Would it be accurate to say that : In essence, a system in QFT is a combination of:
The quantum fields under consideration. The spacetime and symmetry context. The dynamics defined by the Lagrangian or Hamiltonian. Quantization rules and boundary/initial conditions. Together, these aspects define the physical and mathematical structure of the system being studied.
@HerrFeinmann That's nonsense; before lecturing became a thing, and when lecturers are crappy, you were supposed to just go to the uni library and use the many textbooks there. You'll ask around for which books are good, maybe recommended by the lecturers who are crappy, or just flip through a few. Just doing this will help you turn out fine. I mostly did this, and didnt check any website for guidance. Plenty of wonderful books.
It is only when you reach the level of research, i.e. needing to find research papers, that things are much easier now than before.
@HerrFeinmann as someone who prefers self study, I would say it would be extremely tough at least for me without internet and placed like SE, which helped refined my knowledge. And exactly, I have come to know books which I never knew existed otherwise.
@HerrFeinmann 😅
I have read about people who faced such situations and yet later made great contributions. In physics.
Amal Kumar Raychaudhuri, an Indian physicist, is a case in point. The lecturers were not only lackadaisical and indifferent in the classes, they hardly cooperated (in the undergrad classes). Situations didn't improve in masters either. Finally he decided to take matters in his hands and started doing self-study.
Hey @Allie.
He one day went to library and found some GR papers and liked those and yet he had pretty primitive knowledge of the maths involved. He went to the university's applied maths department but long story short, it was a bad idea. Nothing improved. Finally he had to take some odd jobs in experimental stuffs. He did self-study going through papers and it took three years, probably, to finally write that phenomenal paper he is famous for.
I know one of my professors who somewhat did such thing although the magnitude is lesser compared to above.
What about George Green? He was a self made man! And I am not sure but last time I saw, it is still not known how he gained all such knowledge in spite of having formal schooling only at the age of 9.
> When Green was thirty, he became a member of the Nottingham Subscription Library. This library exists today, and was likely the main source of Green's advanced mathematical knowledge.
@HerrFeinmann looking forward to :P (but I probably cannot help you due to my lack of knowledge; let's see)
Hi everyone
@naturallyInconsistent well, depending where you study/have studied "back then", it can make a huge difference. For example, there might be only a very few copies of some books, or only a limited variety, e.g. due to political reasons
so I really think that the internet/www made such things waaaay more "democratic"
and freed students/persons in general from the necessity to visit bad lectures, perhaps even with high study fees, in order to learn physics (well, of course to get a degree you still have to go to uni)
@TobiasFünke miao miao was one of the few holdouts; by the time miao miao was spending time at the library, the transition from paper textbooks to online textbooks was happening. I think that problem is sorting itself out
@User1865345 wow,I've heard of Raychoudhuri before but didn't know he had to struggle this much in life,really inspiring indeed,would always love to hear such stories,thanks for sharing : ))
@TobiasFünke So, I take it you have read the SE question you upvoted. :P A lot of literature (Annett Superconductivity section 6.5, or section 4.5 in the draft, De Gennes, Tinkham as in the question) writes that $$\langle V \rangle = \sum_{\bf{k},\bf{k}'}V_{\bf{k},\bf{k}'} u_k v_k^{}u_{k'}^{}v_{k'}$$ few acknowledge the problem arising with the $k=k'$ term in the sum (as discussed in the question there too), so that the actual result would be
The second is the one you find in literature in the books above, the first is the one OP calculated and so did I. Now, a few books actually mention it, such as Combescot Superconductivity, section 2.4, that the $k=k'$ term gives a correction somewhere else in the energy. So says Mattuck in equation $(15.16)$
@naturallyInconsistent I did mention. I want to calculate the band structure of Bismuth. How do I find the exact k-path for direct band gap calculation
@SamyakMarathe there are tutorials; the bandstructure for Si is a standard tutorial question, full with examples and solutions. You are supposed to just convert that for the Bi one
So, the problem at end (which I haven't explained step by step of course) is that even if some acknowledge that the $k=k'$ terms should not be in the sum, a few step later the do as if it were
It's a big deal because the gap is $\delta\propto\sum_k u_k v^\ast_k$; on the other hand $k, k\neq=k'$ would make it $k'$ dependendent even in the mean field theory :(
@naturallyInconsistent I think anyone who is familiar with these things can understand what I want to know. Either you are not familiar or I don't understand what you want to know. I literally told you that I want to calculate the direct band gap of Bismuth. For this I want the higher-symmetry k points. For this, I want to know the k-path for the R3m symmetry (hexagonal) lattice of bismuth. How do I find the higher symmetry k points.
and with those in hand, how do I use them in BURAI to find the band gap. Whether I should use crystal coordinates (fractional but unscaled) or scaled (with 2pi factor)
@naturallyInconsistent I didn't explain carefully because it is long, but at the beginning you are right $\sum_{k, k', k\neq k'}$ just gets rid of the diagonal terms. When later on I differentiate to minimize, one sum collapses and $k'$ is now fixed and it becomes a sum over $k$ with just that single value $k'$ excluded. I understand that we can't have a proper discussion without me writing down the entire thing
I am a room owner, so if things are kicking off big style you can ping me to start evicting malefactors, but experience suggests this is very rarely necessary.
@HerrFeinmann In one book I've found an explanation: They say that they use the $p\neq p^\prime$ relation also for the $p=p^\prime$ term; they say this is justified by the fact that the contribution of the term with $p=p^\prime$ is negligible compared to the off-diagonal terms
Zur Berechnung von ⟨Hint⟩ (36.3) benutzen wir die Beziehung (36.19). Der Einfachheit halber verwenden wir diese Beziehung auch für p = p'. Dies ist gerechtfertigt, da der Beitrag von den Termen mit p = p' gegenüber den vielen Beiträgen von p ≠ p' vernachlässigbar ist. Wir erhalten dann
but I agree that with this derivation something is weird, i.e. people should really spell out what they are doing.
great :)
but in any case, you can see if you can somewhat verify the explanations given in those two books, and decide if this is reasonable (one should always check such things)
@TobiasFünke In general I agree (I already checked the German book above), in this case it's really about making approximations, so even if I want to have them motivated, it's already good that someone acknowledges them
When half of the specific literature just doesn't give a damn
At least they confirm that my doubt is reasonable
The funny thing is that the MBT/QM books discussing SC marginally seem to be more useful than SC books :P
I'm not sure why it has not yet been pointed out that all known applications of mathematics to explain or predict phenomena in the real world only rely on a very weak part of mathematics. For example, it is well known that ACA (see Reverse Mathematics) suffices for almost all real analysis, and o...
the discussion with the headline "reduced hamiltonian"
should be part of the Bogoliubov chapter, I think
ah lol, yes I have the second edition too haha.
so I meant that on p. 180 they compute the average values, and then on p. 182 (the last paragraph) they start the discussion about the "dubious" terms (if I am not mistaken; the notation is a bit different etc.)
I'm trying to make sense (mathematically) of spinors. We have a complex vector space of spinors with a $Spin(1,3)_C$ action, the complexified double cover of the Lorentz group $SO(1,3)$. Considering only the connected components, $Spin(1,3)^+ \cong SL(2,C)$. Complexifying $SL(2,C)$, we get $\mathfrak{sl}(2, C)_C = \mathfrak{sl}(2, C) \oplus \mathfrak{sl}(2, C)$. From this, how are the representations of $Spin(1,3)_C$ classified and how is $SU(2)$ involved?
@TobiasFünke Okay, I checked. This other book instead does what I was talking about. It cuts out $k=k'$ and then resulting gap is $k'$ dependent because of the summation, even with the $k$ independent coupling
It's a different way to handle it but it's the "original" way I would have
I think that for a theory like BCS which relies heavily on simplifications, only experiments can decide how reasonable these approximations are, so I don't need to be a mathematician about it, so long as someone acknowledges it
@GroveRover this line of argument is the wrong way to go about it. It makes much more sense to argue, following Wiger, Weyl, and so forth, about "what DEs are Lorentz invariant" and work from there. You'll construct from SU(2) up to all possible Lorentzian invariant DEs suitable for use in physics, and obtain spin for free there, including Dirac equation.
@GroveRover hi. i am not familiar with the rigorous details, but the reason SU(2) becomes involved in physics book is that we use linear combinations of the so(3,1) generators to define two sets of su(2) generators
there are some technicalities involved. if we were instead working with SO(4), then we would exactly have so(4)~=$so(3)\oplus so(3)$
In one derivation, you'll understand spinors, helicity, and how the whole thing ties together
it is much more sensible than the usual treatment that just comes out of nowhere
You will, however, need to supplement your reading with the other version; Wigner and Weyl approached them in opposite ways and each elucidation is clear on some stuff and not others. You'll need both.
Especially the part about using two copies of SU(2) being necessary to obtain Dirac equation.
@TobiasFünke Do you wanna know they most disturbing thing? I can mention at least two books that say the wrong average and then tell you to compute it in the exercises
One of them is Annett, the other is QFT for the gifted amateur, that uses Annett as a reference :P
@HerrFeinmann if, starting from their wrong initial bits, you will reliably be able to obtain the wrong average, then, surely, at least that is fine enough? Like, this approximation will reliably get you a close enough answer, so what's the big deal?
@naturallyInconsistent The big deal is that a bona fide computation which is elementary fermion algebra shows that that average is wrong; now, thanks to Tobias I have some more sources pointing it out, but I'm annoyed by those other sources that do not point out the underlying approximation (which is not so obvious). I can forgive those who say nothing, but not those who even present the result and tell me to check it in the exercises :P
It's like saying "the Fourier transform of $1\r$ is $k^5$, compute in the exercises (I'm being a little dramatic, I know)
in the Dirac equation, u need copies of the left handed and right handed representation. this is not the same as two copies of SU(2)
@naturallyInconsistent it is the Lorentz group on a 2D Hilbert space. it is not the same as Lorentz group on one copy of SU(2). This is because the boost isn't a unitary operator on the 2D Hilbert space
@GroveRover You are already almost there: $\mathfrak{sl}(2,\mathbb{C})$ is the complexification of $\mathfrak{su}(2)$ (the latter is the compact form of the former). The finite-dimmensional complex representations of a real algebra and its complexification are the same, so the finite-dimensional representation theory of the Lorentz algebra is equivalently the representation theory of two copies of $\mathfrak{su}(2)$.
@naturallyInconsistent but u said we have two copies of SU(2) in the Dirac equation. what we have is a $(1/2,0)\oplus (0,1/2)$ rep. these are not copies of SU(2)
@naturallyInconsistent ?? I meant that Duncan uses quantum arguments for constructing spinors, which is not what I'm doing, so I'll have a look in the future when I'll be doing QFT
Rn I'll look at it from the angle I was already approaching the issue
im just saying that we have two distinct ideas here. one is so(3,1)=su(2)+su(2) (modulo technicalities). the other is the Dirac spinor = (1/2,0)+(0,1/2). U r mixing these two ideas @naturallyInconsistent
But really this is just the representation theory of $\mathfrak{su}(2)$ (which is the example in every book on Lie theory) and standard results from representation theory
You need roughly 1. Simply connected Lie groups and their algebras have the same representations. 2. Complexifications of Lie groups have the same (complex) representations. 3. $\mathfrak{su}(2)$ is the compact real form of $\mathfrak{sl}(2,\mathbb{C})$, which is a straightforward computation with the generators
@RyderRude Obviously not, since there are no exotic smooth structures on $\mathbb{R}$ but there are continuous functions on $\mathbb{R}$ that are not smooth.
@naturallyInconsistent I want to ask you one thing about what we discussed a while ago. You gave me a general picture of NR CPT -> QM, NR CFT -> NQ QFT, SR CFT -> SR QFT. But in this part, in the below shown text, the book is implying that NR complex scalar field theory (a NR CFT) gives NR QM or simple QM. But how's this case possible?
@ACuriousMind there are no exotic smooth structures on R, but there are incompatible smooth structures on R, right? i mean smooth structures which do not agree on the notion of smoothness? i.e. they disagree in which functions are called smooth?
@ACuriousMind the example i have in mind is : take R as our manifold. define two atlass. one is $f(x)=x$. the other is $f(x)=x, x>0$ and $f(x)=2x, x\leq 0$. these two atlass seem incompatible. as in, the transition map between them is non differentiable
so those atlass disagree on what functions are defined to be smooth
@RyderRude The smooth structures are "incompatible" but the resulting smooth manifolds are diffeomorphic.
There is only a single smooth structure up to diffeomorphism on any manifold of dimension 3 or less, this is a famous result in the topic you could have found easily
@ACuriousMind yes. using diffeormorphism, they are essentially the same
but this result doesn't seem relevant to my initial questions : i have a continuous non differentiable function wrt the former atlast. can it not become differentiable wrt the latter atlas? i am basing this question on just one manifold and two atlases @ACuriousMind
the question I want to ask is if one can always switch atlases like this to make any function differentiable
@ACuriousMind Yes I did. Also an example was there, which I am familiar with, where it considered the klein gordan eq, and in the NR limit one gets the SE. If I use the same language as naturallyInconsistent we are in SR QFT and we consider NR , so we would need to have SR CFT. Or am I wrong?
@HerrFeinmann but if you want, consider to write an answer to the question on the main site...it might help others with similar problem. you could, for example, mention a few references and how they handle this (or not). I really think this would be helpful
@ACuriousMind e.g. take a continuous function on R which is not smooth wrt the atlas f(x)=x. i define this function $g(x)=2x, x\leq 0 $ , =x, x>0$. this wrt to the latter atlas, this function becomes $g(x')=x'$, which is now differentiable
@ACuriousMind this means that it is a non trivial fact that spacetime is a smooth manifold. as in, if we just define fields to be continuous, then smoothness is not guaranteed
@RyderRude I don't know what that means. If you don't assume spacetime to be a smooth manifold, many of the standard constructions (like tangent spaces) don't work at all, so you need it to be to even define anything but scalar fields.
You can restrict this to only demanding $C^k$-structures with the appropriate $k$ for the field equations you want to write if you're willing to do a lot more work for very little gain :P
@ACuriousMind oh. i hadn't considered non scalar fields, yes
i was thinking about an alternative way to view physics : we don't start with smooth structures. we just say that we have fields on spacetime that are continuous, such that there exists a smooth structure wrt which the fields are smooth
then the smooth structure is just for human convenience, as in, it leads to a convenient formulation of the laws of physics
@TobiasFünke I will consider it, maybe when I'm less in a rush. I've been for way too long on this and with the number of different references I have I should turn on Qmechanic mode. Moreover, even if the situation is better, that question asked why on physical grounds. What I have is that some acknowledge that you do because it's negligible. I think I should let it sink in my mind first :P
I vaguely remember that there is like some specific embedding of microbundles for non-smooth manifolds into their smoothed versions, and these embeddings relate to each other depending on the smooth structure you pick
i am thinking about spacetime in the ontological sense. the only structure that matters is the field configurations. if we were just handed down the field configurations by God, we wouldn't even need differential equations to predict anything
so i thought that the only use of the smooth structure is to postulate : "there exists a smooth structure wrt to which fields are smooth and obey xyz diff eqns"
but fields can't even be defined without the smooth structure. so this view of physics is wrong
So, I have history with the Italian editions of L&L being low quality paper, except the first volume. After buying the 8th volume, not only the paper was good, but...
It's $\LaTeX$!
I can't believe they rewrote the entire book in LaTeX
Hi everyone. Could you help me with the following question? Assuming that a circular loop contained in the plane of the sheet is crossed by a magnetic field with lines incoming with respect to the plane of the sheet, which of the following statements is correct:
a) The loop is crossed by an induced current circulating in a clockwise direction if the magnetic field is constant over time
b) The loop is traversed by an induced current only if there is a variation in the electric field flow (this current flows through the loop in a clockwise/anticlockwise direction depending on whether the el…
I would say that if the incoming magnetic field increases, the current circulates clockwise. Then, by contrast, the circulating induced current moves counterclockwise. No?
The things that ACM told you are in the first four chapters of Hall, and the 5-6th chapters are useful to understand better stuff about the exponential map (surjectivity and BCH). The later chapters are about root systems @GroveRover
@fqq 2011, yes still Editori Riuniti. The same as the others outside
More specifically, it has smooth cover and white pages. The density is the same as the first volume (my second volume has an "opaque" cover and old yellow pages, the third the same but with smooth cover and they have both a lower density)
Thinking about it, I believe my whole initial discourse is wrong because I used the wrong definition before for Spin_C. It is not the complexification of Spin, but rather the double cover of SO x U(1). So the Lie algebra of $Spin(1,3)_\mathbb{C}$ is just $\mathfrak{sl}(2, C) \oplus \mathbb{R} \cong \mathfrac{su}(2) \oplus \mathfrac{su}(2) \oplus \mathbb{R}$ and I just need to classify its representations from here, right? @ACuriousMind
Thinking about it, I believe my whole initial discourse is wrong because I used the wrong definition before for Spin_C. It is not the complexification of Spin, but rather the double cover of SO x U(1). So the Lie algebra of $Spin(1,3)_\mathbb{C}$ is just $\mathfrak{sl}(2, C) \oplus \mathbb{R} \cong \mathfrak{su}(2) \oplus \mathfrak{su}(2) \oplus \mathbb{R}$ and I just need to classify its representations from here, right?
@ACuriousMind $Spin(1,3)$ is defined via the Clifford algebra to be a double cover of $SO(1,3)$, while $Spin(1,3)_\mathbb{C}$ is defined as $Spin(1,3) \times U(1) / \sim$ to be a double cover of $SO(1,3) \times U(1)$
that's not what people usually mean when they just talk about classifying "spinors", but note that the additional $\mathrm{U}(1)$ factor doesn't really change the representation theory anyway since Abelian groups have only one-dimensional representations
@ACuriousMind but how do I classify the representations of the whole Spin(1,3) (the "complex" version is the same as you pointed out) it being disconnected?
The connected part is SL(2, C) and ok that's trivial
of the identity I mean
This is probably a simple question but I know very little about representation theory
@GroveRover the full representation is just the representation of the connected part + a representation of the group of connected components. See e.g. this answer of mine where I spell this out
In explicit terms it's just the usual representation theory of SO(1,3) together with two operators of order 2 that represent parity and time reversal
I was studying about coplanar forces from a book called "Mechanics" by M.C. Ghosh. It says,
When the system reduces to a single resultant force $R$, the equation of its line of action can be easily found. Take any point $O$ as base and rect. axes $Ox$, $Oy$. Let the system reduce to forces $X$, ...
