Beyond the CPT theorem

enter site alice 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.

pay day loans near me 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: scams 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?

check your icici home loan status 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.

source link 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.

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