Hermitian conjugate of spinor product (Srednicki ch 35)

[LAHLH]

Hi,

I totally understand why [tex] \chi\psi=\chi^{a}\psi_{a}=-\psi_{a}\chi^{a}=\psi^{a}\chi_{a}=\psi\chi[/tex]. Where the first equality is just convention, the second is anticommutation of the fields, the third is due to [tex] \chi^{a}\psi_{a}=-\chi_{a}\psi^{a} [/tex] because of the [tex] \epsilon^{ab} [/tex].

But now if we look at the herm conj, as in Srednicki 35.26:

[tex] (\chi\psi)^{\dag}=(\chi^{a}\psi_{a})^{\dag} [/tex]

Now this product is just a number, as the indices are completely summed over, so it should be totally legitimate for me to take [tex]\dag[/tex] to be just a regular c.c. *. (c.f. Avodyne's discussion at the end of my spinor indices thread a while back :https://www.physicsforums.com/showthread.php?t=438291 in particular post #28)

[tex] (\chi\psi)^{\dag}=(\chi^{a}\psi_{a})^{\dag}=(\chi^{a}\psi_{a})^{*} [/tex]

Now this is just equal to [tex] (\chi^{a})^{*}(\psi_{a})^{*} [/tex], the dagger or star (which are the same thing on these components) converts these into right handed spinors, so now we have:

[tex] (\chi^{\dag\dot{a}})(\psi^{\dag}_{\dot{a}}) [/tex]

Now using anticommutation of these objects, and then using the suppressing convention for dotted indices:

[tex]- (\psi^{\dag}_{\dot{a}})(\chi^{\dag\dot{a}}) [/tex]
[tex]=- \psi^{\dag}\chi^{\dag} [/tex]

So I have found that [tex] (\chi\psi)^{\dag}=- \psi^{\dag}\chi^{\dag} [/tex]

Contrary to Srednicki, where [tex] (\chi\psi)^{\dag}=+ \psi^{\dag}\chi^{\dag} [/tex]

Could anyone help me understand what has happened here? Thanks

 

[Avodyne]

The fields are operators, not numbers, so hermitian conjugation reverses their order.

 

[LAHLH]

Oh yes, of course, even the components of the spinors are operators. Unlike the components of the Pauli matrix say which are just numbers.

Thanks once again!