddenly the rain and wind was sooooo big

Then at the gym halfway through a massage suddenly the wind was so strong that all the glass windows shook. Everybody was shocked. Maybe except miao miao.

On the way back there was no rain, just droplets blown off the top of buildings and trees. A tree was blown down and crashed on top of a fire hydrant and that meant a spray. The wind was so great that it blew dry my closed umbrella.

Madder

15:44

I am not finding a realiable source where they define a boltzmann weight

Michael

@ACuriousMind

Ah, posting to imgbb one sec

ibb.co/ZzFShqx

Michael

15:59

Please help sir

@ACuriousMind

ACuriousMind

@Michael please don't call me 'sir', and you still haven't explained what the question is, you just posted a screenshot of some text without any other explanation

Michael

Sorry

I'm getting Reynolds number of 1250 to 12500 instead of 10^3 to 10^4

I think it's because of the difference in units, cm and cm^2

@ACuriousMind

ACuriousMind

given that that's an "approximate" sign there, I don't see the problem - you understand that 10^3 is 1000 and 10^4 is 10000, right? that's what you got, just rounded to less significant digits

Michael

Ok, I was just making sure I did the math right

Nairit Sahoo

17:11

I got to know that $\frac{d \langle x \rangle}{dt}=\int J d^3x$ where $J=\frac{i \hbar}{2m}(\psi \nabla \psi*- \mathrm{h.c.})$ is the current density. But I always thought $\vec{j}=\rho \vec{v}$ (atleast this is true in the free particle case and many slowly varying potential cases) I don't see any $\rho$ in the above equation... instead a volume integral?

I know that volume integral over the 0th component of a 4-current gives you the charge. But here the integral is over the spatial components: what does that give you?

ACuriousMind

@NairitSahoo That there is the quantum probability current, not a classical electric current.

Nairit Sahoo

Okay so there is no reason apriori to believe that $j =\rho v$ is gonna be true in every case

@ACuriousMind But is there a way to "obviously" see that the volume integral of the spatial components of current density gives you the $\frac{d \langle x \rangle}{dt}$?

ACuriousMind

@NairitSahoo Did whatever source you learned this from not derive it?

Nairit Sahoo

@ACuriousMind Yes I see the derivation

ACuriousMind

then what is the question?

it's not a particularly complicated derivation

Nairit Sahoo

17:20

Yes it is easy. I am trying to relate this with the notion of "integrals of currents give you charge" picture. Okay, I guess I want to know that what does the spatial components of the current density give you in general?

@ACuriousMind I mean the temporal component give you the charge itself. The spatial components vanish after div. theorem is applied usually as I have seen

But here it doesn't?

ACuriousMind

this volume integral of a current does not usually appear in classical theories

Ryder Rude

@NairitSahoo hi. $\frac{d<x>}{dt} = \frac{d}{dt} \int x \rho (x,t) dx= \int x \frac{d}{dt} \rho dx = - \int x \frac{d}{dx} J dx = \int J dx$ (using integration by parts)

this is for 1 space dimension

Nairit Sahoo

@RyderRude Yeah that's the calculation which I did. But thanks.

Ryder Rude

@NairitSahoo also, there is no prior notion of $v$ in QM. One can take this equation as the definition of $v$

but one has to note that $v$ doesn't refer to the velocity field of a fluid of particles in QM

Nairit Sahoo

17:36

btw another question. When we discuss time evolution of wave packets. We go to Fourier space and then use the $e^{-i Et/ \hbar}$. I thought that we do this because we know how the time evolution operator works on free particle energy eigenstates which are just plane waves. So we expand our wavefunction in terms of the plane waves which is just doing Fourier transform

But my question is: if we are dealing with wavepackets for non-free particles, then we decompose $\Psi=c_n \psi_n$ and then apply the time evol. operator on the $psi_n$ the energy eigenstates for the particular system, right?

We don't do fourier transform in this case, right?

ACuriousMind

I mean you usually don't call them "wavepackets" in that case, but yes

Nairit Sahoo

Is wavepackets usually referred to only in the free particle case?

ACuriousMind

It's less about the "free" part than about the Fourier part - the name "wavepacket" is because you're looking at something of the form $\int f(p) \mathrm{e}^{\mathrm{i}xp}$, i.e. a superposition of plane waves. You can do this regardless of whether something is free or not (it might just be less useful), but "wavepacket" implies you're doing that and not some other decomposition

also "wavepacket" usually implies you're specifically thinking about $f(p)$ that lead to "localized" wavefunction, i.e. something that can conceivably be called a "packet"

Nairit Sahoo

Makes sense. Thanks.

imbAF

18:23

@ACuriousMind I wanted to ask you something regarding the bloch function. In my lecture the solutions are written as $\psi_{\vec k,\nu}(\vec r)$

And $\vec k$ is the band number, a quantum number

I thought of it as being a representation of the wave vector

is that wrong?

The professor said so, and 3 weeks of lecture I had the opposite impression

Silly Goose

consider some distribution with vanishing monopole $q = 0$ but nonzero dipole $\vec{p} \neq \vec{0}$.

Here, we have that $\vec{p}$ is origin independent (passive view) or equivalently invariant under translating the charge distribution around (active view)

this seems to mean that I can then move my dipole anywhere in space and write the dipole field as emanating from that point. however, this seems physically wrong. am i misunderstanding?

like if I have some original coordinate system $S$ with my charge distribution and with some reference object devoid of any electrostatic moments, say an ideal chair.

It seems like I should not be able to move my dipole anywhere I want and say that the field is now emanating from that point.

I guess would we precisely say to leading order actually I can make such a modification. However, the difference is encoded in higher order moments, which in general change due to translating the distributions?

Even so, say I am working with ideal dipoles. Then, there will be no higher order moments. Then, the sort of physical paradox still stands.

ACuriousMind

@imbAF I'm afraid I don't know what the phrase "representation of the wave vector" means. $\vec k$ just is the wave vector of these Bloch waves.

imbAF

but it is not according to him

I will elaborate

because my whole understanding has ben turned upside down

And I need someone to clarify this to me, as I have homework related to this.

If yuu could give me 3 min

Nairit Sahoo

Last question for today: I know there is a lot of discussion on this and I have been reading some. But can somebody just give me a final concise answer to: What is the momentum possibly non-normalizable eigen"state" for particle in a box? (The plane waves are not in the domain of $p$ because it doesn't satisfy the b.c.s for the problem which are $\psi(0)=\psi(L)=0$)

ACuriousMind

@SillyGoose I don't understand this claim - why does it "seem you should not be able to move your dipole"?

@NairitSahoo It's the plane waves. The generalized potentially non-normalizable eigenstates never lie "in the domain" of the operator since non-normalizable functions are not even elements of the Hilbert space.

Nairit Sahoo

18:37

@ACuriousMind While discussing these "eigenstates" which don't belong to the usual Hilbert space, we don't care about the boundary conditions?

Silly Goose

ACuriousMind

@imbAF either you have misunderstood your professor or he's using non-standard terminology. The $k$ is a wave vector, since $\psi_k(r) = \mathrm{e}^{\mathrm{i}kr}u(r)$, it appears exactly in the function of what we usually call a wave vector here.

Silly Goose

here I have a chair as the reference object described above.

imbAF

@ACuriousMind I specifically asked him and he said not, the $\vec k$ is quantum number, related to the bands

I will go to the short version and if you need additional info I can tell you. But please I would like to have your insight about this matter

Silly Goose

physically, $\vec{p}$ and the simply translated $\vec{p}'$ yield distinct dipole fields even though I think they should not

maybe I am making an illogical jump. Does origin independence of the dipole moment of a neutral charge distribution imply that I can equivalently translate that distribution without changing its dipole moment?

ACuriousMind

18:40

@imbAF again, this is either a misunderstanding or non-standard terminology. E.g. the Wiki article also calls $k$ a wave vector without any qualifications. But I don't understand why you consider this question important in the first place, it's just a name.

imbAF

You will understand

Silly Goose

@imbAF your professor might be referring to the fact that $\vec{k}$ is "crystal momentum" and not associated with the actual momentum to do with the momentum operator $\hat{p}$

imbAF

I am associating it with energy, since it's wave vector

But you will see the big picture

ACuriousMind

@imbAF ah, that's the mistake; yes, it is a wave vector in the formal sense of the word, but it is not (at least not generally) directly related to kinematic momentum or energy

@SillyGoose Again, I don't follow you at all here - why do you think they should not?

obviously you have realized this is a physically absurd claim, but you haven't really explained why you think that in the first place

imbAF

@ACuriousMind How, see let me tell you

We started with the example of having a crystal and bombarding it with some particles to which we attach a wave vector

and assuming single collision and elastic collisions, the wave vector just changes direction not value

Silly Goose

18:45

@ACuriousMind hm well there is a problem i was doing yesterday which asks something along the lines of what is the leading multipole term. for instance, put a dipole $p\hat{x}$ a distance $d$ to the right of a conducting surface parallel to the $y$ axis. you can solve this by method of images. so the image + real charge distribution is a dipole $p \hat{x}$ at $x = -d$ and a dipole $p \hat{x}$ at $ x = d$.

imbAF

then we consider $\Delta \vec k$ inside the crystal and the conditions that need to be satisfied for diffraction to occur

So one of the conditions is the brillouin one

Silly Goose

okay then the first question is whether the dipole moment of this real + image distribution vanishes.

imbAF

$\vec k \vec G=\frac{1}{2}|\vec G|^2$

in here $\vec k$ is clearly the wavevector

we did substitute $\Delta \vec k \rightarrow \vec k$

Silly Goose

naively, one thinks that they can freely move each individual dipole due to the origin independence argument and so then the net dipole moment is just $2p \hat{x}$

imbAF

So now, how is that not related to energy?

Silly Goose

18:47

in von Laue scattering that $\vec{k}$ is associated to plane waves of incident (and reflected) light, no? Not to the Bloch electrons in the crystal.

ACuriousMind

@imbAF I don't understand what anything you said is supposed to have to do with the Bloch eigenstates from Bloch's theorem

imbAF

Because

ACuriousMind

those aren't states for some particles/waves you scatter off a crystal, they're eigenstates for the crystal Hamiltonian, usually for the electrons in the lattice or whatever

just because two things are both denoted with the symbol $\vec k$ in different contexts does not mean they have anything to do with each other

imbAF

sure but

one of the conditions that we make use in derivation of the expression for the eigenstates of the translation operator is that $\vec K_i \vec A_j=2\pi\delta_{ij}$

This condition was also present when we considered $\vec k \vec G=\frac{1}{2}|\vec G|^2$

ACuriousMind

@SillyGoose I don't know what you mean here - of course the total dipole moment will change if you move the dipoles relative to each other!

but if you only have a single dipole, you can move it anyway and the total field just remains the field of a single dipole

Silly Goose

18:51

@ACuriousMind hm, that is confusing. i was talking to another student yesterday and they were describing doing this. maybe i misunderstood.

ACuriousMind

this isn't different from you being able to move a single charge anywhere you want and the total electric field just "originating" from that charge but when you have more than one charge you can't just freely move them around

Silly Goose

right if you wanted to leave things physically unchanged you need to translate every object in space

imbAF

If $\vec k$ is not a wavevector in the sense that it is tied to the energy etc, what is it then?

ACuriousMind

@imbAF Again, just because you use similar things in two different contexts doesn't mean they have to be the same

Silly Goose

$\vec{k}$ in the bloch functions do label energy eigenstates (together with the band index). The $\vec{k}$ in von Laue scattering labels the wave number of incident light that you are shooting your crystal with.

imbAF

18:52

I mean that isn't

ACuriousMind

@imbAF because it "looks like momentum" it's usually called crystal momentum. See the Wiki article for some interpretations

imbAF

easy to consider

@ACuriousMind and it is still not related to the energy?

even though, you just called it crystal momentum ?

Silly Goose

you can see the former statement by plugging in the bloch function (without labels) into the Schrödinger equation. you then get a family of schrödinger equations for the periodic function $u$ parameterized by $\vec{k}$. each of these equations will have multiple solutions, which we index by the band index $n$.

ACuriousMind

since it labels energy eigenstates of course it is somewhat "related to the energy"

but it's not ordinary momentum

imbAF

So, for a fixed k-value

ACuriousMind

18:54

you need to accept that we sometimes have mathematical expressions that look very similar to each other whose physical significance is very different :P

imbAF

Ok I will accept this thing

But for a fixed $\vec k$ value, you can vary $\nu$ value

then, according to what the professor said today

for a fixed $\vec k$ you have different energy eigenstates/values, just like when you consider a particle in a potential well

where you have discrete values

Then this leads to the clear statement that $\vec k$ is not energy, as he said and as you are saying

then $\nu$ is the index to show different energy eigenvalues, right?

Silly Goose

if you are talking about bloch functions, the pair $(n, \vec{k})$ labels an energy eigenvalue.

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