Hermitian conjugate of spinor product (Srednicki ch 35)

[LAHLH]

Hi,

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

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

(\chi\psi)^{\dag}=(\chi^{a}\psi_{a})^{\dag}

Now this product is just a number, as the indices are completely summed over, so it should be totally legitimate for me to take \dag 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)

(\chi\psi)^{\dag}=(\chi^{a}\psi_{a})^{\dag}=(\chi^{a}\psi_{a})^{*}

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

(\chi^{\dag\dot{a}})(\psi^{\dag}_{\dot{a}})

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

- (\psi^{\dag}_{\dot{a}})(\chi^{\dag\dot{a}})
=- \psi^{\dag}\chi^{\dag}

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

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

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!