- #1
LAHLH
- 409
- 1
Hi,
Srednicki says in (38.14) that for any combination of gammas [tex] \bar{A}\equiv \beta A^{\dag}\beta[/tex]. This is fine, and I can work at such relations as (38.15), like [tex] \bar{\gamma^{\mu}}=\gamma^{\mu}[/tex] and so on.
We also have for spinors [tex] \bar{u_s}\equiv u^{\dag}_s\beta[/tex], and for the Dirac field [tex]\bar{\Psi}=\Psi^{\dag}\beta[/tex]
Oh and believe that the following law holds [tex] \overline{AB}=\bar{B}\bar{A} [/tex].
Now my questions is, how does Srednicki obtain the first equality in (39.8)? namely
[tex] \left[ \bar{u_s}\gamma^{0}\Psi\right]^{\dag}=\overline{\bar{u_s}\gamma^{0}\Psi\right} [/tex]
I can't see how this follows from any of his prior definitions. We know from the previous chapter that if A is some combo of gammas then [tex] \bar{A}=A[/tex], but no relations concerning [tex]\bar{A}=A^{\dag}[/tex] for certain combinations.
Srednicki says in (38.14) that for any combination of gammas [tex] \bar{A}\equiv \beta A^{\dag}\beta[/tex]. This is fine, and I can work at such relations as (38.15), like [tex] \bar{\gamma^{\mu}}=\gamma^{\mu}[/tex] and so on.
We also have for spinors [tex] \bar{u_s}\equiv u^{\dag}_s\beta[/tex], and for the Dirac field [tex]\bar{\Psi}=\Psi^{\dag}\beta[/tex]
Oh and believe that the following law holds [tex] \overline{AB}=\bar{B}\bar{A} [/tex].
Now my questions is, how does Srednicki obtain the first equality in (39.8)? namely
[tex] \left[ \bar{u_s}\gamma^{0}\Psi\right]^{\dag}=\overline{\bar{u_s}\gamma^{0}\Psi\right} [/tex]
I can't see how this follows from any of his prior definitions. We know from the previous chapter that if A is some combo of gammas then [tex] \bar{A}=A[/tex], but no relations concerning [tex]\bar{A}=A^{\dag}[/tex] for certain combinations.