The h Bar - 2021-11-29 (page 2 of 2)

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3:07 PM

a possibly very naive question about the standard $\mathbf{SU}(2)$ vs $\mathbf{SO}(3)$ thing: we know there is a two-to-one homomorphism $\mathbf{SU}(2)\to \mathbf{SO}(3)$, which is a universal covering etc. Mostly thinking in terms of qubits and Bloch sphere, we also know an "equivariant map" between these two representations:
given $\psi\in\mathbb C^2$, defining $f(\psi)=(2\bar\psi_0\psi_1,|\psi_0|^2-|\psi_1|^2)\in\mathbb R^3$ (there is some abuse of notation here, hopefully you can tell what I mean), and denoting with $\Phi:\mathbf{SU}(2)\to\mathbf{SO}(3)$ the homomorphism, we have $\Phi…

I suppose the question is similar to the more general: is there a general way to find the equivariant map connecting different representations of a group? I found some discussion on math here, but it flies a bit over my head. Except the map we have in this case is not even really an "equivariant map" according to the standard definition: it's nonlinear, and maps between vector spaces over different fields

I was thinking of asking this on math.SE, but figured I'd pass it through here first to possibly get some insight

I think I’m running into a version of the second problem in a HW question I’ll be grading

@glS the Bloch sphere is neither "3-dim" nor real

Namely, finding a nice isomorphism between different two representations of SO(1,2)

(The vector rep and the spinor rep to be specific)

it's the projective space constructed from the Hilbert space $\mathbb{C}^2$ with the linear SU(2) representation, and what it carries is a projective representation of SU(2) induced from that

3:23 PM

I think I know the answer to my question at the level of the algebra (ie generators) but I’d prefer to be able to do it at the group level (ie exponentials)

@ACuriousMind mh. Good point, one shouldn't think of it as having a linear structure. It's just $S^2$. I'm probably getting confused trying to think of these as linear representatoins, which they might be not. When you say "it's the projective space", I suppose you are thinking about the diffeomorphism $\mathbb{CP}^1\simeq S^2$?

that the embedding of the Bloch sphere into $\mathbb{R}^3$ matches the SU(2) projective representation up with the normal action of rotations as SO(3) on $\mathbb{R}^3$ is essentially a happy accident

(or rather, because su(2) and so(3) are isomorphic, you can choose the embedding such that this is the case)

@glS Yes - in my mind, we should start from $\mathbb{C}^2$ as the Hilbert space of states. It's a general fact that quantum states really are rays in Hilbert space, so if you want a space where every point/element corresponds to a unique state, you have to mod that equivalence relation out

that's what gives you $\mathbb{CP}^{n-1}$ as a space for a quantum system with $\mathbb{C}^n$ as its Hilbert space

Hence why we love pretending everything is just a bunch of qubits. Easier to understand what’s happening when you can break it up like that

and for $n=2$ you find that this projective space is isomorphic to $S^2$, which you can embed neatly into $\mathbb{R}^3$

but nothing about that is general - higher projective spaces are not isomorphic to spheres, and they don't embed into $\mathbb{R}^n$ such that the actions of su(2) or whatever match up with the native rotations in these $\mathbb{R}^n$

@ACuriousMind I’ve heard all of this amounts to the simplest example of the Hopf fibration, though I’ll confess that’s mostly just a slogan to me

3:28 PM

the visualization of the Bloch sphere is useful, neat, and very dangerous because it tries to seduce you to forget that QM is weird :P

@ACuriousMind let me take a step back here. We start from projective representations of SU(2), because we want maps between states, which are rays. So we should really just deal with PU(2), which comes with a natural action on $\mathbb{CP}^1$. Now, we know that PU(2)$\simeq$SO(3), which is equivalent to the usual statement about SU(2)\to SO(3). It is now natural to ask the question: what is the correct way to map between the set on which PU(2) acts, to the set on which SO(3) acts?

But apparently you can apply the various Hopf fibrations to different systems to describe the pure states

@Semiclassical sure - the sphere of unit vectors in $\mathbb{C}^2$ is an $S^3$, and we're "modding out the phases" $S^1$, so we have $S^1\to S^3 \to \mathbb{CP}^1\cong S^2$

Right

Hmm, what’s the latex for an embedding (right arrow with curled tail)

$\hook$

$\hookrightarrow$

3:31 PM

Ah, there we go

the answer is the usual Hopf fibration, thought of as a map $\mathbb{CP}^1\to S^2$, but my question is, how do you get there from solely what we know about the relation between the corresponding (Lie) groups?

I'm afraid I don't really understand the question

If you know that $\mathbb{CP}^1\cong S^2$, then you know that there must be a way to line this isomorphism up such that it maps the action of the generators of SU(2) on $\mathbb{CP}^1$ to the action of the generators of SO(3) on $S^2$

Quoting from Wikipedia: “ Similarly, the topology of a pair of entangled two-level systems is given by the Hopf fibration $S^3 \hookrightarrow S^7 \hookrightarrow S^4$”

I’d guess the idea is similar but it seems less trivial

@ACuriousMind mh, yes that's probably the correct angle. Do we know that? Couldn't the actions of PU(2) and SO(3) on respectively $\mathbb{CP}^1$ and $S^2$ in principle be "incompatible" with the Hopf diffeomorphism $\mathbb{CP}^1\simeq S^2$?

I suppose the counting is to start from a generic C^4 vector (eight real numbers) and impose normalization to get S^7

But then the S3 fibers don’t represent something as simple as choices of overall phase

3:40 PM

@glS If all we knew was that they were "actions", then yes, they might be incompatible. But you can just fix any isomorphism $S^2\cong \mathbb{CP}^1$ and look at what the action of SO(3) becomes when concatenated with that

(Source is this paper: arxiv.org/abs/quant-ph/0108137)

it might not "line up" correctly, but the image will be the action of the projective unitary group on $\mathbb{CP}^1$

then all you have to do is "rotate the sphere around" until the actions match up correctly

aren't you essentially saying that being the base spaces isomorphic, any group action on one can be trivially made into a group action on the other? Sure, I agree. My question is essentially if one could have obtained the specific structure of the diffeomorphism $\mathbb{CP}^1\simeq S^2$ only from the condition of compatibility of the actions. If you already know the isomorphism, you can act on either space without problems, fine.

But imagine you didn't know that $\mathbb{CP}^1\simeq S^2$. Could you have found the map starting from how you wanted it to interact with the group actions?

Hmm. Looking at the paper, the S3 for the two-qubit case is somehow “the Bloch sphere of the second qubit”

@glS Yes, but that's only part of what I'm saying - I'm also saying that when you do this, you observe that both group actions aren't just "actions" but projective representations, and since SU(2) representation theory tells us that any two representations (linear or projective) with the same dimensions are isomorphic, that's the case here, too

i.e. you can start from just knowing $\mathbb{CP}^1\cong S^2$ without any constraint on what this does with the action, and get two different projective representations of SO(3) on $\mathbb{CP}^1$ - one from this isomorphism, one from its construction as a projective space

then you figure out the isomorphism $\mathbb{CP}^1 \to \mathbb{CP}^1$ that shows these two representations are isomorphic

and concatenating these two isomorphisms then gives you the equivariant isomorphism $\mathbb{CP}^1\to S^2$ that plays nice with the group actions

@Semiclassical yes, this strikes me very much as an "we wanted to find a way to cram the other Hopf fibrations in here" and less as a natural generalization :P

3:49 PM

Lol

Fair

I'm not making any judgement about how useful this is (I don't know), just the motivation seems a bit...lacking

The idea that the Hopf fibration could give some insight into entangled qubits, just as it did for single qubits, is appealing

But it does feel unmotivated. What’s the physical meaning of the S3 fibers and S4 base

@ACuriousMind what if you didn't know about the isomorphism at all? You only know that PU(2) has a projective representation on $\mathbb{CP}^1$ and SO(3) has a projective representation on $S^2$, and that PU(2)$\simeq$SO(3). Probably assuming the representations to be faithful. Can you find from this that $S^2\simeq\mathbb{CP}^1$?

They do have a story for separable states, where S4 is for one of the qubits and S3 is for the other

And that does make sense in my brain: any separable 2-qubit state can be rotated to having the first state be the north pole

But then they get to the entangled case and I dunno

Maybe the story should be: use separable states to define what the map should do, and use that to figure out what should happen to non-separable states

4:05 PM

@glS yes - both have the same dimension, and there's only a single projective representation of so(3) with any given dimension

but in order to say that $S^2$ "has a projective representation" you'd already have to know it is isomorphic to some $\mathbb{CP}^n$ because what do you mean by "projective representation" if you're not on a projective space??? :P

so that's kind of tautological

to be honest I don't really know much about projective representations; I know the definitions, but I've never seen some actual derivations/examples. For example, that SO(3) has a projective representation on $S^2$ makes sense intuitively and geometrically, but I don't know how to handle it algebraically

@ACuriousMind well, it has a natural real projective structure, no?

really, there aren't that many 2-dimensional manifolds, it's not very hard to figure out that $\mathbb{CP}^1\cong S^2$ :P

@glS yeah but that's not a "projective representation". Real projective representations aren't interesting because you're only quotienting out $\{\pm1\}$, not an entire group of phases U(1)

I really think this is a typical low-dimension accident

wait a minute though. When we consider rotations on the Bloch sphere we are considering "standard" rotations, not "projective ones", no?

yes

$S^2$ is also not a real projective space btw, the real projective space is what you get when you identify antipodes on it

and it's not embeddable in $\mathbb{R}^3$, so be happy that the geometry works out so you get a Bloch sphere and not an unvisualizable Bloch real projective plane ;)

note that in my sketch above I didn't say that the action of SO(3) on $S^2$ was projective, I said that this action becomes a projective representation on $\mathbb{CP}^1$ once you use some iso $\mathbb{CP}^1\cong S^2$

@ACuriousMind blah. So we are saying that the projective representation of SU(2) on $\mathbb{CP}^1\simeq S^2$ just so happens to descend into an action on $S^2$ that matches the standard linear representation of SO(3) on $\mathbb R^3$ when restricted on $S^2$?

what a happy little accident

4:20 PM

yes, but this is probably not as restrictive as you think

consider that there are many, many diffeomorphisms of $S^2$ (to itself) that turn the action obtained from the rotations on $\mathbb{R}^3$ into something that looks very different

but all these other spheres with their "non-standard rotations" still are isomorphic to $\mathbb{CP}^1$ and its action (just concatenate the standard $\mathbb{CP}^1\cong S^2$ and the diffeo $S^2\to S^2$ you used to generate them)

this would only be a mystery if you knew that there are, like, infinitely many smooth actions of $\mathrm{SO}(3)$ on $S^2$ that are not related by automorphism

mh... example?

I don't know if there are, actually

@glS for what I'm saying about generating different SO(3) actions?

yes

btw, do you know a good reference on projective representations and such?

@glS see my Q&A on them and the references in the comments

as for the example - just rotate the sphere by some degree in some direction

then the rotations no longer "miraculously match up", they're mismatched

but the fix is just to undo the rotation and then do the standard diffeomorphism

@ACuriousMind uh, these things are studied in the context of conformal field theories? No wonder I don't know much about it

4:27 PM

if you take a crazier transformation than a rotation, it will be harder to recognize the result on the sphere is still the same SO(3) action, but it'll still work exactly the same

So $\epsilon_{\alpha \beta}\psi^*{}^\alpha \psi^\beta$ is invariant under $SU(2)$

Do I just impose the fermionic commutation relations?

i.e. $\{\hat{\psi}^*{}^\alpha, \hat{\psi}_\beta\} = \delta^\alpha{}_\beta$?

@ACuriousMind so what do you mean exactly? You take an SO(3) rotation $R$, then consider (linear) actions of SO(3) on $\mathbb R^3$ defined as $A.v\equiv ARv$, with $A\in\mathbf{SO}(3)$ and $v\in\mathbb R^3$, where on the RHS we are using the "natural" linear action?

@glS No, I mean - we have our given action $g.x$ of $\mathrm{SO}(3)$ on $S^2$. Now I take any diffeomorphism $f : S^2\to S^2$

and I define an action as $g.x := f(g.(f^{-1}(x)))$

i.e. I define the action by mapping my point back to the "old" $S^2$ via $f$, doing the "normal" action, and then mapping forward again by $f$

this can give you a bunch of very strange group actions if you choose crazy $f$s

but all those actions will still be isomorphic to the old action (precisely via $f$)

so what I'm saying here is: the accident is not that there is one specific group action on $\mathbb{CP}^1$ and one specific group action on $S^2$ and they happen to match up

What would be an amazing accident would be if we had a "natural" choice of how SO(3) acted on $\mathbb{CP}^1$ and $S^2$ and a natural choice of isomorphism $\mathbb{CP}^1\cong S^2$ and this latter iso would happen to map the two actions onto each other

but the latter is not the case - we are free to choose this iso such that the mapping happens

and in order to say how much of an "accident" that really is you'd have to know how many classes of smooth group actions there are on $S^2$ that are not related by diffeomorphism as I describe above

not sure whether that's really a good question (I don't know much about actions in general outside of representation theory), but that would measure how surprised we really should be

4:50 PM

Why is $\psi^\dagger \psi \phi$ bounded below?

@ACuriousMind I'm not sure I follow. Which action are you saying is not "natural" here? SO(3) on $\mathbb{CP}^1$? The isomorphism $\mathbb{CP}^1\simeq S^2$ is fairly natural... although I don't know if you mean something specific by natural. It's not unique I guess, so non-natural in that sense.

@glS I mean the isomorphism - you're free to choose it, there's nothing unique about it

what makes it unique is saying you want to choose the one that maps the actions to each other nicely

are you saying the situation is not too surprising because for whatever choice of diffeo $f:\mathbb{CP}^1\to S^2$ and (projective) action of SU(2) on $\mathbb{CP}^1$, the corresponding action on $S^2$ is bound to look like an SO(3) rotation?

let's use $g : \mathbb{CP}^1\to S^2$. Then $f \circ g$ with $f : S^2\to S^2$ some other diffeo is still a diffeo $\mathbb{CP}^1\to S^2$

as I said above, that choice of $f$ is what you can use to "adjust" how the group action looks on $S^2$

so you start with any $g$, and then you get some "random" action of SO(3) that might or might not look like a rotation, but it's not a priori clear to me whether it is "surprising" that you can find an $f$ so that it becomes the standard rotation

it would be surprising if there are many smooth actions on $S^2$ not related to each other by any $f$, but I feel like there shouldn't be :P

ah yes: "If a compact connected group whose dimension is greater than that of SO(2)$\times$ SO(n-2) acts continuously and effectively on S^n, then this action is equivalent to the the natural [i.e. that induced from $\mathbb{R}^{n+1}$] action of a subgroup of SO(n+1)" (from here) [pdf link])

so indeed, there aren't any actions of SO(3) on the sphere that are not equivalent to the natural one, and this isn't an accident

5:11 PM

mh. So we are considering an action $\varphi(U):S^2\to S^2$ of $U\in\mathbf{SU}(2)$, defined as $\varphi(U)(v)\equiv g(\phi(U)(g^{-1}(v)))$ with $g$ your diffeo, $v\in S^2$, and $\phi$ the standard projective representation, and ask whether this should always amount to an SO(3) rotation?

@glS not always - we ask whether there's a choice of $g$ always such that this is true

If I know $\phi$ is positive, why can't I just have a potential term $\phi$ to the power $1$?

@ACuriousMind right. Choice of $g$ being the standard one, which we know makes it true..

5:26 PM

I feel like we are going in circles here. You are essentially saying it's not surprising because we know there is a choice of $g$ that makes it true, but the question was essentially about whether such a choice should exist in the first place.

We can say that it's not hard/surprising that $\mathbb{CP}^1\simeq S^2$, i.e. that there is such a $g$, fine, but then it's not really obvious that this choice of $g$, which is *fairly* natural, in the sense of not requiring any sort of weird manipulation, should just so happen to be the one corresponding to the SO(3) nice rotations.

@glS whether it should exist in the first place the reference I just posted above addresses

i.e. any action of SO(3) on $S^2$ is equivalent to the normal one

so such a choice must exist

and the relation between what starts as a projective representation and ends up being a sort-of-linear representation (i.e. the restriction of a linear representation on a compact subset) is also a bit peculiar. Is there any generality to it? Do you get the same starting with a general Lie group and considering projective representatoins of its universal cover (or something like that)

I don't know what sort of generalization we would be looking for - the higher projective spaces aren't spheres

@ACuriousMind yes, I meant about the relation between projective representation of the universal cover and restriction of the linear representation of the group.

I'm not sure what sort of relation you see here that could be generalizable

What I'd have tried to generalize is that the $S^n$ carries an action of $\mathrm{SO}(n+1)$ by restriction

and then you have some low-dimensional projective representation of that group on some $\mathbb{CP}^k$

but since $\mathbb{CP}^k\neq S^n$ in any other case, I don't know what we're looking for

5:38 PM

@ACuriousMind something like $f(\varphi(U)([v])) = \varphi'(\Phi(U))(f([v]))$, where $U\in G$, $\Phi:G\to H$ Lie group homomorphism, $\varphi(U)\in\mathrm{Aut}(\mathbb P(\mathcal H))$ for some Hilbert space $\mathcal H$ is the projective representation, $\varphi': H\to\mathbf{GL}(W)$ is the linear representation, $v\in\mathcal H$ and $[v]\in\mathbb P(\mathcal H)$, and finally $f:\mathbb P(\mathcal H)\to W$ where $W$ is some vector space on which $H$ acts linearly

generalising from $G=\mathbf{SU}(2)$, $H=\mathbf{SO}(3)$, $\mathcal H=\mathbb C^2$, $W=\mathbb R^3$, and $f(\mathbb{CP}^1)=S^2$

so on the LHS we have a projective representation, which is however "equivalent" via $f$ to a linear representation, at least when restricted to a subset of the linear space on which it acts

all projective representations are equivalent to some linear representation (of the universal cover/central extensions)

M. Çağlar TUFAN

Hey everybody

I feel you're trying to construct a very unnatural thing here - don't forget that almost always obtains a projective rep from a linear rep to begin with: We usually start with the Hilbert space and a linear rep, and only pass to the projective space in some cases

M. Çağlar TUFAN

I've got a question. An partiacle with mass m in parabolic orbit will have angular momentum of $J=mr^2\omega$ right?

I assume that the angular momentum is conserved so the motion is on a plane, thus position vector and linear momentum vectors are perpendicular to each other.

5:54 PM

@ACuriousMind I've got to read up more about projective representations I think. Is this stuff covered in the Martin Schottenloher's notes linked above? I can't seem to find too much when googling things like "lie groups projective representations", so I'm not sure whether this is studied in some more specific context, so I'm using the wrong keywords

@glS well, the thing is, for "nice" groups there really isn't much more than what I show in the answer to my question: All the projective reps come from linear reps of the universal cover

you won't find much more about that in Schottenloher either because there just isn't more to it

@ACuriousMind oh, I see. I thought the theory was more developed than that. Guess I'll go read up on that one then

I guess this is a topic of interest mostly to physicists then?

where it gets interesting is when your group has non-trivial central extensions by U(1)

then you have to figure out how to actually compute the $H^2(G;\mathrm{U}(1))$ I define in my answer, and that then is a field all on its own - group cohomology

but nice groups like $\mathrm{SO}(N)$ aren't complicated like that

argh, group cohomology, great.. how bloody deep is the rabbit hole?

Feb 22 '20 at 20:50, by ACuriousMind

There are no different rabbit holes, they are all isomorphic to the universal rabbit hole object

6:07 PM

lol

that begs the question, does the universal rabbit hole object have a bottom? Because I'm not sure it does

that's just asking whether the category of rabbit holes has an initial object!

did you just type "bottom" on ncatlat to see whatever came out? Do they plan to just overload any english word into a categorical term?

yes, but I knew it was there because top/bottom are indeed somewhat common terms for this kind of binary like initial/final, 0/1, true/false

@glS haven't they done that already?

note that TeX has $\top$ and $\bot$ built-in, this is some of the less exotic stuff on nlab :P

6:24 PM

yea, fair enough

7:20 PM

what about

ncatlab.org/nlab/show/being

M. Çağlar TUFAN

I've got a question. An partiacle with mass m in parabolic orbit will have angular momentum of $J=mr^2\omega$ right?
I assume that the angular momentum is conserved so the motion is on a plane, thus position vector and linear momentum vectors are perpendicular to each other.

1 hour later…

8:41 PM

Do classical spinors anticommute?

9:02 PM

@DIRAC1930 not necessarily

spin-statistics only applies to relativistic quantum field theories

you can have a classical spinor field that's Graßmann-even, it's not inconsistent

When I write $\epsilon_{\alpha \beta} \psi^*{}^\alpha \psi^\beta$, is that only invariant under $SU(2)$ if they anticommute?

Maybe not, I think they commute at that stage

But just to make sure I'm doing everything correctly, when lowering indices, do I have to write the indices in exactly the order $\epsilon_{\alpha \beta} \psi^\beta = \psi_\alpha$

i.e. I can't lower like this $\epsilon_{\beta \alpha} \psi^\beta = \psi_\alpha$

It only works for the metric being invariant under transposition for the latter case

9:20 PM

@DIRAC1930 Srednicki's book should be good for this sort of questions

it's not very deep but it's nice for index manipulations etc

but yes, if $\epsilon$ is antisymmetric you obviously get a minus sign if you raise/lower indices "the other way"

Okay, and a consequence of this is that $\phi^\alpha \psi_\alpha = - \phi_\alpha \psi^\alpha$

Is that correct?

9:58 PM

yes

10:39 PM

I have a question: It is known that $F=ma$ only holds for constant mass, otherwise one would have to use $$F=\dot p=\frac{\mathrm dm}{\mathrm dt}+ma$$ Is this any different to using $$F=m(t)a$$ where $m(t)$ describes mass as a function of time? And in both cases, is there anything to pay attention to when the force $F$ depends on the mass and thus time for non-constant mass (say gravity, for example)?

The classic example is the rocket equation

When the rocket expels some of its mass during take off

and it can be checked that this is indeed $F = \dot{p}$

@M.ÇağlarTUFAN this is not correct: the velocity vector will generally have both radial and tangential components. What conservation of J does is place a constraint on how the tangential component varies with radius. The only times when the radial velocity vanishes is at closest or farthest approach

You in fact need that radial velocity to not vanish, in order to satisfy conservation of energy

@Slereah I know that, but I wondered whether using some expression $F=ma$ and let $m$ be time-dependent would give the same result

@Jonas $m$ is already time dependent in the equation you wrote

$$F(\vec{x}, t) = \dot{p}(t) = \dot{m}(t) \dot{x}(t) + m(t) \ddot{x}(t)$$

10:57 PM

I remember the rocket equation being a headache to derive b/c it’s all based on this stuff

In particular i forget if it’s easier to make sense of it in the rocket’s frame or an external frame

Rocket’s frame is of course not inertial, but it’s in that frame that the mass is ejected with fixed speed

Can you say that an operator is unitary only on a subset of the states?

11:12 PM

You could certainly write down such an operator

I think I get it now, thanks!

Eg it being unitary on one sub space but not another

Simplest example being an operator with at least one eigenvalue of modulus one

Okay thanks

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