@Mr.Feynman isn't the vector supposed to live in the tangent space of some position right next to the identity? That means it needs to be parallel transported back to identity's tangent space, and that would need the dg? I don't know if this is the case, but it would be what I think it is doing
@fqq Oh I don't know if one "needs" to. But in the context of explaining the representation theory of orbital angular momentum vs spin, all resources I've seen assume such a partition (cf. Hall's QM for mathematicians chapter 17.6)
oh! interesting okay ill checek it out. i thought tikz was just for commutative diagrams but i just realized from ur comment that i use the package tikz-cd which is probably particularly for cd
@naturallyInconsistent what you describe is done by the left translation, which for matrices is that $g^{-1}$. The $dg$ is the identity map in a tangent space
Yes, you have $v_g\in T_g G$. $dg$ takes it into itself and the left translation takes it into $v_e\in\mathfrak{g}$. So we have $$v_g\overset{dg}{\longrightarrow} v_g\overset{\text{left tr.}}{\longrightarrow} v_e$$
I always swap the overset command arguments :P
As you can see the first arrow is apparently useless
@SillyGoose holy carp that is so impossible. Actually, in classical mechanics the configuration space should be position and velocities. Only then can you do the Legendre transformation to give phase space. But I suppose we have been so incredibly inconsistent that nobody knows if configuration space w/o extra specifications include velocities or not
The tangent bundle is for Lagrangian mechanics, the cotangent bundle (phase space) is for Hamiltonian mechanics. Both of them are tensor spaces over the configuration space.
And then there is momentum space, and if that includes either momentum integral or time derivative, then it is suitable to do classical mechanics too. Just VERY weird.
The configuration space only denotes the position(s) of the various particles. A single particle in 3-dimensional space has configuration space $\Bbb R^3$, but fully describing the state of the particles requires knowledge of it's velocity, the information of which isn't provided in it's configuration space alone.
Actually, one sad thing about Goldstein is precisely that it uses old physics notation and thus does not go into all the crazy detail of tangent bundles, cotangent bundles.
If you want to study those, you have to find something more modern than Goldstein, and the quality of the writing can be seriously scary.
There is actually a tonne of headaches with the way we write down classical mechanics, at least in the old notation. There is this amusing book called "Structure and Interpretation of Classical Mechanics", where the authors were busy fighting over how to teach computers to do classical mechanics. They kept "discovering" that old physics notation is full of inconsistencies. No sure how much of "discovering" they would find if they had access to modern maths notation for classical mechanics.
@SillyGoose If you want a proper mathematical treatment: A classic is Arnold's Mathematical Methods of Classical Mechanics, a more recent one is Abraham's and Marsden's Foundations of Mechanics
i have a question about the definition of coordinates as presented in this image. so if the initial mapping $\phi$ goes from a point $P$ on the manifold to $R^n$ and then the projection operator goes from $R^n \rightarrow R$, i am wondering where the identity of our coordinates comes from? for instance $R^3$ can be described by the coordinates $x,y,z$ or $r,\theta, \phi$, but im not seeing how the difference between these comes into this formalism, so i think im misunderstanding something.
@Relativisticcucumber you're not looking at $\mathbb{R}^n$ as some abstract space with different possible coordinates here
the $\mathbb{R}^n$ is literally just the space whose elements are $n$-tuples $(x_1,\dots,x_n)$ of real numbers
that you can also look at $\mathbb{R}^n$ as a manifold in its own right and then think about coordinates on it is perhaps confusing but irrelevant here
@Relativisticcucumber I do not think you can talk about "identity of our coördinates". The choice of the coördinates is already inside the choice of $\phi$ so that the values that appear in $\mathbb R^3$ have to be interpreted as already being of each particular type of coördinates
@ACuriousMind do you know how to, without deferring to the Lie Algebras, precisely describe why the differences in the irreducible (projective) representations of SO(3) for orbital angular momentum vs. spin come about in representation theoretic terms.
Taking the following as axioms: SO(3) properly describes spatial rotations. Spatial rotations are implemented on Hilbert space via representations.
@SillyGoose Orbital angular momentum is by definition $x\times p$. Stone-von Neumann says the only representation of $[x,p] = \mathrm{i}$ is $L^2(\mathbb{R}^n)$, and you can explicitly show that the $L^2$ for $L = x\times p$ on that space only has integer $\ell$.
i.e. the thing here is that we're not starting from "we want to represent SO(3)" for orbital angular momentum, we're starting from wanting to represent the CCR, which is much more restrictive due to SvN
We will check if "boosting then time evolving" and "time evolving then boosting" r the same. Boosting then time-evolving : $(x, p) $ gets boosted to $(x, p+mv)$. Time evolution using $H=\frac{p^2}{2m}$ gives: dp/dt=0 and dx/dt= p/m + v
And this trajectory is exactly what time evolving then boosting gives u
As in start w/ an arbitrary Hilber space H. Consider projective representations of SO(3) over H. Deduce that irreps corresponding to spin and irreps corresponding to oam exist.
So we have shown that boosting then time evolving is the same as time evolving then boosting. This proof works only when our definition of boost is $(x, p) \rightarrow (x, p+mv) $ @naturallyInconsistent
the classification of irreps doesn't tell you that there is such a thing as "orbital angular momentum" that specifically acts only on $L^2(\mathbb{R}^3) \cong L^2(\mathbb{R},\mathrm{d}r)\otimes \bigoplus_{\ell \in \mathbb{Z}} H_\ell$
there's nothing about the $H_\ell$ in there that would identify them as "representations of orbital angular momentum" rather than "representations of spin"
@Relativisticcucumber $\phi_\text{Cartesian}$ maps manifold point $p$ to $(x,y,z)\in\mathbb R^3$ whereas $\phi_\text{spherical}$ maps the same $p$ to $(r,\vartheta,\varphi)\in\mathbb R^3$, different values, same physical point
@naturallyInconsistent Time evolving then boosting gives u this same trajectory : 1. Time evolving gives u the trajectory: $(p/m t, p)$ 2. Boosting this gives u the trajectory : $((p/m + v)t, p+mv) $
so it is impossible to start with an abstract hilbert space and then discover that there are two "types" of projective reps of SO(3) over it .-. @ACuriousMind
@SillyGoose There are two types! Half-spin and integer spin representations. This just isn't directly related to anything about angular orbital momentum.
let's take an easier example: Take a sphere (like the Earth), choose an equator and a zero meridian, then assign to every point the tuple of latitude and longitude. This defines a coordinate chart from patches of $S^2$ to $\mathbb{R}^2$.
Time evolving gave me $(\frac{p}{m} t, p)$. Now i shud boost this, not according to the standard definition, but acccording to my definition if I'm being consistent
there's nothing about this coordinate chart "mathematically" that tells you that the first component in the $(x_1,x_2)\in\mathbb{R}^2$ "is longitude" (or maybe it isn't and I chose the second component to be longitude!)
I think there is really no way to do boosts on phase space alone. "Boosting then time evolving" works out well, but "time evolving then boosting" gets screwed
@ACuriousMind so is it correct to say: the projective irreps of SO(3), i.e. the irreps of SU(2) come in two types. even dimension and odd dimension. the odd dimension ones can be lifted to be irreps of SO(3). Both even and odd dimension irreps of SU(2) correspond to spins. Odd dimension irreps of SU(2) also correspond to orbital angular momentum?
Err I don't know. All the resources I've read only really look at finite dimensional proj reps of SO(3)
and then Hall which is the only book which looks at infinite dimensional reps that ive read only looks at genuine reps of SO(3), not projective
@SillyGoose This is just separation of variables + spherical harmonics: A $L^2$-function $f(x,y,z)$ on $\mathbb{R}^3$ is the sum of separable functions $R(r)Y(\theta,\phi)$ where $r$ is the radial coordinate in $\mathbb{R}^3$ and $\theta,\phi$ are spherical coordinates on $S^2$. Spherical harmonics tell you there's a basis of spherical harmonics $Y^{\ell m}$, where the set of all $Y^{\ell m}$ for fixed $\ell$ spans an irrep $H_\ell$ of SO(3)
@SillyGoose what do you mean by "correspond to orbital angular momentum"?
The whole point of ignoring time in phase space, as opposed to keeping time also in, in contact space, is because we care more about extracting symmetries from studying Hamiltonians, and those symmetries can either be studied at one time slice, or are invariant in time. Then there is no need to keep the time coördinate
@ACuriousMind okay so can i say that the map from the manifold to the tuples is the coordinate or should it be the tuple itself? im a bit confused now on what i should actually refer to as the coordinate in all of this
So the spherical harmonics are one way of writing the irreps of SO(3). which for each $\ell$ correspond to an orbital angular momentum. then, for a given $\ell$ is this irrep of SO(3) isomorphic to an integer spin irrep?
@ACuriousMind but i thought $\phi$ is the coordinate chart and then the mapping that i think is called a coordinate in that picture i sent is the coordinate chart + the projective mapping? maybe im misunderstanding
@RyderRude having x and p, one half is contravariant and the other is covariant. It happens to be Lorentz invariant if you care about the volume element.
@RyderRude You do not ever write wavefunctions in phase space. Density operators can be coerced into phase space form, but when you need to deal with that, you weep.
what is true is that orbital angular momentum as $x\times p$ induces a very particular representation of SO(3) on $L^2(\mathbb{R}^3)$, in particular one where only irreps of integer spin occur. This does not mean that every integer spin rep is somehow related to orbital angular momentum.
(actually, if you really want to court insanity, you can work with wavefunctions that are parametrised by phase space coördinates, with some smearing between position and momentum, and that is truly infernal.)
@ACuriousMind wait sorry just to clarify one thing -- so you cannot have a chart without having coordinates be defined, right? so if you have a chart, you have coordinates established as well?
@ACuriousMind i think i will answer by analogy to spin because i actually don't have any familiarity with orbital angular momentum. in textbook QM, one first deals with the spin operators. for spin-1/2 systems these are the pauli matrices. the pauli matrices generate the 2D irrep of SU(2) (the 2D projective irrep of SO(3)). This is what I mean by the "2D irrep of SU(2) corresponds to spin-1/2"
@Relativisticcucumber If you take a collection of sets $\{X_i\}_{i\in[1,n]}$ by taking Cartesian products to obtain $(X_1,X_2,...,X_n)$ there are $n$ natural projection maps, $\pi_j:(X_1,X_2,...,X_j,...,X_n)\rightarrow X_j$ whose image is the "$n$th" set in the product
It wud be great if the infinite time limit were a fundamental description of reality. Then the path integral never needs to speak of time-dependent wavefunctionals
@SillyGoose Almost half the books cover from orbital angular momentum first, and the other half from Stern-Gerlach first. You should probably not insist upon one specific view after so many days of bashing head at wall
I don't understand where orbital angular momentum is supposed to enter here, we just generically use the name "spin" for the number that labels a representation
@RyderRude Actually, this is the viewpoint taken in QFT---the problem is not that we dont want to be able to discuss QFT in finite times, but that we really do not know the Hamiltonian well enough to discuss that, and are forced to take the infinite time limit. It is a huge limitation, not something to enjoy
However, if you are going to keep going on about infinite time being fundamental or whatnot, I am out of the convo right now. I cannot be wasting my finite time on this.
@naturallyInconsistent I think the usual justification of infinite time is that the interaction region and interaction time r very tiny, so we can pretend that the total volume of the experiment is infinite
This assumes that finite time QFT is fundamental, yes
@ACuriousMind okay so you're saying we cannot assign the label "orbital angular momentum" to an irrep without deferring to the commutation relation argument you wrote above. because a priori the irreps which are generically called "spin" (a mathematical usage of the word spin, not physical) have nothing to do with orbital angular momentum
you can write down theories of three real scalar fields with an SO(3) symmetry that "rotates in field space" but that symmetry is unrelated to the actual rotations of angular momentum
i.e. the thing here is that we're not starting from "we want to represent SO(3)" for orbital angular momentum, we're starting from wanting to represent the CCR, which is much more restrictive due to SvN
we're looking for "projective representations of the CCR", but the SvN theorem says there is only a single one. It is the representation of the CCR that induces a representation of SO(3) here with generators $L_i$ as components of $L = x\times p$.
@naturallyInconsistent @ACuriousMind In scattering experiments, if we pretend a definite product has been obtained prior to measurement according to Born rule probability, do we get any wrong predictions?
I dont think so becuz the energy eigenstates dont interfere
@SillyGoose It isn't! The spin angular momentum operators commute with $x$ and $p$, that's what we mean when we say spin is "intrinsic" - in contrast to orbital angular momentum, spin angular momentum is not just a consequence of the CCR
@RyderRude how far prior? If it is from state preparation, then you will fail to get the oscillations of different energy eigenstates. If it is right before measurement, then you cannot know the difference, and we have to resort to unnaturalness arguments to point out that it should not be taken as a serious contender for explaining our universe.
but this whole business was not to push for a way to view things but to see if there was an explanation of angular momentum that simultaneously produced the existence of orbital ang and spin
@naturallyInconsistent yes, i only meant after asymptotic future evolution but prior to measurement. And yes, the unnaturaless arguments r there for finite time evolution. Then we hav to resort to keeping all the branches of the wavefunction
@RyderRude The unnaturalness argument is that there is no point prior to measurement that is special enough to be selected out as the place where the system decohered / collapsed / measured / etc
@WaveInPlace what do u find weird? To see that zero divergence doesnt mean zero field, u can consider a constant electric field as the simplest example
This constant field is also technically an electromagnetic wave of wavelength and frequency 0
@naturallyInconsistent yes, this is what i mean. With finite time evolution, there is no way to collapse stuff without getting artificial bad-looking laws
Becuz we have to pick a time-point in that finite time evolution to artificially collapse and get rid of the other branches
So we hav to resort to keeping all branches that the Schrodinger eqn gives us
@ACuriousMind so then what allows us to identify the irreps of SU(2) with spin? If there are multiple views to obtain this same result all would be helpful to hear
No, today you are saying that you are seeking "an explanation of angular momentum that simultaneously produced the existence of orbital angular momentum and spin" This is impossible.
"given an arbitrary Hilbert space, there exists a projective rep[...]" is also impossible, but for a totally other reason
"that we identify with the 'spin of the system' and a distinct other proj[...] which we identify with the orbital angular momentum" is possible, just not in the direction you chose.
@WaveInPlace divergence is a linear operation. To describe actual light, u can superpose plane waves. The divergence of the superposition wud also b zero becuz of linearity
@ACuriousMind why is the 2D irrep of SU(2) the group generated by the operators which are physically the spin-1/2 observable operators; emphasis on physically
@WaveInPlace Every property of light can be correctly captured by studying plane waves and then making wavepackets using them to model real-world behaviour of light.
@naturallyInconsistent in any case it still stands that no one has said until today that either 1) or 2) were impossible to do :P
also from today i verbatim said, "so it is impossible to start with an abstract hilbert space and then discover that there are two "types" of projective reps of SO(3) over it .-. ..." signifying that this was still my q
@WaveInPlace The plane waves form a basis for the space of functions; since the Maxwell equations are linear, you can just solve problems for plane waves, and then for any other situation write the actual field as a sum ("superposition") of plane waves and re-use your solution for plane waves
@SillyGoose What do you mean "mathematically model"
@ACuriousMind / @naturallyInconsistent it would work well until you got down below the wavelength/width of a single unit of light. Ie. the boundaries with QM.
also i am not upset about finding out it is impossible; i am countering your claim that i have been purposefully searching to do something impossible :P the statement made by you seems to imply that you have thought it was impossible all along, which you (or anyone else) have not explicitly stated until today. this is what i am claiming
you are correct that at some point the classical model of EM of course stops working and is replaced by a quantum description, but the relation between the QM notion of photons and an EM wave is more complex than "the classical EM wave is composed of tiny units of light with a certain size"
@SillyGoose No, I only realised that you were asking for an impossible thing just now. Buried under all that verbiage prior to this---we have been on this for sooooo longgggg
@WaveInPlace And I was just telling you that if you are thinking of one wavelength of light as being sufficient to define a photon, then you really do not understand quantum theory at all.
@SillyGoose sometimes you'll see people write stuff like $\mathrm{U}(1)_\text{em}$ or $\mathrm{U}(1)_A$, where the subscripts stand for stuff like "electromagnetic" or "axial", i.e. the physical meaning is attached to the group in the notation
i mean i guess this is actually one of the points of groups :P that what you act on doesnt matter; so generically a group should have multiple uses in disparate physics
@WaveInPlace What is there to "figure out"? The Maxwell equations are just the equations of motion (both classically and quantumly), and light does not have charge because it does not interact with itself at the level of the equations of motion: Two EM waves simply pass through each other due to the linearity of the equations.
the point is that the Yang-Mills Lagrangian produces a non-linear self-interaction term (and hence a "charge") for non-Abelian but not for Abelian gauge groups
hmm i suppose this all makes sense. when you go from operators to groups, there is a loss of information as you jump to groups. so one cannot go from groups to operators :P
in the language of representation theory, charges are associated with non-trivial representations of a symmetry group, and the gauge field always transforms in the adjoint representation - and the adjoint representation of Abelian groups is trivial.
Again, you cannot think about the quantum description just by trying to think of light as chunks of photons or whatever. The quantum description is proper QFT, and you can also show there that - in pure EM without matter - photons will not interact
@SillyGoose No. But you can do Wigner and Weyl, and then you should be happy with angular momentum, because all the important questions you want to ask of it, is quite neatly sorted.
@SillyGoose that's likely a reference to the spin-statistics theorem, which explains why particles with certain spin have certain statistics - in non-relativistic QM there is no reason why fermions should be half-integer spin objects
@SillyGoose What do you mean by "represent orbital angular momentum"? Again, orbital angular momentum is just $x\times p$, we're representing $x$ and $p$, not "orbital angular momentum"
@naturallyInconsistent that's why I said without matter!
you need fermions to get the non-linear box diagarms
@WaveInPlace this is false in the integral PoV too. e.g. u can consider a field that is "created" by a point charge and u can choose ur surface such that the charge point is outside of the enclosed volume. The field inside the volume is non-zero but Guass laws gives zero divergence becuz no charge is enclosed
@SillyGoose Again, why not? If you pick the $\ell=2$ subspace, which is thus finite dimensional, why can it not have a working orbital angular momentum operator on it?
@WaveInPlace But you can enclose multiples of units of light, one single unit of light, or part of a unit of light, and they will always give you zero in the divergence integral. What are you talking about?
hmmm i see okay so the irreps are finite dimensional which i think is a theorem somewhere :P. but the representation of the CCR itself must be over an infinite dimensional space?
@SillyGoose Yes, the CCR, due to SvN, and thus if you want to accurately capture the entire spectrum of orbital angular momentum, yes you have to have infinite Hilbert spaces.
@naturallyInconsistent, look at the figure at the top of this wiki page. If you enclose less than a full wavelength in a surface the flux is imbalanced. Ie. you have a net charge.
If the wave is a plane wave that's not true, obviously.
Most importantly, QFT allows photons to exist on their own. So the situation is the same. Photons dont need electrons anywhere in the universe to exist. The free field solutions are non zero @WaveInPlace
The classical theory allows EM field to exist on its own. QFT allows the quantised EM field to exist on its own
There is not much difference. Its just that the free field solutions can b non zero
There is smthing similar that shows up in General relatovity too where u can hav free solutions only in higher dimensions
I think we lose gravitational waves in 2+1 GR. Is this tru @ACuriousMind
@naturallyInconsistent, put a cylinder/torus around the wave, with its centre along the z axis. If you enclose only one of the blue curves (ie. half a wavelength) there is more flux in the +X than -X.
Oh, the situation is at least internally self-consistent in GR. If you say you have so and so number of dimensions, the degrees of freedom are also that many. One is not so fortunate in condensed matter physics. 1D stuff can vibrate in all 3D. Sickening, I tell you!
@WaveInPlace There is no possible configuration of the shape of the box you choose that can violate the mathematically proved theorem that it will always be zero.
@naturallyInconsistent What do you mean, "also that many"? The formula for the d.o.f. of a gravitational wave in $d$ dimensions is $\frac{(d-2)(d-1)}{2} - 1$
@naturallyInconsistent, it's only zero if the E field does not vary with x, as in the original linked example (where it moved along x and was constant in y). That's kind of my issue. To match experiment the field must vary with x/have boundaries.
Can someone help me understand why do we need $dg$ at all in the way he arrives to it? It maps a vector to itself, so why would I compose $(L_{g^{-1}})_*$ with it instead of just having $\Omega=(L_{g^{-1}})_*$?
@WaveInPlace And I told you 1) those that have boundaries can always be built up from infinite plane waves without boundaries 2) even if you take the case with boundaries and thus the fields are not infinite plane waves, the waves will then curve in such a way to keep this strict zero intact. There is just no way there will be a problem.
@Mr.Feynman Okay, second attempt with less strange notation: If you just write $g^{-1}$ it wouldn't be clear what you're acting on; writing $\mathrm{d}g$ makes it clear this is an action of $g^{-1}$ on the tangent bundle, and $\mathrm{d}g$ is a bit of a strange notation for the identity on the tangent bundle. The reason for this notation is that for any Lie group morphism $f : H\to G$ you have that $f^\ast \omega = f^{-1}\mathrm{d}f$, where $\omega$ is the Maurer-Cartan form on $G$.
In particular this is true when $f$ is an embedding into $\mathrm{GL}(n)$.
so writing $\omega = g^{-1}\mathrm{d}g$ is just a special case of this general fact for $g : G\to G$ the identity map instead of some non-trivial $f$.
> The reason for this notation is that for any Lie group morphism $f : H\to G$ you have that $f^\ast \omega = f^{-1}\mathrm{d}f$, where $\omega$ is the Maurer-Cartan form on $G$
I'll look into that. Does this result have a name that you know?
@Mr.Feynman You need a $\mathfrak g$-valued $1$-form. On $T_eG$, it will be the identity map (viewing it as a $(1,1)$-tensor and then translating into a vector-valued $1$-form). Think of $dg$ as being like $dx$ on $\Bbb R^n$. The first structure equation on $\Bbb R^n$ writes $dx = \sum\omega^i\otimes e_i$. This is also writing the identity map as a $(1,1)$-tensor.
@Mr.Feynman Yes, for $v\in T_gG$, $\Omega(v) = L_{g^{-1}*}(v)$ is correct. But you want a differential form so that you can differentiate and wedge it :)
You missed my final point. How will you take the exterior derivative of that form? Or wedge it with another $\mathfrak g$-valued form? If I write $\Omega = g^{-1}dg$, then I can easily apply the rules of differential calculus/differential forms.
the analogy to $\mathrm{d}x$ is a good point, though - consider that we also have on $\mathbb{R}$ that the identity on tangent space expressed as a differential form is $\mathrm{d}x$
so this $\mathrm{d}g$ is just a generalization of that
Yes, that analogy made me understand what the $dg$ stood for in the first place. On the other hand, putting it there felt as ad hoc as saying e.g. $f\circ\mathrm{id}$ instead of $f$ because the differential of $L_g$ is per se $\mathfrak{g}$-valued form
