The h Bar

Relativisticcucumber

00:00

@HerrFeinmann when will u change ur name back? it disturbs me

00:25

@Relativisticcucumber Well, the relevance is this: The EM wave that we are used to in vacuum is transverse, no doubt about that, but in a waveguide, you can have it just be a metallic box-like thingy with vacuum in it. Intuition says that we should just have transverse waves in them too, but instead, it earns an extra longitudinal part in some modes. If you cannot be sure that waves in vacuum are transverse, you cannot be sure that it will also be so in materials.

imbAF

I was trying to calculate the Hamiltonian for the free scalar field by using the hamiltonian density and something takes place depending on the variables I choose.
The Hamiltonian is $H=\int \frac{1}{2}(\dot \phi^2 + (\nabla \phi)^2+m^2\phi^2)$.
If I choose the arguments to be $x^\mu$ and $k^\mu$ I can't get the desired solution of the type $\omega (a^\dagger a + a a^\dagger)$.
But If I consider $x^\mu$ and $k^'mu$ and $k^\mu'$ everything works

Why is that?

Relativisticcucumber

00:43

@naturallyInconsistent every time i start to think i understand smth XD

hm ok i will look into that / think ab that

one of my profs also told me smth amazing today -- i hope it's true. he said there can be instances where a material completely changes it's lattice type and that this is some kind of phase transition. i was really curious about it

naturallyInconsistent

@Relativisticcucumber yeah, it hurts so much

@Relativisticcucumber oh yes and it is quite interesting too. But you should have seen it back in first year thermal physics, the kind of stuff that they introduce the CO_2 phase diagram with critical point, triple point, and then many ice phases

Relativisticcucumber

i never took a thermal physics 0.o maybe like general chemistry?

i did take stat mech tho -- that was a bad time but im ready to go ham on it next semester now that i wont be forced to do schroeder

i think this thing of interest is maybe even smth that carbon does w diamond and graphite?

if you wait a really really long time maybe? i am very curious about this process

Silly Goose

01:00

why is the most general solution not integrating over $\omega$ and making the amplitudes $\omega$-dependent?

oh the cucumber answered my question i think

naturallyInconsistent

01:18

@Relativisticcucumber yes, it is incredibly slow for diamonds in RTP to convert to graphite. So negligibly slow that we wont live to see any appreciable conversion

@SillyGoose $\omega$ and $\vec k$ need to satisfy the dispersion relation, which in this case fixes the velocity, both group and phase, to be the spacetime conversion factor. You can do it in two ways; this way with one less integral (because, really, you have one less degree of freedom to integrate over, precisely because of the dispersion relation), or if you want to make it manifestly Lorentz invariant, integrate 3+1D with a Dirac delta enforcing the dispersion relation.

Silly Goose

@naturallyInconsistent ah i see

danke

Sir Cumference

01:35

You know, it's really hitting me how much rarer snow has become where I live

Like it's mid december and we're only at 15°C here. I feel like it definitely would've been freezing by now when I was a kid

Not sure if it's actually a result of climate change or e.g. confirmation bias though

Anyone else notice something similar tho?

naturallyInconsistent

02:30

Climate change is not just about it getting hotter, it is also about getting erratic. When I was a kid, the climate pattern is extremely regular, and we can count on having drought in summer and flooding in winter. Last decade, we had a flooding in summer that was the first recurrence of having flooding in the city for 30 years. It is beyond ridiculous.

qwerty

we had random very hot days/spells (over 30°C) at the end of winter this year

then went cold again

very much not "normal"

Silly Goose

i randomly found this differential geometry book on my laptop "Smooth Manifolds and Observables" by Nestruev. Anyone ever heard of it?

qwerty

02:53

no, unusual title?

Aditya Krishna Panickar

Doubt regarding general theorems of 1D Potentials and how particle on a circle seems to violate it...

For 1D potentials, degenerate energy levels do not exist for normalizable energy eigenstates(i.e. bound states). The fact proved was quite general and was proved assuming only that the wavefunctin vanish at infinity.

However, this result does not apply no particle on a circle. Why so?

For a particle on a circle, what does it mean to vanish at infinity exactly? The real line itself is wrapped around, so what does it mean to vanish at inf, because what exaclty infinity here?

Ok! We are here describing the real number line as the circle, with x+L equivalent to x. But, still, how do we solve the main dilemma

Ok, if we are "projecting" (sort of) the particle on a circle case to the real line, how exactly do we distinguish a wave in the clockwise and counterclockwise direction(they are moving definitely, as they are also momentum eigenstates! )

Herr Feinmann

09:14

@Relativisticcucumber I tried some time ago but it said that I can change once a month

Probably the month has expired but I'm too lazy for that

naturallyInconsistent

mewww

(gtg)

Herr Feinmann

@naturallyInconsistent nI, the RPA is driving me crazy 😭

Tobias Fünke

morning

@HerrFeinmann Random Phase Approximation?

nooo, keep the name. it is perfect

Arjun

09:31

@AdityaKrishnaPanickar For non-degeneracy in 1D(Real potentials)it is not necessary that all wavefunctions vanish at infinity,it is sufficient(although not necessary) if all wavefunctions vanish at a given point on the real line.Now for a particle on a circle this is not necessarily true,all that is known is $\psi(x+L)=\psi(x)$ ,this condition by itself does not guarantee non-degeneracy.

@AdityaKrishnaPanickar x lies in (0,L] what is infinity here?

Herr Feinmann

@TobiasFünke Yes

I'm reading about an elementary version of it but the digrams don't really make sense for me

Arjun

09:47

@AdityaKrishnaPanickar That's trivial to see no? You have two eigenstates for each energy E(>0),when you "unwrap" it you'll have $e^{\pm ikx}$,k>0,the natural choice would be the plus one going along positive x and minus along the negative x,now regarding clockwise or counter clockwise when you wrap it into a circle is immaterial imo,since it will change based on which way you view the circle,if you look at it one way and it is say,clockwise you can flip it and it will change to counter-clockwise.

Tobias Fünke

10:22

@HerrFeinmann in which context?

Herr Feinmann

@TobiasFünke It's actually nothing big. It is the computation of the linear response to and external potential, causing the effective potential screening. My source seems to deal with it using Hartree approximation but I think there are diagrams missing

I think that some also call this Lindhard dielectric function

Tobias Fünke

I see

Herr Feinmann

It's probably too vague like this, isn't it?

Tobias Fünke

;d a bit, but I could guess the context

Which source do you study? Is it research or course work?

Herr Feinmann

Course work :P
Source is Sec. 3.8 of Rickayzen, *Green's Functions and Condensed Matter*

Tobias Fünke

10:37

aha

Herr Feinmann

This is the thing that I'm not really sure about, figure (b). The Hartree approximation that I know looks like

Tobias Fünke

I don't have the book right now at hand, but IIRC I've skimmed through it

Herr Feinmann

Tobias Fünke

hmhm I don't know by heart. But are you sure both are really the Hartree approximations (and not e.g. one Hartree and the other Hartree-Fock or so)?

Herr Feinmann

My only guess is that we don't make that resummation because we only want first order stuff, but it's a bit of a leap in the dark

@TobiasFünke Yes, because the book states it

The Fock approximation would involve the self-interaction diagram too

Aditya Krishna Panickar

11:25

ok, that makes sense. There is no vanishing condition here, and the proof of no degenracy requires vanishing functions and bounded derivatives. Makes sense.

In general 1D potentials, At least for normalizable(bound) states, functions definately vanish at infinity, which sufficently ensures non-degenracy

Arjun

@AdityaKrishnaPanickar yupp non-degeneracy is not necessarily true and I'm not entirely sure of what you meant in your second question

Aditya Krishna Panickar

But for the circle, we do not consider the entire real line anyways, so normalizability doesn't guarantee vanishing conditions.

Arjun

Entire real line?Could you be more precise for a particle in a circle it can only be in the interval (0,L]

Aditya Krishna Panickar

My 2nd question is ignorable

Arjun

ok

Aditya Krishna Panickar

11:33

What I meant is precisely that, since we only consider [0,L). So normalizability does not guarntee vanishing.

Unlike "true" 1D potentials, where wavefunction extend to infinity, so, normalizability guarntees vanishing conditions.

Arjun

@AdityaKrishnaPanickar normalizability doesen't guarantee vanishing at infinity,physical sensibility guarantees it.There could be non-vanishing wavefunctions in the L^2 Hilbert space.

Aditya Krishna Panickar

Oh, did not know that.

Arjun

@AdityaKrishnaPanickar physics.stackexchange.com/a/332040/304878 See this

Aditya Krishna Panickar

Thanks. Greatful for not having dumb pathologies

in physics

BTW, this is totally unrelated question regarding the Casmir Effect...

Arjun

what is that? I'm sorry I didn't study it yet!

Aditya Krishna Panickar

11:49

Well, the fact that Brownian motion and the ability of matter to have thermal properties(temp), was suggestive of matter being atomic, or having microscopic degrees of freedom.

The Casmir effect also seems to be an analogue of brownian motion, and the fact that spacetime also has "temp". Could this hint toward the fact that spacetime will also have microscopic degrees of freedom, or "atoms" of spacetime so to speak

This is heavily inspired by T.Padmanabhan's interview. I swear I absolutely will not turn to HEP or Cosmology, but his interview was captivating for sure.

Interview with Thanu Padmanabhan

►

Arjun

Oh! I've watched that interview

Aditya Krishna Panickar

Lolol

ACuriousMind

@AdityaKrishnaPanickar the real Casimir effect has little to do with the vacuum, see this answer of mine

Aditya Krishna Panickar

Thank you @ACuriousMind

Arjun

@ACuriousMind Hi,does one need to know qft to understand it?

Herr Feinmann

13:09

I don't think ACM will like that question :P

Slereah

why do you want to understand QFT without knowing it

Herr Feinmann

13:57

@Slereah I think Arjun just means if you need some kind of prior knowledge to understand it

Slereah

Plenty

Reading

important skill

knowing numbers

probably other things

I guess I should link the thing

webspace.science.uu.nl/~hooft101/theorist.html

Herr Feinmann

14:25

Prior knowledge of QFT itself...

It's a self-consitency question

Slereah

How would you understand it without knowing it

Herr Feinmann

14:42

I don't know

Slereah

Is it like a zen riddle

Herr Feinmann

Just self consistent

Silly Goose

15:28

When does one usually learn about renormalization group?

and why is it referred to with “group”? There is a stack answer which states that it is somehow relates to a semigroup but even then is not a group.

Shashaank

Hello Guys

Could someone explain Schrodinger's cat thought experiment?

I don't understand how will the cat be both alive and dead, and how is the atom a mix of both stable and emitting states?

Arjun

15:42

@Slereah lmao I didn't realise my question could be interpreted that way lol,I meant if one needs to know qft to understand casimir effect and his answer on the same XD

naturallyInconsistent

@SillyGoose after learning renormalisation, and never a moment earlier.

@SillyGoose seriously, this is the abuse of language from physicists that you want to die on?

@Shashaank this isn't the problem. How much else of physics you have learnt, however, is.

(actually, more like how much maths)

Silly Goose

@naturallyInconsistent even though the names semi-group and group are similar, they are (in my opinion) in a physics context extremely different, no? Inverses and the identity are quite central ideas when talking about physical transformations in usual contexts.

so i am not sure how one (or why one) would ever conflate the word "group" with "semi-group"

naturallyInconsistent

I have not denied your correctness. In fact, I'm specifically trying to tell you that it is yet another abuse of language; why, after all that you have already been through, would you not be expecting abuses of language?

Silly Goose

well more like i don't understand how one could have ever stared at a semi-group and called it a group

unless the naming was a political move to make it more catchy because what physicists know what a semi-group is (other than open system theorists maybe)

naturallyInconsistent

@SillyGoose It is extremely obvious that that particular semigroup does not have inverses, and so it cannot possibly cause confusion in practical use, and on top of that, semigroup is such a mouthful

Relativisticcucumber

15:55

@ACuriousMind do you know what is the purpose of the rule that you can only change ur username once a month?

@Arjun im just curious -- what does this mean

i have zero knowledge to provide regarding qft lol

naturallyInconsistent

@Relativisticcucumber well, did you read ACM's answer? Especially the link to arXiv notes by Jaffe?

ACuriousMind

16:12

@Relativisticcucumber see the discussion at meta.stackexchange.com/q/29966/263383

Shashaank

16:25

@naturallyInconsistent I didn't learn much of quantum physics, I know classical physics though and I am trying to learn Modern Physics, I know concepts like superposition though, yet I am not able to understand how will a a quantum particle emit and not emit radiation at the same time?!

naturallyInconsistent

@Shashaank how much linear algebra do you know?

Shashaank

pretty much, I had taken an intermediate course on Linear Algebra

I know subspaces, nullspace and dimension and some more concepts too

naturallyInconsistent

If you already know about eigenvectors, then you can go straight to Feynman lectures volume 3 and flip directly to the chapter on the Ammonia Maser

Shashaank

But I actually am a chemical Engineer major, so I don't know much of physics, but I wanna learn...

@naturallyInconsistent yeah I will look into it

naturallyInconsistent

you wont actually need to know much

you will, however, need the postulate: the result of an actual measurement with an operator, that can be made to appear as a matrix, will only ever be one of the eigenvalues of the matrix.

and when you measure an eigenvalue, the state of the quantum system is then given by the eigenvector if the eigenvector is unique.

Shashaank

16:31

Is there no other way to explain superposition ( say I wanted to explain these concepts in a fun way to my younger brother, Cannot I do that )

naturallyInconsistent

You can easily explain superposition in the classical sense.

It just doesn't have the kick that quantum superpositions have.

Relativisticcucumber

@ACuriousMind this led me to meta.stackexchange.com/questions/19478/the-many-memes-of-meta/… which is the best thing ever

naturallyInconsistent

I have, however, no difficulty explaining quantum superpositions to kids. Even at age 12 they can understand, since I'll just explain matrices along with it.

Shashaank

Yeah, because I am a beginner myself, I'll try understanding the concept first lol

Thanks @naturallyInconsistent

Can you guys suggest some physics books?!

naturallyInconsistent

17:01

I just did

Herr Feinmann

17:16

@naturallyInconsistent just please don't use chairs :P

naturallyInconsistent

@HerrFeinmann why not? They can sit nicely on them whilst myow myow's exposition blows their minds

dont want them to get physically hurt from falling down

Herr Feinmann

17:33

Ok, have them sit on quantum chairs

Relativisticcucumber

17:46

@Shashaank what do u want to learn

if you know linear algebra and ODE and basic physics (kinematics, maybe some basic electromagnetism, etc.) then I think you can just jump into quantum. sakurai is a great book for most of what it covers. can be challenging at first, but once you get a grasp of the notation, it's a good study imo.

nI might say that i shouldnt recommend sakurai to someone who doesnt have a grasp on lower level physics -- to each his own XD

Silly Goose

I second Sakurai for at least chapters 1, 2, and 4. I would recommend learning the "theory of angular momentum" from someone who has an appreciation for representation theory instead of out of any of the canonical texts. I would suggest leaving Sakurai behind after the aforementioned content and finding other, more suitable textbooks for further study.

naturallyInconsistent

@HerrFeinmann all chairs in the real world are quantum in nature...

@Relativisticcucumber on the contrary, a person who does chem eng, is well-placed to understand quantum theory.

Tobias Fünke

in case anyone is interested in my opinion: I don't like Sakurai at all

but I know it is a widely used textbook, especially in the US

Silly Goose

18:02

@TobiasFünke What textbook did your courses follow? Cohen?

Relativisticcucumber

@naturallyInconsistent yay agreed!

Tobias Fünke

nah, none specific

Silly Goose

I think Sakurai's chapter 1 is pretty good. I think it is otherwise not a great textbook.

naturallyInconsistent

@TobiasFünke miao miao prefers Ballentine, if following the same route, that is so much now the convergent paradigm for teaching QM.

Relativisticcucumber

@naturallyInconsistent i was also thinking quantum/CM is probably the most useful thing in physics for someone doing chem

naturallyInconsistent

18:04

But Feynman Hibbs is really the kind of route miao miao would prefer much more

@Relativisticcucumber yes~

ok, it really is beyond regular sneeppuu time so tada

Silly Goose

Oh yes I forgot about Ballentine. I guess Ballentine probably has better introductory chapters. The presentation of perturbation theory in both Ballentine and Sakurai is pretty bad though (in my opinion).

Tobias Fünke

yes, ballentine is good. also isham, Sobrino is nice too (but very unknown, I believe)

A.Z. Capri's book (and the corresponding exercise + solution book) is nice as well

@SillyGoose well, yes, it is rather short in Ballentine, no? But in e.g. time-dependent perturbation theory he makes one or two comments rarely found in other books, if I remember well

Galindo Pascual is a classic too

Silly Goose

@TobiasFünke the time-independent section is short indeed. i'll have to check the book to see what you mean by the one or two comments. but i would not be surprised i guess as i generally have more trust in ballentine not sweeping things under the rug

Tobias Fünke

yep

e.g. regarding the term "transition probability"

Silly Goose

separately, why is electrical conductivity what is focussed on (e.g. in my EM and condensed matter courses)? I have not heard of magnetic conductivity, for instance.

Relativisticcucumber

18:19

@SillyGoose i already asked this in cm lecture

like why we do microscopic models for electricity and not magnetism

i think the answer was "magnetism is hard"

why is $\nabla$ called nabla and del

Silly Goose

@Relativisticcucumber lol

ACuriousMind

@Relativisticcucumber see Wiki

Tobias Fünke

@Relativisticcucumber agreed on lol

What exactly do you mean? Of course you can (and do) study magnetic properties of materials, using e.g. quantum mechanics

ACuriousMind

@SillyGoose electric circuits are everywhere (in our modern world) and conductivity/resistance are fundamental concepts to understand those, but how many magnetic circuits have you seen? :P

Tobias Fünke

e.g. in Nozières and Pines they study "spin susceptibility" --just a quick search through that book yielded that result

Relativisticcucumber

18:26

@ACuriousMind inductors use magnetism, no?

ACuriousMind

@Relativisticcucumber did I claim they don't?

Herr Feinmann

@naturallyInconsistent Gott würfelt nicht

@TobiasFünke I heard about them today when I checked a book of theirs published in 1966 D:

Slereah

Magnetic resistance is useful for some things

Tobias Fünke

GMR...

Slereah

ie hard drives are protected from magnetic fields with magnetic resistant metal

Relativisticcucumber

18:29

@ACuriousMind well i thought ur point is that circuits require a knowledge of electricity but not magnetism

Tobias Fünke

@HerrFeinmann :d yeah the books are quite old (but gold)

Herr Feinmann

I was just checking the references to find something about that RPA thing :P

And since Rickayzen is from the 80's...

ACuriousMind

@Relativisticcucumber no, my point is that electric resistance is very important to understand circuits, magnetic resistance not so much

that doesn't mean they don't involve magnetism in some form

Tobias Fünke

@HerrFeinmann can you specify again the question? or is it really only about the hartree diagram thing?

Relativisticcucumber

@ACuriousMind hm i see

Silly Goose

18:33

separately, why do we allow for dielectric functions and conductivities to be complex? Naively, we decree that all fields are real valued. Thus, you can work with complexified fields for convenience, but you need to take the real part at the end of your computation. You also need to take the real part of the fields if you do multiplication operations between two complex-valued fields (e.g. $\vec{E} \times \vec{B}$). However, this rule does not apply to dielectric functions or conductivity. Why?

Relativisticcucumber

@SillyGoose can i hazard a guess

Tobias Fünke

@Relativisticcucumber go ahead

Silly Goose

For instance, in my EM course we describe atomic polarizability (microscopic electric susceptibility) as $\vec{p}(t) = 1 + 4\pi \alpha(\omega) \vec{E}$ where $\vec{p}(t) := \vec{p}_\omega e^{-i\omega t}$ and $\vec{E} := \vec{E}_\omega e^{-i\omega t}$ and $\alpha(\omega)$ is generically complex-valued.

Relativisticcucumber

if we dont take the real of electric fields, we get nonphysical results. if we take the real of conductivity, we get nonphysical results XD

my nonphysical i mean like matching known data

Silly Goose

but conductivity can be measured, right?

Tobias Fünke

18:36

well, do you know the meaning of the real and imaginary parts of $\epsilon$?

Relativisticcucumber

@SillyGoose im p sure the im part of conductivity is crucial to get the proper physics

but i see ur point

Silly Goose

the real and imaginary parts of $\epsilon$ can be used to characterize wave phenomena upon examining the waves

Relativisticcucumber

someone also asked this in lecture

"what does the im part mean / is it 'real'"

Silly Goose

but that is assuming the prescription of allowing $\epsilon$ to be allowed to be complex in the first place

Tobias Fünke

@SillyGoose well, yes. I don't know if it can be done "purely" real, but for sure it is convenient, don't you agree?

Silly Goose

18:38

it is convenient, but ad hoc.

Tobias Fünke

The imaginary parts are related to losses

Relativisticcucumber

@TobiasFünke i think the ratios/signs of the im versus real components characterize your physical regime, right?

Silly Goose

i do not doubt that the phenomena predicted by a complex dielectric function is not real. i doubt that a real (pun intended) explanation is being provided to theoretically justify letting $\epsilon$ be complex.

Tobias Fünke

which propagates further to response functions etc; see fluctuation dissipation theorem etc

Relativisticcucumber

@SillyGoose isnt all of enm based off of experiments -- i mean do we need a theoretical justification?

Silly Goose

18:39

@Relativisticcucumber for a subject this old, i would expect a reasonable justification

Relativisticcucumber

@SillyGoose to me experimental justification is reasonable

this reminds me of when i was a freshman and i asked my physics prof why $F_g = \frac{GmM}{r^2}$

his reply was "what do you mean why" with a facial expression of a broken man

back to studying for me ciao ciao

Tobias Fünke

haha

see you around

Silly Goose

well i am not questioning the fundamental laws here :P

i mean for instance, there have at least been several attempts (I do not know the validity of) to justify the necessary use of complex numbers for quantum mechanics. and the inadequateness even of natural extensions (e.g., to quaternionic quantum mechanics).

this is what I mean by theoretical justification. is it really necessary (mathematically, or mathematical physically, or whatever) to use a complex dielectric function?

Tobias Fünke

well, you have a quantity which you fourier transform, no?

so I would expect it to be a complex function in general

Herr Feinmann

@TobiasFünke The question is quite specific to the way the section of that book handles it and since it is much longer than a yes/no question, are you sure it won't bother you? :P

Tobias Fünke

18:48

@HerrFeinmann Well, I cannot promise that I can help you. But perhaps it also helps you when you write it down. So you decide...I don't want to waste your time

Silly Goose

@TobiasFünke hm well this seems like a step towards the answer i am looking for. if it is true that $\epsilon(t)$ is always real and $\epsilon(\omega)$ is complex merely as a mathematical artifact of fourier transforming, then I could buy into this.

Tobias Fünke

"In physics, dielectric dispersion is the dependence of the permittivity of a dielectric material on the frequency of an applied electric field. Because there is a lag between changes in polarisation and changes in the electric field, the permittivity of the dielectric is a complex function of the frequency of the electric field. Dielectric dispersion is very important for the applications of dielectric materials and the analysis of polarisation systems." this is what Wikipedia writes

Dielectric

In electromagnetism, a dielectric (or dielectric medium) is an electrical insulator that can be polarised by an applied electric field. When a dielectric material is placed in an electric field, electric charges do not flow through the material as they do in an electrical conductor, because they have no loosely bound, or free, electrons that may drift through the material, but instead they shift, only slightly, from their average equilibrium positions, causing dielectric polarisation. Because of dielectric polarisation, positive charges are displaced in the direction of the field and negative charges...

Herr Feinmann

@TobiasFünke No, not at all. Even if you didn't answer I would have a "well-formulated" question written down. I will type it up later, possibly in a different room to avoid blocking the chat and I will tell you (not in a hurry :P)

Tobias Fünke

@HerrFeinmann ok!

Silly Goose

well i mean i see that mathematically in a simple model of dielectric functions the complex-valuedness of $\epsilon(\omega)$ comes from introducing a first time derivative term (i.e., drag) + using complexified dipole moment and etc.. but i think that just pushes the question to why allow the equation of motion of bound electrons to be complex.

Tobias Fünke

18:51

@SillyGoose yeah, I don't know actually

2

Q: What is the complex conductivity of a metal?

I was told that it is possible to describe frequency dependent properties of metals in terms of the equations of motions of the charge carriers. In order to do so the Drude model can be exploited and the following relationship can be derived: $$\sigma _n = \frac{Nq^2}{m(\tau^{-1} - i\omega)}$$ ...

ACuriousMind

@TobiasFünke indeed the Fourier transform of a real function is real only if the function is even, i.e. $f(x) = f(-x)$

Tobias Fünke

@ACuriousMind yes

and e.g. the susceptibility vanishes for $t<0$, and admits a real and imaginary part

the reason is causality

ACuriousMind

yes, exactly - you have $\epsilon(t < 0) = 0$ and so its Fourier transform cannot be real if it's non-zero

Tobias Fünke

and since you can relate this to $\epsilon$, I think it is indeed natural to get a complex valued dielectric function

yes!

Silly Goose

in any case, maybe i will focus my observation + question. Seemingly, we decree that because the electric field is observable, it must be real-valued! But conductivity is also observable, yet we allow it to be complex valued. Furthermore, it seems that actually we can even measure the real part and imaginary part of conduction so that observable things need not be real-valued at all! So why should the electric field be real-valued...and etc.

Tobias Fünke

18:57

what you call "conductivity" might not be what others call conductivity, I guess. care for the language

on a general note: you can always put an $i$ in front of the scale of your measurement device ;) or use ANY arbitrary (distinguishable) symbols... also in QM. the point we want real numbers is that we can compare them quantitatively and do statistics

we do not need to work with self-adjoint operators. in the standard QM framework, we just need a complete projection family

Silly Goose

i think your remark about $\tilde{\epsilon}(\omega)$ (tilde for emphasis) being generically complex was illuminating. I guess I should investigate further into what $\epsilon(t)$ generically looks like.

Tobias Fünke

19:20

@ACuriousMind actually you must be careful here

the dielectric function itself does not obey the Kramers-Kronig relation

i.e. it is not a proper response function

but I think the change is trivial and does not really invalidate the argument. If you have e.g. $\epsilon=1+\chi$ (symbolically), the the KK relations should hold for the real and imaginary parts of $\epsilon-1$...in any case, for me the most intuitive quantity is $\chi$, and that is a response function, obeys KK, and is complex.

if anyone is interested, a good starting point seems "On an admissible sign of the static dielectric function of matter" by Dolgov et al. 1981

Tobias Fünke

19:42

sorry, the one paragraph is very imprecise if not wrong. What is true is that $1/\epsilon(\omega)$ (say, in a homogeneous medium) is a (causal) response function, i.e. obeys the Kramers-Kronig relations, in contrast to $\epsilon(\omega)$ itself (see mentioned paper above). But I think the argument is still valid, i.e. the inverse dielectric function must be complex in general, for the mentioned reason of causality

More Anonymous

Is the charge conservation violated in general relativity given that volume is not constant? But then I realized that this can be fixed if I only allow Dirac delta functions as my charge density. Is this the correct way to go about it?

@Slereah I think I asked you this before but I didnt consider only allowing dirac delta functions as a fix

Slereah

I mean total charge is fixed?

More Anonymous

@Slereah I mean charge density is fixed but then I integrate it with volume and its not?

Slereah

Isn't it

More Anonymous

Volume changes in GR no?

Slereah

19:51

Yeah but as far as i know the total charge is conserved?

More Anonymous

By using dirac delta functions?

Slereah

Just in general

More Anonymous

I know theres a conservation of charge density which is partial_\mu j^\mu = 0

imbAF

Can anyone help me with something. I have an exercise homework where I consider the $\phi^4$ field. I am asked to calculate the unperturbed and perturbed hamiltonian. I was able to solve and find the unperturbed one. Now when it comes to the perturbed hamiltonian one has $H_I=\frac{\lambda}{4!}\int \phi^4 d^3x$.
Now if you express the field via the ladder operators, one has 16 possible combinations of them. And by using the integration over position, one get's delta functions of the form $\delta(\vec k_1 + \vec k_2 + \vec k_3 + \vec k_4)$ and any other possible combination. Is there a short…

More Anonymous

But if I go in the rest frame of j^\mu and integrate with volume the charge is not constant

Slereah

19:53

well first bad news it's $\nabla_\mu j^\mu$

More Anonymous

fair

imbAF

And how does one solves things like $\int d\vec k_1d\vec k_2d\vec k_3d\vec k_4 \delta(\vec k_1 + \vec k_2 + \vec k_3 + \vec k_4)$ ?

Slereah

arxiv.org/pdf/2005.13233

this seems relevant

More Anonymous

Hes using a killing vector :/

I mean this means I can construct spacetimes where charge cannot be defined since it will have no killing vectors

according to this paper

Silly Goose

is there an intuition behind why for high frequency EM waves: above the plasma frequency waves are transparently propagated and below the plasma frequency waves are totally reflected?

this "all or nothing" behavior is quite interesting but I cannot think of intuitively why it should be true

imbAF

20:10

How would one solve $\int d\vec k_1d\vec k_2d\vec k_3d\vec k_4 \delta(\vec k_1 + \vec k_2 + \vec k_3 + \vec k_4)$ ? ?

Silly Goose

@imbAF have you tried doing the integrals one at a time

imbAF

Yes

You end up with $\int d\vec k_1d\vec k_2d\vec k_3 a_{k_1}a_{k_2}a_{k_3}a_{k_123}$

That doesn't bring you anywhere

Silly Goose

what is the context that you are seeing this integral in

imbAF

I had an assignement. I consider the $\phi^4$ field. And my task is to find the Hamiltonian of it

the hamiltonian is

Tobias Fünke

interestingly, there seem to be many books stating the exact wrong thing I've done (but I messed up with notation additionally lol)...anyway, a good discussion of all of this can be found in the book of Vignale, and Pines and Nozières. non-trivial and interesting, indeed

imbAF

20:16

$H=H_0+H_I$ where H_0 is the unperturbed hamiltonian and $H_I$ the perturbed one @SillyGoose

So I was given the lagrangian density, which means:

$H_0=\int \mathcal{H}_0d^3x$ where $\mathcal{H}_0$ is the hamiltonian density, which we can calculate from the relation that it has with the lagrangian density $\mathcal{L}_0$ unperturbed

I found that

The problem is that now I have the following:

$H_I=\int \mathcal{H}_Id^3x=-\int \mathcal{L}_I d^3x=\frac{\lambda}{4!}\int\phi^4 d^3x$

Now if you write: $\phi(x)=\int e^{ikx}a^\dagger(\vec k)+ e^{-ikx} a(\vec k) \frac{d^3k}{(2\pi)^2 2\omega}$

you would,as you suspect get 16 combinations of ladder operators

and one term would look like:

$\int a_{\vec k_1}a_{\vec k_2}a_{\vec k_3}a_{\vec k_4}e^{-ix(\vec k_1 + \vec k_2 + \vec k_3 +\vec k_4)}\frac{d^3k_1}{(2\pi)^3 2\omega_{k_1}} \frac{d^3k_2}{(2\pi)^3 2\omega_{k_2}}\frac{d^3k_3}{(2\pi)^3 2\omega_{k_3}}\frac{d^3k_4}{(2\pi)^3 2\omega_{k_4}}$

Now how do you go about it?

qwerty

21:20

@MoreAnonymous to define a density you need a volume in the first place. as slereah was saying charge is fixed not charge density

I think the link slereah gave is talking about how you would define, knowing that, an arbitrary (covariantly) conserved current in a general spacetime

qwerty

21:38

@MoreAnonymous section 5 discusses the case where there's no killing vector

this paper seems a bit counter to my intuition too though. maybe I will have a look when I have time. I would have thought something like MTW would cover this so I would check there as well.

More Anonymous

Thanks I'll have a look

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