## Geometric Quantization (Part 5)

Now let’s start looking at some examples of the adjoint functors introduced in Part 4: quantization and projectivization. It’s really the examples that bring the subject to life. They give new insights into hoary old topics in physics, and also raise some puzzles about the relation between classical and quantum mechanics.

I’ll start with the classical spin-j particle and its quantization. I recently discovered through conversations on Twitter how few physicists have heard of the classical spin-j particle. They all know that the quantum spin-j particle has a Hilbert space $\mathbb{C}^{2j+1}$, an irreducible representation of $\mathrm{SU}(2).$ But the corresponding classical system whose quantization gives this Hilbert space seems remarkably little-known, especially given how simple it is. So, I’ll describe it and its geometric quantization slowly and carefully, before feeding it into our functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

which does this quantization quickly and automatically.

The space of states of the a classical spin-j particle is the sphere, $S^2.$ The radius of the sphere is the total angular momentum of the particle, namely j. A point on this sphere represents a possible angular momentum vector of the particle: a vector $\vec{J} \in \mathbb{R}^3$ with length j. There are 3 important observables in this system: the components of the angular momentum, $J_x, J_y$ and $J_z.$

To do classical mechanics with this system we want a symplectic structure on the sphere. A symplectic structure determines Poisson brackets of observables, and we want to make sure that

$\{J_x, J_y\} = J_z, \quad \{J_y, J_z\} = J_x, \quad \textrm{ and } \quad \{ J_z, J_x\} = J_z$

There’s a unique symplectic structure $\omega$ that does the job. A symplectic structure on the sphere is just a nowhere vanishing 2-form (since such a 2-form is automatically closed and nondegenerate), and you can almost guess $\omega$ just from the fact that it had better be rotation-invariant, so it has to be some multiple of the usual area 2-form on the unit sphere. If you figure out which multiple gives the above Poisson brackets, you see that $\omega$ must be $j$ times the area 2-form for the unit sphere. With painful explicitness:

$\omega = j \sin \theta d\theta \wedge d\phi$

This is a bit peculiar: you might expect $j^2$ here instead of $j,$ since the area of a sphere of radius $j$ in Euclidean space is proportional to $j^2.$ But that’s not what the calculation gives! If you want some more justification for the Poisson brackets we’ve chosen, which force this symplectic structure, you can think of the sphere as a coadjoint orbit of the group $\mathrm{SU}(2),$ and use the fact that any such coadjoint orbit gets a Poisson structure, which happens to give the above Poisson bracket formula in this example. Or, you can use dimensional analysis to see $\omega$ must be proportional to $j,$ since they both have units of action. But let’s not get into that here.

To geometrically quantize our symplectic manifold using Kähler quantization, we need to equip it with lots of extra structure, as explained in Part 2. For starters, we need to give it a complex structure. There’s a unique one invariant under rotations, namely the one that makes our sphere into a copy of the Riemann sphere $\mathbb{C}\mathrm{P}^1.$ There is then a unique Kähler structure on $\mathbb{C}\mathrm{P}^1$ whose imaginary part is our symplectic structure $\omega.$ The real part is a Riemannian metric: the usual metric on a round sphere of radius $\sqrt{j}$ in Euclidean space. (Again, that square root looks peculiar, but that’s what we get.)

Next we need to choose a hermitian line bundle over the sphere, equipped with a hermitian connection whose curvature is $i \omega.$ By the miracle of algebraic topology, this exists precisely when the integral of $\omega$ over the sphere is $2 \pi$ times an integer. Since this area is $4 \pi j,$ this forces $j = 0,1/2,1,3/2,\dots.$ So, while we can work with the classical spin-$j$ particle for any real $j \ge 0,$ we can only quantize it when $j$ takes the usual integer or half-integer values!

Let’s focus on the spin-1/2 particle for a while. In this case we can easily describe the relevant line bundle over our sphere. It’s just the bundle I’ve been calling $L:$ the dual of the tautological line bundle on $\mathbb{C}\mathrm{P}^1.$ The holomorphic sections of this bundle are just linear functions $\psi \colon \mathbb{C}^2 \to \mathbb{C}.$ So, the space of sections is ${\mathbb{C}^2}^\ast,$ or just $\mathbb{C}^2$ if we identify this space with its dual using the usual inner product.

That’s good! We got the right answer! But as you can see, the process seems rather long and tortuous.

If we use our functor

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}$

everything goes a lot faster, mainly because all the choices we had to make are built into the definition of $\texttt{Class}.$ An object in this category, you’ll remember, is a linearly normal subvariety $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ for some arbitrary $n.$ When we quantize it we get $\texttt{Q}(M) = V$ where $V \subseteq \mathbb{C}^n$ is the smallest linear subspace with $M \subseteq PV.$

In other words, quantization ‘flattens out’ or ‘linearizes’ $M,$ replacing this possibly quite interesting projective variety by the smallest vector space whose projectivization contains this variety.

For the spin-1/2 particle, we take $M = \mathbb{C}\mathrm{P}^1.$ And we get $\texttt{Q}(M) = \mathbb{C}^2,$ since this is the smallest subspace of $\mathbb{C}^2$ whose projectivization contains $\mathbb{C}\mathrm{P}^1.$ Voilà!

That was pretty trivial. But it was trivial for an interesting reason. Remember from Part 4 that we have an adjoint functor going back from the quantum to the classical:

$\texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

This sends any subspace $V \subseteq \mathbb{C}^n$ to its projectivization $P V.$ Moreover, we have

$\texttt{Q} \circ \texttt{P} = 1_{\texttt{Quant}}$

In words: if we quantize the projectivization of a quantum system we get that quantum system back.

And that’s what we’re doing in the case of the spin-1/2 particle! The space of states of the classical spin-1/2 particle, $\mathbb{C}\mathrm{P}^1,$ was the projectivization of $\mathbb{C}^2.$ So when we quantize it, we just get $\mathbb{C}^2$ back.

That’s not how it will work for the spin-1 particle, or any higher-spin particle. The spin-j particle still has $\mathbb{C}\mathrm{P}^1$ as its classical state space, but when we quantize it we get $\mathbb{C}^{2j+1}.$ That’s what I’ll talk about next time.

Now, experts may be yawning at this point, because they already know how to handle such a higher-spin particle! We simply choose a different Kähler structure on $\mathbb{C}\mathrm{P}^1,$ and a different line bundle over it. Namely, we rescale the Kähler structure we’ve already been talking about to make the total area of the sphere be $4 \pi j.$ If j = k/2, this means we have to multiply our earlier Kähler structure by k. And to get a line bundle with a connection whose curvature is the imaginary part of this rescale Kähler structure, we just take the kth tensor power of our line bundle $L.$

The holomorphic sections of this new line bundle $L^{\otimes k}$ are homogeneous polynomials of degree k on $\mathbb{C}^2.$ The space of all these has dimension $2j+1,$ and this is the right space of states for quantum spin-j particle.

All this is great, and it’s an illustration of a theme I’ll eventually want to talk about much more: when you’re doing the Kähler quantization of some physical system, you can always multiply your Kähler structure by a natural number k, and take the kth tensor power of your line bundle, and get a new system!

What does this mean physically? I’ll give away part of the answer now: it corresponds to dividing Planck’s constant by k! You see, angular momentum is really measured in units of Planck’s constant, so using this procedure to go from the spin-1/2 particle to the spin-5/2 particle (for example) is the same as dividing Planck’s constant by 5.

However, in the setup described last time, we had no explicit choice over what Kähler structure or line bundle to use on our subvariety $M \subseteq \mathbb{C}\mathrm{P}^{n-1}.$ It inherited those structures from $\mathbb{C}\mathrm{P}^{n-1}.$ So to change those structures, we need to embed $M$ in a different way, perhaps into a different projective space.

This may seem a bit silly, and in a way it is—but not completely.

As I mentioned last time, we are describing the geometry of varieties extrinsically, through their embedding in projective space. This is a bit old-fashioned compared to the more modern intrinsic viewpoint. At some point I’d like to switch to an intrinsic approach. However, the interplay between the intrinsic and extrinsic approaches is a time-honored theme in algebraic geometry, and we should exploit it in geometric quantization!

In particular, even if we start with an abstract variety not embedded in any projective space, once we choose a line bundle $L$ over it we can often embed in a projective space built from $L.$ By definition, we can do this whenever $L$ is a ‘very ample’ line bundle. Sometimes $L$ isn’t very ample but $L^{\otimes k}$ is whenever k is large enough; then we call $L$ ‘ample’. The study of ample line bundles is a big deal in algebraic geometry.

So, it’s not so silly to start our research on geometric quantization by taking our varieties to come equipped with an embedding into a projective space, and then later think about how we get such an embedding if we don’t already have one.

So, next time I’ll geometrically quantize the spin-j particle by taking the space of states of a classical spin-j particle, $\mathbb{C}\mathrm{P}^1,$ and embedding it, not in itself, but in some higher-dimensional projective space. The math involved is well-known; the interesting part to me is its physical interpretation.

Part 1: the mystery of geometric quantization: how a quantum state space is a special sort of classical state space.

Part 2: the structures besides a mere symplectic manifold that are used in geometric quantization.

Part 3: geometric quantization as a functor with a right adjoint, ‘projectivization’, making quantum state spaces into a reflective subcategory of classical ones.

Part 4: making geometric quantization into a monoidal functor.

Part 5: the simplest example of geometric quantization: the spin-1/2 particle.

Part 6: quantizing the spin-3/2 particle using the twisted cubic; coherent states via the adjunction between quantization and projectivization.

Part 7: the Veronese embedding as a method of ‘cloning’ a classical system, and taking the symmetric tensor powers of a Hilbert space as the corresponding method of cloning a quantum system.

Part 8: cloning a system as changing the value of Planck’s constant.

### 9 Responses to Geometric Quantization (Part 5)

1. This has been a great series of posts that leads to a lot of interesting questions. I’m still a grad student so maybe you’ll forgive any naive ones.

One question is about G/P and the corresponding geometric objects, how do they fit into this picture? For example, in Lie type A we have G/B where B is the stabilizer subgroup of a full flag. Then with partial flags we obtain various G/P. These can be interpreted as specifying certain geometric figures and their incidence relations within projective geometry. All of these G/P are projective varieties and so we can apply the functor you defined. But what happens in other types? And, in both type A and other types how do the geometric objects correspond to aspects of the quantized systems?

Another question I had concerned varying the projective embeddings. For the Grassmannian we have the Plucker embedding, and this suggests considering taking various exterior powers and then projectivizing, to obtain other projective embeddings. You already mentioned that the cartesian product of vector spaces comes with the Segre embedding into the tensor product. Are there any other sources of projective embeddings to keep in mind? As we change our projective embedding, what properties of the quantized system change and how?

Again, thanks for the posts, I think they have helped my understanding a lot and even better raise interesting questions to think about!

• John Baez says:

Your questions are not naive at all. Mostly you’re raising issues that I’d love to talk about in future posts if I have time. (The range of issues I’d like to discuss keeps expanding faster than I can write.)

One question is about G/P and the corresponding geometric objects, how do they fit into this picture? For example, in Lie type A we have G/B where B is the stabilizer subgroup of a full flag. Then with partial flags we obtain various G/P. These can be interpreted as specifying certain geometric figures and their incidence relations within projective geometry. All of these G/P are projective varieties and so we can apply the functor you defined.

I should probably explain for those not in the know. Here $G$ is a complex simple Lie group, $B$ is a Borel subgroup, and $P$ is parabolic subgroup. The spaces $G/P$ are known as as generalized flag varieties. They are compact manifolds, but there are well-known ways to put Kähler structures and compatible line bundles on them, so we can hit them with Kähler quantization—and by varying our choice of $P$ and the extra data, we can get all the irreducible representations of $G$ this way!

The buzzword for this whole network of ideas is ‘Borel–Weil–Bott theorem’, and all mathematicians should learn this stuff.

As you note, when $G$ is of type $\mathrm{A}$ (that is, $\mathrm{SL}(n,\mathbb{C})$ for some n), these generalized flag varieties $G/P$ are spaces of ‘figures’ like points, lines, planes, etc. in projective geometry. When $G$ is of some other type, these generalized flag varieties are spaces of figures in other kinds of geometry, like < a href = “https://cameroncounts.files.wordpress.com/2015/04/pps1.pdf”>’polar geometry’. This is a whole huge fascinating story in itself. But what it means for us here is that all these kinds of geometry can be quantized!

What does that actually mean? I’m not completely sure except in examples like $G = \mathrm{SL}(2,\mathbb{C}),$ where quantized incidence geometry plays a big role in ‘loop quantum gravity’ and ‘spin foam models’—attempts to study geometry quantum-mechanically. It would be fun to generalize those ideas to other groups in a very systematic way.

Today I just did the smallest bit of this enormous story, taking $G = \mathrm{SL}(2,\mathbb{C})$, taking $P = B$ to be the subgroup of upper triangular matrices in $\mathrm{SL}(2,\mathbb{C}),$ and getting $G/B = \mathbb{C}\mathrm{P}^1.$

Another question I had concerned varying the projective embeddings. For the Grassmannian we have the Pl¨cker embedding, and this suggests considering taking various exterior powers and then projectivizing, to obtain other projective embeddings.

At the end of this post I mentioned that next time I’d be quantizing the spin-j particle for j > 1/2 by looking at embeddings of $\mathbb{C}\mathrm{P}^1$ in higher-dimensional projective spaces. This is a bit like what you’re talking about. Maybe later we can look at other examples. Right now I’m mainly interested in examples that give me ideas about physics, but probably every example would give me ideas about physics if I were smart enough.

I feel I didn’t really answer your questions, but that’s probably because I think it’s the most fun to look at specific examples and draw conclusions from what happens, and I plan to do more of that in future posts. Stay tuned!

• Hi Alex. One thing you might be asking is, if John is studying subvarieties of projective space, why would someone else be studying subvarieties of other $G/P$, in particular other Grassmannians?

The simplest answer to that is, projective spaces (all put together) are the classifying space for line bundles, whereas $k$-Grassmannians (all put together) are the classifying space for $k$-plane bundles.

I don’t know how to make that desirable to a classical mechanic. On a line bundle, we can put a connection, take its curvature, and get a (possibly) symplectic $2$-form. On a $k$-plane bundle, we can put a connection, take all its Chern-Weil forms, and get lots more forms. But I don’t know whether a physicist would say “hey, my phase space actually had all these other $2j$-forms on it, that I didn’t know what to do with.”

Alternately, if $M$ carries a $k$-plane bundle $\mathcal V$, we can look at the antitautological bundle $\mathcal O(1)$ on $\mathbb P \mathcal V$, push it down to $M$, and get $\mathcal V$. So in that sense, we can work with the projectively embedded $\mathbb P\mathcal V$ instead of the Grassmannianly embedded $M$. At that point, the physicist’s question becomes “is my phase space actually a projective bundle over something smaller, and maybe that’s the right space to work with instead”? Again, I don’t really know examples of that.

• I realize I was misreading your question, which is not about generalizing the ambient projective space to a different ambient $G/P$, but about what the physics should be associated to $G/P$ as a phase space.

I think the easiest point of view is to see $G/P = K/L$ as a symplectic reduction of $T^* K$, where $K$ is the maximal compact of $G$ complex (so $G$ is $K$‘s complexification). Start with a free particle traveling on $K$. By using the left or right $K$-action to move the particle back to the origin, one can derive two $\mathfrak k$-valued “momentum vectors”, and from there, take their $K$-invariant functions. (The $K$-invariants of the left $\mathfrak k$-momentum determine those of the right one, and vice versa, a classical version of the Peter-Weyl theorem.)

If you fix the value of these invariants (constant on the $K\times K$-orbits), you get a subset of the phase space $T^* K$, which you can then divide by the right $K$-action. I’m not sure how best to describe the resulting phase space physically, other than as a free particle on $K$, with a fixed kind of momentum, up to the right action.

The dumbest example would be about a free particle on the round $3$-sphere in $\mathbb R^4$. It’s got a $4$-momentum, an antisymmetric $4\times 4$-matrix. The $SO(4)$-invariant momentum to compute here (note that $S^3\times S^3 \twoheadrightarrow SO(4)$!) is the kinetic energy $|\vec v|^2$. So assume the energy is fixed, use the right $S^3 = SU(2)$ action to move the particle back to the origin $e$, and the remaining freedom is a point in $T^*_e(S^3)$ with fixed $|\vec v|^2$.

2. Lurking physicist says:

Regarding the prefactor of $j$ instead of $j^2$ in the symplectic form $\omega$, a physicist’s low-brow way of seeing it is to recall the angular variable $\phi$ and the angular momentum $L_z = j\cos\theta$ form canonical conjugate pair, and hence the prefactor of $j$.

I am one of those physicists who did hear about the classical spin-$j$ particle though I’ve never learned symplectic geometry. A few years back, I found an interesting connection, perhaps well known among mathematicians, between the classical spin-$j$ particle and George Marsaglia’s method of picking random points in $S^2$ with uniform probability density. In the latter, Marsaglia employed the following transformation: $x = 2x_1\sqrt{1-x^2_1-x^2_2}$, $x = 2x_1\sqrt{1-x^2_1-x^2_2}$, $y = 2x_2\sqrt{1-x^2_1-x^2_2}$, and $z = 1-2(x^2_1+x^2_2)$, where $(x,y,z)$ are Cartesian coordinates of a point on $S^2$, and $(x_1,x_2)$ are Cartesian coordinates of a point on a unit disk. This transformation is a symplectic map and thus volume preserving. It then follows that a uniform probability density on the unit disk, through this transformation, generates a uniform probability distribution on $S^2$.

3. Classically, if we take a vector of length 1/2 in $\mathbb{R}^3$ and triple it we get a vector of length 3/2. This gives a map from classical states of the spin-1/2 particle to classical states of the spin-3/2 particle. In other words: a map from the sphere of radius 1/2 to the sphere of radius 3/2. Simple! But this is not a symplectic map, since as we saw last time, we are giving the latter sphere a symplectic structure that’s 3 times as big. So, it’s not a valid classical process. Still, it’s a perfectly fine way to get our hands on lots of states of the classical spin-3/2 particle! All of them, in fact.

4. Bruce Bartlett says:

Can geometric quantization of the spin j particle give us a geometric way of understanding why the square of the total angular momentum,

J_x^2 + J_y^2 + J_z^2

is quantized to be j(j+1) and not j^2? That is the central mystery to me which I would love to see a geometric explanation for. The radius of the sphere of the spin 1/2 particle is somehow supposed to be Sqrt(3)/2 and not 1/2…. even though when you measure J_z you can only ever get +- 1/2. I would love to see a geometric quantization explanation for that.

• John Baez says:

I could argue that once you get geometric quantization to cough up the irreducible representations of SU(2) as sections of equivariant line bundles over the sphere, you’re done. Then the Casimir will act as an operator on any of these irreducible representations, and it’s bound to have the right eigenvalue: $j(j+1).$

But you seem to be wanting to see a sphere of radius $\sqrt{j(j+1)}$ pop out of the geometry, before you talk about any operators and their eigenvalues. That would be cool, but I don’t think that’s going to happen.

This reminds me of some approach where the answer comes out to be $(j+\frac{1}{2})^2.$ The Duflo isomorphism is an isomorphism of algebras going from the Ad-invariant part of the symmetric algebra of a Lie algebra to the center of its universal enveloping algebra, and we can use this to quantize the Casimir of SU(2) and get

$(j+\frac{1}{2})^2 = j(j+1) + \frac{1}{4}$

this is tantalizingly close to being ‘correct’.

It might be easier to get a sphere of radius $j + \frac{1}{2}$ to come out of geometric quantization than a sphere of radius $\sqrt{j(j+1)}.$ Then the slight error could be a source of lifelong frustration.

• Bruce Bartlett says:

Ok – thanks, I will ponder this.

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