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 \(\Theta^\prime\).
