The Three-Fold Way (Part 1)
Posted by John Baez
It’s a wonderful fact that nature is described using complex Hilbert spaces. We can take a beam of electrons and split it. If we do it right, each electron goes both ways! Then we can insert a tightly wound coil of wire between the two beams. By running some current through this wire, we can make a magnetic field that’s mostly trapped inside the coil. By this method, we can multiply the part of the electron taking one route by $i$, as compared to the part that takes the other route. And we can check that this is true by studying the interference patterns that appear as the beams recombine! Indeed we can do this for any complex number on the unit circle, say $exp(i \theta)$.
But what’s so great about the complex numbers? You can set up a theory of Hilbert spaces based on any normed division algebra. And as you’re undoubtedly sick of hearing, there are three choices: the real numbers $\mathbb{R}$, the complex numbers $\mathbb{C}$, and the quaternions $\mathbb{H}$. So mathematically, at least, there are three possible kinds of quantum mechanics!
Only three? There could be more, but Solèr’s theorem picks out these three from among a vast set of alternatives, based on some simple axioms about how infinite-dimensional Hilbert spaces should work.
What about the finite-dimensional case? The Jordan–von Neumann–Wigner theorem classifies the possibilities in an approach based on algebras of observables. The Koecher–Vinberg theorem starts from seemingly different assumptions, but leads to the exact same conclusions. Both these theorems leave room for some exotic possibilities involving octonions and spin factors — but the overall message of all these results seems to be: real, complex and quaternionic quantum mechanics are equally good.
However, for some reason — or perhaps no good reason — nature is best described by complex quantum mechanics. We can take an electron and multiply it by $i$, so real quantum mechanics is out. But we can’t multiply it by $j$ or $k$ — or at least that’s what everyone says. So quaternionic quantum mechanics is out too, apparently.
This has led people to look for mathematical ways in which complex quantum mechanics is ‘better’ than the real or quaternionic theories. Lucien Hardy proved a nice result along these lines:
- Lucien Hardy, Quantum theory from five reasonable axioms.
But I want to tell you about two others.
One of the greatest discoveries in physics is Noether’s theorem, which sets up a one-to-one correspondence between observables and one-parameter groups of symmetries. In its original form, this theorem holds in a particular approach to classical physics. But by now it’s become a general idea that applies to quantum physics as well. Energy corresponds to time translation; momentum corresponds to translation in space; angular momentum corresponds to rotation — these have become a basic part of our understanding of the world, transcending the details of any particular theory we have of the world. If the world has some symmetry, we expect there to be a corresponding observable.
In quantum mechanics, observables are self-adjoint operators. These correspond in a one-to-one way to continuous one-parameter groups of unitary operators, thanks to Stone’s theorem. So, Stone’s theorem is the quantum version of Noether’s theorem.
One problem with real and quaternionic quantum theory is that this idea breaks down, or at least becomes more subtle. Let’s see how.
Suppose $\mathbb{K}$ is any associative normed division algebra: in other words, $\mathbb{R},\mathbb{C}$ or $\mathbb{H}$. I’ve told you how to define $\mathbb{K}$-Hilbert spaces, and — you can take my word for this — a lot of what you know about complex Hilbert spaces works for real and quaternionic ones, too.
For example, let $H$ and $H'$ be $\mathbb{K}$-Hilbert spaces. We say a linear operator $T: H \to H'$ is bounded if there exists a constant $K \to 0$ such that $\|T v\| \le K \, \|v\|$ for all $v \in H$. We define the adjoint of a bounded operator $T: H \to H'$ in the usual way: $\langle u, T^\dagger v \rangle = \langle T u, v \rangle$ for all $u \in H$, $v \in H'$. It is easy to check that $T^\dagger$, defined this way, really is an operator from $H'$ back to $H$.
Let’s define a $\mathbb{K}$-linear operator $U: H \to H'$ to be unitary if $U U^\dagger = U^\dagger U = 1$. I should warn you that when $\mathbb{K} = \mathbb{R}$, people say usually say ‘orthogonal’ instead of ‘unitary’ — and when $\mathbb{K} = \mathbb{H}$, people sometimes use the term ‘symplectic’. But let’s use the same word, ‘unitary’, for all three cases.
We define a one-parameter unitary group to be a family of unitary operators $U(t) : H \to H$, one for each $t \in \mathbb{R}$, such that $U(t + t') = U(t) U(t')$ for all $t, t' \in \mathbb{R}$. Let’s say this group is continuous if for each vector $v \in H$, $U(t)v$ depends continuously on the parameter $t$.
To avoid distractions, let me assume now that $H$ is finite-dimensional. In this case every operator is automatically bounded. And in this case, every continuous one-parameter unitary group can be written as $U(t) = \exp(t S)$ for a unique operator $S$. It is easy to check that $S$ is skew-adjoint: it satisfies $S^\dagger = -S$. Conversely, any skew-adjoint $S$ gives a continuous one-parameter unitary group by the above formula. This is a version of Stone’s theorem that applies to real and quaternionic Hilbert spaces as well as complex ones.
However, in quantum theory we usually want our observables to be self-adjoint operators, obeying $A^\dagger = A$.
Now, when $\mathbb{K} = \mathbb{C}$, we can write any skew-adjoint operator $S$ in terms of a self-adjoint operator $A$. This gives the usual correspondence between one-parameter unitary groups and self-adjoint operators, which we know and love from ordinary complex quantum mechanics.
How do we do it? Easy: we set $S(v) = A(v) i .$ Huh? Remember, we’re taking $\mathbb{K}$-Hilbert spaces to be right $\mathbb{K}$-modules: that’s why I’m writing the $i$ on the right here. It’s just an arbitrary convention, and right now I’m beginning to regret this convention: this equation looks dorky. But when I use the other convention, I also regret that. It’s no big deal, just a small annoying itch.
But never mind that — here’s the main point:
When $\mathbb{K} = \mathbb{R}$ we have no number $i$, so we cannot express our skew-adjoint $S$ in terms of a self-adjoint $A$. Nor can we do it when $\mathbb{K} = \mathbb{H}$. This time the problem is not a shortage of square roots of $-1$. Instead, there are too many—and more importantly, they do not commute! We can try to set $A(v) = S(v)i$, but this operator $A$ will rarely be linear. The reason is that because $S$ commutes with multiplication by $j$, $A$ anticommutes with multiplication by $j$, so $A$ is only linear in the trivial case $S = 0$.
Let’s do the calculation in detail, in case my verbal argument wasn’t clear. We have $A(v) = - S(v) i$ so $A(v j) = - S(v j) i = - S(v) j i = S(v) i j = - A(v) j$ In short, when we pull a $j$ out, we get an unwanted minus sign. So, $A$ is linear only when $S = 0$.
That’s one problem. A second problem, special to the quaternionic case, concerns tensor products of Hilbert spaces.
In ordinary complex quantum theory, when we have two systems, one with Hilbert space $H$ and one with Hilbert space $H'$, the system made by combining these two systems has Hilbert space $H \otimes H'$. Here the tensor product of Hilbert spaces relies on a more primitive concept: the tensor product of vector spaces.
Remember, our $\mathbb{K}$-vector spaces are one-sided $\mathbb{K}$-modules, which I arbitrarily took to be right $\mathbb{K}$-modules in some vain attempt to avoid as many silly-looking equations as possible. In an algebra class, you learn that you can tensor two one-sided modules of an algebra and get another such module if your algebra is commutative. But the quaternions are not!
Unfortunately, this issue is not addressed head-on in Adler’s book on quaternionic quantum theory:
- Stephen Adler, Quaternionic Quantum Mechanics and Quantum Fields, Oxford U. Press, Oxford, 1995.
The tensor product of bimodules of a noncommutative algebra is another bimodule over that algebra. So, when tensoring two quaternionic Hilbert spaces, Adler essentially chooses a way to make one of them into a bimodule… without being very explicit about this.
Alas, there is no canonical way to make a quaternionic Hilbert space $H$ into a bimodule. Indeed, given one way to do it, you can get lots of new ones by introducing a new way to do left multiplication by scalars while keeping the right multiplication unchanged: $(x v)_{new} = \alpha(x) v , \qquad (v x)_{new} = v x ,$ where $v \in H$, $x \in \mathbb{H}$ and $\alpha$ is an automorphism of the quaternions. Every automorphism is of the form $\alpha(x) = g x g^{-1}$ for some unit quaternion $g$, so the automorphism group of the quaternions is $SU(2)/\{\pm 1\} \cong SO(3)$. Thus, we can ‘twist’ a bimodule structure on $H$ by any element of $SO(3)$, obtaining a new bimodule with the same underlying quaternionic Hilbert space. If I remember correctly, this idea is lurking in Adler’s work. I just wish the underlying math had been explained in a way that I could understand better.
When Toby Bartels did some work on quaternionic functional analysis, he used bimodules explicitly, right from the start:
- Toby Bartels, Functional analysis with quaternions.
More recently, Chi-Keung has investigated these questions in more depth. His paper provides a nice overview of many different approaches I hadn’t known about:
- Chi-Keung Ng, Quaternion functional analysis.
But he points out an interesting fact: if we take a quaternionic Hilbert space to be a bimodule of the quaternions (equipped with an $\mathbb{H}$-valued inner product satisfying some sensible conditions), then the category of quaternionic Hilbert spaces is equivalent to the category of real Hilbert spaces! So, at least from a category-theoretic perspective, it seems like we’re back to real quantum mechanics!
The reason is not hard to see, if you know a little algebra. It’s easier to see for vector spaces, rather than Hilbert spaces. A bimodule of the quaternions is the same as a left $\mathbb{H} \otimes \mathbb{H}^{op}$ module, where $\mathbb{H}^{op}$ is the quaternions with the multiplication turned around. But because $\mathbb{H}$ is a $\ast$-algebra, $\mathbb{H}^{op} \cong \mathbb{H}$ Moreover, $\mathbb{H} \otimes \mathbb{H}$ is isomorphic to $M_4(\mathbb{R})$, the algebra of $4 \times 4$ real matrices. So, the category of $\mathbb{H}$-bimodules is equivalent to the category of $M_4(\mathbb{R})$-modules.
But it’s well-known that $M_4(\mathbb{R})$ is Morita equivalent to $\mathbb{R}$, meaning that their categories of modules are equivalent!
Thus, the category of $\mathbb{H}$-bimodules is equivalent to the category of real vector spaces.
In short: so far it’s not looking too good for real and quaternionic quantum mechanics. In both of these, we lack the correspondence between observables and symmetries that we’re used to. And in the quaternionic case, we can’t describe systems built from two parts… unless we pull a trick that basically takes us back to real quantum mechanics!
But in the posts to come, I’ll propose a way to tackle these problems. The trick is to treat real, complex and quaternionic quantum theory as part of a single unified structure: the ‘three-fold way’. This nice thing is that the necessary math is already known — it’s been sitting there for many years, patiently waiting for us to appreciate it.
Re: The Three-Fold Way (Part 1)
You need to find a way of algebraically playing with some “I” (squared to -1, maybe +1 as well) that doesn’t appear ad hoc and carries some sensible interpretation.
That would be Geometric/Clifford algebra, then?