The h Bar

02:27

@Slereah ... have we misinterpreted the saying all along, and that it is instead that the Turtle family inherited the problem from Hercules? :P

@Slereah Why the self-abuse?

naturallyInconsistent

02:44

@Sanjana I know yall had a nice convo about this whole thing, so I just wanna inject some general side remark. The space of possible theories to study is so huge, that we kinda have to try and limit ourselves into whatever subspace is most likely to get interesting / useful results. To that end, we want to consider only theories at least giving us SR behaviour, i.e. Lorentz invariance, and spin-statistics theorem asserts that if you have any form of macro-causality,

then the full spin-statistics theorem with micro-causality also has to hold. Macro-causality is extremely intuitive and agreeable with experimental measurements, so we have no reason to expect a violation of spin-statistics theorem. This leads us to consider only local theories. Of course, we have examples of interesting non-local theories, but it would be difficult to know what kinds of non-local theories are worth exploring.

Now, we turn back to a bit about the SR stuff. ACM told you earlier during the convo that if you do not really care about SR, then in NR, you could consider functions of space that are also, incidentally, a function of time, that is unrelated to space. The manifold that such theories work on, separates apart, so that it is more natural to speak of just the 3x3 stress tensor, not a 4x4 stress-energy tensor.

As for why I linked back to your follow-up question rather than the opening question, is because this is something I can add: Since you have already seen tensor theory, diff geo and especially GR uses it, you should note that an important physical ingredient as to why we care about only tensor fields and not arbitrary matrix functions of spacetime, is that only tensor fields have the chance to be physically meaningful. Especially under rotations and boosts,

non-tensorial quantities could be found to change the direction that they point to in physical spacetime. This makes them physically meaningless. After careful study, it becomes clear that only tensors and spinors can be suitable ingredients in any physics theory.

@Mr.Feynman Actually, we kinda need to see Equation (11-31); the statement as written is visibly bollocks, since Equation (11-33) only asserts a zero for when the Dirac delta distribution isn't zeroed, and we cannot read from this alone that the bracketed term isn't also zeroed by happenstance in the many cases the Dirac delta distribution is zero. But I suppose the ensuing convo helped you out.

naturallyInconsistent

03:08

@Relativisticcucumber It is extremely annoying that, now, we have the picture quotes of Equation (2.89) and of Equation (2.91), but missing the bit in the middle to judge for ourselves what the argument actually is. Please quote more next time. However, it seems like your problem here is, again, a lack of a good foundation in basic SR do's and dont's. In particular, the whole "all spacelike separations can reduced to same-time" and "no Lorentz boost can reverse time ordering".

Sanjana

05:25

@naturallyInconsistent actually I wasn't confused about the "tensor" part but the "Lorentz" part of it. But yeah, now I understand.

@naturallyInconsistent Yes, true.

naturallyInconsistent

@Sanjana But you said something about Jacobian of GCT. That is not necessarily a tensor, and if not, would then not have physical meaning.

Silly Goose

shweggums

naturallyInconsistent

schweggums?

Silly Goose

SWAG GUMS

Relativisticcucumber

@naturallyInconsistent indeed that was part of the problem. but one more thing learned now :)

Sanjana

05:29

@naturallyInconsistent Oh I meant that, instead of the stuff being Lorentz cov. what would go wrong if I impose covariance under general coordinate transformations instead...I didn't bring extra jacos, I was just speaking of the two jacos which appear in the transformation law of any 2-tensor such as the stress tensor!

naturallyInconsistent

@SillyGoose still not idea what that means but sure, schweggums if you like them so much

Silly Goose

@naturallyInconsistent it means the hills have hair

naturallyInconsistent

@Sanjana I roughly knew what you mean, but I am pointing out that Jacobians are n-densities and not n-forms and that might be an issue. Yes, it is interesting to want to see what would GCTs bring to the discussion. However, we care about GCTs for getting GR, i.e. if you want to do this, it is a matter of trying to get to quantum gravity, and we all know that there is, as yet, no suitable theory of quantum gravity.

@SillyGoose googled all those terms and still made no sense to meow

Sanjana

What does the student mean here when he says the "spinor part gets thrown out" and the following explanation by Prof. Shiraz Minwalla? Relevant conversation from 1:01:10 to 1:02:00.

I understand that the gamma matrix projects the pure spinor part, but why is it zero?
I know this formally from varying the action. But what's the reasoning given here---He says that the tensor product has two representations and suddenly sets the RHS to be zero...what's going on!

@naturallyInconsistent Hmm hmm

Relativisticcucumber

05:49

time ordering operators does not actually change anything -- it's just a rearrangement, right? so is the feynman propagator the same thing as the "general" propagator?

british alligators be like "im a propuh gatuh"

naturallyInconsistent

@Relativisticcucumber lol

@Relativisticcucumber no, it actually does change the results, since the commutators are where the interesting physics are happening.

Relativisticcucumber

@naturallyInconsistent but dont we time order by using commutation relations to shift operators over? so if its mathematically equivalent how can it change the result?

naturallyInconsistent

I think you are getting it backwards. We start with the Dyson series, which is explicitly using the time-ordering operator in its expansion, and then use Wick's theorem to convert that to commutators and normal ordering; the normal ordered part is vanquished by the vacuum, and thus all the results are in the commutators.

Relativisticcucumber

bah i have not seen the dyson series yet. i will look it up

bolbteppa

@Sanjana It's just saying that the vector spinor $\psi_{\mu \alpha}$ is irreducible, it does not reduce from a spin 3/2 spinor to a spin 1/2 component on projecting down to $\gamma^{\mu} \psi_{\mu \alpha} = \hat{\psi}_{\alpha}$, this component is always zero, thus the spinor part of the vector-spinor gets thrown out and we have an irreducible representation of the Poincare group

naturallyInconsistent

06:03

@Relativisticcucumber You kinda should already have. It is the unitary time evolution operator in interacting QFT.

Relativisticcucumber

@naturallyInconsistent oh yeah this is free field still

in the notes im currently on

i saw wicks theorem in srednicki interacting tho

naturallyInconsistent

Then I am confused as to why Tong is covering time ordering in a situation that is way before its intended use-point. Especially since there is only one place that time-ordering is used, and it is precisely in the perturbation expansion of the unitary time evolution operator

Sanjana

@bolbteppa Thanks. A not really well defined follow-up question: If I would want to project the vector part instead, that would be also zero, right---and with what should I have projected then?

Relativisticcucumber

@naturallyInconsistent let me look at the next set and see if that clarifies it

naturallyInconsistent

@Relativisticcucumber unlikely; it would have to take a long time to get from free fields to interacting fields

Relativisticcucumber

06:15

but the next set is interacting XD

with the time evolution you mentioned above

naturallyInconsistent

Well, it will still take a long exposition just to get to and understanding of the Dyson series and so on

bolbteppa

There's nothing else you can do right

Relativisticcucumber

@bolbteppa ?

bolbteppa

Should use the index $i$, $\gamma^i \psi_i = 0$, because you are in $SO(D-2)$

and he explains the count in the next minute or so

Regarding the Maharishi comment in the video:

> Q: This ad in Time magazine. It's a two-page ad. Cost $71,800. (It says) the unified field is "described by the supergravity theory of quantum physics as a super-symmetric (perfectly balanced), non-Abelian (self- interacting) field of pure intelligence, which generates the fundamental particles and forces of nature through its infinitely self-referral dynamics at the Planck scale of nature's functioning, (10 to the minus 33 power at CM or 10 to the minus 43 power sec.) giving rise to the infinite diversity of the universe."

Relativisticcucumber

06:35

well still is the definition of the feynman propagator just that the fields are time ordered?

Mr. Feynman

@Relativisticcucumber it is the vacuum expectation value of a time ordered product of two field operators

Relativisticcucumber

@Mr.Feynman and propagator in general is just vacuum expectation value of a product of two field operators, right? no order specifications?

or actually of any number of fields probably?

Mr. Feynman

In QFT one typically uses that as a shorthand for Feynman propagator

Relativisticcucumber

oh, so propagator and feynman propagator are the same?

Mr. Feynman

Then in some cases other Green's functions are useful (such as the one you mention without time ordering)

But those are not called propagators as far as I know

@Relativisticcucumber yes

Wish me luck, my mission is hard

Relativisticcucumber

06:39

hm?

Mr. Feynman

Infiltration

naturallyInconsistent

I would have to point out that the retarded propagator is also commonly considered, if only for pedagogy's sake.

nickbros123

07:01

What does site say about questions like this physics.stackexchange.com/q/792155/346785

Slereah

07:58

@nickbros123 Probably more of a question for the Academia SE

Jagerber48

08:29

@ACuriousMind @ACuriousMind the other day we were discussing spinor dimensionality in the context of your comment on this answer: physics.stackexchange.com/a/791633/128186. You mentioned that Dirac spinors are elements of the unique irreducible representation of the Clifford algebra and that the dimension of that representation is $2^{\lfoor d/2\rfloor}$.

Perhaps the answer is pointing out that the even grade sub algebra of the Clifford algebra form A spinor space, not necessarily the spinor space that corresponds to the irreducible representation of that same Clifford algebra...

ACuriousMind

@Jagerber48 I mean, yes, that's possible - that's why I'm asking in my comment in what sense this is equivalent. However, if this is just "some" spinorial representation, then I find it pretty disingenuous to pretend that this is an answer to your question, because the spinors that matter most in physics are the Dirac and Weyl spinors

naturallyInconsistent

I would have to point out that we really only care about the $d=3+1$ case and not any higher dimensions; and using GA to represent spinors simply happens to work, so why not use it? I am not too bothered that there are potentially extra degrees of freedom we can throw away.

Silly Goose

i cannot find an ArXiv, but this "essay" on quantum geometry seems interesting: journals.aps.org/prl/abstract/10.1103/PhysRevLett.131.240001

ACuriousMind

@naturallyInconsistent I mean with that attitude you can just skip the whole theory altogether and just write down the action of $\mathrm{SO}(3,1)$ on the Dirac spinors $\mathbb{C}^4$ without doing any representation theory at all

no need for any "theory of spinors" at all in that case

Jagerber48

@SillyGoose that was published yesterday :o <24 hrs ago

naturallyInconsistent

08:44

@ACuriousMind I don't think that is equivalent. The expression of spinors using GA perfectly captures the rotation behaviour that we want spinors to have, in a way that is natural to spacetime, including the half-angles. I don't see why having some slight extra degrees of freedom is detrimental to an understanding.

ACuriousMind

@naturallyInconsistent why do you "want" spinors to have that rotation behaviour?

naturallyInconsistent

@Jagerber48 it is typical for stuff to be published on arXiv first. This is definitely true of APS; nowadays they prefer for you to publish to arXiv, and then ask APS to absorb the manuscript into APS so that they would fix things from there

Jagerber48

Yes, I know, just surprising something relevant is so hot off the press.

ACuriousMind

To me, the Dirac spinor as the irrep of the Clifford algebra is straightforwardly motivated by Dirac's original derivation of the $\gamma$-matrices: The square root of the KG equation needs objects that fulfill the relations of the generators of the Clifford algebra, and we ask: What kind of object can fulfill this? The answer is the representation theory of Clifford algebras: Only a vector of dimension $2^{\lfloor d/2\rfloor}$

naturallyInconsistent

@ACuriousMind I'm not sure what your question is? Spinors have that rotation behaviour basically by definition? It was first experimentally discovered that spin-half's rotation properties are so-and-so, and then we figured out a way to make that make sense in some general framework.

ACuriousMind

08:49

@naturallyInconsistent you've got your history wrong: Dirac found Dirac spinors while trying to build a relativistic wave equation, not while trying to explain any "half-spin" phenomena

Jagerber48

@ACuriousMind well no.. there are vectors of higher dimension that would satisfy the equation as well.

Silly Goose

a recount from dirac himself on what ACM just mentioned: youtube.com/watch?v=Ci86Aps7CMo

Jagerber48

The point is those vectors aren't the vectors of lowest dimension that satisfy the equation.

naturallyInconsistent

@ACuriousMind Pauli spinors predated Dirac spinors

ACuriousMind

@Jagerber48 but by rep theory they're all just "multiples" of the Dirac spinor

Ryder Rude

08:52

note that the objects that GA calls spinors are actually operators that act on spinors

naturallyInconsistent

I mean, Dirac knew to try to express Dirac gamma matrices in terms of Pauli matrices. He cannot possibly have not known about Pauli spinors beforehand.

Jagerber48

I'm listening to spinor videos by Eigenchris on youtube. He indicates that "Hestenes" spinors are elements of the even-grade subalgebra of Clifford algebras but that "standard" spinors are elements of left-ideals of Clifford algebras and that the two are isomorphic (I think?) by an application of a projection operator.

naturallyInconsistent

Oh yay Eigenchris!

Jagerber48

@ACuriousMind Sure, but perhaps the perspective is that the story about Clifford algebras (even grade sub algebras or minimal left ideals) answer the question "what is a spinor" and then the physics question about: What is the smallest dimension spinor that satisfies this equation, is an "implementation detail" from the perspective of mathematics?

Ryder Rude

but the number of dimensions is different!

Jagerber48

08:55

@RyderRude Between the even-grade sub algebra and the minimal left ideal?

Ryder Rude

idk what to say. 2 component spin-1/2 spinors can be identified with Pauli matrices

so there is some map at least here

Jagerber48

See youtu.be/… for the eigenchris video comparing the two definitions of Spinors. Not sure how it ties into Dirac/Weyl/Pauli spinors, but he does explore those definitions earlier in the video series, possibly earlier in the same video.

@RyderRude right, so by these Clifford algebra definitions the spin-1/2 Pauli spinors arise from Clifford algebras on vector spaces with dimension 3, like Cl(3, 0).

Ryder Rude

in QM, spinors r vectors on which elements of Clifford algebra can act

but GA wants to call elements of the Clifford algebra as spinors

does this map offer any intuition tho

Silly Goose

I thought in QM spinors are talked about when 1) we have an irrep of a Spin group, 2) elements of the representation space of certain irreps are spinors

Slereah

The Clifford algebra can act on itself

As it is an algebra

Ryder Rude

09:05

it seems to offer some intuition in a 2 component spinors case

ACuriousMind

@Jagerber48 Physically, I can motivate why you'd want the Clifford $\gamma^\mu$ acting on something - cf. Dirac equation. I have no idea why I'd look at the even subalgebra and expect those objects to appear in physics.

Slereah

For instance rotations on vectors are simply defined as members of the algebra acting on itself as $x v x^{-1}$

ACuriousMind

I don't understand what's supposed to be easier or more elegant about looking at this subalgebra instead of just answering the question "What can the $\gamma^\mu$ from the Dirac equation actually act on?"

Ryder Rude

in case of the 2 component spinor, the complex valued vector is hard to identify with a direction, while the corresponding Pauli matrix can directly be identified with a direction

ACuriousMind

and then it doesn't even result in the Dirac spinor we actually use in the general case

Ryder Rude

09:07

so the latter is in some sense more intuitive

ACuriousMind

@SillyGoose The context here is this question and its answers

Ryder Rude

but a spinor is a state vector. it directly encodes probabilities. the probabilities r not obvious in the Clifdord algebra mapping

so it brings some intiuition and removes the more important intuition

Slereah

That's because spinors are not wavefunctions

The states of fermions aren't "spinors"

They're some Hilbert space vectors that transform under the spinor rep

naturallyInconsistent

Arrgh, I got dragged away for something. @ACuriousMind Note that Dirac knew that it would be extremely important for Dirac spinors to have the $g=2$, which was ad hoc added to Pauli matrices just to get agreement with experiment. Dirac even said that he wanted to bask in the joy of discovering his equation, and postponed until the next day to perform the calculation that naturally $g=2$ for Dirac spinors.

Ryder Rude

let's also call the sub-algebra a spinor. it transforms the spinor way. and we still have a state vector for probabilities (which also transforms the spinor way)

so the GA spinor does not replace the state vector

ACuriousMind

09:14

@naturallyInconsistent So? That doesn't change that the fundamental idea of Dirac spinors is "something that is acted upon by $\gamma^\mu$" not "anything in a half-spin representation"

Ryder Rude

we cant get rid of the state vector. even in algebraic QM without a Hilbert space, there is a state

it has its separate use

naturallyInconsistent

@ACuriousMind That's because that is the opposite to the actual logical flow. It is that, under GA (and really this Hamilton's quaternions), we found that the natural way to describe rotations is a rotor, that has the half-angle property, and then later on we note that this half-angle property is pretty much an essence of spin-half, so we then have a reason to try to merge the two concepts.

Ryder Rude

what the even sub algebra can do is give some geometric picture of spinor rep

ACuriousMind

Note that the even subalgebra is not even something that is straightforwardly acted on by $\gamma^\mu$ since the product of something even and $\gamma^\mu$ is odd

@naturallyInconsistent I find nothing "natural" about this

I've never understood people's obsessions with quaternions either

Ryder Rude

this half angle thing also shows up in quaternions. i dont think it has to do with spin

i mean that formula which has a half angle on the left and right

naturallyInconsistent

09:18

@ACuriousMind The thing I was replying to was you trying to point out that Dirac worked out Dirac spinors by building a SR wave equation. I replied that Pauli spinors predated Dirac spinors. It is clear that it is a mix of both, but I don't see why we have to be so fixated on Dirac spinors as the only version to pay attention to.

@ACuriousMind It is natural as being the correct generalisation of rotations in complex numbers to higher dimensions.

Jagerber48

@ACuriousMind my beef with the emphasis on "the things that something acts on" is like this. Someone asks what are A? and the response is "A are the things B act on". But this feels like putting the cart before the horse and you do this whole runaround with representation theory, and the question never really gets answers. Hence why I'm seeking a representation free answer to "what are spinors", noting that we can give a representation free answer to "what are vectors".

Ryder Rude

note that quaternions cannot do a general 4D rotation. they r a special case of 4D rotation matrices

the correct generalisation of rotation to any dimension is matrices

ACuriousMind

@naturallyInconsistent Pauli spinors are just Dirac spinors one dimension lower - the $\sigma^i$ are the $\gamma^i$ for 3d

@Jagerber48 What exactly is the problem here? I say "Dirac spinors are objects on which the $\gamma^\mu$ can act so that their Dirac equation can square to the KG equation". You ask "But what are they?". I say "$2^{\lfloor d/2\rfloor}$-dimensional complex vectors, and I can prove this is the unique choice". What is missing?

naturallyInconsistent

@ACuriousMind But the point of bringing this up is that we knew the rotational behaviour of spin-half stuff by considering Pauli spinors before Dirac came by with Dirac spinors. This rotation behaviour is identical to the GA rotors, so that there is a reason to try to merge the two; at some point it was realised that we can absorb Dirac spinors into the same mathematical framework. Why is this a problem?

Ryder Rude

@Jagerber48 in that case, u shud think of the double cover of the rotation group. it is a Lie group, completely abstract, representation independent

ACuriousMind

09:25

@naturallyInconsistent My problem is that you're not "absorbing Dirac spinors", it's just an accident that this works in low dimensions, while the standard theory of Dirac spinors gives you a theory of spin-1/2 fermions in arbitrary dimension

that's my problem with the dimensional mismatch: yes, the concepts match in low dimension, but that's because there's not much "space" in low dimensions to get anything else

Ryder Rude

i think spinors can b identified with the Lie algebra on that double cover manifold. so it's just the tangent space at identity

Jagerber48

@ACuriousMind I guess you've explained to me what are the things that solve the Dirac equation (this is where the $\gamma^{\mu}$ matrices come from). But why are you calling them spinors? (the same things that are involved in the non-relativistic description of electrons, or in the precession of tops etc.)? Calling them spinors implies they are part of a more general framework, so the question remains, what are spinors?

naturallyInconsistent

@ACuriousMind But again, this is a vacuous objection. We don't really care about Dirac spinors other than the $d=3+1$ specific case; just like how Dirac complained about Feynman's Lagrangian path integral version of QFT being restricted only to Lagrangians with up to 2 derivatives---that is more than enough to get correct answers.

Jagerber48

Like, the definition you gave above is: spinors are the things which make sense in the Dirac equation, but why do those things also make sense in other contexts?

naturallyInconsistent

@ACuriousMind I think this is somewhat talking past each other. I think Jagerber48 is trying to get a realist exposition of what spinors are, whereas you are giving him abstract mathematical definitions.

ACuriousMind

09:29

@Jagerber48 You have that the commutators $[\gamma^\mu,\gamma^\nu]$ fulfill the commutation relations of the $\mathfrak{so}(p,q)$ algebra (this is a general fact about Clifford algebras), so every such spinor automatically carries with it a notion of how it transforms under rotation

Jagerber48

Well abstract mathematical definitions are fine as long as they avoid representation theory (per the rules of the question)

ACuriousMind

and I call them "spinors" because you can show that they have half-spin under those transformations

naturallyInconsistent

@Jagerber48 Well, then ACM already satisfied your requirements, I think.

Jagerber48

I'm missing it in some of the fast flying text, but what was the definition that fulfils the requirements?

ACuriousMind

All of this is representation theory!

Ryder Rude

09:30

u can think of spinors as the tangent space of the double cover SO(3) group

this is representation independent

ACuriousMind

When I say "spinors are objects the $\gamma^\mu$ act on", I'm saying "spinors form a representation of the Clifford algebra"

really, demanding to "avoid representation theory" is to me about as meaningful to tell us to avoid multiplication when writing down formulas

as soon as you have a group of transformations (like rotations) that act on vectors (objects you can sum), you have representations

you can contort yourself to avoid the modern phrasings of representation theory, but you'll still do stuff described by it

Ryder Rude

in textbooks, when anyone writes "spinors" they wud always b talking about a vector on a Hilbert space, not some abstract algebra object that transforms like spinors @Jagerber48

Jagerber48

Well, the point in my question is that, I don't feel like you need representation theory to understand vectors. Given that, please provide an understanding of spinors free of representation theory, or, if that's impossible, explain why the difference

ACuriousMind

@Jagerber48 A spinor is a vector with extra structure

it's not some sort of other entity

it's a vector together with a specific way of transforming under rotations, which we call "half-spin"

Ryder Rude

@Jagerber48 in the abstract, u can hav objects that transform like spinors

naturallyInconsistent

09:34

Meanwhile in GA language, we have something more realist: A quantum particle represented by a spinor field rotates the standard basis tetrad to align with the energy-momentum flow, rotating with the angular-momentum of the quantum particle.

Ryder Rude

but in physics, spinors r state vectors

ACuriousMind

@naturallyInconsistent Do you have a reference/explanation for this claim?

Jagerber48

RIGHT, but how do the rules for how objects "transform under rotations" get set. I HATE the "spinors are things that transform like spinors". You don't need to resort for this for vectors. You have a differentiable manifold and vectors are elements of the tangent space. You can derive how vectors transform when you make a change of coordinates. In particular rotations if you wish. That is, after defining vectors as elements of the tangent space you can DERIVE how they transform.

You don't have to stipulate how they transform ahead of time.

Ryder Rude

note that u wud never get any definition of anything related to a spinor without involving the rotation group

abstract or otherwise

Jagerber48

You don't need the rotation group to get a Clifford algebra. Just a vector space with a metric

naturallyInconsistent

09:36

@ACuriousMind Don't Hestenes's many papers on it explicitly constructs this? That's the whole point of identifying Dirac spinors with GA rotors

Ryder Rude

@Jagerber48 the rotatiom grp preserves the metric

bolbteppa

@Jagerber48 I added an example of $SO(2)$ spinors at the end of my answer which illustrates explicitly what the other answer meant by the 'even subalgebra' in that case

ACuriousMind

@Jagerber48 You are completely correct that spinors in this sense are different from vectors!

in fact, there are manifolds where you have vectors but cannot have spinors

bolbteppa

@Jagerber48 An intrinsic definition of a spinor is that it is a(n element of a) minimal left ideal in a Clifford algebra

ACuriousMind

because in order to have spinors, you need to have a spin structure (which is essentially a choice of spin representation at each point) and there are topological obstructions to that

everyone agrees that spinors need this extra structure; you may "hate" that they are not as intrinsic as vectors, but that's just how it is

@naturallyInconsistent I was hoping to get a more precise answer than "anything by Hestenes" :P

naturallyInconsistent

09:39

@ACuriousMind I'm sure this is the same shit that relates to vierbeins.

Ryder Rude

vectors r defined without metric. there r no spinors without metric/rotation group

bolbteppa

If you think of a Clifford algebra as a bunch of matrices, then a minimal left ideal is a matrix with one non-zero column (basically a matrix acting like a vector) that is acted on from the left by the matrices in the Clifford algebra, sending elements of the ideal into one another (i.e. vectors into vectors)

naturallyInconsistent

@ACuriousMind What about the book by Doran and Lasenby? It had the same construction there too.

ACuriousMind

@naturallyInconsistent oh no, don't tell me this is just a garbled retelling of the idea of a spin connection

bolbteppa

This is just hiding the fact that spinors live in a representation of a Clifford algebra, phrasing it in a new language, you can't avoid this idea

naturallyInconsistent

09:40

With the same mysterious $\beta$ degree of freedom; they write some stuff on it, but I never understood that

bolbteppa

@Jagerber48 A spinor is also technically also just a vector in a vector space (that gets acted on by Clifford algebra elements, or equivalently an element of a minimal ideal which is ultimately the same thing), so you should be happy calling a spinor a vector and moving on right? Obviously not, the important thing is that it gets acted on by a Clifford algebra

Jagerber48

Do any of you follow what I'm trying to paraphrase from Eigenchris about the minimal left-ideal construction of spinors compared to the Hestenes even-grade algebra approach?

Eigenchris says the minimal left-ideal coincides with the standard notion of spinors. I don't know enough about the standard notion of spinors to know if I agree with that

naturallyInconsistent

@ACuriousMind Well, I'm not fully sure of this myself, but when I asked friends who know more about spinors in GR why we need vierbeins, and he was like "you just need?". Later on, someone explained to meow that it was because the metric doesn't care which way is positive and which way is negative, so we kinda need a square rooted object that smoothly connects the choice of directions across tangent spaces

Jagerber48

But both the minimal left-ideal and even-grade algebra constructions satisfy what I'm looking for, so if either of them are passable I'm satisifed.

In both of these cases spinors are particular elements of the Clifford algebra

naturallyInconsistent

@Jagerber48 Meanwhile I do not know enough of minimal left-ideals to check if that is correct. He is, however, correct about the standard notion of spinors. It is very standard to Weyl and Wigner construction of spinors.

ACuriousMind

09:44

@Jagerber48 yes, that is true- simply by virtue of, you guessed it, representation theory: The ideal is in particular a representation of the algebra, and being minimal, it is an irrep. But the irrep of the Clifford algebra is unique (in even dimensions), so they are the same

Ryder Rude

note that a better definition of a spinor is just the projective reps of a rotation group. this does not involve any Clifford algebra

naturallyInconsistent

lol, and now we have yet another invocation of vectors that subtly ignores the physics.

Ryder Rude

Clifford algebra reps often give u decomposable reps from the previous definition

Jagerber48

@ACuriousMind ok, if you agree that elements of minimal left ideals coincide with the standard notion of spinors, then can you work on the dimensions in that case? Does the minimal left ideal thing give e.g. $2^{\lfloor d/2 \rfloor}$?

Ryder Rude

e.g. for Dirac eqn, it gives a direct sum rep

ACuriousMind

09:47

"minimal left ideal" is just a more ring-theoretic way to say "let's decompose this algebra into its irreps under its action on itself"

I'm afraid this is just representation theory in a trenchcoat :P

@Jagerber48 it has to, because it's an irrep of the Clifford algebra!

bolbteppa

@Jagerber48 I even added a comment explaining how you can see these two perspectives in the $SO(2)$ example very explicitly, it's better to study a very specific example if you're confused about all this stuff

Jagerber48

@bolbteppa thanks for all the effort in your answer, unfortunately it is very heavy for me to follow. I think you possibly assume a lot of background knowledge I don't have, but I'm not sure.

It also seems to heavily center on representation theory which I'm specifically trying to avoid (maybe this is an impossible goal)..

@ACuriousMind I'm pondering what you're saying about the relationship between minimal left ideals and representation theory.

Ryder Rude

the Clifford algebra definition works with any metric in any dimension, but Urs says spinors only exist in dimensions which can be identified with complex numbers, quaternions, octonions

Jagerber48

Like... I don't define vectors by saying "vectors are the things that matrices act on"

Ryder Rude

anyone know what Urs was talking about?

i will try to find it. do u recognise any such connection with complex numbers, quaternions and octonions? @bolbteppa

Clifford algebra exists in any dimension

bolbteppa

09:55

Clifford 'numbers' can be used to generate the real, complex, quaternion, and octonion numbers in a unified form

naturallyInconsistent

@Jagerber48 If you don't mind that you would have to completely abandon standard notation, then maybe "spinors are a type of rotor as in GA" and "rotors rotate (and Lorentz boost) vectors on spacetime" would be tolerable, even if they might not map exactly onto the same thing?

Ryder Rude

yes. but spinors r more directly related to Clifford algebras rather than these numbers. Clifford algebras exist in any dimension @bolbteppa

ACuriousMind

@Jagerber48 An algebra $A$ is in particular a vector space. The map $A\mapsto \mathrm{GL}(A), a\mapsto a\cdot$ defines a representation of $A$ on itself by left-multiplication. An "ideal" $I$ is a subvectorspace $I$ such that $AI = I$, so in rep theory language $I$ is a subrepresentation. The ideal is minimal when the subrepresentation is irreducible.

Ryder Rude

but Urs was talking about some connection with these numners

ACuriousMind

it's just two different terminologies for the same idea

naturallyInconsistent

09:58

@Jagerber48 Oh, no, that is not what that particular thing meant. After learning quite a bit of GR, this definition is actually somewhat smart. I've ended up coming to the realisation that all physics theories should start with a set of postulates that sound somewhat circular, but are not.

Jagerber48

@ACuriousMind I see

bolbteppa

@Jagerber48 I see, I am guessing it is heavy because you need to study the unitary group $U(n)$ and its Lie algebra, and the idea of constructing representations of a Lie algebra might be difficult. Technically my answer explains everything as it goes it's not missing anything, but I can see it might be hard without some familiarity, my references explain things in more detail from first principles if that will help

Jagerber48

@naturallyInconsistent Not sure exactly what you mean here, but this sounds exactly like the Hestenes thing and it sounds like that is controversial as to whether that corresponds exactly to (is isomorphic to?) standard spinors or not

Ryder Rude

urs talks about it here physicsforums.com/insights/…

bolbteppa

Honestly it may take months to get this simple point, the eigenchris stuff is probably only going to leave you more confused at the end of all this

Ryder Rude

10:01

apparently, Urs does not mean general spinors

Jagerber48

@naturallyInconsistent Not sure what you mean here. Starting with postulates that are circular sounds terrible. If they only sound circular and aren't actually circular that's ok I guess.. if they can be understood in a non-circular way.

Ryder Rude

he means something called "exceptional spinors" which only exist in these dimensions

these r related to quaternions, octonions

he says "the theory of spinors is most natural in these dimensions"

bolbteppa

It really looks unmotivated to call (left Weyl, assuming even dimensions) spinors the even subalgebra of a Clifford algebra, however if you take the approach in my answer it becomes motivated and natural

The second sentence of the spinor GA paper the answer linked to says

> However, when spinors are defined to be elements of the even subalgebra of some real geometric algebra, the gap between algebraic, geometric, and physical methods is closed.

a paper is not going to casually define spinors in this way if it is in any way controversial

naturallyInconsistent

@Jagerber48 Yes, it is the Hestenes thing. I am fully aware that all this bickering will give you a bad impression of it, but I vastly prefer having a direct physical interpretation that makes sense.

bolbteppa

Honestly it's just confusing because it is in geometric algebra language which very few people know, I don't either and I don't want to

naturallyInconsistent

10:07

@Jagerber48 Yes, only sounding circular. A lot of people don't realise this, but even in Newtonian mechanics we define forces using masses and accelerations, and then define masses using weight, which is a force. This particular circularity is broken by operational definitions, but it still sounds circular.

bolbteppa

I would bet serious internet points that GA cannot give a more intuitive explanation of spinors than my answer (assuming one is familiar with groups like $SU(n)$ and $SO(n)$ and their lie algebras, and the idea of wanting to construct representations of Lie algebras, indeed 'fundamental representations' is usually pretty advanced but I introduced it in a super simple way in the answer)

Indeed the spinor GA paper just does the usual stuff most physics spinor papers do, introducing random things that magically work with no intuition

Jagerber48

@bolbteppa are you able to weigh in on the dimensionality issue that ACM raises?

bolbteppa

A simple way to see there is no issue is that the even subalgebra of the Clifford algebra of $SO(2n)$, when you apply these elements to the vacuum $|0>$, reduces to the $SU(n)$ creation operators acting on the vacuum, thus we have reduced counting the no. of components to counting the number of $SU(n)$ creation operators acting on the vacuum, so the dimensionality goes as I explained in my answer, I'm sure there's another way directly from combinatorics I can't remember

This is not obvious without knowing to invoke $SU(n)$ and it's confusing to think in terms of the Clifford algebra directly I would like the OP to answer how one does the count directly as well

Jagerber48

10:32

In the standard formulation, how are spinors related to the spin group Spin(p, q)?

They are elements of representations of the spin group?

Slereah

It's the vector space associated to its canonical representation

bolbteppa

The spin group is the group generated by $\frac{1}{2}\gamma^{ab} = \frac{1}{4}[\gamma^a,\gamma^b]$, where the $\gamma^a$ satisfy the Clifford algebra $\{\gamma^a,\gamma^b\} = 2 \eta^{ab}I$ where $\eta^{ab}$ has signature $(p,q)$, where the spin group elements are $e^{(i/2)\omega_{ab} \gamma^{ab}}$, and spinors are the things these elements act on, in other words, spinors live in a representation of the Clifford algebra elements $\gamma^a$, again you can't avoid representation theory

Giorgi Lagidze

hi all. had a question about taylor. Assume lagrangian is such as $L(\sigma(q, a), \frac{d}{dt}(\sigma(q, a), t)$. Then the article says that taylor expansion of $L$ around point 0 is: $L(q, \dot q) + a\frac{\partial L}{\partial a}\Bigr|_{a=0} + ...$ and then it proceeds to say that when $L'$ and $L$ are differed by total time derivative, $a\frac{\partial L}{\partial a}\Bigr|_{a=0} = \frac{d\Lambda}{dt}$.

What we did here is we neglected high orders and said that difference between lagrangians is only given by $a\frac{\partial L}{\partial a}\Bigr|_{a=0}$ while in reality, it's more correct to say: $a\frac{\partial L}{\partial a}\Bigr|_{a=0} + a^2\frac{\partial^2 L}{\partial a^2}\Bigr|_{a=0}$.

I am wondering how exact this is. I get that since $a$ is infinetisemal, approximation would be super good, but would it be "perfect" ? we got 2 choices to represent lagrangian change.

$a\frac{\partial L}{\partial a}\Bigr|_{a=0}$ or $a\frac{\partial L}{\partial a}\Bigr|_{a=0} + a^2\frac{\partial^2 L}{\partial a^2}\Bigr|_{a=0}$. The 2nd way is more correct for sure, but since article does first option, what's the "error" approximation ?

Jagerber48

The Clifford algebra somehow gets this navel gazing effect where the Spin(p, q) group lives within the Clifford algebra (as the even graded elements, right?), but because it's a Clifford algebra, it can act on itself. So, if you like, it can be the space where the representation lives. So the Clifford algebra can be both the Spin group and it's representation, the spinors.

I'm curious if people find Eigenchris's mapping between the even grade subalgebra and minimal left ideals, facilitated via a projection, to be compelling: youtu.be/tASmO3NE4IQ?t=1640. I wonder if the dimensionality only works out for small $d$, or if it generalizes to all dimensions in the same way he shows, or in a more complicated way.

See also Sec VII. https://=arxiv.org/pdf/math-ph/0403040.pdf

bolbteppa

10:53

I even did something very similar to that in my SO(2) example, ultimately yes that is what you're going to have to do, but you're still left asking why spinors are minimal left ideals or why they are the even subalgebra of a Clifford algebra

ACuriousMind

@Jagerber48 it makes perfect sense to me, but it shows what I claimed all along: Hestenes' definition is not the same as the usual definition of Dirac spinors

in low dimensions, the notions match because there is no "room" for the even subalgebra to be actually bigger than the minimal left ideal so the kernel of that projector on the even subalgebra is trivial, but in higher dimensions they diverge, and the minimal left ideal is increasingly smaller than the even subalgebra

Jagerber48

"it" being the video or the article? or both? At least the article seems to be making some claim that there is a correspondence that can be made between the two notions.

In any case, where I'm at now: I *think* I'm satisfied with the minimal left ideal definition, and others seem to be satisfied with that one as well. I'm grappling with if I'm ok with the fact that it seems to be representation theory in a trenchcoat.

I like the GA definition more in terms of even sub-algebras, it's very straight forward. But if it doesn't have a simple correspondence to "usual" spinors then it's annoying…

ACuriousMind

@Jagerber48 the video

Jagerber48

Yeah right, the video shows that in low dimensions the correspondence is straightforward. But it's not clear from the video what happens in higher dimensions.

bolbteppa

It's not giving a Dirac spinor in even dimensions it's giving a Weyl spinor

ACuriousMind

11:09

@Jagerber48 In higher dimensions what you get is that the projector will sometimes project out a full even element like $\gamma^i\gamma^j$ to 1 - it still gives a correspondence between the even subalgebra and the minimal ideal, but it's not an equivalence

Jagerber48

I see.. is it correct to say it is a one-to-many correspondence?

ACuriousMind

I mean I'd just say that the projector is not injective

Jagerber48

Yeah

ACuriousMind

but sure you can invert it to get a "one-to-many" map from the ideal to the even subalgebra

Jagerber48

And there are more even-grade-elements than there are elements of minimal left ideals

bolbteppa

11:19

Sorry in the SO(2) example it does give a Dirac spinor...

Ryder Rude

12:20

@GiorgiLagidze their expression is correct upto first order. urs upto the second order

if e is infiniesimal, we only take first order becuz e^2=0

Ryder Rude

12:36

but realistically, Lagrangians with translational symmetry r only first order in $x$ at most. First order corresponds to the pseudo force case

so u can ignore the other terms for realistic Lagrangians

there r fictional lagrangian counter examples like $x^2\dot {x}$, which hav translational symmetry but r second order in $x$

Giorgi Lagidze

@RyderRude yes, but derivation the article is doing is for lagrangians that could be anything, so they still use only the first order. because e is infinetisemal, approximation of only first order will be pretty good, but it won't be 100% perfect which would make the later calculations incorrect.

$a \frac{\partial L}{\partial a} = \frac{d\Lambda}{dt}$ (option 1)
$a \frac{\partial L}{\partial a} + a^2 \frac{\partial^2 L}{\partial a^2} = \frac{d\Lambda}{dt}$ (option 2)

for all lagrangians, option 1 might not give the exact calculations which would cause the conservation law to be derived in…

Ryder Rude

@GiorgiLagidze what is the article trying to derive?

Giorgi Lagidze

noether's theorem (jhwilson.com/blog/2022/Noether-classical) @RyderRude

Ryder Rude

oh

ACuriousMind

@GiorgiLagidze You need to stop thinking about infinitesimals as "approximations". They are the physicists' way to talk about very well-defined mathematical notions (but those notions are not the same in every context). In the context of Noether's theorem, what we have is a continuous flow $q_\epsilon(t)$ with $q_0(t) =q(t)$, and we are interested in $L(q_\epsilon(t)) - L(q(t))$ being a total time derivative $\dot{F}_\epsilon$.

The expansion in the infinitesimal and dropping terms of higher order is equivalent to taking $\frac{L(q_\epsilon(t)) - L(q(t))}{\epsilon} = \frac{\dot{F_\epsilon}}{\epsilon}$ and sending $\epsilon\to 0$.

because only terms of linear order in $\epsilon$ in the numerators will survive, you can drop any higher order terms without making an error

Giorgi Lagidze

12:53

@ACuriousMind if lagrangian was such that it contains $q^2$, dropping higher order term would be incorrect

ACuriousMind

@GiorgiLagidze no, it would not be

again, what we're really looking at when we talk about infinitesimals here is that difference quotient in the limit $\epsilon\to 0$

in that context it is entirely correct

Ryder Rude

@GiorgiLagidze u can show that dL/d(elsilon) at epsilon =0 must be a total derivative for all Lagrangians.

@ACuriousMind this is the proof @GiorgiLagidze

@GiorgiLagidze u can try $L=(x+\epsilon)^2\dot{x}$ and check dL/depsilon at epsilon=0

Giorgi Lagidze

yes, that's what I am doing now

ACuriousMind

No one is claiming that dropping the higher order terms is "correct" in the sense that that would accurately reflect what a finite transformation does. The claim is that this infinitesimal version vanishing/being a total derivative is in fact equivalent to the finite version vanishing/being a total derivative.

This is a specific instance of the general correspondence between Lie algebras and Lie groups (the finite transformation is the Lie group element, the infinitesimal one the Lie algebra one)

Giorgi Lagidze

13:10

okay, it doesn't exactly get the function back. assume $L = \frac{1}{2}m\dot q^2 - mgq^2$ and $L' = \frac{1}{2}m\dot q^2 - mg(q+a)^2$ and we want to get taylor expansion for $L'$. So as you say, $L(0) + a\frac{\partial L}{\partial a}\Bigr|_{a=0}$ is enough without higher order terms. If you calculate this taylor now, you get $L(q, \dot q) -2mgqa$ which is $\frac{1}{2}m\dot q^2 - mgq^2 - 2mgqa$ and this is not the same as $L'$. So taylor expansion we did for $L'$ doesn't give us back the $L'$

Ryder Rude

@ACuriousMind this @GiorgiLagidze

Giorgi Lagidze

I actually am doing the same thing as limit( i am doing the derivative and evaluate it at $a=0$ so I am not coming up with some strange ideas, it's the same thing), the whole point of taylor is to approximate the function

Ryder Rude

the only claim is that, if L(a)-L(0) is a total derivative for all a, then dL/da at a=0 is a total derivative

Giorgi Lagidze

I'm not sure I understand what you meant @RyderRude

does my example make sense to you ?

Ryder Rude

no, u r claiming that L(a)-L(0) requires all derivatives at 0 which is correct

Giorgi Lagidze

13:14

exactly, so why can we drop the higher order terms :P

Ryder Rude

the claim.in the proof is that dL/da at a=0 is a total derivative. note that this total derivative is not the same total derivative as L(a)-L(0)

ACuriousMind

@GiorgiLagidze but what is the problem? $2mgq$ is not a total derivative, so $q+a$ is not a (quasi-)symmetry for this $L$

also, taking translations for this is about the most confusing example you could have chosen because the infinitesimal and the finite translation look exactly the same

Ryder Rude

u can check it with the fictional translationally invariant Lagrangian $x^2 \dot{x}$

ACuriousMind

this is exactly the one example where you cannot really see a meaningful difference between infinitesimal and finite, and the infinitesimal version just looks as if it drops random terms

Giorgi Lagidze

Ah, I think I get what the proof is saying. It doesn't say that taylor expansion for $L$ exactly matches the function without higher order terms

Ryder Rude

13:16

yes

it only says that dL/da at a=0 is some total derivative. This is not the same total derivative as L(a)-L(0)

the latter requires all derivatives of $L$

@ACuriousMind the two total derivatives r related by this derivation @GiorgiLagidze

Giorgi Lagidze

exactly. That was my confusion. can't there be a scenario where difference between lagrangians include $a$ in it (in this case, $\frac{\partial L}{\partial a}$ would end up without including a and equality would be wrong)

Ryder Rude

in L(x)=2x , the difference includes a . what's the problem again?

ACuriousMind

@GiorgiLagidze Equality of what

Giorgi Lagidze

no, that $a$ goes outside

ACuriousMind

23 mins ago, by ACuriousMind

No one is claiming that dropping the higher order terms is "correct" in the sense that that would accurately reflect what a finite transformation does. The claim is that this infinitesimal version vanishing/being a total derivative is in fact equivalent to the finite version vanishing/being a total derivative.

^again, no one is claiming the infinitesimal version is equal to the finite one

just that for the purposes of invariance, it in fact suffices to look at the infinitesimal version

and, again, the broader mathematical underpinning of this is the idea that finite transformations forming a Lie group have infinitesimal versions forming a Lie algebra, and you can show in full generality that invariance under the Lie algebra implies invariance under the Lie group and vice versa

I already pointed out a few days ago that this is the proper setting for Noether's theorem, and that her original formulation was in fact very generally about Lie groups acting on action-like integrals

Giorgi Lagidze

13:29

I have no idea what Lie group is, so maybe that's why it's confusing me, but in an intuitive way, I understand

ACuriousMind

sure, the usual physics texts do not explain in great detail why the "infinitesimal trick" works, but it does work. You can either accept that or you have to learn the actual mathematical theory :P

Giorgi Lagidze

so, if $a$ was not infinetisemal, what wouldn't work in this proof ? taylor expansion still would be the same thing

ACuriousMind

What proof?

Ryder Rude

i dont think the derivation uses infinitesimals

Giorgi Lagidze

@ACuriousMind ibb.co/d2tP3DY

it does say that it uses infinetisemal :P

Ryder Rude

13:32

in the non total derivative case, the derivatioj is just : dL/dx dx/da + dL/dv dv/da=0. this uses no infinitesimals

ACuriousMind

@GiorgiLagidze If you don't have $\alpha$ infinitesimal, then you cannot neglect the higher order terms there!

Giorgi Lagidze

i know about that. i am just only interested in this specific derivation jhwilson.com/blog/2022/Noether-classical

Ryder Rude

in the total derivative case, u just put a total derivative to the RHS. and then it's the same derivation

ACuriousMind

and then you can't conclude that the first order term should be a total derivative

Ryder Rude

i think what we need to prove is that d/da (d/dt (f_a(t)) at a=0 must be a total derivative. then we can proceed with the usual proof

this proof shudnt require infknitesimals

Giorgi Lagidze

13:34

@ACuriousMind can you think of a lagrangian such as change ends up total time derivative, but $\frac{\partial L}{\partial a}\Bigr|_{a=0} \neq \frac{d}{dt}\Lambda$ ?

ACuriousMind

@GiorgiLagidze the question makes no sense

as I said, there is a fully general correspondence between the infinitesimal and finite transformations, and invariance under one implies invariance under the other

Giorgi Lagidze

I am asking for a specific example of lagrangian such as ibb.co/d2tP3DY fails

ACuriousMind

there is none

but the fully general argument (instead of handwaving about infinitesimals) requires Lie theory

Giorgi Lagidze

because you said that if $a$ was not infinetisemal, this wouldn't work, that's why I asked for the counterargument

but seems like there's none and i should be understanding this generally

ACuriousMind

Well, the argument as written wouldn't work

because if you don't omit the higher orders, you'd just get that $\sum_n \alpha^n\frac{\partial^n L}{\partial\alpha^n}$ is a total derivative

Giorgi Lagidze

13:38

exactly

that's what i was feeling would be the most general, most correct way

but since $a$ is infinetisemal, only first order is enough

ACuriousMind

the magic of Lie theory is precisely that the first order vanishing/being a total derivative suffices

Giorgi Lagidze

should I learn about lie theory to understand this ?

ACuriousMind

I can't tell you if you "should"

Slereah

Can't you

It would be a good idea

Giorgi Lagidze

better question would be is it something that can be learned in a day? basically I am only interested why the following proof works for infinetisemal but not for finite in general

ACuriousMind

13:40

@GiorgiLagidze certainly not

Giorgi Lagidze

got it, but at least that's the start

ACuriousMind

and proper Lie theory will require other math you likely do not have if you haven't even heard of "Lie groups" at all

Giorgi Lagidze

without lie theory, i won't understand what I am asking, right ? :D

ACuriousMind

however, there are really 2 options here: Either you get comfortable with handwaving about infinitesimals, or you need to learn a bunch of math

Giorgi Lagidze

none of them is a good option lol

ACuriousMind

13:42

not learning a bunch of math but going into existential crisis mode every time a physicist pulls out an infinitesimal is the worst of both worlds :P

Giorgi Lagidze

I mean, since I am learning all this, either today or after a month, i am 100% sure

there will be another subject that pops up the infinetisemal

and the same argument

one way or another, i need to learn it at some point

I will ask 2 more questions if you don't mind. maybe i will understand these. i understand extremizing the action for a case when we have S[q] and S[q+k(t)] such as q+k(t) is a different path but its starting and end point is the same

this is how we derive EL

but here - ibb.co/vxFyDXf

it also extremizes the action but a(t) is not the same as $a(t_0) = 0$ and $a(t_1) = 0$

because of that, extremizing action concept still holds ?

mainly, object moves from one initial "fixed" location to another final "fixed" location. and q(t) + a(t) is a totally different path that don't have the same initial/end locations at $t_0$ and $t_1$

so is it still correct to say that $s[q(t) + a(t)]$ action should still be bigger than $s[q(t)]$ or in other words, the same in first order ?

Shing

hey, I am studying QFT, a bit confused, would anyone be kind enough to explain to me: why the creator operator is equal to partial derivative with respective to annihilation operator? (in a lecture note)

I made some calculations with $[a,a^\dagger]=1$, all led to contradictions. I suspect creator operator is equal to partial derivative with respective to creation operator instead.

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