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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?

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The Dirac equation is invariant under C,P,T even separately, so of course it is invariant under CPT in this form as well. However, the true general CPT invariance shouldn't be tested on one-particle equations or classical fields such as Dirac's equation. The general argument why CPT works only makes sense on the full multiparticle quantum field theory. –  Luboš Motl Jul 23 at 14:57

2 Answers 2

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.

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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).

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How can it be shown the representation invariant under this representation? \left(\frac{n}{2}, \frac{m}{2}\right) \oplus \left(\frac{m}{2}, \frac{n}{2}\right) –  user55944 Jul 30 at 12:43
    
@user55944 : I have wrote the general way of proof into the answer. –  Andrew McAddams Jul 30 at 13:30
    
How can you show that the weak interaction is parity violating theory according to this representation? –  user55944 Sep 13 at 10:58
    
It's easy because it consists projectors on states $\left(\frac{1}{2} , 0 \right)$ and $\left( 0 , \frac{1}{2}\right)$ unequally. For example, W-boson interaction doesn't consist right-fermions interaction. So after making parity transformation of lagrangian we won't get initial lagrangian. –  Andrew McAddams Sep 13 at 11:05

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