Sunday, 1 November 2020

quantum field theory - CPT transformation for bilinears


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?



Answer



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,


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


And,


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



$\gamma^0_{ij} = \gamma^0_{ji}$ 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


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...