CPT invariance of Dirac equation

Question

We know that Dirac equation is

\begin{equation} ( i \partial _\mu \gamma ^\mu - m ) \psi ~=~0. \end{equation}

How can we show that Dirac equation is invariant under CPT transformation?

Answer 1

In order to show this, figure out how the transformations $C, P, T$ act on each element in the equation individually.

Pay special attention to $C$. That's a tricky one! For a start on C you can check out this physics SE post.

Answer 2

Here is one tricky way to show that the Dirac representation is invariant under C,P,T-transformations. The free Dirac theory refers to the direct sum $\left( 0 , \frac{1}{2}\right) \oplus \left( \frac{1}{2}, 0 \right)$ of the Lorentz group irreducible spinor representations. As it can be shown, the representation $\left(\frac{n}{2}, \frac{m}{2}\right) \oplus \left(\frac{m}{2}, \frac{n}{2}\right)$ is always invariant under C, P, T transformations. So you don't need to find the explicit form of $C, P, T$-matrices for the Dirac theory if you know the spinor representation of the Lorentz group.

The short proof of the invariance of $\left( \frac{m}{2} , \frac{n}{2}\right) \oplus \left( \frac{n}{2}, \frac{m}{2} \right)$ under discrete transformations of the Lorentz group

The proof, of course, is very formal. In a few words,

spinor representation of the Lorentz group involves two Casimir operators, $$ C_{1} = M_{ab}M^{ab}, \quad C_{2} = M^{\dot {a} \dot {b}}M_{\dot b \dot a}. $$ Here the generators of spinor representation of the Lorentz $M_{ab}, M_{\dot a \dot b}$ are connected with $M_{\mu \nu}$ (the vector generator of the Lorentz group) by the definite relation (here it's not important what exactly is this relation): $$ \tag 1 C_{1}\left( \frac{n}{2}, \frac{m}{2}\right) = -\frac{n(n + 2)}{2}\left( \frac{n}{2}, \frac{m}{2}\right), $$ $$ \tag 2 C_{2}\left( \frac{n}{2}, \frac{m}{2}\right) = -\frac{m(m + 2)}{2}\left( \frac{n}{2}, \frac{m}{2}\right). $$ Then we may introduce the general operators of $T, P, C$-inversion by their (anti)commutators with the $M_{ab}, M_{\dot {a}, \dot {b}}$ (which can be done by their clear (anti)commutators with $M_{\mu \nu}$). Finally, $$ \hat{T}C_{1} = C_{2}\hat {T}, \quad \hat {P}C_{1} = C_{2}\hat {P}, \hat {T}C_{2} = C_{1}\hat {T}, \quad \hat {P}C_{2} = C_{1}\hat {P}. $$ and the similar for $C$-inversion.

So when acting on $(1)$, $(2)$ by $C, P$ or $T$ operators, we will get (for example, by $\hat {P}$) $$ \tag 3 C_{1}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right) =-\frac{m(m + 2)}{2}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right), $$ $$ \tag 4 C_{2}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right) =-\frac{n(n + 2)}{2}\hat {P}\left( \frac{n}{2}, \frac{m}{2}\right). $$ So we see, that $\hat {P}$ acting on $\left( \frac{n}{2}, \frac{m}{2}\right)$ changes it to $\left( \frac{m}{2}, \frac{n}{2}\right)$, so in general (except the case $m = n$) $\left( \frac{n}{2}, \frac{m}{2}\right)$ isn't invariant under $C, P, T$-transformations. But the direct sum $\left( 0 , \frac{1}{2}\right) \oplus \left( \frac{1}{2}, 0 \right)$ (in particular) is invariant, because $$ \hat {P}\left(\left( \frac{n}{2}, \frac{m}{2}\right) \oplus \left( \frac{m}{2}, \frac{n}{2}\right) \right) = \left( \frac{m}{2}, \frac{n}{2}\right) \oplus \left( \frac{n}{2}, \frac{m}{2}\right) $$ (nothing have changed).