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6:52 AM
@BalarkaSen The nuclear spectral theorem
$$\forall \phi, \psi \in \Phi, \langle \phi, \psi \rangle = \int \xi_\lambda(\psi) \overline{\xi_\lambda(\psi)} d\mu(\lambda)$$
Where $\xi_\lambda$ are the generalized eigenvectors of some self-adjoint operator $A$ such that $A^\times \xi_\lambda(\phi) = \lambda \xi_\lambda (\phi)$
I mostly use that book on the topic
7:14 AM
oh wait, was that the one
can't find it in there
here mb
it's theorem 3.3.12
8:05 AM
@Slereah Yeah I'm reading it from Gelfand, Generalized functions vol 4
the OG
Thanks for the references! Seems good
2 hours later…
10:27 AM
A: Intuition on spectral theorem

Disintegrating By PartsAlmost everything about this subject was derived in the opposite order of what you have been taught. That's why it is difficult to answer your question. The infinite-dimensional case was studied for functions before the finite-dimensional case, and well before the notion of a vector space. Orth...

I don't think I've ever seen a qft book mention the phrase rigged Hilbert space come to think of it
I assume it is roughly similar, except much worse
If you work in the Fock space I'm guessing the appropriate Gelfland triple is just the Fock construction of the one particle rigged Hilbert space
This proof of the spectral theorem looks pretty good
A: Proof of Spectral Theorem

MickGThe main reason for posting this was to answer it, thus collecting all this stuff in a single place for future reference -- and present too. The first item on this proof is that a linear operator on a finite-dimensional complex vector space admits an upper triangular representation. This is prov...

If you're using like $L^2(\mathcal{S}(\mathbb{R}^3))$, then who knows
there's a lot of (...)
But I guess it's hard for the handful of people at nlab to describe all of physics
Sometimes I want to contribute but then I remember that I am not a category wizard
1 hour later…
11:53 AM
Is spin a relativistic phenomenon? Why does it show up on its own from the Dirac equation?
What do you mean "on its own"?
@Charlie I am reading from a textbook that derived Dirac equation starting by writing a general linear equation and then forcing it to obey the relativistic dispersion relation. And some more steps and we can show that the state vector must have at least four components. There are two sets of energy eigenstates with the same energy eigenvalue and we associate one set with the particle and the other with its antiparticle. 1/2
We then try to find a quantum number to distinguish the degenerate states which happens to be spin. So isn't it like spin just showed up out of nowhere? Whether it's spin or something else, the derivation predicted that there is a two-fold degeneracy. 2/2
In other words, I am trying to pinpoint (some intuitive physics explanation) what exactly caused this degeneracy.
Spin doesn't show up on its own like this in non-relativistic QM and it's often patchworked to accommodate spin I think. Hence, I was suspecting that its origins has something to do with relativity because that's the only additional physics assumption (if you exclude the baseline QM stuff) that goes in the derivation of Dirac equation.
12:12 PM
@Yashas have a read of this for a general discussion
Also, the elementary spins give rise to magnetic moments, and relativity stitches electricity and magnetism together. So I am speculating that it has something to do with the charge?
You can have spin in non-relativistic QM. But since the representation theory of the Lorentz group is related to that of SU(2) you are going to find that relativistic theories usually involve some discussion of spin
su(2), the Lorentz algebra and the Clifford algebras are all involved in building the Dirac equation and they are all related in various ways
2 hours later…
1:52 PM
@bolbteppa This is how you prove the infinite-dimensional spectral theorem as well
Decompose the space into cyclic subspaces, at the cost of spectrum of the operators restricted to each cyclic subspaces overlapping, and then decompose each cyclic piece into eigenspaces, and rearrange
1 hour later…
3:11 PM
Can you get very far with a version of GR that doesn't use the Levi-Civita metric?
I guess right out of the gate you get weird geodesics so maybe not
3:24 PM
what do you mean by "getting far"?
(also, I assume you meant Levi-Civita connection)
@Yashas hi. What prerequisites should I cover before writing a rudimentary yolo alg?
I guess making some accurate predictions, also yeah lol
i am reading a book by mohammed engeldy
nvm about it, just a thought
Spin is not related to either QM or relativity
but it is more important in those theories, yes
3:27 PM
Is it possible to have QM and SR without spin?
Depends what you mean by that
@satan29 Writing YOLO algorithm? That doesn't sound quite right. YOLO is a neural network architecture. You need to be comfortable with some neural network framework (PyTorch is easy to work with but YOLO can be trained using Darknet repository too which is what most people prefer I believe). You need to know basics of deep learning. Convolutions, residual connections, etc. and a general idea of object detection networks.
You could have only scalar theories, sure
yeah I thought that might be the answer :P
ah that's true
but you still have spin there, it's just zero
Spin is just the result of the rotation group having a cover
3:28 PM
@satan29 Have you done a course on deep learning?
@Charlie Well I mean, that's like asking me if you can have a field without bunnies
if there weren't, there would be zero bunnies
@Yashas no, though I have started reading up on the basics
i understand a fait bit about MLP now
The real question is what is the spin of the bunny field
"Spin" is a pretty vague word in physics anyway
there's like 10 different things called spin something
If you mean the quantity then it stems from Noether's theorem
@Charlie There are plenty of alternative theories like Einstein-Cartan that use a connection with torsion, but really you have to be specific what you mean by "GR with a different connection" because for instance in the Palatini formulation of GR, the connection is a dynamical variable and it is an equation of motion following from the EH action that the connection is LC, so in that formulation "choosing a different connection" is not even possible on its own
3:30 PM
@satan29 Yeah, so you need to have pretty strong basics in deep learning to train a YOLO model. MLPs are very primitive and unsuitable for image processing tasks. Convolution is everything in deep learning models for computer vision. You also need to know basic image processing and basics of OpenCV.
With Noether's theorem, there's a Noether current associated with rotation
@Yashas i see
you can split that current in two parts : one part that depends in the momentum (that's angular momentum), and the rest
the rest is the "spin"
oh that's kind of a cool idea to have the connection be dynamic
only to find that the "on-shell" connection is torsionless
although of course, we don't call it spin in classical mechanics
the only common example in classical mechanics is that the spin of EM fields is the polarization
part of the angular momentum of light is due to it
@Charlie it's torsionless on its own, though
it may acquire torsion through other terms
3:34 PM
hmm kind of like interactions?
Basically, yeah
spinor fields will induce torsion if you allow the connection to run wild
@Charlie it's that formulation that makes GR look closer to other gauge theories - the Christoffels are a gauge field, so to speak, see also this answer of mine
So (part of) the argument why GR isn't really a gauge theory would be the fact that measuring quantities in different coordinate systems (corresponding [in some cases] to different frames) are physically distinct situations?
Whereas in the true gauge theory different gauges are indistinguishable in every physically measurable way
@Charlie I would not talk about "coordinate systems" in this case, but it is simply a fact that any non-scalar is not "gauge invariant" under the local $\mathrm{GL}(n)$ transformations.
but it is a fact that we can observe non-scalars, like measure an electric field, which is a vector
oh yeah, sure
3:45 PM
so the usual slogan "the observables are the gauge-invariant functions" in this case fails to really do justice to what we mean by "observable"
I'm not sure how to resolve this, really. GR likes to pretend to be a "normal" theory with an action but it somehow isn't like the others :P
It is a complex issue
Famously you can make classical mechanics diffeomorphism invariant
and you can just have a relativistic action where the metric isn't dynamics, it would be diffeomorphism invariant but sans gravity
But I have seen some people argue that the metric tensor is Special in some way with diffeomorphism invariance
I forget the argument, though
As usual nlab probably has the solution but mortals have yet to read it
"The configuration space is not the naive quotient of fields by diffeomorphisms as above, but is the homotopy quotient, or action groupoid"
I am already feeling my soul leave my body
"diffeomorphism ghosts"
"This ambient structure on the spacetime breaks its general diffeomorphism invariance and hence the effective resulting theory on this background is not generally covariant (a special case of the general phenomenon of spontaneous symmetry breaking)."
So maybe it is???
But overall I feel like the statements "Spacetime is a manifold and you can do diffeomorphisms on it" and "Gravity is a thing" are independent
"The statement of general covariance is that the distinct configurations of the gravitational field form the set $Riem(\Sigma)/Diff(\Sigma)$."
If that's your definition of general covariance, then yes, I would say that implies GR
or something close to it, anyway
4:09 PM
Slereah's relationship with nLab should be the focus of a biopic
should it
nLab is just TVTropes for theoretical physicists
not much new
3 hours later…
6:46 PM
I was mentioned
2 hours later…
9:08 PM
@RyanUnger you were.
3 hours later…
11:42 PM
I have been studying a bit of category theory for the first time in the last couple of weeks. It's cool, but when I read something on nLab I feel like I understand something if and only if I already knew it (in non-categorical language maybe)
@Charlie that's not fair :) everything has spin if you count having spin zero. but I guess you could still have bound states with integer non-zero angular momentum, so composite particles with spin. But at least the free KG field is QM+SR without spin

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