@naturallyInconsistent i have arrived at dyson's formula :o
in proving dyson's formula, tong says "observe that under the T sign, all operators commute since the order is fixed by the T sign". what does he mean by this?
@SillyGoose If a matrix $C$ is Hermitian then it is diagonalizable. This is easier to state and check than "If a matrix $C$ is a sum of two simultaneously diagonalizable matrices then $C$ is diagonalizable".
Some updates on my spinor question. - I started a bounty on it https://physics.stackexchange.com/questions/791628/can-spinors-be-explained-or-understood-without-group-or-representation-theory - I asked Eigenchris about his projection correspondence between even grade subalgebra and minimal left ideal spinors and he said he didn't think about the correspondence in higher dimensions, he mainly wanted to show the two agree for low dimensional cases like Pauli spinors. - I think I do prefer the minimal left ideal definition to the representation definition (even though they're very similar) bec…
@Jagerber48 I think I "feel" the objection you have, but there is really no requirement on physical (or mathematical) theories to be somehow parsimonious with their objects. If introducing an additional vector space "out of the blue" (and it's not really out of the blue as I've argued via the Dirac equation which needs something on which the $\gamma$s can act) leads to a good theory, that's fine. There's no reason to demand a theory should not define additional objects
(and I don't think you could even coherently define what that means)
But don’t you see the distinction with vectors? With vectors you have something concrete you can dig your teeth into. Spinors as representations are like shadows of ghosts. This leads to so much confusion, this is on top of the fact that quantum mechanics already feels like chasing ghosts. In EM and gravity you have concrete fields you can imagine. You wonder “are spinors weird because quantum
is weird?” But no, spinors exist outside of quantum mechanics.
@Jagerber48 but the reason they are physically relevant is because of quantum mechanics!
the half-spin representations are relevant because representations of symmetry groups in physics only need to be projective representations in QM
and your precious construction of specifically spin-1/2 spinors in low dimensions via the Clifford algebra does not construct the general n/2-spin objects you get when coupling n objects of spin 1/2
you do not escape the theory of projective representations with this, you only delay it in one specific instance
The GA is not precious to me if that’s what you mean. It’s just an example of a concrete construction. You’ve convince me the relationship with “standard” spinors is at best very non trivial.
@Jagerber48 my point is that even if this construction worked and was perfectly equivalent, you would still be much poorer in understanding spinors than via general representation theory. What do you do with coupling of angular momenta and Clebsch-Gordan coefficients in this framework?
If I've constructed my spin-1/2 objects via rep theory, then when someone asks "oh, what about the spin of a composite system of n of those?" I just get out the toolkit of decomposing tensor representations into irreducible representations. If you're somehow tied to imagining spinors as being tied specifically to the even subalgebra or left ideal in a Clifford algebra I don't see what you're going to do here
Physicists tried to resist Wigner's introduction of group and representation theory into physics when it started - they even called it the "group plague" (Gruppenpest), but in the end Wigner won because it was just so much more efficient than anything else. I feel we're just retreading this early resistance to abstract mathematics here.
Abstract math is ok and good to support physics and improve insight and intuition. But eventually it gets hard to see where the math makes contact with reality and it feels like you’re just doing math and not physics. It’s this way with representation theory for me.
@Obliv Come to think of it, I'm sure I have had some of that, but not it alone. I should try that next time. But that is never for work/studying. It is for shakies ass. Or rather, to get others to do it with meow; I'm pretty weak and would quickly drop asneeppuuu.
Instead, at work I drink tea. Boiled leaf juice. No additives other than what is already on the leaves (which could include smoking, honey, whatever else they add)
@ACuriousMind Why not adopt a standard that would never get into the quagmire? Consistently $\mathbb Z^+$ for without zero, $\mathbb Z^+_0$ for with zero, same with $\mathbb R$, and then nobody would complain that natural numbers is only natural with zero or without zero.
@Obliv just the negatives will do. As you write it here, you have repeats.
@Jagerber48 is abstract math ever useful for anything other than predictive value? I don't want to be a heathen but I feel like analytical solutions or "nice" looking models are secondary to predictive power. If your solutions are ugly or numerically approximated, so be it?
@naturallyInconsistent what kind of tea leaves do you use? I'm thinking of switching over to tea if it makes me less jittery lol.
@Obliv There are so many! I'm actually leaving the choice to the secretary; she is so good at finding tasty stuff at cheap prices. However, whether you like which tea at all is a matter of preference and acquired tastes, so it is probably not smart to dictate what it is you should be trying at such an early stage. Instead, look for a variety, so that you can quickly eliminate those you don't like and go for what you like.
I definitely would try a bunch of stuff first. I'm not a snob but I don't like cheap stuff generally, like instant coffee (unless it's iced or lots of creamer). I wanted to try some good oolong tea at some point but the prices are kinda high for the good stuff :P
can I guess where you're located @NaturallyInconsistent
welp at least I have it narrowed down to countries where people drink tea.. I'm guessing europe then because of timezone difference not being that substantial
you are a cat, that much is certain.
Where does there exist talking cats? Or talking geese maybe you are where @SillyGoose is
yeah it's like light/moderate racism, which is funny in itself I think because it seems so unaware of it.. could be genius or really dumb but it doesn't change the effect :P
well maybe you're right it might not be that subtle xD
if we have that fields are time ordered, we have $T\phi(x)\phi(y)$ and the time ordered version is $T\phi(x)\phi(y) = \phi(x)\phi(y)$ and we know that $[\phi(x),\phi(y)] = \phi(x)\phi(y) - \phi(y)\phi(x)$, then we can say $T\phi(x)\phi(y) = [\phi(x),\phi(y)] + \phi(y)\phi(x)$ [...]
It cannot possibly be self-unaware. They have Chinese speaking people on their set, and the subtitles are specifcally chosen to keep the Google Translate mistakes in, and it keeps going between Thai, Vietnamese, Japanese, Korean, etc
[...] so then the confusion i have is that tong defines $\Delta(x - y) = [\phi(x),\phi(y)]$ and the feynman propagator as $\Delta_F(x - y) = \langle 0 \vert T\phi(x)\phi(y) \vert 0 \rangle$ but then says the $T\phi(x)\phi(y) = :\phi(x)\phi(y): + \Delta_F(x - y)$. this would imply that $\Delta_F(x - y) = \Delta(x - y)$, [...]
[...] but this doesnt seem right given their defs? and also how do we know that the remaining difference between time ordered and $\Delta_F(x - y)$ is always normal ordering of the fields?
@Relativisticcucumber Equations (2.87) and (2.88) define $\Delta$ as a commutator. It is not a propagator. Equation (2.90) defines a propagator $D$, and Equation (2.93) defines $\Delta_F$; these 3 functions are thus defined for you and you should see that they are all different. Normal ordering is covered in Section 3.3; it is totally different from the commutator alone.
okay i have narrowed the confusion more. in 2.90 $D(x - y)$ is defined to be an expectation value but in 2.92, the $D$'s are the elements of a commutator. this seems to be in contradiction? @naturallyInconsistent
@Relativisticcucumber You are absolutely wrong on this. In Equation (2.92), which, as you can compare with Equations (2.87) and (2.88), you can see that what it is doing is that it is trying to express the commutator $\Delta$ as a subtraction of $D$ as defined in Equation (2.90)
wait i am confused as well. the propagator is a map from a pair of spacetime points to a real number $D: M \times M \rightarrow \mathbb{R}$. Here, the commutator ($\Delta$ in Tong) of quantum fields is (essentially) a map from a pair of operators to another operator $[\cdot, \cdot]: \mathcal{A} \times \mathcal{A} \rightarrow \mathcal{A}$.
although from your comment it seems like maybe there is a typo in Tong in defining $\Delta$
@SillyGoose The commutator operation is a map from a pair of operators to operators. The commutator of quantum fields happens to be proportional to the identity operator.
@SillyGoose It is not necessary; it happens to be a "c-number" function.
where "c-number" is just complex number, and the reason why they are, is because they are proportional to the identity operator, and we conventionally factorise out the identity operator when that is possible.
Yes, but remember, we started the whole discussion on imposing only the equal-time commutation relations. Turns out, if you define this properly, the SR invariance of the rest of the theory enforces the relationship you are finding out via these propagator business everywhere else in spacetime.
i don't get how strictly speaking the commutator of operators can result in a non operator. i can understand if we are implicitly taking the expectation value of the vacuum state (which does get rid of the identity operator)
@SillyGoose Remember that in QM, $[L_x,L_y]=iL_z$ has an operator output, whereas $[x,p]=i\hslash$ is actually $[x,p]=i\hslash\mathbb I$ that we always ignored the identity operator?
@Relativisticcucumber Lorentz boosts never interfere with time ordering, so this is not a problem?
I would have to point out to you that there are a TONNE of places in both maths and physics that a "special case" formula, that ostensibly seems to be a specialisation of a more general formula that you might see later on, is actually equivalent to the more general formula.
The "greater generality" is only in the usage; the mathematical content is identical because you can use the restricted special case to derive the general case.
@SillyGoose It is some multiple of the Dirac delta distribution, not zero. It is only zero when spacelike-separated. The fact that it is a Dirac delta distribution is actually a source of the infinities that renormalisation is used to conjure away.
(Note that I am fully aware that, since same time, all separations are space-like separations. I am just trying to help be extremely clear.)
But Equation (2.79), which is the full expression of the commutation relations at equal time, automatically fixes the entire spacetime's worth of commutation relations.
It is extremely annoying, because you are putting in all these effort to understand this part, (as I also did back then), and in the end we will basically throw all these away and work with something completely else. In the interacting case, we cannot assert that these expansions would continue to make sense; we just pretend and hope they do. It just so happens that the results seem to make sense.
@SillyGoose Compare the computation in Equation (3.33) with the definitions in Equations (2.69) and (2.70) to see the correctness of the statement after Equation (3.33) that the commutator happens to be a "c-number" function; you might find it helpful to just directly compute the commutator in that notation, and notice that the "double creation/annihilation" operator parts cancel out, leaving the pure function. Proportional to identity operator. @Relativisticcucumber
(again, I can help you with the understanding of these things, but I really think the effort won't likely be worth it. However, I totally agree with your gut feelings that, if you don't go through some of these processes, you would keep feeling that everything is on a house of cards and that will totally hinder further learning.)
@Jagerber48 I have broken my answer up into sections, it addresses everything you asked about and more, e.g. no other answer is going to be able to explain why $SU(n)$ has no (or only) spinors for example, I've given links and references to sections which you can use for more background if you need it
This dimensionality criticism is not correct, I have no idea where this 'complex dimension $2^{d-2}$' thing is coming from but it's not correct, presumably the discussion below equation (2.6) of the linked GA paper explains why the dimensionality is correct from a pure GA perspective
@RyderRude Why and how? My answer makes it clear for $SU(n)$ at least
The problem is it might take the reader a year to make sense of my answer if they don't have the background, in a years time you'll see why any other approach is just a dead end
I defined the Lie algebra of the Unitary Group, you don't even need to know anything about the unitary group really, just notice that operators defined as $\hat{K}^a_b = |a><b|$ satisfy $\hat{K}^a_b \hat{K}^c_d = |a><b|c><d| = \delta_b^c \hat{K}^a_d$, thus I can write down what $[\hat{K}^a_b,\hat{K}^c_d]$ is, and so I get a Lie algebra. I can now try to find representations of this Lie algebra.
I can then re-write the exact same Lie algebra by re-defining $\hat{K}^a_b = |a><b|$ as $\hat{K}^a_b = b^{a \dagger} b_b$ using fermionic oscillators
@bolbteppa It is a standard result that the even subalgebra of $\mathrm{Cl}(q,p)$ is isomorphic to $\mathrm{Cl}(q-1,p)$. Since a Clifford algebra of dimension $d = q+p$ has real dimension $2^d$, this means the even subalgebra has real dimension $2^{d-1}$ (which, if for some reason we have a complex structure on this, yields complex dimension $2^{d-2}$)
Now I can look for vector representations of this Lie algebra. We usually just define $[\hat{K}^a_b , T^c] = \delta^c_b T^a$ as a vector representation. I just showed how to do this explicitly using Fermionic oscillators.
@ACuriousMind I'm not sure I understand what's going on on this point, the criticism may be valid and I'm misunderstanding it, I really doubt the paper is wrong over something apparently simple, and I believe my answer explains how to do whatever they are doing with no ambiguity, we could easily construct an example e.g. in 12D to check everything explicitly that might be high enough $D$ for things to potentially diverge
but yes, their "semi-spinors" are what everyone else calls a Dirac or Weyl spinor (whatever the irreducible representation of Spin(p,q) with half-spin in that dimension is)
I just find this terminology deeply silly and the repeated claims about "our spinors are so much simpler" when you don't even use the word for the same thing as everyone else disingenuous, but it is not technically wrong. It's not the first time I say something like that after looking into some weird GA claim
@ACuriousMind I think sections 2.6 and 2.7 of my answer explain why it does not give a Weyl spinor it actually gives a Dirac spinor. On a basic level, $\gamma^{m_1 .. m_k}|0>$ are the tensor reps, writing the $\gamma^m$'s in terms of oscillators, since $\gamma^c = b^{c \dagger} + b^c$ and $\gamma^{n+c} = i(b^{c \dagger} - b^c)$ (for $c \leq n$ in $SO(2n)$), only the term with all creation operators remains.
Without an $i$ in front, we get a $SU(n)$ tensor living in one Weyl spinor, but with an $i$ in front, we thus get the second copy of a Weyl spinor, so it gives a Dirac spinor.
Their definition of intrinsically just using the $\gamma^{m_1 .. m_k}$'s without the vacuum means their 'spinor' includes all the contributions with annihilation operators which is god damn ugly to say the least, I'm not sure if it's wrong though
I tried hard to bash GA but in this specific case it's hard to criticize GA because Clifford algebras are unavoidable, however the main thrust of GA is still irrelevant
Below (2.17) in the GA paper, they are basically saying that a spinor is defined as an object multiplied from the left by the spin group, and because direct multiplication of the $A^{ab}$'s on a state $|0>$ adds two creation oscillators to this turning it into $|ab>$, and because when we go to $J^{ij}$ we can also get $i|ab>$ treated as a distinct state,
we can ignore the spin group and just focus on the big bag of tensors that gets sent into itself under the spin group, treating these as basis elements gives us a Dirac spinor. However you need to know what a spin group and a matrix representation of it is to do this, and you're sneakily making a choice of one of the two actions I discuss in my post 'because it's natural because GA = cool', it begs the question why are we even working in a Clifford algebra in the first place
@Jagerber48 see the past few comments above
(GA says we're working in a Clifford algebra because GA = cool, we can re-do everything we do with vectors using matrices, how novel! By this logic, Dirac randomly pulling out a Clifford algebra when deriving the Dirac equation should be sufficient and everybody should completely understand spinors...)
> In what is titled On Non-Existence or On Nature, Gorgias develops three sequential arguments: first and foremost, that nothing exists; second, that even if existence exists, it is inapprehensible to humans; and, third, that even if existence is apprehensible, nevertheless it is certainly not able to be communicated or interpreted for one's neighbors
What if our modern day 'ancient geniuses' are all just the previous generations Debate Guys and we lost the more advanced understandings people had back in the day that they just never wrote down :\
@Slereah We all know people like that, and those of us who are not equally demented, tend to understand how detrimental they are to the functioning of society.
Because $x^{\mu} \to X = x^{\mu} \gamma_{\mu}$, everything you do with a vector surely has a complicated matrix analog no matter how unintuitive, things like dot products have analog's somewhere in the matrix description, and higher rank tensors can also be written as matrices of this form, also anti-symmetric tensors have volume interpretations, so conflating it all and putting it into matrices = Geometric Algebra is miraculous!
@ACuriousMind So is it okay if I identify it with $\frac{1}{\gamma^\mu p_\mu-m}$ in momentum space? In the calculations I am familiar with the creation operator comes to the right, and the annihilation to the left like $\langle c_k c_p^\dagger \rangle$, not the other way round...In the way it is stated here, won't the annihilation operator just annihilate the vacuum on it's right?
@ACuriousMind But if the fermions still obey the Dirac equation, would the answer to my first question be correct? (Some authors actually used results from this paper in the relativistic case)
how are fermions on a lattice supposed to "obey the Dirac equation"?
and again, the propagator you cite is a VEV in relativistic QFT, here's you're looking at EVs of Slater determinants, those are two completely different contexts
@ACuriousMind Somebody else proved that the results hold true when they take the continuum limit.
"An expression for the reduced density matrix ρV in terms of the two point correlators is known for free Dirac fields" and then they cite this paper which starts with the screenshot I attached above.
I cannot divine from a small snippet what's going on
but the relationship between lattice fermions and continuum fermions is rather subtle (cf. fermion doubling), you can't just "take the continuum limit" and hope that does the correct thing
But really there _is_ no information other than that! The original paper is https://arxiv.org/abs/0903.5284 which mentions that the two point correlator of Dirac fermions is that thing in the lattice case (maybe the authors have taken care of the subtleties, maybe not idk...they don't seem to mention these at all!)
it is not at all unusual to find that the specific claim something is being cited for does not appear in the citation, or is itself cited from elsewhere. whether this is because the authors considered the missing steps obvious, made a mistake or are trying to lie to you is usually impossible to tell :P
@ACuriousMind btw it's in section 2 of the link I sent. They say "An expression for the reduced density matrix ρ$_V$ in terms of the two point correlators is known for free Dirac fields. This follows from the fact that the expectation values of polynomials on the fields located in $V$ computed with the help of ρ$_V$ must obey Wick’s theorem (here the global state is the vacuum)."
@Sanjana I mean if you just need the factorization, then what does the order matter? The same proof that shows that $\langle (c^\dagger)^n c^n\rangle$ factorizes should also show that $\langle c^n (c^\dagger)^n\rangle$ factorizes
@ACuriousMind I need to know whether the order in the snippet actually gives the $\frac{1}{\not{p}-m}$ because authors of the other paper used the snippet's result and directly put that! I doubt it.
user587860
12:55
If $V, W$ are irreducible representations of a Lie algebra $\mathfrak{g}$ that doesn't lift to its Lie group $G$, then is it also true that $V\oplus W$ doesn't lift to $G$?
@Supersymmetry Yes. The condition that the representation doesn't lift is that for $\tilde{G}$ the universal cover, $\pi_1(G)\subset \tilde{G}$ is not represented trivially, i.e. $\rho_V(\pi_1(G)) \neq \{1\}$ and likewise for $W$ (I'm using $\rho_V$ forthe representation map on $V$). This immediately implies $\rho_{V\oplus W}(\pi_1(G))\neq \{1\}$ by definition of the direct sum.
user587860
@ACuriousMind Thanks, I thought along the same lines, but I wanted to make sure.
@Slereah If it is regarded as a complete classical theory of gravitation which should include both geodesic equation and EFE, then EFE contradicts geodesic equation...
Yeah...maybe. I mean atleast MTW pg. 186 puts it that way. But it's nowhere near any "formal axiomatization" to be sure. Plus you can always add some corrections to FP theory to make it consistent...
@Slereah At an "informal" level...there might be many things---point particle electrodynamics with backreaction, expecting effective field theories to be UV finite upto all orders...all historical artifacts...
Recently, I came to know about paraconsistent systems: where you allow some inconsistencies to some extent and disregard others...inconsistent but unabsurd
@Slereah in order to be formally inconsistent, the theory would have to be given in a formal way - how many historical physical theories can you name for that that's the case?
@Sanjana I am extremely sceptical of such a claim. It is often the case that, for condensed matter physics, the Coulomb potential of the nuclei needs to be explicitly handled. Instead, what is more likely is that you are dealing with the linearisation approximation of the Dirac cones of graphene or something like that, in which case the Dirac equation approximately holds near the Fermi surface.
@Slereah The construction seems to be wrong, or trivial. I mean, if the vector a is to be interpreted as from the centre of circle to any arbitrary point outside the circle, and if the "unit contour line" is really supposed to be some distance along the vector a from circle to the finally constructed line, then yes, this is trivial: the vector a forms the secant, and the finally wanted distance is the cosine, so if the secant is multiplied by k, then the cosine is divided by k. Trivial result.
If, instead, the finally drawn chord is what is wanted, then it is double of the sine, and this entity is a nonlinear function of the length of the secant.
@naturallyInconsistent afaik the authors of the paper are high energy theorists and might not be aware of these subtleties or even the ones mentioned by ACM, or they are intentionally taking a very special case after explicitly mentioning the other cases don't work.
I would not expect HEP theorists to make elementary mistakes like these, tbh. It is like having titans making mistakes on basic mechanics.
still, the probability is non-zero
Anyway, @Slereah, I am perfectly fine with the secant v.s. cosine analogy method to construct forms, but then one would have to question them on the length unit that makes up the radius of the unit circle.
Not to mention what happens if the metric is not diagonal. Those of us who have to deal with condensed matter, know that hexagonal lattices are common and nice.
The non-trivial case of the metric is in the case where it is a conic section
The elements of the metric tensor correspond to the various properties of your conic
The Minkowski metric would be a boring unit hyperboloid for instance
> At one time it was often called the ' Eudemian summary ', on the assumption that it was an extract from the great History of Geometry in four Books by Eudemus, the pupil of Aristotle.
Another bibliography descent ending in a dead end
damn book is lost
> Eudemus of Rhodes (circa 350 bce–290 bce) is known to have written three works on the history of mathematics: History of Arithmetic, History of Astronomy, and History of Geometry. In fact, each is now lost and only known today from references to the works by others whose writings did survive
@naturallyInconsistent The continuum limit is a topic that is notoriously difficult and also underdiscussed in standard hep-th QFT approaches. I think the problem is that we pretend the lattice->continuum limit is simple because that's the way we "define" the QFT path integral, usually.
But in practice matching lattice theories with continuum theories is surprisingly difficult (which probably reflects on some level that the limiting procedure that is supposed to define the path integral does not actually work)
@ACuriousMind I'd point out that it was incomprehensible when Feynman introduced it, the cavalier way he does, way back in all of his presentations on QFT, and it is equally incomprehensible in the other expositions.
@Sanjana are we talking about Casini and Huerta? I'd bet they are more aware of subtleties of the continuous limit more than anyone in this chat. What claim do you have a problem with exactly? The sentence you mentioned is just a passing remark
