Beyond the CPT theorem

By Bryan 18 May 2013 quantum gravity, quantum theory

Virtually all known laws of physics are invariant under the CPT transformation — that is, the combined operations of Charge conjugation (C), Parity or “mirror flipping” (P), and Time reversal (T). What that means is the following. Start with a trajectory $\psi(t)$ through state space, which represents some possible way for a system to change over time according to the known laws of physics. Now transform that trajectory, by reversing the order of all the states, and then applying C, P and T to each of them:

$\psi(t) \mapsto CPT\psi(-t).$

$CPT$ -invariance means that the resulting trajectory $CPT\psi(-t)$ will also be a possible according to the laws. One can check that this is equivalent to the statement that $CPT$ commutes with the Hamiltonian, $[CPT,H]=0.$

Why is $CPT$ so often symmetry? There is a theorem that explains it: if we characterize quantum field theory in a very plausible and general way (such as by the Wightman axioms or Haag axioms), and in particular assume that it admits a unitary representation of the Poincaré group, then $CPT$ -invariance is guaranteed. This result is called the CPT theorem. See Borchers & Yngvason for a very readable proof in the Haag framework.

Ok, that’s the background for today. Now:

Question: as we move forward, and begin to adopt theories that go beyond the standard model of particle physics, will we continue to have a CPT theorem or something like it?

It is widely believe that the CPT may fail in generic extensions of quantum theory. In particular, the requirement of a unitary representation of the Poincaré group is pretty strong, and may not hold in the kind of general context of interest in quantum gravity. Just search for CPT-violation on the arxiv to see what I mean.

But there is a sense in which something “like” a CPT theorem probably will hold in physics beyond the standard model. That sense is this: every unitary dynamics admits infinitely many “time reversing transformations” (i.e., time reversal plus some other linear symmetries) under which the dynamics is invariant. Here’s a more careful statement of this fact.

Fact. Let $\mathcal{H}$ be a separable Hilbert space with a unitary group $U_t = e^{-itH}$ describing the quantum dynamics, and let $T$ be the (antiunitary) time reversal operator. Then there exists a unitary operator $\Theta$ such that the dynamics is $\Theta T$ -invariant, in that $[\Theta T,H]=0$ .

Think of $\Theta$ as some kind of generalized symmetry transformation, similar to $CP$ , but something else entirely. It is in this sense that this fact expresses something like the $CPT$ -theorem, although unlike the $CPT$ -theorem the mathematics is completely trivial.

There are two steps to seeing why this “Fact” is true. The first is to observe that, for every self-adjoint operator $H$ , there is something called a conjugation operator $K_H$ such that $[K_H,H]=0$ . Here’s how it’s defined. The self-adjoint operator $H$ comes with its own basis set for the Hilbert space, ${v_1,v_2,v_3,\dots}$ . That’s because of the spectral theorem. So for every vector $\psi$ in the Hilbert space there are complex constants $c_i$ that allow you to write that vector,

$\psi = c_1v_1 + c_2v_2 + c_3v_3 + \cdots.$

The conjugation operator $K_H$ is just the operator defined by conjugating all the complex constants of a vector written in the $H$ -basis,

$K_H\psi = c_1^*v_1 + c_2^*v_2 + c_3^*v_3 + \cdots.$

So that’s pretty easy. And it’s easy to check that $K_H$ satisfies some special properties: it is antilinear $K_H(a\psi+b\phi)=a^*K_H\psi+b^*K_H\phi$ , antiunitary $\langle K_H\psi,K_H\phi\rangle = \langle \psi,\phi \rangle^*$ , and it commutes with the $H$ that we used to define it $[K_H,H]=0$ . We will use all of these properties in the next step.

The second step to seeing why our “Fact” is true is to recognize that if $K_1$ and $K_2$ are any two antiunitary operators, then they are related by a unitary operator, $K_2 = UK_1$ . It’s a nice exercise to check for yourself that this is true, but if you get stuck, try here.

Since time reversal $T$ and the conjugation operator $K_H$ for the Hamiltonian are both anitunitary, this means that $K_H$ is related to $T$ by some unitary operator $\Theta$ :

$K_H = \Theta T.$

So, there is always a unitary operator $\Theta$ such that $\Theta T$ commutes with the Hamiltonian $H$ .

Above, I said there were actually infinitely many such operators. Puzzle: Can you work out why?

If you’re impatient, here’s the reason. Let $f:\mathbb{R}\rightarrow\mathbb{C}$ be a function (a Borel function if we’re being pedantic), and let $f(H)$ be the corresponding Hilbert space operator as a function of the Hamiltonian $H$ . (For example, if $f(x)=x^2$ , then $f(H)=H^2=H\circ H$ .) Every such function $f(H)$ commutes with $H$ . And we already know that $K_H$ does as well. So their composition commutes with $H$ as well:

$0 = [f(H)K_H,H] = [f(H)\Theta T,H] = [\Theta^\prime T, H],$

where $\Theta^\prime = f(H)\Theta$ . Since there are infinitely many such functions, this means that there are infinitely many such operators $\Theta^\prime$ .

Soul Physics

Philosophical Foundations of Physics

Beyond the CPT theorem