## From Classical to Quantum and Back

30 January, 2019

Damien Calaque has invited me to speak at FGSI 2019, a conference on the Foundations of Geometric Structures of Information. It will focus on scientific legacy of Cartan, Koszul and Souriau. Since Souriau helped invent geometric quantization, I decided to talk about this. That’s part of why I’ve been writing about it lately!

I’m looking forward to speaking to various people at this conference, including Mikhail Gromov, who has become interested in using category theory to understand biology and the brain.

Here’s my talk:

Abstract. Edward Nelson famously claimed that quantization is a mystery, not a functor. In other words, starting from the phase space of a classical system (a symplectic manifold) there is no functorial way of constructing the correct Hilbert space for the corresponding quantum system. In geometric quantization one gets around this problem by equipping the classical phase space with extra structure: for example, a Kähler manifold equipped with a suitable line bundle. Then quantization becomes a functor. But there is also a functor going the other way, sending any Hilbert space to its projectivization. This makes quantum systems into specially well-behaved classical systems! In this talk we explore the interplay between classical mechanics and quantum mechanics revealed by these functors going both ways.

For more details, read these articles:

• 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.

• ## Classification Problems in Symplectic Linear Algebra

21 January, 2019

Check out the video of Jonathan Lorand’s talk, the second in the Applied Category Theory Seminar here at UCR. It was nicely edited by Paola Fernandez and uploaded by Joe Moeller.

Abstract. In this talk we will look at various examples of classification problems in symplectic linear algebra: conjugacy classes in the symplectic group and its Lie algebra, linear lagrangian relations up to conjugation, tuples of (co)isotropic subspaces. I will explain how many such problems can be encoded using the theory of symplectic poset representations, and will discuss some general results of this theory. Finally, I will recast this discussion from a broader category-theoretic perspective.

Check out the video of his talk above, and also his talk slides. It was nicely edited by Paola Fernandez and uploaded by Joe Moeller.

Here are some papers to read for more detail:

• Jonathan Lorand, Classifying linear canonical relations.

• Jonathan Lorand and Alan Weinstein, Decomposition of (co)isotropic relations.

Lorand is working on a paper with Weinstein and Christian Herrman that delves deeper into these topics. I first met him at the ACT2018 school in Leiden, where we worked along with Blake Pollard, Fabrizio Genovese (shown below) and Maru Sarazola on biochemical coupling through emergent conservation laws. Right now he’s visiting UCR and working with me to dig deeper into these questions using symplectic geometry! This is a very tantalizing project that keeps on not quite working…

## Geometric Quantization (Part 8)

21 January, 2019

Puzzle. You measure the energy and frequency of some laser light trapped in a mirrored box and use quantum mechanics to compute the expected number of photons in the box. Then someone tells you that you used the wrong value of Planck’s constant in your calculation. Somehow you used a value that was twice the correct value! How should you correct your calculation of the expected number of photons?

I’ll give away the answer to the puzzle below, so avert your eyes if you want to think about it more.

This scenario sounds a bit odd—it’s not very likely that your table of fundamental constants would get Planck’s constant wrong this way. But it’s interesting because once upon a time we didn’t know about quantum mechanics and we didn’t know Planck’s constant. We could still give a reasonably good description of some laser light trapped in a mirrored box: there’s a standing wave solution of Maxwell’s equations that does the job. But when we learned about quantum mechanics we learned to describe this situation using photons. The number of photons we need depends on Planck’s constant.

And while we can’t really change Planck’s constant, mathematical physicists often like to treat Planck’s constant as variable. The limit where Planck’s constant goes to zero is called the ‘classical limit’, where our quantum description should somehow reduce to our old classical description.

Here’s the answer to the puzzle: if you halve Planck’s constant you need to double the number of photons. The reason is that the energy of a photon with frequency $\nu$ is $h \nu.$ So, to get a specific total energy $E$ in our box of light of a given frequency, we need $E/h \nu$ photons.

So, the classical limit is also the limit where the expected number of photons goes to infinity! As the ‘packets of energy’ get smaller, we need more of them to get a certain amount of energy.

This has a nice relationship to what I’d been doing with geometric quantization last time.

I explained how we could systematically replace any classical system considering by a ‘cloned’ version of that system: a collection of identical copies constrained to all lie in the same state. The set of allowed states is the same, but the symplectic structure is multiplied by a constant factor: the number of copies. We can see this as follows: if the phase space of our system is an abstract symplectic manifold $M$ its kth power $M^k$ is again a symplectic manifold. We can look at the image of the diagonal embedding

$\begin{array}{rcl} \Delta_k \colon M &\to & M^k \\ x & \mapsto & (x,x,\dots, x) \end{array}$

The image is a symplectic submanifold of $M^k,$ and it’s diffeomorphic to $M,$ but $\Delta_k$ is not a symplectomorphism from $M$ to its image. The image has a symplectic structure that’s k times bigger!

What does this mean for physics?

If we act like physicists instead of mathematicians for a minute and keep track of units, we’ll notice that the naive symplectic structure in classical mechanics has units of action: think $\omega = dp \wedge dq$. Planck’s constant also has units of action, so to get a dimensionless version of the symplectic structure, we should use $\omega/h.$ Then it’s clear that multiplying the symplectic structure by k is equivalent to dividing Planck’s constant by k!

So: cloning a system, multiplying the number of copies by k, should be related to dividing Planck’s constant by k. And the limit $k \to \infty$ should be related to a ‘classical limit’.

Of course this would not convince a mathematician so far, since I’m using a strange mix of ideas from classical mechanics and quantum mechanics! But our approach to geometric quantization makes everything precise. We have a category $\texttt{Class}$ of classical systems, a category $\texttt{Quant}$ of quantum systems, a quantization functor

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

and a functor going back:

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

which reveals that quantum systems are special classical systems. And last time we saw that there are also ways to ‘clone’ classical and quantum systems.

Our classical systems are more than mere symplectic manifolds: they are projectively normal subvarieties of $\mathbb{C}\mathrm{P}^n$ for arbitrary n. So, we clone a classical system by a trick that looks more fancy than the diagonal embedding I mentioned above: we instead use the kth Veronese embedding, which defines a functor

$\nu_k \colon \texttt{Class} \to \texttt{Class}$

But this cloning process has the same effect on the underlying symplectic manifold: it multiplies the symplectic structure by k.

Similarly, we clone a quantum system by replacing its set of states $\mathrm{P}(\mathbb{C}^n)$ (the projectivization of the Hilbert space $\mathbb{C}^n$) by $\mathrm{P}(S^k(\mathbb{C}^n)),$ where $S^k$ is the kth symmetric power. This gives a functor

$S^k \colon \texttt{Quant} \to \texttt{Quant}$

and the ‘obvious squares commute’:

$\texttt{Q} \circ v_k = S^k \circ \texttt{Q}$
$\texttt{P} \circ S^k = v_k \circ \texttt{P}$

All I’m doing now is giving this math a new physical interpretation: the k-fold cloning process is the same as dividing Planck’s constant by k!

If this seems a bit elusive, we can look at an example like the spin-j particle. In Part 6 we saw that if we clone the state space for the spin-1/2 particle we get the state space for the spin-j particle, where $j = k/2.$ But angular momentum has units of Planck’s constant, and the quantity j is really an angular momentum divided by $\hbar.$ So this process of replacing 1/2 by k/2 can be reinterpreted as the process of dividing Planck’s constant by k: if Planck’s constant is smaller, we need j to be bigger to get a given angular momentum! And what we thought was a single spin-1/2 particle in the state $\psi$ becomes k spin-1/2 particles in the ‘cloned’ state $\psi \otimes \psi \otimes \cdots \otimes \psi.$ As explained in Part 6, we can reinterpret this as a state of a single spin-k/2 particle.

Finally, let me point out something curious. We have a systematic way of changing our description of a quantum system when we divide Planck’s constant by an integer. But we can’t do it when we divide Planck’s constant by any other sort of number! So, in a very real sense, Planck’s constant is quantized.

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.

## Geometric Quantization (Part 7)

8 January, 2019

I’ve been falling in love with algebraic geometry these days, as I realize how many of its basic concepts and theorems have nice interpretations in terms of geometric quantization. I had trouble getting excited about them before. I’m talking about things like the Segre embedding, the Veronese embedding, the Kodaira embedding theorem, Chow’s theorem, projective normality, ample line bundles, and so on. In the old days, all these things used to make me nod and go “that’s nice”, without great enthusiasm. Now I see what they’re all good for!

Of course this is my own idiosyncratic take on the subject: obviously algebraic geometers have their own pefectly fine notion of what these things are good for. But I never got the hang of that.

Today I want to talk about how the Veronese embedding can be used to ‘clone’ a classical system. For any number k, you can take a classical system and build a new one; a state of this new system is k copies of the original system constrained to all be in the same state! This may not seem to do much, but it does something: for example, it multiplies the Kähler structure on the classical state space by k. And it has a quantum analogue, which has a much more notable effect!

Last time I looked at an example, where I built the spin-3/2 particle by cloning the spin-1/2 particle.

In brief, it went like this. The space of classical states of the spin-1/2 particle is the Riemann sphere, $\mathbb{C}\mathrm{P}^1.$ This just happens to also be the space of quantum states of the spin-1/2 particle, since it’s the projectivization of $\mathbb{C}^2.$ To get the 3/2 particle we look at the map

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

You can think of this as the map that ‘triplicates’ a spin-1/2 particle, creating 3 of them in the same state. This gives rise to a map between the corresponding projective spaces, which we should probably call

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

It’s an embedding.

Algebraic geometers call the image of this embedding the twisted cubic, since it’s a curve in 3d projective space described by homogeneous cubic equations. But for us, it’s the embedding of the space of classical states of the spin-3/2 particle into the space of quantum states. (The fact that classical states give specially nice quantum states is familiar in physics, where these specially nice quantum states are called ‘coherent states’, or sometimes ‘generalized coherent states’.)

Now, you’ll have noted that the numbers 2 and 3 show up a bunch in what I just said. But there’s nothing special about these numbers! They could be arbitrary natural numbers… well, > 1 if we don’t enjoy thinking about degenerate cases.

Here’s how the generalization works. Let’s think of guys in $\mathbb{C}^n$ as linear functions on the dual of this space. We can raise any one of them to the k power and get a homogeneous polynomial of degree k. The space of such polynomials is called $S^k(\mathbb{C}^n),$ so raising to the kth power defines a map

$\mathbb{C}^n \to S^k(\mathbb{C}^n)$

This in turn gives rise to a map between the corresponding projective spaces:

$P(\mathbb{C}^n) \to P(S^k(\mathbb{C}^n))$

This map is an embedding, since different linear functions give different polynomials when you raise them to the k power, at least if $k \ge 1.$ And this map is famous: it’s called the k Veronese embedding. I guess it’s often denoted

$v_k \colon P(\mathbb{C}^n) \to P(S^k(\mathbb{C}^n))$

An important special case occurs when we take $n = 2,$ as we’d been doing before. The space of homogeneous polynomials of degree k in two variables has dimension $k + 1,$ so we can think of the Veronese embedding as a map

$v_k \colon \mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^k$

embedding the projective line as a curve in $\mathbb{C}\mathrm{P}^k.$ This sort of curve is called a rational normal curve. When $d = 3$ it’s our friend from last time, the twisted cubic.

In general, we can think of $\mathbb{C}\mathrm{P}^k$ as the space of quantum states of the spin-k/2 particle, since we got it from projectivizing the spin-k/2 representation of $\mathrm{SU}(2),$ namely $S^k(\mathbb{C}^n).$ Sitting inside here, the rational normal curve is the space of classical states of the spin-k/2 particle—or in other words, ‘coherent states’.

Maybe I should expand on this, since it flew by so fast! Pick any direction you want the angular momentum of your spin-k/2 particle to point. Think of this as a point on the Riemann sphere and think of that as coming from some vector $\psi \in \mathbb{C}^2.$ That describes a quantum spin-1/2 particle whose angular momentum points in the desired direction. But now, form the tensor product

$\underbrace{\psi \otimes \cdots \otimes \psi}_{k}$

This is completely symmetric under permuting the factors, so we can think of it as a vector in $S^k(\mathbb{C}^2).$ And indeed, it’s just what I was calling

$v_k (\psi) \in S^k(\mathbb{C}^2)$

This vector describes a collection of k indistinguishable quantum spin-1/2 particles with angular momenta all pointing in the same direction. But it also describes a single quantum spin-k/2 particle whose angular momentum points in that direction! Not all vectors in $S^k(\mathbb{C}^2)$ are of this form, clearly. But those that are, are called ‘coherent states’.

Now, let’s do this all a bit more generally. We’ll work with $\mathbb{C}^n,$ not just $\mathbb{C}^2.$ And we’ll use a variety $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ as our space of classical states, not necessarily all of $\mathbb{C}\mathrm{P}^{n-1}.$

Remember, we’ve got:

• a category $\texttt{Class}$ where the objects are linearly normal subvarieties $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ for arbitrary $n,$

and

• a category $\texttt{Quant}$ where the objects are linear subspaces $V \subseteq \mathbb{C}^n$ for arbitrary $n.$

The morphisms in each case are just inclusions. We’ve got a ‘quantization’ functor

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

that maps $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ to the smallest $V \subseteq \mathbb{C}^n$ whose projectivization contains $M.$ And we’ve got what you might call a ‘classicization’ functor going back:

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

We actually call this ‘projectization’, since it sends any linear subspace $V \subseteq \mathbb{C}^n$ to its projective space sitting inside $\mathbb{C}\mathrm{P}^{n-1}$.

We would now like to get the Veronese embedding into the game, copying what we just did for the spin-k/2 particle. We’d like each Veronese embedding $v_k$ to define a functor from $\texttt{Class}$ to $\texttt{Class}$ and also a functor $\texttt{Quant}$ to $\texttt{Quant}.$ For example, the first of these should send the space of classical states of the spin-1/2 particle to the space of classical states of the spin-k/2 particle. The second should do the same for the space of quantum states.

The quantum version works just fine. Here’s how it goes. An object in $\texttt{Quant}$ is a linear subspace

$V \subseteq \mathbb{C}^n$

for some $n.$ Our functor should send this to

$S^k(V) \subseteq S^k(\mathbb{C}^n) \cong \mathbb{C}^{\left(\!\!{n\choose k}\!\!\right)}$

Here $\left(\!{n\choose k}\!\right)$, pronounced ‘n multichoose k’ , is the number of ways to choose k not-necessarily-distinct items from a set of n, since this is the dimension of the space of degree-k homogeneous polynomials on $\mathbb{C}^n.$ (We have to pick some sort of ordering on monomials to get the isomorphism above; this is one of the clunky aspects of our current framework, which I plan to fix someday.)

This process indeed defines functor, and the only reasonable name for it is

$S^k \colon \texttt{Quant} \to \texttt{Quant}$

Intuitively, it takes any state space of any quantum system and produces the state space for k indistinguishable copies that system. (If you’re a physicist, muttering the phrase ‘identical bosons’ may clarify things. There is also a fermionic version where we use exterior powers instead of symmetric powers, but let’s not go there now.)

The classical version of this functor suffers from a small glitch, which however is easy to fix. An object in $\texttt{Class}$ is a linearly normal subvariety

$M \subseteq \mathbb{C}\mathrm{P}^{n-1}$

for some $n.$ Applying the kth Veronese embedding we get a subvariety

$v_k(M) \subseteq \mathbb{C}\mathrm{P}^{\left(\!\!{n\choose k}\!\!\right)-1}$

However, I don’t think this is linearly normal, in general. I think it’s linearly normal iff $M$ is k-normal. You can take this as a definition of k-normality, if you like, though there are other equivalent ways to say it.

Luckily, a projectively normal subvariety of projective space is k-normal for all $k \ge 1.$ And even better, projectively normal varieties are fairly common! In particular, any projective space is a projectively normal subvariety of itself.

So, we can redefine the category $\texttt{Class}$ by letting objects be projectively normal subvarieties $M \subseteq \mathbb{C}\mathrm{P}^{n-1}$ for arbitrary $n \ge 1.$ I’m using the same notation for this new category, which is ordinarily a very dangerous thing to do, because all our results about the original version are still true for this one! In particular, we still have adjoint functors

$\texttt{Q} \colon \texttt{Class} \to \texttt{Quant}, \qquad \texttt{P} \colon \texttt{Quant} \to \texttt{Class}$

defined exactly as before. But now the kth Veronese embedding gives a functor

$v_k \colon \texttt{Class} \to \texttt{Class}$

Intuitively, this takes any state space of any classical system and produces the state space for k indistinguishable copies that system that are all in the same state. It has no effect on the classical state space $M$ as an abstract variety, just its embedding into projective space—which in turn affects its Kähler structure and the line bundle it inherits from projective space. In particular, its symplectic structure gets multiplied by k, and the line bundle over it gets replaced by its kth tensor power. (These are well-known facts about the Veronese embedding.)

I believe that this functor obeys

$\texttt{Q} \circ v_k = S^k \circ \texttt{Q}$

and it’s just a matter of unraveling the definitions to see that

$\texttt{P} \circ S^k = v_k \circ \texttt{P}$

So, very loosely, the functors

$v_k \colon \texttt{Class} \to \texttt{Class}, \qquad S^k \colon \texttt{Quant} \to \texttt{Quant}$

should be thought of as replacing a classical or quantum system by a new ‘cloned’ version of that system. And they get along perfectly with quantization and its adjoint, projectivization!

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 the symmetric tensor powers of a Hilbert space as the corresponding way to clone a quantum system.

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

## Unsolved Mysteries of Fundamental Physics

2 January, 2019

In this century, progress in fundamental physics has been slow. The Large Hadron Collider hasn’t yet found any surprises, attempts to directly detect dark matter have been unsuccessful, string theory hasn’t made any successful predictions, and nobody really knows what to do about any of this. But there is no shortage of problems, and clues. Watch the talk I gave at the Cambridge University Physics Society for some ideas on this! Warning: this is for ordinary folks, not experts.

There are some squeaky sounds on the video at first, but they seem to go away pretty quick, so hang in there! You can also see my talk slides here:

and click on the links for extra information.

## 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.

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.

## 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.

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.