Combined PCT Symmetry

Recall the parity transformation $\vec{x}\rightarrow \vec{x}^\prime = -\vec{x}$,

$$P\psi(\vec{x},t) = \psi^\prime(\vec{x}^\prime,t) = e^{i\phi}\gamma^0\psi(\vec{x},t) ,$$

$$P\phi(\vec{x},t) = \phi^\prime(\vec{x}^\prime,t) = \phi(\vec{x},t) ,$$

$$P\vec{A}(\vec{x},t) = \vec{A}^\prime(\vec{x}^\prime,t) = -\vec{A}(\vec{x},t) .$$ (5.262)

The parity transformation leaves the Dirac equation and all physical observables unchanged.

We now combine all three symmetries, $P$, $\mathcal{C}$ and $\mathcal{T}$

$$P\psi(\vec{x},t) = e^{i\phi}\gamma^0\psi(\vec{x},t) = \psi^\prime(\vec{x}^\prime,t) ,$$

$$\mathcal{T}\psi(\vec{x},t) = T\psi^*(\vec{x},t) = i\gamma^1\gamma^3\psi^*(\vec{x},t) = \psi^\prime(\vec{x},t^\prime) ,$$

$$\mathcal{C}\psi(\vec{x},t) = C\gamma^0\psi^*(\vec{x},t) = C\overline{\psi}^T(\vec{x},t) = i\gamma^2\gamma^0\overline{\psi}^T(\vec{x},t) .$$ (5.263)

to obtain

$$\psi_{P\mathcal{CT}}(x^\prime) \equiv P\mathcal{C}\mathcal{T}\psi(x) = P C\gamma^0 (\mathcal{T}\psi(x))^* = P C \gamma^0 T^* \psi(x)$$

$$= (e^{i\phi}\gamma^0) (i\gamma^2\gamma^0) \gamma^0 (i\gamma^1\gamma^3)^* \psi(x) = -i e^{i\phi} i\gamma^0\gamma^2\gamma^1\gamma^3 \psi(x)$$

$$= ie^{i\phi} (i\gamma^0\gamma^1\gamma^2\gamma^3) \psi(x) = i e^{i\phi} \gamma_5 \psi(x) ,$$ (5.264)

with $x_\mu^\prime = -x_\mu$.

Since $\psi_{P\mathcal{CT}}(x^\prime)$ can represent a positron, we see that it is an electron moving backward in space-time and multiplied by $ie^{i\phi}\gamma_5$. Thus positrons are negative-energy electrons running backward in space-time. This is the basis of the Stückelberg-Feynman form of positron theory.

For a free-particle spin-momentum eigenstate $\psi(x)$ and negative energy, we see

$$\psi_{P\mathcal{CT}}(x^\prime) = ie^{i\phi}\gamma_5 \left( \frac{-\not{p} + m}{2m} \right) \left( \frac{1 + \gamma_5\not{s}}{2} \right) \psi(x)$$

$$= \left( \frac{\not{p} + m}{2m} \right) \left( \frac{1 - \gamma_5\not{s}}{2} \right) ie^{i\phi}\gamma_5 \psi(x)$$

$$= \left( \frac{\not{p} + m}{2m} \right) \left( \frac{1 - \gamma_5\not{s}}{2} \right) \psi_{P\mathcal{CT}}(x^\prime).$$ (5.265)

Therefore a positron wave function is a negative-energy electron moving backward in time, multiplied by $ie^{i\phi}\gamma_5$.

For an arbitrary solution in the presence of electromagnetic forces, the negative energy eigenvalue equation is

$$[\vec{\alpha}\cdot(-i\vec{\nabla} - e\vec{A}) + \beta m + e\phi ]\psi = -E\psi.$$ (5.266)

Carrying out the $P\mathcal{CT}$ transformation gives

$$[\vec{\alpha}\cdot(-i\vec{\nabla}^\prime + e\vec{A}^\prime(x^... ...e) ]\psi_{P\mathcal{CT}}(x^\prime) = +E\psi_{PCT}(x^\prime) .$$ (5.267)

Notice that since

$$P\phi(\vec{x},t) = \phi(\vec{x},t) ,$$

$$P\vec{A}(\vec{x},t) = -\vec{A}(\vec{x},t) ,$$

$$\phi^\prime(t^\prime=-t) = +\phi(t) ,$$

$$\vec{A}^\prime(t^\prime=-t) = -\vec{A}(t) .$$ (5.268)

Therefore $A_\mu^\prime(x^\prime) = +A_\mu(x)$.