The n-Category Café

Last time we talked about Jordan algebras and their role in the foundations of quantum mechanics. But the formalism of Jordan algebras seems rather removed from the actual practice of physics. After all, physicists hardly ever take two observables $a$ and $b$ and form their Jordan product ${1\over 2}(a b + b a)$. As I hinted last time, it is better to think of this operation as derived from the process of squaring an observable, which is something physicists actually do. But still, I can’t help wondering: can we see the classification of finite-dimensional formally real Jordan algebras, and thus the special role of normed division algebras, as arising from some axioms more closely tied to quantum theory as physicists usually practice it?

One answer involves the duality between states and observables. To understand this, you need to know about ‘mixed states’. A ‘state’ describes your knowledge about a physical system. If you know as much as possible we call it a ‘pure state’, but more generally, you may not know as much as possible, and then we speak of ‘mixed states’. It’s usually best to consider pure states as a special case of mixed states.

How does this play out in ordinary quantum theory? If a quantum system has the Hilbert space $\mathbb{C}^n$, observables are described by self-adjoint $n \times n$ complex matrices: elements of the Jordan algebra $\mathrm{h}_n(\mathbb{C})$. But matrices of this form that are nonnegative and have trace 1 also play another role. They are called density matrices, and they describe mixed states of our quantum system. The idea is that any density matrix $\rho \in \mathrm{h}_n(\mathbb{C})$ lets us define expectation values of observables $a \in \mathrm{h}_n(\mathbb{C})$ via $$\langle a \rangle = \mathrm{tr}(\rho a) .$$ The map sending observables to their expectation values is real-linear. The fact that $\rho$ is nonnegative is equivalent to $$a \ge 0 \; \implies \; \langle a \rangle \ge 0$$ and the fact that $\rho$ has trace 1 is equivalent to $$\langle 1 \rangle = 1 .$$

We might call this relationship between states and observables ‘state-observable duality’. By this, we are not merely referring to the fact that states live in the dual of the vector space of observables: that much is obvious, given that the expectation value of an observable should depend linearly on that observable. The nontrivial thing is that we can identify the vector space of observables with its dual: $$\begin{aligned} \mathrm{h}_n(\mathbb{C}) &\stackrel{\sim}{\longrightarrow}& \mathrm{h}_n(\mathbb{C})^* \\ a &\mapsto& \langle a, \cdot \rangle \end{aligned}$$ using the trace, which puts a real-valued inner product on the space of observables: $$\langle a, b \rangle = \mathrm{tr}(ab) .$$ Thus, states can be identified with certain special observables!

All this generalizes to an arbitrary finite-dimensional formally real Jordan algebra $A$. Every such algebra automatically has an identity element. This lets us define a state on $A$ to be a linear functional $\langle \cdot \rangle : A \to \mathbb{R}$ that is nonnegative: $$a \ge 0 \implies \langle a \rangle \ge 0$$ and normalized: $$\langle 1 \rangle = 1 .$$ But in fact, there is a one-to-one correspondence between linear functionals on $A$ and elements of $A$. The reason is that every finite-dimensional Jordan algebra has a trace $$\mathrm{tr} : A \to \mathbb{R}$$ defined so that $\mathrm{tr}(a)$ is the trace of the linear operator ‘multiplication by $a$’. Such a Jordan algebra is then formally real if and only if $$\langle a, b \rangle = \mathrm{tr}(a \circ b)$$ is a real-valued inner product. So, when $A$ is a finite-dimensional formally real Jordan algebra, any linear functional $\langle \cdot \rangle : A \to \mathbb{R}$ can be written as $$\langle a \rangle = \mathrm{tr}(\rho \circ a)$$ for a unique element $\rho \in A$. Conversely, every element $\rho \in A$ gives a linear functional by this formula. While not obvious, it is true that the linear functional $\langle \cdot \rangle$ is nonnegative if and only if $\rho \ge 0$ in terms of the ordering on $A$. More obviously, $\langle \cdot \rangle$ is normalized if and only if $\mathrm{tr}(\rho) = 1$. So, states can be identified with certain special observables: namely, those observables $\rho \in A$ with $\rho \ge 0$ and $\mathrm{tr}(\rho) = 1$.

In short: whenever the observables in our theory form a finite-dimensional formally real Jordan algebra, we have state-observable duality. But what is the physical meaning of state-observable duality? Why in the world should states correspond to special observables? A state is a way for your system to be; an observable is something you can measure about it. They seem quite different!

Here is one attempt at an answer. Every finite-dimensional formally real Jordan algebra comes equipped with a distinguished observable, the most boring one of all: the identity, $1 \in A$. This is nonnegative, so if we normalize it, we get an observable $$\rho_0 = \frac{1}{\mathrm{tr}(1)} \, 1 \in A$$ of the special kind that corresponds to a state. This state, say $\langle \cdot \rangle_0$, is just the normalized trace: $$\langle a \rangle_0 = \mathrm{tr}(\rho_0 \circ a) = \frac{\mathrm{tr}(a)}{\mathrm{tr}(1)} .$$ And this state has a clear physical meaning: it is the state of maximal ignorance! It is the state where we know as little as possible about our system — or more precisely, at least in the case of ordinary complex quantum theory, the state where entropy is maximized.

For example, take $A = \mathrm{h}_2(\mathbb{C})$, the algebra of observables of a spin-$\frac{1}{2}$ particle. Then the space of states is the so-called Bloch sphere, really a 3-dimensional ball.

The ball is convex, and for a good reason Suppose I flip a coin, don’t show you the result, and tell you “I made the particle’s state be $a$ if the coin landed heads up, and $b$ if it landed tails up”. Then the mixed state that describes your knowledge is halfway between $a$ and $b$. More generally any convex linear combination $p a + (1-p)b$ of mixed states is another mixed state, where the probability $p$ is between $0$ and $1$. That’s what we mean by saying the ball is convex. Indeed, for any formally real Jordan algebra, the space of states is convex!

So, on the surface of this ball are the pure states: the states where you know as much as possible. For any point on this surface, there’s a state where you know the electron’s spin points that way. At the center of the ball is the state of maximum ignorance. This corresponds to the density matrix $$\rho_0 = \left(\begin{aligned} \frac{1}{2} &\; & 0 \\ 0 & & \frac{1}{2} \end{aligned}\right)$$ In this state, when I ask you about the particle’s spin along any axis, all you can say is that there’s a $\frac{1}{2}$ chance that it’s pointing one way, and a $\frac{1}{2}$ chance of it pointing the other way.

Now, back to the general theory:

$A$ acts on its dual $A^*$: given $a \in A$ and a linear functional $\langle \cdot \rangle$, we get a new linear functional $\langle a \circ \cdot \rangle$. This captures the idea, familiar in quantum theory, that observables are also ‘operators’: they act on states. And state-observable duality means we can get any state from the state of complete ignorance by act on it with a suitable observable. After all, any state corresponds to some observable $\rho$, as follows: $$\langle a \rangle = \mathrm{tr}(\rho \circ a)$$ So, we can get this state by acting on the state of maximal ignorance, $\langle \cdot \rangle_0$, by the observable $\mathrm{tr}(1) \rho$: $$\langle \mathrm{tr}(1)\rho \circ a \rangle_0 = \frac{\mathrm{tr}(1)}{\mathrm{tr}(1)} \mathrm{tr}(\rho \circ a) = \langle a \rangle .$$ So, we see that the correspondence between states and special observables springs from two causes. First, there is a distinguished state, the state of maximal ignorance. Second, any other state can be obtained from the state of maximal ignorance by acting on it with a suitable observable.

While these ideas raise a host of questions, they also help motivate an important theorem of Koecher and Vinberg. The idea is to axiomatize the situation we we have just described, in a way that does not mention the Jordan product in $A$, but instead emphasizes:

• state-observable duality,
• the fact that ‘positive’ observables, namely those whose observed values are always positive, form a cone.

To find appropriate axioms, suppose $A$ is a finite-dimensional formally real Jordan algebra. Then seven facts are always true.

First, the set of positive observables $$C = \{a \in A \colon a \gt 0\} .$$ is a cone: that is, $a \in C$ implies that every positive multiple of $a$ is also in $C$. Second, this cone is convex: if $a,b \in C$ then any linear combination $p a + (1-p)b$ with $0 \le p \le 1$ also lies in $C$. Third, it is an open set. Fourth, it is regular, meaning that if $a$ and $-a$ are both in the closure $\overline{C}$, then $a = 0$. This last condition may seem obscure, but if we note that $$\overline{C} = \{ a \in A \colon a \ge 0 \}$$ we see that $C$ being regular simply means $$a \ge 0 \; and \; -a \ge 0 \quad \implies \quad a = 0 ,$$ a perfectly plausible assumption.

Next recall that $A$ has an inner product; this is what lets us identify linear functionals on $A$ with elements of $A$. This also lets us define the dual cone $$C^* = \{ a \in A \colon \forall b \in A \; \; \langle a,b \rangle \gt 0 \}$$ which one can check is indeed a cone. The fifth fact about $C$ is that it is self-dual, meaning $C = C^*$. This formalizes the notion of state-observable duality!

The sixth fact is $C$ is also homogeneous: given any two points $a,b \in C$, there is a real-linear linear transformation $T : A \to A$ mapping $C$ to itself in a one-to-one and onto way, with the property that $Ta = b$. This says that cone $C$ is highly symmetrical: no point of $C$ is any ‘better’ than any other, at least if we only consider the linear structure of the space $A$, ignoring the Jordan product and the trace.

From another viewpoint, however, there is a very special point of $C$, namely the identity $1$ of our Jordan algebra. And this brings us to our seventh and final fact: the cone $C$ is pointed, meaning that it is equipped with a distinguished element (in this case $1 \in C$). As we have seen, this element corresponds to the ‘state of complete ignorance’, at least after we normalize it.

In short: when $A$ is a finite-dimensional formally real Jordan algebra, $C$ is a pointed homogeneous self-dual regular open convex cone. All the elements $a \in C$ are positive observables, but certain special ones, namely those with $\langle a, 1 \rangle = 1$, can also be viewed as states.

In fact, there is a category of pointed homogeneous self-dual regular open convex cones, where:

• An object is a finite-dimensional real inner product space $V$ equipped with a pointed homogeneous self-dual regular open convex cone $C \subset V$.
• A morphism from one object, say $(V,C)$, to another, say $(V',C')$, is a linear map $T : V \to V'$ preserving the inner product and mapping $C$ into $C'$.

Now for the payoff. The work of Koecher and Vinberg, nicely explained in Koecher’s Minnesota notes:

• Max Koecher, The Minnesota Notes on Jordan Algebras and Their Applications, eds. Aloys Krieg and Sebastican Walcher, Lecture Notes in Mathematics 1710, Springer, Berlin, 1999.

shows that:

Theorem: The category of pointed homogeneous self-dual regular open convex cones is equivalent to the category of finite-dimensional formally real Jordan algebras.

This means that the theorem of Jordan, von Neumann and Wigner, which we saw last time, also classifies the pointed homogeneous self-dual regular convex cones!

Theorem: Every pointed homogeneous self-dual regular open convex cones is isomorphic to a direct sum of those on this list:

• the cone of positive elements in $\mathrm{h}_n(\mathbb{R})$,
• the cone of positive elements in $\mathrm{h}_n(\mathbb{C})$,
• the cone of positive elements in $\mathrm{h}_n(\mathbb{H})$,
• the cone of positive elements in $\mathrm{h}_3(\mathbb{O})$,
• the future lightcone in $\mathbb{R}^n \oplus \mathbb{R}$.

Some of this deserves a bit of explanation. For $\mathbb{K} = \mathbb{R}, \mathbb{C}, \mathbb{H}$, an element $T \in \mathrm{h}_n(\mathbb{K})$ is positive if and only if the corresponding operator $T : \mathbb{K}^n \to \mathbb{K}^n$ has $$\langle v, T v \rangle \gt 0$$ for all nonzero $v \in \mathbb{K}^n$. A similar trick works for defining positive elements of $\mathrm{h}_3(\mathbb{O})$, but we do not need the details here. We say an element $(x,t) \in \mathbb{R}^n \oplus \mathbb{R}$ lies in the future lightcone if $t \gt 0$ and $t^2 - x \cdot x \gt 0$. This of course fits in nicely with the idea that the spin factors are Minkowski spacetimes. Finally, there is an obvious notion of direct sum for Euclidean spaces with cones, where the direct sum of $(V,C)$ and $(V',C')$ is $V \oplus V'$ equipped with the cone $$C \oplus C' = \{(v,v') \in V\oplus V' \colon \; v \in C, v' \in C' \} .$$

In short: self-adjoint operators on real, complex and quaternionic Hilbert spaces arise fairly naturally as observables starting from a formalism where nonnegative observables form a cone, and we insist on state-observable duality.

There is a well-developed approach to probabilistic theories that works for cones that are neither self-dual nor homogeneous: see for example the work of Howard Barnum and his coauthors. This has already allowed Barnum, Gaebler and Wilce to shed new light on the physical significance of self-duality. But perhaps if we think more about state-observable duality we can better understand its meaning… and thus the appearance of normed division algebras in quantum physics!

Finally, as Urs Schreiber pointed out, it is worth comparing state-observable duality to the ‘state-operator correspondence’. This was made popular in the context of string theory, but it really applies whenever we have a $C^*$-algebra of observables, say $A$, equipped with a state $\langle \cdot \rangle : A \to \mathbb{C}$. Then the Gelfand-Naimark-Segal construction lets us build a Hilbert space $H$ on which $A$ acts, together with a distinguished unit vector $v \in H$ called the ‘vacuum state’. The Hilbert space $H$ is built by completing a quotient of $A$, so a dense set of vectors in $H$ come from elements of $A$. Thus again, some observables give us states. In particular, the vacuum state $v$ comes from the element $1 \in A$.

This is reminiscent of how in the Jordan algebra framework, the state of maximal ignorance comes from the element $1$ in the Jordan algebra of observables. But there are also some differences: for example, the Gelfand-Naimark-Segal construction requires choosing a state, and it works for infinite-dimensional $C^*$-algebras, while our construction works for finite-dimensional formally real Jordan algebras, which have a canonical state: the state of maximum ignorance. Presumably both constructions are special cases of something more general.