In the page 5 of the document 'CPT Symmetry and Its Violation' by Ralf Lehnert (https://core.ac.uk/download/pdf/80103866.pdf), appears a discussion about how the spin-statistics theorem applies to the CPT theorem proof. It is said that for 2 spinors $\chi, \psi$, CPT transformations looks like:

$$ \bar{\chi}\psi \rightarrow -\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T} = \dots = (\bar{\chi} \psi)^\dagger $$

Nevertheless, from the left hand side of the first equal symbol I derive,

$$ -\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T} = (-\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T})^{\dagger\ *} $$

Since a bilinear and its transpose is the same thing. Now I'm going to use introduce inside bracket the conjugation operation represented by $*$. Then,

$$ (-\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T})^{\dagger\ *} = -(\chi^\dagger \gamma^0 \psi)^\dagger = -(\bar{\chi}\psi)^\dagger $$

So, my result has different sign from the one in the document. It is no conflict with the usual CPT result that says $\bar{\psi}\psi \rightarrow \bar{\psi}\psi$ since you can choose $\chi = \psi$ and due to anti-commutation of the 'bar' fields with fields you get precisely that result. Otherwise, it would be, $\bar{\psi}\psi \rightarrow -\bar{\psi}\psi$

Am I right or I'm loosing something?


2 Answers 2


I believe you're correct. Another reference here gives the identity $$CPT: \qquad \bar{\chi}\psi \rightarrow \bar{\psi}\chi$$ which is $$ -(\bar{\chi}\psi)^\dagger = \psi^\dagger \gamma_0 \chi = \bar{\psi} \chi$$

The error in the reference you are working from is in the second-to-last equality. They have used the fermion anticommutation relation in the last equality, but ignored it in the second-to-last, writing $$-\chi^{\dagger T \dagger} \gamma^{0*} \psi^{\dagger T} = -(\psi^T \gamma^{0T} \chi^{T\dagger})^\dagger$$

But this should not have a negative sign after the equality. In fact, the whole derivation is quite circular and inconsistent, they have simply attempted to apply two transpositions inconsistently to summon a change of sign. The trick is that for Grassmannian operators, the usual identity $(AB)^T = B^T A^T$ needs to be amended to $(AB)^T = -B^T A^T$. Otherwise the following is inconsistent: \begin{align} 1)& \qquad (\chi^T \psi)^T = \chi^T \psi\\ 2)& \qquad \chi^T \psi = -\psi^T \chi\\ \implies 3)& \qquad (\chi^T \psi)^T = - \psi^T \chi = \chi^T \psi \end{align}

  • $\begingroup$ Thank you for your answer. If you like, you could take a look to my new related post physics.stackexchange.com/questions/476870/… where I try to prove CPT theorem by using what we discuss here. I'm kind of stuck $\endgroup$
    – Vicky
    Apr 30, 2019 at 3:55
  • $\begingroup$ Ok, I think that I was wrong and I'm going to write an answer to explain it $\endgroup$
    – Vicky
    May 1, 2019 at 13:21

In the text, you can see that the CPT transformation can be written as

$$ \bar{\chi}\psi \rightarrow -\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T} = -(\psi^T \gamma^0 \chi^{T\ \dagger})^\dagger $$

If you go on with that expression,

$$-(\psi^T \gamma^0 \chi^{T\ \dagger}) = -(\psi^T \gamma^0 \chi^*) = -(\psi^T \gamma^0 \chi^{T\ \dagger}) = -(\chi^\dagger \gamma^0 \psi)^T $$


$$ -(\chi^\dagger \gamma^0 \psi)^T = -(\psi^T \gamma^0 \chi^*) = -\psi_i(\gamma^0)_{ij}\chi^*_j = +\chi^*_j(\gamma^0)_{ji}\psi_i = -\chi^\dagger\gamma^0\psi = \bar{\chi}\psi $$

$\gamma^0_{ij} = \gamma^0_{ji}$ and since $\gamma^0_{ii} = 0$ you can use without Dirac deltas the anti-commutation between $\chi$ and $\psi$ even if $\chi = \psi$

So under CPT,

$$ \bar{\chi}\psi \rightarrow (\bar{\chi}\psi)^\dagger $$

The key is not to consider that transpose or adjoint introduces sign. It's just as simple as if $A, B$ are fermion fields, then

$$ (AB)^T = B^T A^T,\quad (AB)^\dagger = B^\dagger A^\dagger \tag{A}$$

The second one comes from the definion of adjoint operator, i.e., if ${\cal O}$ is an operator, its adjoint ${\cal O}^\dagger$ is given by

$$ \langle f|{\cal O}g \rangle = \langle {\cal O}^\dagger f|g \rangle $$

So, if ${\cal O} = AB$ you have that,

$$ \langle f|ABg \rangle = \langle {A}^\dagger f|Bg \rangle = \langle B^\dagger A^\dagger f|g \rangle $$

The first one of Eq. (A) it's now a corollary that comes from the definition of adjoint as transpose plus complex conjugation.

A last remark is that it's NOT true that $(\bar{\chi}\psi)^T = \bar{\chi}\psi$, so in general

$$ (\bar{\chi}\psi)^T \neq \bar{\chi}\psi $$

This is due to $\bar{\chi}\psi$ is not a number, it's an operator and it's not true, in general, that an operator and its transpose is the same thing. I write this because I've seen it in other post related to similar questions about transposition and adjoit of bilinears, and I think that I have already proved it to be wrong in this answer. I recommend to visit Transposition of spinors


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