physics | Azimuth

Geometric Quantization (Part 6)

1 January, 2019

Now let’s do some more interesting examples of geometric quantization using the functor described in Part 4. Let’s look at the spin-j particle with j > 1/2.

To be specific, let’s consider the spin-3/2 particle. There’s nothing special about the number 3 here: everything I’ll say can be generalized. But the number 3 will give me a nice excuse to show you a picture of a curve called the ‘twisted cubic’.

We can build a spin-3/2 particle from three spin-1/2 particles, all having angular momenta pointing in the same direction. Classically this procedure amounts to tripling a vector, but quantum-mechanically it’s related to cubing. This is a bit mysterious to me, but let me explain.

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.

Quantum-mechanically the state space for the spin-1/2 particle is $\mathbb{C}^2.$ We can think of a guy in here as a linear functional on the dual ${\mathbb{C}^2}^\ast.$ We can cube this and get a homogeneous polynomial of degree 3 on ${\mathbb{C}^2}^\ast.$ The space of such polynomials is called $S^3(\mathbb{C}^2),$ and this is the space of states of the quantum spin-3/2 particle. So, we get a map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Simple! This describes the process of ‘triplicating’ the state of spin-1/2 particle and getting a spin-3/2 particle. But this is not a linear map, so it’s not a valid quantum process. As the saying goes, “you can’t clone a quantum”. Still, it’s a perfectly fine way to get our hands on lots of states of the quantum spin-3/2 particle! But not all of them, as we’ll soon see.

Geometric quantization should reconcile and combine classical and quantum mechanics. How can we do this here?

It’s pretty simple. We projectivize the map

$\text{cubing} \colon \mathbb{C}^2 \to S^3(\mathbb{C}^2)$

Cubing sends any line through the origin in $\mathbb{C}^2$ into a unique line through the origin in $S^3(\mathbb{C}^2).$ So, we get a map from $\mathbb{C}\mathrm{P}^1$ to the projective space of $S^3(\mathbb{C}^2).$ And we define the image of this map to be the space of classical states of the spin-3/2 particle!

Let’s see what it looks like. I’m thinking of an element $(a,b) \in \mathbb{C}^2$ as a linear functional

$f(x,y) = ax + by$

If we cube it we get this homogeneous polynomial of degree 3:

$f(x,y)^3 = a^3 \, x^3 + 3a^2b \, x^2y + 3ab^2 \, xy^2 + b^3 \, y^3$

If we take $x^3, x^2y, xy^2$ and $y^3$ as our basis of the homogeneous polynomials of degree 3, we can identify $S^3(\mathbb{C}^2)$ with $\mathbb{C}^4.$ So we get

$\begin{array}{clll} \text{cubing} \colon & \mathbb{C}^2 &\to& \mathbb{C}^4 \\ & (a,b) & \mapsto & (a^3, 3a^2b, 3ab^2, b^3) \end{array}$

If we projectivize this map we get a map between projective varieties, which I’ll call

$P(\text{cubing}) \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^3$

The image of this map is the space of states of the classical spin 3/2-particle! It’s a copy of $\mathbb{C}\mathrm{P}^1$ sitting inside projective 3-space. But because cubing a nonlinear map, this copy will be twisted: not a ‘line’ but a ‘curve’. People call it the twisted cubic.

Just for kicks, let’s take a closer look at it. Let’s use homogeneous coordinates and write $[a,b]$ for the point in $\mathbb{C}\mathrm{P}^1$ corresponding to a nonzero vector $(a,b) \in \mathbb{C}^2.$ Similarly, any point in $\mathbb{C}\mathrm{P}^3$ can be written as a nonzero 4-tuple of complex numbers with a bracket around it. We get

$\begin{array}{cccl} P(\text{cubing}) \colon & \mathbb{C}\mathrm{P}^1 &\to& \mathbb{C}\mathrm{P}^3 \\ & [a,b] & \mapsto & [a^3, 3a^2b, 3ab^2, b^3] \end{array}$

so the twisted cubic is

$\{ [a^3, 3a^2b, 3ab^2, b^3] : \, (0,0) \ne (a,b) \in \mathbb{C}^2 \} \subset \mathbb{C}\mathrm{P}^3$

This is hard to visualize, so we can work with a copy of $\mathbb{C}^3$ that’s dense inside $\mathbb{C}\mathrm{P}^3,$ namely the set where $a = 1.$ The portion of the twisted cubic sitting in this set is

$\{ (b, 3b^2, b^3) : \, b \in \mathbb{C} \} \subset \mathbb{C}^3$

This is still hard to visualize, so we can restrict to the reals and think about the curve

$\{ (b, 3b^2, b^3) : \, b \in \mathbb{R} \} \subset \mathbb{R}^3$

This is also called the twisted cubic! People often rescale the $y$ axis to get rid of the number 3 here.

This version of the twisted cubic is still a bit hard to visualize, but it’s the intersection of two very nice surfaces: the surface $y = x^2$ and the surface $z = x^3.$ So, the twisted cubic is the black curve:

in this nice picture uploaded to ResearchGate by Alexander M. Kasprzyk.

Okay, back to serious business! Let’s call the twisted cubic $C_3 \subseteq \mathbb{C}\mathrm{P}^3.$ It’s exactly the sort of thing we can geometrically quantize using our functor

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

The reason is that it’s a projective variety and it’s linearly normal. Moreover when we quantize $C_3$ we get $\mathbb{C}^4,$ which is just what we want for the spin-3/2 particle!

$\texttt{Q}(C_3) = \mathbb{C}^4$

Why? Remember, $\texttt{Q}(C_3)$ is defined to be the smallest linear subspace $V \subseteq \mathbb{C}^4$ such that $C_3 \subseteq P(V).$ But $C_3$ twists around so much that the smallest $V$ that works is all of $\mathbb{C}^4.$

Now let’s take stock of where we are and draw some general conclusions from what we’ve seen in this example. We now have a firm grip on the space of quantum states of the spin-3/2 particle:

$S^3(\mathbb{C}^2) \cong \mathbb{C}^4$

and also the space of classical states of the spin-3/2 particle, the twisted cubic:

$C_3 = \{ [a^3, 3a^2b, 3ab^2, b^3] : \, (0,0) \ne (a,b) \in \mathbb{C}^2 \}$

Now for something cool: the latter sits inside the projectivization of the former! This is obvious, but it has a very nice physical meaning. While I’ve been calling $\mathbb{C}^4$ the space of quantum states of the spin-3/2 particle, it is very reasonable to argue that quantum states are actually points of its projectivization, $\mathbb{C}\mathrm{P}^3.$ I will skip the argument, which is old, famous and convincing:

• Wikipedia, Projective Hilbert space.

What matters for us here is that classical states of the spin-3/2 particle give some of these quantum states—far from all, but some of the nicest ones! They are the states obtained by ‘cubing’ a state of a spin-1/2 particle. In other words, they are the states where we can think of our spin-3/2 particle as made of three spin-1/2 particles with their spins perfectly aligned.

It’s nice to say this with a bit more physics jargon. These quantum states coming from classical ones are called ‘coherent states’ . A general quantum state of the spin-3/2 particle is a ‘quantum superposition’ of these coherent states. By this, I mean that the smallest linear subspace $V \subseteq \mathbb{C}^4$ for which $P(V)$ contains the twisted cubic is all of $\mathbb{C}^4.$

And while we’ve seen this in a particular example, it’s a completely general feature of our setup! Remember, projectivization

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

is the left adjoint of quantization

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

This means that for any $M \in \texttt{Class}, V \in \texttt{Quant}$ we have

$\texttt{Q}(M) \subseteq V \quad \iff \quad M \subseteq \texttt{P}(V)$

We get something interesting if we take $V = \texttt{Q}(M)$ here. Since $\texttt{Q}(M) \subseteq \texttt{Q}(M),$ we get

$M \subseteq \texttt{P} (\texttt{Q}(M))$

Category theorists call this inclusion of $M$ in $\texttt{P} (\texttt{Q}(M))$ the ‘unit’ of our pair of adjoint functors. It says how classical states sit inside the projectivization of $\texttt{Q}(M).$ If we call points of $\texttt{P}(\texttt{Q}(M))$ quantum states, those in $M$ are called the coherent states.

Moreover every quantum state is a ‘quantum superposition’ of coherent states! In other words, the smallest linear subspace $V \subseteq \texttt{Q}(M)$ for which $M$ sits inside $\texttt{P}(V)$ is all of $\texttt{Q}(M).$

Why is this true? It’s just the definition of $\texttt{Q}(M)!$

So, I hope you see how much physics is packed into these adjoint functors $\texttt{Q}$ and $\texttt{P}.$ I’ll do more examples next time. In the meantime, you can read more about this approach to the spin-3/2 particle here:

• Dorje C. Brody and Lane P. Hughston, Geometric quantum mechanics.

I was greatly inspired by this article!

• 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.

11 Comments | mathematics, physics | Permalink
Posted by John Baez

Geometric Quantization (Part 5)

30 December, 2018

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.