@rob I made a query to list all such questions, if it helps

so that's ~6,200 questions out of ~192,000 questions that have accepted answer score < top answer score

Can a gauge theory be quantized in such a way that the mediators of the interaction are fermions?

Anyone here specializing in Optics and Photonics?

I have a really basic question but it's eating up my headspace

Hey @JohnRennie I don’t know what it is but congratulations

@cOnnectOrTR12 Hi :-)

Congratulations? What did I do?

evening @JohnRennie

@antimony Hi :-)

@ManasDogra I guess?

Via supersymmetry for a start

Gravitinos and photinos are fermions

@ManasDogra What is your definition of "a gauge theory"?

If you have a traditional gauge field $A$ that's a 4-vector like the 4-potential of ED, then you have no choice during quantization - quantizing it will give you a massless vector boson, that's not a choice, it's just how quantization works

I remember that Feynman has a whole argument as to why gravity can't be mediated by fermions

In a supersymmetric theory do the selectrons etc act as a gauge field?

I'm sure nlab has a whole section as to what kind of higher gauge fields they are

Let's see

@JohnRennie the superpartners of the bosonic gauge field form a "gauge superfield" with the bosonic gauge field if you will

Supergravity is from the super Poincaré group it seems

Ah, so it's the photinos, winos, etc that form a gauge field not the bosons that are the superpartners of the standard model fermions?

I would guess so yes

@ACuriousMind Maybe by modifying the way of quantizing..including some anticommutators instead of commutators?

The gauge fields depend on the tangent bundle of the principal bundle

@JohnRennie yes, one might say that both the gauge boson and the gauge fermion "mediate the gauge interaction" but a supersymmetric gauge theory tend to behave rather differently from the usual gauge theories we're used to

@Slereah Can you give a reference?

@ManasDogra the spin-statistic theorem says you can't do that

when you have a vector field, you have to use bosons

@ManasDogra It's in the first chapter of his lectures on gravity

I suspect this is too complicated for me to really grasp :-)

@ManasDogra only in the context of supersymmetry as mentioned above

@ACuriousMind Hmm...I really need to learn about this then :)

thanks everyone

note that this doesn't get rid of the gauge bosons - it just adds "gauge fermions"

Photino gaugino--all of these sounds so cute..but then when you look at the math :(

@ACuriousMind I was just speculating whether people have tried something like this..and as always they already have :)

rather than gauge fermions, I find it much more interesting that every massless vector boson must almost always be a gauge boson (see physics.stackexchange.com/a/265646/50583 and Weinberg's book for details), i.e. even though gauge symmetry is just an artifact of the way we describe our system, there isn't really a consistent way in ordinary QFT to describe massless vector bosons without getting a gauge symmetry "for free".

what of the higgs

oh vector

nvm

but then again, how many massless vector boson lagrangians are there?

@ManasDogra the "almost" is just me hedging me bets :P

I give an account of the basic idea of Weinberg's proof at the end here

you won't see this often outside of Weinberg because you need to follow Weinberg's idiosyncratic approach to QFT of starting from the particles and constructing the fields instead of starting from the fields + their Lagrangian to even get the idea to do this :P

@ACuriousMind Will give it a read at night.

Didn't see Weinberg's QFT book.. Saw the GR one and it's almost non-geometric there

So I can feel how the QFT book is like..

Weinberg's QFT book (actually, books, it's several volumes) is really good and really interesting but it absolutely should not be the first QFT book you read

this approach is so different to the "default" that it is much easier to first understand how everyone else does it and then appreciate the very distinct insights one gets from doing it like Weinberg

Hmm... I am having a hard time in reading even Peskin Schroeder which seems like an easy read to everyone else around me :(

he's also infamous for the cluttered notation

which is mostly because he's really pedantic about what depends on what in which way, which is both a blessing and a curse

Well I'm glad someone does it at least

it can be useful to have

as I say - it's really good, and it does a lot of stuff very explicitly that other people just claim or sweep under the rug

@NiharKarve It's generally a good idea to log into SEDE (it's a separate login) before composing queries

that way the attribution is clearer :-)

Also, it's a good idea to give it a good title ;-)

@NiharKarve CC @rob One technical comment: the query needs to disregard cases where OP self-accepted their own answer, since those are already not pinned to the top. (example.)

@NiharKarve Thank you!

@EmilioPisanty Good catch, this eliminates around 1,000 entries

here is the revised edition, replete with title and name :)

*replete -> complete

@NiharKarve almost!

that version discards threads where the accepted answer has a null OwnerUserId (because the owner account has been deleted)

here is an improved fork

I've also added a column indicating dupe/closed status

which feels like relevant info

CC @rob

@EmilioPisanty with physics.stackexchange.com/questions/81190/whats-inside-a-proton being a good example, easily findable as it is the 1st place by score difference.

then again my query only returns ~1800 threads, so there's something funky going on. Let me check.

It's definitely the dupe checker

removing it bumps it from yours's 5279 to 5505 (so there's some 230 threads with missing AA owners)

OK, done

5505 entries, with closed/dupe checker

great!

Hi, I have been reading a bit on the attempted derivations of the Born rule lately and a basic point that confuses me about the motivation is the following: people contrast the Born rule with the unitary time-evolution of the Schrodinger equation and say, in the Everettian tradition, that it ought to be the case that the Born somehow emerges from this unitary time-evolution of the system as a whole.

However, the whole reason as to why we say that the time-evolution in quantum mechanics ought to be unitary in the first place is because of Wigner's theorem as far as I understand. Now, Wigner's theorem is simply derived from the postulate that a symmetry would preserve inner-products up to an overall phase. And the physical motivation for this is that the inner products are what give us probabilities pertaining to empirical outcomes vis-a-vis the Born rule!

So, the Born rule is the basis of unitary evolution in the first place, so it seems rather hopeless/incoherent to take the unitary evolution as physically unique/meaningful in the first place if you don't have the Born rule.

I am not saying that there is no tension between the unitary evolution and the Born rule, of course, the Born rule is non-unitary. But, importantly, it is also the reason we give special physical importance to unitarity in the first place. And I do feel the discomfort of having a separate axiom for what happens when you measure something. But I just think that one would also need to justify the unitary evolution without referring to the Born rule if one is to attempt deriving the Born rule.

Those who solve integration manually are gigachads

The guy who generated dot code using AI.

@DvijD.C. "without the Born rule" - what do you really have to work with?

if you don't have the Born rule, you have nothing that tells you how the formalism connects to experiment

Yeah, I don't know, just the axiom that if a state is an eigenstate then the measurement would yield the corresponding eigenvalue? It connects it to some of the experiments.

Hartle's derivation kind of attempts to do this, I guess.

@DvijD.C. if you have that, then you know observables need to be self-adjoint (in order to have real eigenvalues), and then the Schrödinger equation gives you unitary time evolution since the Hamiltonian in there is self-adjoint

I.e. I think the real root of "time evolution is unitary" is "the Hamiltonian is self-adjoint", not some argument about Wigner's theorem

Hmm, I see, interesting. I always thought of the Schrodinger equation as secondary, coming from Wigner's theorem.

Well, that's the problem with "motivations" for axioms - there's no uniquely correct one ;)

but the Schrödinger equation in the end is just the statement "the Hamiltonian generates time translations", which you can motivate purely by analogy to classical Hamiltonian mechanics

@ACuriousMind Yeah, that's true.

@ACuriousMind I have been attracted to this view of arranging unitary transformations at the core and seeing Hermitian operators being observables coming from them being generators of these unitary transformations. IIRC Weinberg kind of voices this line of thought in his introductory chapter on QM in Volume I.

Yeah, OK, I had probably exaggerated in my mind Weinberg's rather modest assertions in favor of my view.

Heh. I mean, the two are equivalent by Stone's theorem - every one-parameter unitary transformation has a self-adjoint generator, and every self-adjoint operator generates a unitary transformation.

It's kinda pointless to ask "which comes first" - you always get both!

Yeah, true, I guess I think sometimes looking at a set of axioms in one version over the other can clarify some physical point. But then I often get carried away and start thinking that that version is more fundamental.

@ACuriousMind know any good sources for that, by any chance? i need to cut my teeth on that

also, i'm here mostly because the Mathematics chat is giving me headaches and I need a respite from non-physicists presuming to know what relativity means

^^ +1. Yeah, a version which only talks about observables to start with does sound therapeutic.

@Semiclassical unfortunately not, most of my knowledge of that is...osmotical in that I couldn't tell you where I got it from :/

for context, see chat.stackexchange.com/rooms/36/mathematics

^^ I guess Heisenberg's matrix mechanics has a similar appeal but its reputation has always convinced me to not bother trying to understand it, at least in the original version :|

which has somehow devolved into "silly scientists with their godless materialism"

@ACuriousMind drat

one bit I've been curious about lately is how one gets to the density matrix formalism

because on one hand it's largely a story that deals with operators alone

@Semiclassical entanglement - if you want to describe the state of a subsystem of an entangled state, you need to use a mixed state/density matrix

obviously the operators are themselves basis-dependent, so in that sense you don't really get rid of the underlying appeal to vector space, but in a practical sense you're only using operators

neverthless, the axioms I know of still take state vectors as axiomatic rather than density matrices

@ACuriousMind well, that's sorta the point: you usually start from axioms that deal with pure states, and then justify generalizing to mixed states

but i don't know if i've seen an axiomatic account that starts from mixed states to begin with

ah, that would be the $C^\ast$ algebra approach again - states are linear functionals on the observables (the functional represents the expectation value) and the pure states are "extremal" in a certain sense in this space of states

yeah, that's what i want

so far the closest I've come to that on my own is the following

Suppose we start with a Hilbert space $\mathcal{H}$. Then the dual space of continuous linear functionals $\mathcal{H}^*$ is itself a Hilbert space. Moreover, the two are isomorphic per the Riesz representation theorem. (One always has $g(v)=\langle u,v\rangle$ for some $u$, basically)

now we can form the set of all operators from $\mathcal{H}$ to itself, with elements $|u\rangle \langle v|$. We can give that an inner product as $\text{tr}[A^\dagger B]$, and using that we can get to the Hilbert-Schmidt class of operators which themselves form a Hilbert space

Using the Riesz representation theorem on this Hilbert space, we find that continuous linear functionals on operators are always given by $E[A]=\text{tr}(M^\dagger A)$

for some operator $M$. if we insist that this assignment is real for Hermitian $A$ and positive whether $A=B^\dagger B$ (so positive definite) then this gives positive hermitian $M$

in other words, a density matrix

where I get hung up: this story works just fine when we're dealing with finite-dimensional Hilbert spaces. but i know juuuuust enough functional analysis to expect Weird Shit (TM) happens when they're infinite-dimensional

plus the above starts from a Hilbert space of state vectors and proceeds to a Hilbert space of Hilbert-Schmidt operators. i like the geometry of that, but it all derives from the original Hilbert space.

@Semiclassical Isn't the point of the Hilbert-Schmidt condition precisely to tame this?

because the weird thing in the infinite-dimensional case is that the space of operators from $H$ to itself is not $H\otimes H^\ast$, but the space of Hilbert-Schmidt operators is.

yeah, that's a better way to put it

what I've left out in the above is the role of "completion"

there's no need to worry about completing the tensor product (w/r/t Hilbert-Schmidt norm) to get a Hilbert space when it's finite dimensional

I think where I start to lose the plot is that, from what little I know of C^* algebras

they're in general Banach spaces, not Hilbert spaces

so you lose that inner product

and that's basically because you're completing w/r/t to the operator norm rather than the H-S norm

(all H-S operators are bounded but not all bounded operators are H-S)

@ManasDogra "need of photon due to gauge invariance" is a red herring - why would you have gauge invariance if not because you want to describe EM/photons?

and I see little difference between the conservation laws argument for neutrinos and "directly detecting" muons or tauons

the only reason we have to use the indirect conservation law argument is because neutrinos are hard to detect directly

there are infinitely many possible consistent QFTs - "hardcore theory" cannot tell you which one to choose

of course, there many other indirect "tells" that muons and tauons exist - once you know about the electron neutrino and figured out how to detect it, you can wonder why there seem to be fewer of them than expected coming from the sun - the answer is neutrino oscillation, and hence you might postulate the muon/tauon just from the neutrino evidence

but in the end you need some experimental evidence to pin it on

@ACuriousMind anyways, if you do come across a good C^* algebra primer i'd want to see it

yeah, I should probably look for one

@ManasDogra From en.wikipedia.org/wiki/Muon The eventual recognition of the muon as a simple "heavy electron", with no role at all in the nuclear interaction, seemed so incongruous and surprising at the time, that Nobel laureate I. I. Rabi famously quipped, "Who ordered that?"