@imbAF If it happens a few times, rarely, over a long period of time, sure, people wont hold it against you. But you have this thing happening so often, it will piss people off even if it is not on purpose.
@imbAF If the professor said this, he is wrong. It is far more likely that there is a miscommunication somewhere, and that what is actually wanted to be conveyed, is that $\nu$ is the band number, for any fixed wavevector $\vec k$
That $\nu$ is the same band index that @SillyGoose called $n$
@naturallyInconsistent do you agree that serious study can't be done in group(s)? Unless all members of the group are creating something together. Here study means like what we usually do like thinking on problem, understanding chapters. Yeah, discussing with someone else, after studying, who has that knowledge helps.
@SillyGoose Consider a charge distribution with compact support. This makes for the easiest way to talk about the multipole expansion. The compact support means that there is some way to define the "centre of mass" of this charge distribution. If you rigidly translate the whole charge distribution around, it should be utterly expected that, the multipole moments relative to this "centre of mass" should never change.
Instead, the interesting statement is that when you rigidly move the charge distribution around, its multipole expansion relative to the origin, whatever is your arbitrary origin, will only change from the 2nd non-vanishing multipole moment onwards.
The constancy of the first non-vanishing multipole moment is actually pretty intricate (and miao miao only learnt about this from the following wonderful answer, thanks to you asking)
The multipole moments of a distribution are independent of origin if all the lower terms are zero. I can explicitly verify this statement by hand up to the quadrupole level, but is there any straight forward way of proving this statement?
@GrowingMind I cannot interpret posting this when the chat clearly has plenty of messages from mere hours ago as anything but trolling. Maybe it befits you to be silent in here for a whole week. You've repeatedly come in here and made bizarre grandstanding claims about yourself (most notably you "understand[ing] Newtonian mechanics way better than anyone else"), take some time to reflect on how this is alienating to other users here.
@LuckyChouhan It's very important to learn to communicate about what you learn, which you cannot do alone. Also, one of the best ways to find gaps in your understanding is to try to explain something to someone else. If you are in a situation where you can talk to other students about what you're trying to learn, do so.
"I'm a serious student so I spend all my time reading books on my own" is not only likely to make you personally miserable, it will also not lead to the effects you're hoping for
@LuckyChouhan I think we all mean different things with "learning together". Sure, I agree that learning the very basics, i.e. the "language" is probably not ideal, as different people have different ways of learning and it takes different times. Some may prefer giving a quick read first and expanding later, others may prefer to waste their lives reading the first equation and so on. Nevertheless, what I think is very important, is having debates about your doubts with other students or with Professors
So, to wrap it all up, I find that "studying together" is very important, but we all have a different idea of it and a different way to do so. In my case, it is debating specific points, like I sometimes do here, instead of studying the whole topic with others. But you never know what works best for someone else
it's just, like any other no true scotsman fallacy, basically a gatekeeping/purity statement. the reality is that everyone uses mixed modes of learning, some people have preferences, and learning both how to work independently and in groups is important -- and distinguishing when it's appropriate to do either. independent study not somehow more serious or moral.
For courses with weekly exercises we usually had to hand them in in groups of two or three. This meant we had to talk about the exercises - and by extension the course material - and I thought this was overall a pretty good system (even though once or twice I got stuck with partners I didn't particularly like :P)
the more advanced courses usually made the exercises optional - but by then most people had understood what they were useful for after being forced to do them for the first few semesters and continued doing them
@RyderRude What experiment? To date, there is zero experimental evidence for the existence of any supersymmetry partner particles. en.wikipedia.org/wiki/Superpartner
What would you call the kind of human relationship you have with people exclusively online? This applies to forum, discord servers or... Here. For example, many users always hang out here and we don't discuss only physics, but also joke and so on. What would you call this?
The word "friend" may seem extreme to some and it has varying degrees of, uhm, intensity? Acquaintances? Online acquaintances? Fellows?
@ACuriousMind i think this system only works well if the majority (or a partner on average) is interested in physics and sufficiently competent, which is perhaps not the norm:P
I do not understand what Schwartz is trying to say in this section. (My understanding): The end result that is wanted is some operator that relates the free to interacting vacuum. I am not understanding the logic that Schwartz comes to the result.
@naturallyInconsistent Hm okay I think this is helpful
(there is also a known erratum in 7.53; I think the sign of the argument of the exponential should be opposite)
Alternatively, does someone have a suggestion for a resource that has a nice section on this time ordering, normal ordering, Wick's theorem, etc. business as is being developed here?
@PM2Ring i meant the experiments showing the non existence of supersymmetry upto LHC scales. this much be shown in string theory via spontaneous symmetry breaking
I guess maybe first I am not understanding what one should expect. Is it false that in general the vacuum of the corresponding free QFT evolves into the vacuum of the interacting QFT?
@SillyGoose By which time evolution would that happen? The vacuum of a theory is by assumption Lorentz invariant, so in particular invariant under time evolution (=time translation) of that theory
if we are to do this hocus pocus dance of asymtotically the theory is the free QFT and somewhere in the middle it is the interacting QFT. it seems one would naively just write down schwartz's final statement
@SillyGoose Ah, I think you're again confused about the relationship between the free and interacting theories: The setup of QFT is not such that there is an adiabatic switching on of interactions. It's not that the theory "is free" in the asymptotic past in the sense that somehow the Hamiltonian would change from the free to the interacting one,
it's the scattering states in the asymptotic past are free in the sense that you can pretend they are the particle states from the corresponding free theory.
no one is "turning off the interaction" in the Hamiltonian itself in the asymptotic past/future, the theory itself does not "become free" in any meaningful sense
(this gets even more confusing when some sources do introduce an adiabatic switching function $f$, but in the end they always have to take the limit $f(x) \to 1$ everywhere, i.e. remove the switching and get the "interaction always on" theory as a limit)
am i understanding correctly: you are saying that the theory is always the interacting QFT. we just assume that the incoming and outgoing scattering states look exactly like a momentum, spin, etc. eigenstate (or wave packet) of the corresponding free QFT?
In the context of testing axial compression strength, what does the force-displacement graph tell us? Is displacement the bending or deformation of something like a long bone before it breaks?
well, sometimes it's an accurate description of what you're modeling - if you start with some particles in a box and then turn on a magnetic field or whatever, you are switching on the interaction literally
and sometimes it's just a mathematical trick, but as I just said, if you use it as a trick you have to remove the switching at the end via a limit
(or take some other route that results in predictions that are independent of the specifics of the switching function)
and then also the wrap up the previous comment: so is the assumption being made is that yes, we can write a state out of interacting QFT eigenstates that looks exactly like the states of the free QFT?
the claim is that, in scattering, the in and out states of scattering separate well enough before/after the scattering process itself that they can be treated as states of the free theory
this is part of the assumption that Weinberg calls the cluster decomposition principle
but this statement seems dubious. it is conceivable that the interaction allows for more freedom of what possible on-shell (?) states the system can take on. I.e., an interacting system can produce a state that a free system could not (physically).
@SillyGoose yes, which is why careful sources (such as Reed and Simon which you have already tried to read) mention the additional assumption of asymptotic completeness
@ACuriousMind ACM, just to make sure that I'm connecting the dots correctly: this is the same thing we say in the LSZ when we consider the field as asymptotically free but with a field renormalization factor, isn't it?
@SillyGoose It is the same as what you said, since asymptotic completeness means all states of the theory are superpositions of the "free" Fock states, i.e. it is not the case that the "interacting system can produce a state that a free system could not" (which is what you asked about)
@Mr.Feynman I'm not sure what you mean by "the same" but of course it is all related
I just found this blog post, which gives an interpretation of virtual particles I haven't seen before.
Consider a 1D system of springs and masses, where the springs are slightly nonlinear. A "real particle" is a regular $\cos(kx-\omega t)$ wavepacket moving through the line, where $\omega$ satis...
Returning to this vacuum question. Schwartz motivates the operator $\mathcal{U}(t, t_0) = e^{iH(t-t_0)}S(t,t_0)$ where $S$ is the Schrödinger picture time evolution operation of the interacting QFT as relating the field evolving in the FQFT (equivalently, the field in the IQFT in the interaction picture) to the field in the IQFT in the Heisenberg picture.
So why would one not naively just say that $\mathcal{U}(t, t_0) \lvert 0 \rangle \sim \lvert \Omega \rangle$.
here, I guess I do not understand the "More precisely..." so I believe in a free scalar QFT $a_p^{(h)}(t) = a_p e^{-i\omega(t-t_0)}$ via the usual commutation relations. Okay. Then what is this statement about the vacuum (even if it does not have zero energy) being annihilated far in the past? Perhaps a prequestion: when would the vacuum not have an energy that can be set to $0$?
I think I must have some misunderstanding about the vacuum here. In the free QFT, we have that $H_0 \lvert 0 \rangle = \lambda$ and $a_p^{(s)} \lvert 0 \rangle = 0$. Then, $a_p^{(s)}$ will annihilate $e^{-iH_0(t-t_0)}\lvert 0 \rangle = e^{-i\lambda(t-t_0)}\lvert 0 \rangle$ irrespective of the time $t$ and irrespective of the value of $\lambda$.
I gotta admit that this section of Schwartz was never very clear to me. Not that you should consider that too much. Even if I made some progress, the interacting/free vacuum thing is the most difficult thing I ever had to deal with and I still struggle to understand
My current POV is that it's all handwavy bullshit to set up the perturbation expansion
part of me is thinking to just skip to the examples of applying the rules. if this theory is not fleshed out anyways...there is a sense of asking what's the point :P. but there is also wanting to know where the theory really fails
@SillyGoose The claim about the free theory is indeed trivial. That's not the point. The point is that at the point in time where you have that the $a_p$ are both the operators of the free and the interacting theory, $S\lvert \Omega\rangle$ is annihilated by them just like $\mathrm{e}^{\mathrm{i}H_0(t-t_0)}\lvert 0\rangle$ and so they are both the vacuum at that point and this finally relates $\lvert 0\rangle$ and $\lvert \Omega\rangle$.
you seem to be putting too much emphasis on discerning some hidden meaning in the intermediate steps here when all Schwartz wants is to be able to write down (7.53) :P
@ACuriousMind yeah, it is true. What I told here was from my experience and here group study means when more than 2-3 classmates (or friends) try to study together the topic on which none of them know anything. And I feel that studying in such groups is distraction. (Do you think so ?) And yeah whatever you said is true once we know what we're going to talk and have some understanding of that.
@SillyGoose Yes, and that's the entire derivation :P Now you just use that the vacuum for a given set of $a_p$ is unique, so this means the two states being annihilated here are multiples of each other, and you've got (7.53)
@LuckyChouhan I find the idea that "none of them know anything" confusing: Surely they are trying to revise the content of a lecture, or work through a chapter of a book of which they've read the previous chapters or something like that?
Obviously putting N people together who don't know anything is a bad idea for any value of N, but people who don't know anything are quite rare :P
I actually have way more qm and QFT books on my physical shelf than any other topic say GR, class mech etc lol because i kept on buying more books in undergrad in hopes that the next one would be more illuminating
I have a question about the solutions to the Dirac equation. I considered $u_+(p)$ which is the spin up solution for the particle and I wanted to gain the spinor for the anti-particle. So I used the charge conjugation matrix $C=i\gamma^2\gamma^0$. The result I got gives me one minus. Which when compared to the spinor for the antiparticle with spin down, which would be the opposite of the positive spin up particle, it has an additional minus. Is that supposed to happen?
@ACuriousMind yeah, you know what I think is when I have any (math/programming) problem to do, I try to think about that problem first myself without anyone's help, and most of the time I find that "it takes time to get to the solution, and sometimes I get solutions different from given in book or somewhere else and it really helps me" but on the contrary when I discuss with someone (assumes s/he has that knowledge) I find I don't have anything to say on that problem
because I didn't spend time on that problem, so it helps me to talk to others when I have done the problem or thought what I could, then it is nice to discuss with other to know how they'll approach the problem.
Is it correct that if I use the charge conjugation matrix on a spinor solutin to the D.E the result will be the spinor of the anti particle but multiplied with a minus 1?
@imbAF There are so many conventional choices involved in these things that it's impossible to answer whether a sign is correct or not from the details you've given. The only way is to ask a person who is familiar without your exact conventions.
@ACuriousMind I cross checked it 5 times and I always have a minus appearing. But when compared to the spinor of the anti-particle, (because all 4 spinors are given, the expressions) there is no minus
So I fear I might be wrong, but perhaps something else was off that I wasn't considering
Also, I am wondering why Schwartz uses a different constant of proportionality for $\langle \Omega \lvert = \mathcal{N}_f \langle 0 \lvert \mathcal{U}_{\infty 0}$ shouldn't $\mathcal{N}_f = \mathcal{N}_i^*$?
@SillyGoose depends on whether you assume the in and out Fock spaces are the same space, it appears he does not if he does not identify the two constants
I considered u_+ and v_+ to show that if you use the conjugation matrix you get the other. But I get the minus, which doesn't appear in the expression for v_+
I am asking because $U_{0-\infty} := U(0, -\infty) = \lim_{t \to -\infty} U(t_0, t)$, where the right most side is what we started with. this seems to be implicitly setting $t_0 = 0$.
it's perfectly possible that Schwartz simply uses $U_0$ as a shorthand for $U_{t_0}$. Again, you read the book, you have to know whether he assigns any special significance to $t_0 = 0$
@RyderRude Considering that we have a good understanding how the body works overall, with the exception of the brain. We don't have fully knowledge of it
guy who said "i think we should not compete": Let's rank incommensurable achievements on a linear scale, thereby giving the impression science is a competition!
Which is one of the big 3 mysteries, together with how did everything begin, where do organisms come from and how and where and how do humans come from, but this last one has been answered
I remember I read somewhile ago a test/experiment the japanese scientists did
the claim that evolutionary theory is "not as great" of an achievement as QCD is bizarre (the other way around it would be just as bizarre) but if you absolutely want to play this silly game, xkcd already settled this in favour of the biologists ages ago.
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@imbAF The observation that individual people compete for fame, money and other things within most fields of human activity does not imply we have to conceive of the activities themselves as competitions
@ACuriousMind You can't separate the activity from the competition. If nearly all the members act in a certain way, then that is part of the activity itself
@imbAF I don't think you're getting my point. I'm saying that aspirationally, science should not be a competition. Yes, the way it is practiced in the here and now, it usually is. But we should not do silly things like Ryder here and start ranking which field is the "smartest" or which achievements the greatest in casual discussion, because that implies we think of it as a competition even ideally.
Darwin made a great hypothesis for his time. his ideas are also outdated now. but in the end, i dont think his work was as hard to come up with as Newton's work
i guess it requires huge creativity to come up with those hypotheses
@RyderRude What's your purpose in saying this? Why do you feel the need to compare Darwin and Newton in the first place? Why is this kind of comparison at all interesting or important? Both made significant historical contributions to science.
Why do you want to rank them as if you're a little kid trying to figure out who the strongest superhero is?
yes. i think my perspective is skewed toward physics because physics uses a lot of math. and i biasedly think that doing math is harder than writing things in english @imbAF
@imbAF that sounds like a good summary. there is also the problem of consciosuness, but there is doubt over whether or not such problems can even be addressed
e.g. there is this "why is there something rather than nothing" question, which almost everyone agrees cant be addressed
So, sure thing anime and manga got me interested but then it turned out that I was just interested in Japanese culture per se. That being said, in my physics era I would engage in the study of many different things at once; after the burnout, a void is left and I need to learn something, be it the lore of some show/book or a language or music theory or whatever it is
@imbAF I don't know how you could arrive at your second interpretation. When an exercise tells you "Derive the equation for X: <an equation for X>" you're asked to derive the equation for X written in the exercise, not some other thing the exercise doesn't mention.
@ACuriousMind I don't know if the same takes place in Heidelberg, but where I am, you need to collect points in order to be able to enter the exam. So, I want to be certain about stuff, just so I won't risk losing points
Specially when the exercises are written in english, but after being translated
@LuckyChouhan I have only now logged on the site - not directly on the chat, seeing this ping. My favorite way of learning is reading books, possibly one I like more to lay the foundations and other books to fill the gaps or to see if they say something in a way I prefer. I also enjoy spending a lot of time on things that I know but I don't understand why the book would say differently...
Good lectures can be a replacement for a book, sometimes
We had an introductory lecture on 2nd quantization, and there was a point in the lecture that I didn't understand. The lecturer said: We assume that we have a set of orthonormal single particle states that we can write as $|\nu\rangle$ $\nu=1,n$. We consider $c^\dagger_\nu$ a creation operator in these states. Then the Hamiltonian in quadratic form can be given in these fermion operators: $H=\sum_{\mu,\nu=1}^n h_{\mu,\nu}c^\dagger_{\mu}c_{\nu}$ where $h_{\mu,\nu}$ are the matrix elements of the Hamiltonian between the respective basis state
And then, he says: if we consider $\gamma^\dagger_\alpha$ the the creation operators for the orthonormal eigenstates $|\alpha\rangle$ of H, one can write the hamiltonian as:
$H=\Sum_{\alpha=1}^nE_\alpha\gamma^\daggner_\alpha\gamma_\alpha$, where $E_\alpha$ is the eigenvalue of $|\alpha\rangle$
My question is: How does the Hamiltonian changes form
when in both cases (of writing it), we are considering ladder operators of some basis vector
You should check single particle operators in 2nd quantization. Have you wondered what $h_{\mu\nu}$ above was? $h_{\mu\nu}=\langle \mu\lvert H\rvert\nu\rangle$
So if you consider the eigenstates of $H$, you have $h_{\alpha\beta}=E_\alpha\langle\alpha\lvert\beta\rangle=E_\alpha \delta_{\alpha\beta}$ @imbAF
The multipole moments of a distribution are independent of origin if all the lower terms are zero. I can explicitly verify this statement by hand up to the quadrupole level, but is there any straight forward way of proving this statement?
