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00:41
hey what happened to lucabtz
also -- in a&m, they say that they are going to improve upon the free electron model by including potential terms that incorporate the electrons influencing each other (eq 17.2) then they say that, in an independent electron pictures, the contribution of an electron level to the charge would be what is given in eq 17.5. so isnt this kind of contradictory? are we moving away from free electron or not? these are the assumptions that give rise to the hartree equations
01:02
i am noticing a pattern in physics
anything created by landau is evil
 
3 hours later…
04:03
@Relativisticcucumber It is independent particle approximation; they are free as in not having correlations, such that you don't insist upon using coördinates that enforce correlations, but you may have the antisymmetry in Slater determinant, and in DFT you ignore even the Slater determinant. The Coulomb interaction is too strong, though, so we must not be free to the point of ignoring the other electrons.
@Relativisticcucumber That's going overboard. Condensed matter physics is just difficult, no matter who it is that is attacking the problem. Don't attack the messenger who tried to make things nice for you.
@SillyGoose One of the first things that is done when moving from scalar QFT to spinor QFT, is to show that the spinor propagator is a spinor part modification of the scalar propagator, so that most of the machinery is not wasted.
 
4 hours later…
08:24
nice
08:54
@Relativisticcucumber have you heard of the term "mean field theory"? -- Hartree and Hartree-Fock can be seen as such
> Asymptotic scattering theory works well for predictions for many experimental scenarios, especially in particle physics, but does not cover all cases of interest. Rel
ativistic quantum information is a feld in which fnite time processes that occur in a local laboratory environment are theoretically and experimentally investigated
this shows that the S-matrix theory view of physics is wrong
I think some books, including Weinberg's, try to push an idea that S-matrix is the only observable and that fields are merely computation tools
this view was alright in the 70s but it is wrong today, in light of new experiments
@TobiasFünke hi. glad to see u here
@RyderRude good morning!
@TobiasFünke gm :)
when people do scattering in lab, they are ultimately measuring some local observables. i think the S-matrix is an approximate description of those local measurements
09:44
user image
5
Give a Crystalstructure. If we have a general shape for the crystal, say FCC. (Eg: NaCl)
Do the individual atoms, form also a FCC? ( in the case of NaCl, both Na and Cl if considered separetly, form a FCC-structure). Considered spearetly meaning, that if you ignore the other points in the strucutre and just look at them. Is this true generally?
@Mr.Feynman you misspelled "miao" :p
You gave me a mild heart attack: I read the meme like 5 times before sending it :P
@Madder Sometimes yes, sometimes no. For NaCl the Na atoms form an FCC lattice and so do the Cl atoms so yes you can pick any atom as your lattice points.
But there may be structures where the primitive lattice contains two or more of the same atom.
In that case you can pick one of the atoms in the primitive cell as the lattice points, but not all the atoms in the primitive cell.
10:01
@JohnRennie Would you be so kind in providing me an example? i find it hard to imagine, since if i am picking a Basis for my lattice. The translational symmetry needs to apply to each component of my basis. So i do not understand how it is possible for an Atom to not create the same structure as the whole Crystal.
@Madder Suppose we have an element A that has a simple cubic structure with one atom of A per cell. Then all the atoms of A are at lattice points.
But now suppose the cell contains two atoms of A, one at (0,0,0) and the other at (0.1, 0.1, 0.1)
(this doesn't seem very likely but this is just a hypothetical example)
Does this make sense so far?
Sure
We can call the two atoms A and A' e.g. A can be the one at (0,0,0) and A' the one at (0.1, 0.1, 0.1)
Now, if we take just the A atoms then they form the lattice, and likewise we could take just the A' atoms and they would form the lattice.
But if we take both the A atoms then they don't form the lattice.
Can you see that?
10:17
As far as i understood, if i take multiple atoms, it is basically the process of forming a Basis. So instead of two points, i "join" them together into one. But this point would have the exact symmetry as the lattice. I will draw it.
So my new Basis. Which is both of these atoms, do form the lattice?
Yes, in 2D a square lattice with one atom per cell would be like this:
And all the atoms are at lattice points.
What I'm saying is suppose we had two atoms per cell arranged like this:
@Mr.Feynman lmao
Where all the atoms are the same element.
Now if we tried to take all the atoms as lattice points we'd end up with some weird lattice like this:
@TobiasFünke what topics are your main interest in physics?
10:29
@RyderRude I work, roughly speaking, in condensed matter theory. So of course I find this very interesting. But I also enjoy topics outside my field. Overall, I'd say quantum theory and to some extent the underlying mathematics, many body theory, perturbation theory, statistical physics...
@TobiasFünke really interesting. Condensed matter theory is really cool. Hope to learn it
I'm interested in measurement problem and quantum gravity rn
@RyderRude yeah, I agree. It is a huge field, though, with many many subfields and corresponding methods.
@RyderRude This is what I've assumed from your latest questions ;)
@TobiasFünke xD
@TobiasFünke yes. Many things like supersymmetry and maybe AdS/CFT stuff may arise in cond matter theories. So i think it is richer than hep
i think i read supersymmetric models arise in cond matter
anyons too
@RyderRude About this I know barely anything to be honest... Perhaps one day I can learn about these things ^^
@TobiasFünke :)
in this paper, they talk about faster than light signalling in QFT link.springer.com/article/10.1007/s10701-024-00756-8
u might find it interesting
it is to rule out a certain kind of measurements which they name "impossible measurements"
cuz FTL is impossible, so they must rule out these measurements
and it may have consequences to the measurement problem. but it is not immediately about the measurement problem
there is something I can't understand about the paper
10:38
but wouldnt the second atoms you are drawing be also present to the right and under or are they only present in one place? that does not define a crystal.
In your example, if we extend the atoms floating nearby to nearby atoms, you can just take them as a basis. i do not understand frankly what you try to imply.
If A and C are spacelike, is it possible that there exists a B such that AB and BC are both time-like?
i think this is literally impossible
the paper relies on a weaker version of this idea
Do you mean AC is a spacelike interval?
@RyderRude Thanks, I'll have a look.
i think the reason the paper has "impossible measurements" is that they're using a weaker version of this idea
10:40
Yes, if B = (0,0) A = (1,x) and C = (1,-x) for some x < 1 then BA and BC are timelike and AC is spacelike.
@JohnRennie sorry i meant specifically in the situation where B is in the future of A and C is in the future of B
A = (0, x) C = (0, -x) and B = (1, 0)
@JohnRennie but C is not in the future of B
Oops, I misread your post ...
oh :P
i think the idea is impossible. If AB is timelike and BC is timelike and B is future of A, and C is future of B, then C must be future of A
i think AC must be time like
10:45
Yes, imagine it as the lines AB then BC drawn as two vectors. If the gradient of AC is less than 1 (spacelike) then either AB or BC must also have a gradient < 1.
@JohnRennie thanks. This is great way to visualise
this means the reason the paper has impossible measurements is that they used a weaker version of this idea
the weaker version seems possible to me
it says there are three regions A, B, C.
B is partially in the future of A and C is partially in the future of B
and A and C are spacelike
this idea seems possible cuz of "partially"
using this weaker idea, they get a measurement scenario which allows FTL signalling
and they proceed to conclude that these r impossible measurements cuz they allow FTL
 
2 hours later…
12:48
@ACuriousMind Is the long ass definition of covers for smooth spaces to prevent the inclusion of spaces that aren't paracompact
@Slereah which definition do you mean?
Differentially good open covers
@Slereah What's long about it? I only know "It's a cover $\{U_i\}$ such that all finite intersections are either empty or diffeomorphic to Cartesian space", which I wouldn't call "long ass" :P
Longer than "open covers" :p
question is more why that
From what I can find it is maybe due to non-paracompact spaces making everything complicated
I think the root cause here is that you want to be able to talk about parititions of unity
because they feature prominently in gluing many local constructions on charts to global objects
13:01
Does it also forbid non-Hausdorff manifolds?
Which most certainly do not have partition of unity
Wish they would just say things clearly :V
You know what does say things clearly tho
This extremely good commentary : samuel-lereah.com/Work/…
soon to be sold in all major book outlets
Can someone read these two things iw rote and tell me i fi made a mistake?

" Def: Unit-cell: a region that fills the point space without any overlapping by translation only. It is spanned by the lattice vectors.
Theorem: There are exactly 7 of those. They all Parallelepipeds. They are called Crystal Systems. "
I am trying to give precise definitions of these concepts. Sadly most books are very vague about it.
I should probably add also without gaps. But thats implied.
Apparently one issue of large categories is just being too big to have good functors to Set
Which makes sense I guess
13:44
hallo
@TobiasFünke All this time, I had thought your profile picture was you XD
@SillyGoose Hallo ;) haha no, fortunately (?) it is not me lol
and here I thought we had the Tobias Fünke on the site
If one considers two states $|\psi_1\rangle$, $|\psi_2\rangle$, which can be expressed as linear combination of the eigenstates of the Hamiltonian, would it be incorrect to say that the expectation value $\langle \psi_1|H|\psi_1\rangle=\langle \psi_2|H|\psi_2\rangle$ ?
@ACuriousMind Sorry, but I am not that cool :(
I have never watched the show associated with this "Tobias Fünke"
13:51
@imbAF depends on the specific situation. In general the answer is "nope", though
The Book defines a primitive unit cell as a parallelpiped spanned by lattice vectors, and then shows this example for "different choices of unit cells". (c) is clearly not spanned by any vectors!! In (a) and (d) The vectors would point at empty space. Which is a contradiction to the definition for a lattice vector...
@ACuriousMind is there a Tobias Fünke?
@imbAF What is the relative clause doing there? All states can be expressed as linear combinations of eigenstates of the Hamiltonian. We've been over this.
Just to avoid misunderstandings, ACM. I'm NOT the real Richard Feynman.
I'm the complex Richard Feynman.
13:53
@imbAF You can ask yourself the simplest case. is $\langle n_1 \lvert H \lvert n_1 \rangle = \langle n_2 \lvert H \lvert n_2 \rangle$ in general? (nondegenerate eigenstates are themselves trivial linear combinations of eigenstates).
I realized that that is not the case
@Mr.Feynman don't make me feel old :P
@imbAF good. There is no reason to expect that equality would hold, indeed.
@Mr.Feynman you had such a good chance there to say imaginary...
@ACuriousMind If it helps, I'm just out of this world :P
13:54
@Mr.Feynman well, there is THE Tobias Fünke and then there are people like me lol
2
@ACuriousMind I considered that option, actually. I thought it wouldn't be as good :P
@TobiasFünke are you guys a legion or what? :P
@Mr.Feynman I am not allowed to talk about this
3
@TobiasFünke do you have a single or set of recommended condensed matter textbooks (for learning the basics at a graduate level)?
i am sitting in on some lectures that use ashcroft and mermin but its treatment seems a bit unmodern
@SillyGoose You have to be a bit more specific: Do you mean condensed matter theory? If so, what exactly: General topics? Many-Body Theory?Spectroscopy? Solid state physics? --Let me know and I'll check what I know
general topics of condensed matter theory, yes
13:58
Ok. You know already QFT, correct?
some QFT, yes
@TobiasFünke lmao
Or maybe it doesn't work
ugh
Altland and Simon (Condensed Matter Field Theory). Chaikin and Lubensky (Principles of condensed matter physics), but this one I only skimmed through for some certain chapters. The books by Nozières and/or Pines are classic. A solid solid state book is Grosso Parravicini, and I think Ashcroft Mermin is fine too and covers many topics.
These are the first which come into my mind, but I will think about it a bit more. As I mentioned "condensed matter" is a very huge field, and with many sub-disciplines...
thanks :D
i am at a junction between considering condensed matter theory more seriously and some particle theory stuff :P
14:05
No problem. You might want to check the table of contents and let me know if this is what you are looking for, or something completely different
I see
Grosso-Parravicini has such a cumbersome notation :P
They sometimes write the position label in the ket :(
yeah, true... I mean... I saw so many strange notational things in my field that not much can shock me anymore lol
Condensed matter people seem to have some form of guilty/twister pleasure in abusing notations
@SillyGoose "Berry phases in electronic structure theory" by Vanderbilt is a classic, too -- if you are interested in these kinds of things. IIRC, it should be available for free (the pdf).
I am intrigued by the topological and geometric business in CMT...I'll take a lookzie
14:22
Lol I meant *twisted above
CMT seems pretty cool because of what a hodge podge of things it uses
@SillyGoose yes indeed ;)
the freedom afforded in making "judicious" assumptions is also kind of freeing hehehe
@SillyGoose check also this post
52
Q: Books for Condensed Matter after Ashcroft/Mermin

leongzWhat are some good condensed matter physics books that can fill the gap between Ashcroft & Mermin and research papers? Suggestions for any specialized topics (such as superconductivity, CFT, topological insulators) are welcomed.

14:47
I understand that nowadays nobody cares about Landau poles but I want to know why it was even considered serious historically? I mean, we find the one loop beta function under the assumption that higher loops are suppressed i.e. the coupling is small and then suddenly we are in awe that the coupling turns infinite at a finite energy?!
14:58
@NairitSahoo With one-loop you get - as you know - a $\log$ in the denominator
With higher loops resummation would yield $\log\log E$ terms
So the resummations turns out to be actually useful in the regime in which only the leading term matters, i.e. when $\log(E/\mu)\gg1$, having called $E$ the energy scale and $\mu$ the renormalization scale (e.g. the mass)
Once you accept that the addition of higher loop to the mix doesn't affect the qualitative behaviour at high energies, you should see why we are concerned with a divergent prediction: the coupling constant eventually gets so big that perturbation theory is meaningless, so you can't trust a calculation which uses the beta function produced by perturbation theory
15:13
@TobiasFünke Altland and Simon is great, but it is soooo NOT an introductory book suitable at the same level A&M is...
@naturallyInconsistent True that
I have used both recently, as far as I remember they're not just at two different level but also the topics are wildly different
I don't think A&M even uses second quantization
It does not
Usually an introductory textbook on condensed matter physics should introduce the physics with as little of QFT as it can
15:17
I checked it recently and found that they deal with phonons classically first and then directly using quantum statistics
quantum and stat therm is unavoidable if it wants to treat phenomena, but QFT is right out
So it's second quantization in a very broad sense :P
As of now I haven't found a book dealing with phonon-phonon interactions with Feynman diagrams
A&M is a solid state book..
phonon-phonon interactions? Isn't the free or independent particle approximation going to ignore that?
Where have I mentioned that I want free phonons? :P
Maybe I have used improper terminology
I mean: when you write the ion-ion potential and expand wrt equilibrium positions, the zeroth order term is the static potential, the first order term vanishes and the second order term is the term that gives rise to free phonons
You can decide to quantize the third order terms and treat it perturbatively, though.
15:21
No, it is that usually you don't have phonon-phonon interactions, but rather you have electron-phonon interactions, which can then produce the phonon-phonon interactions ALA QFT
@Mr.Feynman That's usually called anharmonic
Oh ok, you mean effective interactions. No, I'm talking about direct interactions
@naturallyInconsistent Yes, that
I dont think people tend to treat direct phonon-phonon interactions. They consider anharmonicity a different way
but I suppose yes, you can handle them by having direct phonon-phonon interactions
Mhh, how so? Isn't it a perturbation of the phonon hamiltonian?
You would get a $\sim (a_k+a_{-k}^\dagger)^3$ (where the power is a lazy way to mean factors with different momenta and $a$ is the destruction operator of phonons
Well, you've definitely seen $\phi^3$ QFT and seen that even that would require non-trivial renormalisation
This was briefly done in a lecture and they later stated it could be used to compute the phonon-phonon self-energy (you get diagrams like the gluon sunsets of QCD), but we did nothing excplicitly
15:31
Yes, if you have anharmonicity, then just as you have the horrible renormalisation in QFT, you would have to do that for the phonons and obtain an effective field theory
Then, since we're at it, I'll ask again a question I asked here in the past.
Oct 28 at 14:36, by Mr. Feynman
So, in the electron-phonon hamiltonian, the electron-nucleus potential is not the same function as the nucleus-nucleus potential, right?
Oct 28 at 14:37, by Mr. Feynman
The nucleus-nucleus potential should be a Lennard-Jones like potential, harmonic to a first approximation
Oct 28 at 14:38, by Mr. Feynman
While the electron-nucleus potential is just the usual Coulomb potential, so the first order linearized term is non-zero
I ask that because sometimes I read about screening also for the latter
15:47
@Mr.Feynman The electron-nucleus potential is very different from the nucleus-nucleus potential. The Born-Oppenheimer approximation says that you fix the nuclei, so that the electrons feel attracted to the nuclei. But even here, it is not the full Coulomb potential, but usually a pseudo-potential that takes the repulsion of the core electrons into account. Then the nuclear-nuclear potential is the leftover attraction after the electrons have done their thing.
15:59
@naturallyInconsistent Okay, I just care to know because the first order term is non-zero
@Mr.Feynman Which first order term?
16:30
Hello Everyone...
@TobiasFünke Dozens of you, I hear
@naturallyInconsistent the one giving rise to the electron phonon interaction when you linearize which is of course non-zero
16:54
@Mr.Feynman electron-phonon coupling is always a thing treated late in cond matt, because it is very complicated...
@Mr.Feynman Hmm that's what I was saying: nobody should have ever taken Landau poles seriously :/
But anyway, I have another question: does anybody know any theory with no RG fixed points?
@GrowingMind hello
@NairitSahoo I think I gave you a reason why it was serious
@Mr.Feynman You said that the perturbation theory breaks down at $\lambda \to \infty$. My objection is that it is obvious that it does, because we are not even considering the higher order loops to make these conclusions
How does people even say that a theory is strongly coupled or weakly coupled in IR or UV if they do not calculate the higher order beta functions?
(Okay, I sound like I am against the traditional mainstream viewpoint: but I am not... I am trying to understand how people makes these conclusions)
17:11
@NairitSahoo and I said that higher order loops wouldn't change much
They don't affect the qualative behaviour at high energy
@NairitSahoo remember that the RG group and non perturbative methods were developed almost 50 years later
When the Landau pole first appeared, QED=perturbation theory
@Mr.Feynman If higher order loops wouldn't change much how can we take coupling $g \to \infty$
I am sorry but I read and reread and reread your words for idk how many times but it still doesn't make any sense to me
17:29
@NairitSahoo the point is that perturbation theory doesn't always work
@NairitSahoo I don't understand what argument you think you're making here: One can, at any order, perfectly well imagine a running coupling that doesn't become infinite - and indeed Landau poles are absent from the Standard Model!
@ACuriousMind I am considering theories which have Landau poles from the beginning. In such theories why was Landau poles a big deal if we naively take the infinite coupling limit which we are not allowed to begin with because we are using one loop beta function and not the exact one?
I don't understand what you mean by "big deal"
One response to Landau poles was indeed that because they were derived by the very perturbation theory whose breakdown they signaled, they weren't really more concerning than just a large but finite coupling
alas, later non-perturbative methods confirmed that Landau poles signaled a problem with the theory at high energies at least in some cases - the now-known triviality of $\phi^4$ in the continuum limit of lattice models is a reflection of Landau's "renormalized charge going to zero" idea of resolving the pole
As I said, when Landau et al said that QFT didn't work, perturbation theory was the only paradigm
17:57
I am not familiar with every branch of physics out there, some of you have more ground covered, so here the question.

When we describe stuff like electrons or EM as waves. Do we really mean, these are waves, their nature and "true" self is a wave, or is the wave just a mathematical model, that we use to predict their behavior. does this question even make sense?
Does renormalization group in the context of physics have a well-defined (in the mathematical sense) meaning?
This has long been something that confused me. Especially when we deal with concepts such as diffraction and so on.
@SillyGoose yes (but it's a semi-group in some contexts - you can go forward to lower cutoffs but not backwards, i.e. the inverses are not there)
@Madder this question is essentially the question of "scientific realism" vs "anti-realism" and much in between, roughly speaking. I don't think there is a clear answer, since the answer depends on how you "see the world".
@Madder What is the difference between "just a mathematical model" that correctly predicts the world and a "true self"?
18:02
@TobiasFünke If you are telling me , that there is no answer, then for me, it is an answer. Also thank you for mentioning these terms, since i was not familiar with them.
@ACuriousMind The moment i sent the message, i asked myself the same question!
Then i thought of this!
Some mathematical models, could paint a false picture, but make true predictions.
So if i imagine a ball as a mathematical manifold. Then i am not wrong, but at the same time, i am not right. It is not that the concept of a "manifold" something intrinsic or known to the cosmos. It is just there. (yes i am awre of space time manifolds)
But maybe that is nonesensical...
check also "underdetermination of theory" in this context...in general, any good intro book on philosophy of science should cover all of this questions at least in a broad sense
I have one question about parity in general. If a function has negative parity, does it mean that the integral is zero? Or an answer cannot be given unless the boundaries of integrations are set?
As natural scientist/physicist, you are interested in creating models which correctly describes/predicts/explains certain (natural) phenomena and experiments. Whether or not the concepts of your model/theory are "real", is, as mentioned, to a very large extent a matter of personal beliefs/interpretation. Physics itself does not tell you if waves "are real" or that the gravitational force is "real" etc...
@imbAF How could the question possibly have an answer without knowing the bounds of integration? A continuous function for which all integrals are zero is zero itself.
Yeah, that is how I feel. I have to show what I said above
But no boundaries of integrations are set
So, that is exactly my question. How can I show that, if I have no boundaries of integration
18:10
@TobiasFünke Yea sure i agree with that, since there seems to be no objective way to determine that. But i was not sure, since no one really mentions those details.
@ACuriousMind I think in every sensible context (?), after all you're condensing the same observables into less degrees of freedom. You are losing something
@Madder as I said: Philosophy of science/physics books will discuss all these things. Interesting subject...
@Mr.Feynman I'm not an expert on RG so I'm hedging my bets :P
@imbAF I suspect you have once again misunderstood the context. People tend to just write $\int f\mathrm{d}x$ without explicit bounds when the bounds are supposed to be clear from context (often the integral is then "over the entire space", whatever the "space" in the context is).
@ACuriousMind No space in particular is mentioned. You can argue, that because this is related to condensed matter theory, then the space is that of a crystal, which can be 1D etc. But no specific claim is made about it
I am pretty positive about this
But I want to show you that that is the case. Unless I can't see it
@ACuriousMind I want to know if it's the case. That I misunderstood the first part of the exercise
I think the sub exercise is pretty general in what is saying and asking.
By what I just said, the integral there is over the whole of $\mathbb{R}^3$, since your function is a function on $\mathbb{R}^3$.
18:21
Ok
 
1 hour later…
19:24
Carl Friedrich GauB
@ACuriousMind I'm not either but as I understand most of my QFT professors are experts in the Wilsonian RG so they brainwashed me with that stuff :P
Before accepting EFT I was a purist :P
@Mr.Feynman I have a question
@ACuriousMind my demise was partly your doing. Since this conversation
permalink
I have to search again, damn me
Ah, I remember "permalink" :P
good times
Dec 5, 2022 at 13:41, by Feynman_00
The only thing that makes me sad about QFT is the "effective field theory" stuff
@imbAF ask the question
@user430580 you have sinned beyond mortal imagination
19:33
It was more about calculating $\langle \psi |H| \psi \rangle$ , where $|\psi\rangle=\sum_\nu c_\nu |\nu\rangle$ ($|\nu\rangle$) eigenstates of H. In the end I got $\langle \psi |H| \psi \rangle=\sum_\nu |c_\nu|^2 E_\nu$. But I got confused a bit
@Mr.Feynman :P
And thought of asking where I was doing something along the way
Okay
Type the specific part because I'm busy fighting right now, so I'm on my phone
@Mr.Feynman have you found EFT useful
I don't find science useful
See my dang profile bio :P
19:37
i would like to know exactly what sort of calculations are done to analyze the data that something like CERN produces :P. it seems impossible to just use textbook QFT to model tons of uncontrolled scattering and decay events.
@Mr.Feynman when i say useful i mean useful for learning more physics or broadening one's view of physics heh
Well, for pure QED process there shouldn't be problems as far as I know. When there is also QCD things get nasty with form factors and so on
@SillyGoose Why would that seem impossible? You have to think in terms of an inexhaustible resource of grad students who can do annoying calculations :P
But I'm not an expert at all
Oh, she meant practically too long. I thought she was hinting at some more complex calculation
there's papers that just consist of people computing pages and pages of Feynman diagrams
@imbAF while I resume my fight I'll wait for you to type something
19:40
I did
I only wanted to know
whether my derivation was accurate
@Mr.Feynman do u still think @SillyGoose is a girl (i.e. is that who the 'she' refers to)
I got confused at some point, but then I realized how $|\psi\rangle$ can be expressed
Additionally I wanted to know, does parity cause a change of boundaries on an integral
in a certain way?
@ACuriousMind that was one of my master thesis
Just doing cross sections for pion scattering
Above, I posted an exercise that I am trying to solve. And I am thinking of playing with the boundary of the integral @Mr.Feynman
hm but are these cross sections phenomenologically derived?
i mean it seems utterly intractable to derive from first principles a cross section for two beams of billions of protons colliding into one another
19:42
What are you talking about?
@imbAF you said you had doubts about that quantum average
@SillyGoose You don't compute a cross section for "billions of particles colliding with billions of particles"
@ACuriousMind no, YOU don't compute a cross section for "billions of particles colliding with billions of particles"
omg jk lolol
Of course he doesn't, stop slandering :P
the assumption is that because from the viewpoint of the particles the beams are still kind-of sparse, you just compute the scattering results for one particle colliding with one other (or perhaps sometimes two or three or whatever) and then you just multiply by the luminosity of the beam
19:44
@Mr.Feynman yes whether I made the correct derivation of $\langle \psi|H|\psi\rangle$
seemingly there is some assumption of independence if we are to reduce this many body problem to a single body problem (or less)?
@Mr.Feynman u avoid my inquiry sir
@ACuriousMind is this assumption really valid?
It works pretty well, given that we are able to fit the CERN data to the theoretical cross sections derived that way
I don't mean just from agreement with experimental data. i mean from a thinking the assumption through sort of perspective. because one can make simplifying assumptions that capture broad strokes of what is going on while just being physically unreasonable.
19:46
@imbAF so you're asking if the result is correct. It is.
like is the fact that these are high energy beams some sort of justification for this assumption? if we do such scattering at lower energies can we see a breakdown and etc.?
last time I was on here a month ago you were talking about donkeys in religion.
@Mr.Feynman Ok
nw its CERN and beams colliding?
@SillyGoose I don't see an obvious reason to doubt it - the time scales on which these collisions happen are extremely short and at extremely small distances (the particles really have to hit each other almost head on), if you do some back-of-the-envelope calculation for the probability that when one particle hits another there's a third within relevant distance for the typical density of these beams, you should find something very small
2
19:48
@somehuman we talk about weird shoot here
don;t say that
Can I say "sh*t"?
yeah, censoring shit as shoot is weird :P
I felt like a well brought up child
Of a noble family
for saying it or not?
19:52
For saying "shoot"
Dagnabbit
Darn
@Slereah sacre bleu
@imbAF which book do you follow?
is there a sort of fundamental theorem for variations? if i say that $\delta \int dr f(r) = f(r)$, then im able to make sense of a given theorem, but im not sure this is right
@TobiasFünke None. I follow the notes given by the professor. Which are taken from various books Peskin, Schwarz etc. But my plan is to read entire Schwarz once the current semester is over
and linear algebra done right
19:58
@Relativisticcucumber what is delta?
@TobiasFünke i have never really understood
@imbAF Sorry, I cannot follow: Do you study QFT? Because what you've asked is basically QM 101
i thought it means "variation"
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