Skip to main content
not sure what CPT is supposed to satisfy
Source Link
Ryan Thorngren
  • 8.1k
  • 1
  • 27
  • 58

C, P, and T need not all exist in a quantum field theory, and they may not even be unique. Only CPT is guaranteed in a general unitary QFT. In the standard model for instance, $CP$ and $T$ are not symmetries but their composition is.

A simple example, consider a 2-component real fermion $\psi$ in 1+1D. The massless free Lagrangian for this field is $$i \psi^T \gamma^0 \gamma^\mu \partial_\mu \psi.$$ There are two choices of time reversal symmetry: $$\psi(x,t) \mapsto \pm \gamma^0\psi(x,-t),$$ and a mass term $$i\psi^T \gamma^0 \psi$$ breaks either one. Parity also has a choice $$\psi(x,t) \mapsto \pm \gamma^1\psi(-x,t)$$ and is also broken by a mass term. Meanwhile there are no gauge charges, so we can choose $C$ to act trivially and $CPT = PT$ is a symmetry even with a mass term. We can also choose $C$ to act by the chiral symmetry $$\psi(x,t) \mapsto \pm\gamma^2\psi(x,t)$$ and get another "CPT" transformation which is a symmetry of the massless model but not a symmetry of the massive model.

So you see that there are lots of symmetries that we can call CPT, the "CPT theorem" just says that no matter how we modify this theory, there will be some anti-unitary symmetry $S$ (sometimes realized literally as C times P times T but not always). $S$ is required to satisfy $S^2 = 1$ as well as the Reeh-Schlieder theorem, which says that if $\mathcal{O}(x)$ is a point operator, then $$\langle \mathcal{O}(x) S \mathcal{O}(x) S^{-1} \rangle \ge 0,$$ with 0 iff $\mathcal{O}(x)$ is a contact operator (ie. it has zero separated correlation functions, but can have delta function correlators). The existence of such an $S$ is equivalent to unitarity.

C, P, and T need not all exist in a quantum field theory, and they may not even be unique. Only CPT is guaranteed in a general unitary QFT. In the standard model for instance, $CP$ and $T$ are not symmetries but their composition is.

A simple example, consider a 2-component real fermion $\psi$ in 1+1D. The massless free Lagrangian for this field is $$i \psi^T \gamma^0 \gamma^\mu \partial_\mu \psi.$$ There are two choices of time reversal symmetry: $$\psi(x,t) \mapsto \pm \gamma^0\psi(x,-t),$$ and a mass term $$i\psi^T \gamma^0 \psi$$ breaks either one. Parity also has a choice $$\psi(x,t) \mapsto \pm \gamma^1\psi(-x,t)$$ and is also broken by a mass term. Meanwhile there are no gauge charges, so we can choose $C$ to act trivially and $CPT = PT$ is a symmetry even with a mass term. We can also choose $C$ to act by the chiral symmetry $$\psi(x,t) \mapsto \pm\gamma^2\psi(x,t)$$ and get another "CPT" transformation which is a symmetry of the massless model but not a symmetry of the massive model.

So you see that there are lots of symmetries that we can call CPT, the "CPT theorem" just says that no matter how we modify this theory, there will be some anti-unitary symmetry $S$ (sometimes realized literally as C times P times T but not always). $S$ is required to satisfy $S^2 = 1$ as well as the Reeh-Schlieder theorem, which says that if $\mathcal{O}(x)$ is a point operator, then $$\langle \mathcal{O}(x) S \mathcal{O}(x) S^{-1} \rangle \ge 0,$$ with 0 iff $\mathcal{O}(x)$ is a contact operator (ie. it has zero separated correlation functions, but can have delta function correlators). The existence of such an $S$ is equivalent to unitarity.

C, P, and T need not all exist in a quantum field theory, and they may not even be unique. Only CPT is guaranteed in a general unitary QFT. In the standard model for instance, $CP$ and $T$ are not symmetries but their composition is.

A simple example, consider a 2-component real fermion $\psi$ in 1+1D. The massless free Lagrangian for this field is $$i \psi^T \gamma^0 \gamma^\mu \partial_\mu \psi.$$ There are two choices of time reversal symmetry: $$\psi(x,t) \mapsto \pm \gamma^0\psi(x,-t),$$ and a mass term $$i\psi^T \gamma^0 \psi$$ breaks either one. Parity also has a choice $$\psi(x,t) \mapsto \pm \gamma^1\psi(-x,t)$$ and is also broken by a mass term. Meanwhile there are no gauge charges, so we can choose $C$ to act trivially and $CPT = PT$ is a symmetry even with a mass term. We can also choose $C$ to act by the chiral symmetry $$\psi(x,t) \mapsto \pm\gamma^2\psi(x,t)$$ and get another "CPT" transformation which is a symmetry of the massless model but not a symmetry of the massive model.

So you see that there are lots of symmetries that we can call CPT, the "CPT theorem" just says that no matter how we modify this theory, there will be some anti-unitary symmetry $S$ (sometimes realized literally as C times P times T but not always).

Source Link
Ryan Thorngren
  • 8.1k
  • 1
  • 27
  • 58

C, P, and T need not all exist in a quantum field theory, and they may not even be unique. Only CPT is guaranteed in a general unitary QFT. In the standard model for instance, $CP$ and $T$ are not symmetries but their composition is.

A simple example, consider a 2-component real fermion $\psi$ in 1+1D. The massless free Lagrangian for this field is $$i \psi^T \gamma^0 \gamma^\mu \partial_\mu \psi.$$ There are two choices of time reversal symmetry: $$\psi(x,t) \mapsto \pm \gamma^0\psi(x,-t),$$ and a mass term $$i\psi^T \gamma^0 \psi$$ breaks either one. Parity also has a choice $$\psi(x,t) \mapsto \pm \gamma^1\psi(-x,t)$$ and is also broken by a mass term. Meanwhile there are no gauge charges, so we can choose $C$ to act trivially and $CPT = PT$ is a symmetry even with a mass term. We can also choose $C$ to act by the chiral symmetry $$\psi(x,t) \mapsto \pm\gamma^2\psi(x,t)$$ and get another "CPT" transformation which is a symmetry of the massless model but not a symmetry of the massive model.

So you see that there are lots of symmetries that we can call CPT, the "CPT theorem" just says that no matter how we modify this theory, there will be some anti-unitary symmetry $S$ (sometimes realized literally as C times P times T but not always). $S$ is required to satisfy $S^2 = 1$ as well as the Reeh-Schlieder theorem, which says that if $\mathcal{O}(x)$ is a point operator, then $$\langle \mathcal{O}(x) S \mathcal{O}(x) S^{-1} \rangle \ge 0,$$ with 0 iff $\mathcal{O}(x)$ is a contact operator (ie. it has zero separated correlation functions, but can have delta function correlators). The existence of such an $S$ is equivalent to unitarity.