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01:11
@imbAF reverse the argument. Start from Weyl spinors and build the Dirac spinor such that it automatically solves the Dirac equation by definition.
01:45
Hey everyone
What is everyones profession?
02:34
@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 $n$
03:10
@SillyGoose I think you are so confused that you are not even able to articulate the confusion.
@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)
8
Q: How does one prove that the lowest-order nonvanishing multipole moment of a charge distribution is independent of the origin, for arbitrary $\ell$?

user157588The 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?

03:47
@LuckyChouhan That is manifestly false. The entire enterprise of science is itself a gigantic version of group study.
The textbook isn't even written. We only just get to do experiments are then argue with each other
Hello Everyone...
I think the hint that it's an ill posed statemnt is the need to use a hedge word like "serious" study. very "no true Scotsman".
then you just get to keep moving the bounds of whatever you decide is "serious" until it's a tautology that serious study is independent study
Also, the "discussing with someone else, after studying, who has that knowledge helps" is also a tautology.
04:35
@Michael *coughs..jobless(3rd yr undergrad)
 
2 hours later…
06:24
hi
 
2 hours later…
08:51
@Michael developer here.
 
1 hour later…
10:00
When i don't have any questions. This group is almost silent whole day or week....
10:31
What?! We write here everyday
@Mr.Feynman it aint written to be reflective of reality.
just... I have no words.
I'm bewildered
I'll need someone to break it down for poor me
The solipsism is so strong it's gone into delulu-land
10:46
@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.
@ACuriousMind trolling attributes a level of self-awareness I honestly don't think he has
I've just learned that Growind Mind is 123
@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.
@Mr.Feynman new name sounds suspiciously familiar
Never Mind
10:53
lol
"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
@ACuriousMind damn, you went mama bear here :P
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.
Yeah, definitely personal. For example, if I have people around I can't study. I have to chat
Instead, if I have a doubt or a specific point to address, it's different
"Never attribute to malice that which is adequately explained by stupidity."
11:03
Socrates joined the chat
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)
Yeah, exercises are a different beast
But they are not a thing in Italy, at least in most universities :P
Like, you only have the exam and doing exercises is up to you. Get a book and do them
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
I'm not talking about just optional: I'm talking about not having exercises handed to us in most cases
not even for the intro courses?
11:19
Susskind says that they need to remake string theory without supersymmetry...
he also says that no figured out how to spontaneously break this symmetry....
did no one figure out the mathematical side or just that no one figured out how to make it agree with experiment
11:35
symmetry breaking seems like a non perturbative idea. so maybe no one has the mathematics for it in string theory
but i think they probably have some calculation about the energy scale of it. one cant just say "symmetry breaking" as a magic word
i will check Polchinski's second book
11:54
@ACuriousMind Sorry, I'm back (hello back). Probably just for the analysis course, for sure not in the physics courses I had :P
Mathematicians are more consistent with it
12:34
@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
A norm is an additional structure, you should not assume it
@SillyGoose ...if you can't assume that the majority of physics students are interested in physics something has gone wrong much more fundamentally :P
i agree
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?
12:52
@ACuriousMind in Soviet Russia physics is interested in you
@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
@SillyGoose which step do you not understand?
Polchinski says this cant be shown in string theory cuz its a non perturbative phenomenon
Polchinski aims to show this in the Effective QFT
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
12:56
@Mr.Feynman online discussion partners
@SillyGoose Oh God no that section
@Mr.Feynman LOL
my condolences
hm but the free and the interacting vacuums are invariant under different time translations, or is this false?
i didnt understand why Schwartz wants to define particle state using the full Hamiltonian's vacuum instead of the free Hamiltonian's vacuum
13:00
@SillyGoose ...that is why I appended "of that theory" to my statement :P
Schwartz is extremely handwavy
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
the free vacuum is invariant under the free time evolution, the interacting vacuum under the interacting time evolution
This is not Schwartz's problem. It's just the nasty interacting vacuum
This question of mine and links therein may help
I should actually update the references of the question
@Mr.Feynman people use friend more or less loosely dependent on context for irl relationships too. putting labels on things is a tricky game
13:02
@Mr.Feynman thanks!
@qwerty I like putting labels on things, until it's my turn to label
@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?
Hello,

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?
@SillyGoose Yes
13:07
what the...that is much simpler than the alternative (to state and conceptually digest; perhaps not computationally)
because, physically, when the states/particles are far apart it should not matter very much that they can in principle interact
so why do people do this adiabatic switching? i just read a paper which presents 1D quantum mechanical scattering like this
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?
@SillyGoose I'm not sure what the phrase "eigenstates" does there
13:10
or i guess using some basis of the interacting qft hilbert space
or more abstractly using some states of the interacting qft hilbert space (independent of basis)
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
like if I start with a separable state I can only produce bipartite entanglement via an interaction.
Ok, I updated the references
13:15
hm well I will have to take a look again.
asymptotic completeness is just the claim that the spaces of scattering states already span the whole space
@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?
hm but this is different I think than what I mean. although, I might be wrong in my interpretation.
oh hm maybe I am getting confused by this adiabatic switching picture again.
@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
@SillyGoose Related, with a fine answer by ACM, but also take a look at the other answers:
37
Q: Can virtual particles be thought of as off-shell Fourier components of a field?

knzhouI 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...

13:18
@ACuriousMind I mean, I know you were talking about adiabatically turning on perturbations and so. I meant related, yes
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 $\mathcal{U}(t, t_0) \lvert 0 \rangle \sim \lvert \Omega \rangle$.
(FQFT := free QFT; IQFT := interacting QFT)
Note that there are >500 responses to Strassler's blog post profmattstrassler.com/articles-and-posts/… I haven't read all of them. ;)
maybe more clearly, Schwartz seems to motivate this $\mathcal{U}(t,t_0)$ as relating the heisenberg picture FQFT to the heisenberg picture IQFT.
So it can be thought of as some sort of map between the FQFT and the IQFT
or i could have totally misunderstood as well
$\mathcal{U}$ is just the usual interaction picture time evolution operator
Beware that it only takes that form when the initial time is the time at which the interaction picture and Schroedinger picture coincide
13:30
@SillyGoose Incidentally, you're missing a $_0$ in the hamiltonian in the exponential here
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 $0$?
@Mr.Feynman oh oopsies
@Mr.Feynman in the approach using interacting vacuum, would one also have to modify the Interaction Hamiltonian?
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 $\lambda$.
Yes, that is indeed true in free theory
or is schwartz saying that the $\lim t \to -\infty$ is a special case of this more general result
13:40
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
I mean, it's not a joke. Some people really see it that way
if you say enough words the audience will nod and move on
More like, this one is one of those handwavy "proofs" in physics to justify the result you already know
You could just postulate 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
13:42
Don't honk and calculate
I wonder what @bolbteppa would say about a formal matter like this. He hasn't logged in a while, hope he's doing well
@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 $\lvert \Omega\rangle$.
bolbteppa would say all is revealed in L&L volume 21 page 67 (joking).
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
well i was under the impression that these intermediate steps would be relevant for proving 7.53 :P
okay from schwartz's steps this is what I directly get
(and both are equal to $0$)
@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.
13:48
@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)
@Mr.Feynman yeah,
._.
bruh
I'm not entirely sure what emotion that is :D
@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
@Mr.Feynman your opinion on this topic resonates at least with me, :') What is your favorite way of learning? I mean books, blogs, videos.
it is the emotion that i should devote part of my life to writing a good qft textbook so that none shall suffer through these throes again
13:56
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
this was before pdf copies were easily accessible
hm but shouldn't the final general result be that $\lvert \Omega \rangle = \mathcal{N}_i \mathcal{U}_{0-\infty} \lvert 0 \rangle$ without the limit?
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.
14:01
@ACuriousMind And yeah, I was talking about this type of study lol, I haven't done but saw many :)
@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 $\mathcal{N}_f = \mathcal{N}_i^*$?
oh oops wait I messed up these formulae
@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
@SillyGoose use mathb.in to recheck your mathjax :')
14:06
I meant to ask: shouldn't the final general result be $\lvert \Omega \rangle = \mathcal{N}_i \mathcal{U}_{t, t_0}^\dagger \lvert 0 \rangle$.
@ACuriousMind are you familiar with this convention ?
I'm not hunting a potential sign error
I am uncertain if what I wrote is actually true, or if the formula is only valid in the limit written by Schwartz.
@SillyGoose only in the limit, because you have to do this at the point in time where the free and interacting operators are equal
at all other times the argument doesn't work
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_+
14:09
@SillyGoose I just remembered once John Green said I'm uncertain about being uncertain.
Hm I guess I don't see how this limit is encoding that we are at the time $t_0$. originally, this fact was encoded in the annihilation operator used.
Okay wait I think this is my question.
1) is Schwartz setting $t_0 = 0$?
it's the time at which the free and interacting operators are equal
I don't know whether Schwartz chooses this to be $t_0 = 0$
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 $t_0 = 0$
oh oops yes that is just $t_0$
15:01
do u think the best biologists have to be just as smart as the best physicists
i think physics has math in it while biology doesnt
or at least not as much math as physics
Considering that the human body is finite while the universe not really
I will side with the physicist on this
but also, human body is a complex system. more complex than anything physicists study
Not sure about that
like, physicists only study simplified systems
I say that's a stupid question that fundamentally misunderstands the nature of intelligence.
15:05
i also think that biologists and physicits are utilising different kids of intelligence. so maybe biology atttacts smart people of a different kind
@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
but i dont see e.g. coming up with an evolution hypothesis to be as great of an achievemet as e.g. coming up with QCD
I think we know about human biology to a satisfactory level
@imbAF but biology also encompasses theories about evolution and origins of life
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!
2
15:07
@imbAF and i dont think we understand everything else except the brain..
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
@ACuriousMind i said it because thats a principle i live by. but that doesnt mean i dont recognise that some people can be smarter than others
to try and prove how our world jumped from inorganic to organic
i just meant that we shouldnt worry about not being smart too much
And they were testing different proteins and such that would "consume" other types of matter
But, I don't remember it right now, it wasn't a satisfactory experiment for some reason
@ACuriousMind within a field, science is a competition
e.g Newton vs Leibniz on calculus. And many more others
15:10
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.
2
@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 i do think that experimental biology takes just as much intelligence..
but, e.g. theoretical evoltutionary hypotheses sound like stories. while theoretical physicists have to do a lot of math
"Never confuse education with intelligence, you can have a PhD and still be an idiot." Feynman
e.g. there was this "selfish gene" hypothesis that was once regarded as a cornerstone of the field
@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
but it's outdated now. it was just a story
15:16
And how said activity would look like, without that element, nobody knows
@SineoftheTime that is true .
@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.
Well.. Ideally, I would agree with you that the primary focus should be in understanding the universe, or nature in general
And I also agree that comparing members of different fields of study isn't the right thing to do
since intelligence manifests itself in different forms
e.g I,personally, consider Mozart one of the smartest humans that has ever lived. But not comparing him with others in other fields
Just his activity alone, makes me say that
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
maybe it's not commensurable
But you can't say that cuz those are two different scenarios
It's like saying that Bolt is better than Phelps for some reason
But that doesn't make much sense, since they compete it totally different sports
How do you translate capabilities in running into swimming? You can't do it
15:25
@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
@RyderRude Can you compose the same way Mozart did? Can Einstein do it?
Does that make you and eistein stupid ?
@ACuriousMind "my team is best team wooohhh"
@ACuriousMind i dont think there is any rational purpose... upon pondering on ur question
@imbAF i regard music as an extremely subjective thing
But that is not the point
15:27
@qwerty indeed, that's what this usually boils down to :P
@ACuriousMind Goku, that is the only safe and right answer
fortunately I only have passing knowledge of Dragonball and cannot judge that statement ,P
@imbAF i get your point. different abilities are non commensurable
@ACuriousMind Consider it an Axiom :D
also, different fields just atttact people with different abilities
15:30
@ACuriousMind the wonder is why do you even entertain that such openly unreasonable behaviour happen under your watch
Would the 3 most important questions in life be:
1. How did everything came to be?
2. Abiogenesis
3. How did humans came to be.(Answered)
At least the one whose growing is manifesting mostly in the ego happened when you werent yet here.
@ACuriousMind I would be mad at you if you didn't compensate that with massive knowledge in physics
By the way, new DB Daima episode today. Stay tuned
@Mr.Feynman is this the story of why you learn Japanese?
To compensate my meager knowledge in physics? :P
15:38
@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
@Mr.Feynman Hm? I have encylopedic knowledge of my favourite works of fiction, too, I'm just not into anime :P
@Mr.Feynman AOT?
@ACuriousMind I know that, but as a anime/manga guy I have to express my disagreement :P
@ACuriousMind what is your encyclopedic knowledge of Planescape
15:49
Though they do appreciate how much he improved the heating system for the flame pit.
5
Tell us of the Blood War
we don't have that kind of time :P
before we get to that we'd have to discuss the general setup of the Great Wheel cosmology
It's not rocket science
@PM2Ring that would be more appropriate on all souls day rather than all Saints Day👍
rocket surgery, perhaps?
🧐
@PM2Ring lol
@PM2Ring what about cartesian demon
"we could give them a completely normal experience"
16:04
we all have our own personal demons to deal with
yes. in cartoons, they often show an angel and a demon inside the character's head
that is a metaphor of the personal demon idea
bringing back a wandering attention would be a common demon
ymmv
16:24
@ACuriousMind Too bad we never got to visit the realms of law, we could have met those guys
good boys and their floating orbs
ah, the cop golden retrievers
I want to visit Mechanus rly
just hang out with a bunch of geometric shapes
you're probably a tad too chaotic to fit in well there
Maybe meet the big guy
you're not even a platonic body
16:28
Only the low level drones are platonic solids
the higher ups look like uuuuh
They look like something
I'm sure they're just platonic solids in higher dimensions
He does not radiate perfect geometric order
our feeble minds just cannot comprehend their beauty
Perhaps add that as a foot note
He is such a silly guy
16:31
I love the 1st and 2nd editon art
yeah the modern day modrons are less fun designs
They just look like robots
they look like extras from Machinarium
well, to be fair they were always supposed to look like that, I guess - Nodrom in Plancescape doesn't look far off from the modern version
yeah but I like the evil geometry design
Reminds me of the Xipehuz
16:47
Anyone know what this identity is called:
\[
\bar{u}_s(k) \gamma^\mu u_{s'}(k) = \bar{v}_s(k) \gamma^\mu v_{s'}(k) = 2 k^\mu \delta_{ss'}.
\]
17:00
@Slereah can I write GR completely using gamma matrices?
The geometry side of it
I mean you can write the metric with gamma matrices
So I guess
There are formalism for doing GR in spinor terms
Hmm... Now if only i could figure what would be the point of doing that :P
@Slereah why does the normal formaism fail
?
It doesn't fail, some people just think it has benefits for some theorems?
 
2 hours later…
18:35
Do you guys think that numerical/computational techniques have made dedicating time to doing certain computations analytically an outdated practice?
no, why would they?
usually you use numerical techniques when you can't solve a problem analytically
19:23
Also it's hard to be sure that you can solve a problem numerically efficiently
You have to prove convergence and such
There are words for people who believe that you can just throw numerical simulations at anything and it is engineers
20:22
Can someone help me with exercise 1?
In the sense
that I am not sure what I am asked to do
should I derive the expression given below, or an expression for as to how u(x) looks like?
@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.
That would just be a plug in in the TISE. And would give me that expression
That seems to easy, to be what the exercise is asking
A plug in of $\psi(\vec x)$
20:37
It's what the exercise is asking.
@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
 
1 hour later…
21:46
@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
Blogs, not much, unless I'm desperate
22:03
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
What is the difference that causes this change?
 
1 hour later…
23:25
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
Goodnight people
I have wondered
But why $E_\alpha$ and not $E_\beta$?
23:52
i officially survived all of my first set of grad midterms
not sure what the outcome is but its over and im still alive
the most satisfying thing ever: closing all of the tabs that were for studying for an exam
congrats
well done!
Relaxation time

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