Is Weinberg missing a \gamma_5 in his mass parameter redefinition?

[ChrisVer]

I'm having problem in deriving 23.6.11 from Weinberg's-Quantum Theory of fields...

We have: [itex] \psi_f \rightarrow \exp (i a_f \gamma_5) \psi_f[/itex], f denoting the flavor.

Then for the mass term lagrangian he writes:

[itex] L_m = - \frac{1}{2} \sum_f M_f \bar{\psi}_f (1+ \gamma_5) \psi_f - \frac{1}{2} \sum_f M^*_f \bar{\psi}_f (1- \gamma_5) \psi_f[/itex]

With [itex]M_f[/itex] the mass parameters. He says that by making a transformation of the fields as above, the mass parameter will be redefined:

[itex]M_f \rightarrow M_f \exp (2i a_f)[/itex]
However I think he is missing a [itex]\gamma_5[/itex]?

Because the first for example term:

\begin{multline}

\\

-\frac{1}{2} \sum_f M_f\psi^\dagger_f e^{-i \gamma_5 a_f}\gamma_0 (1+ \gamma_5) e^{i \gamma_5 a_f} \psi_f=\\

\approx -\frac{1}{2} \sum_f M_f\psi^\dagger_f (1-i \gamma_5 a_f)\gamma_0 (1+ \gamma_5) (1+i \gamma_5 a_f) \psi_f=\\

=-\frac{1}{2} \sum_f \psi^\dagger_f \gamma_0 M_f (1+i \gamma_5 a_f) (1+i \gamma_5 a_f) (1+ \gamma_5)\psi_f=\\

=-\frac{1}{2} \sum_f \bar{\psi}_f M_f (1+i 2 \gamma_5 a_f) (1+ \gamma_5)\psi_f

\end{multline}which leads in the redifinition of M:

[itex]M_f \rightarrow M_f \exp (2i a_f \gamma_5)[/itex]

Any help?

 

[UVCatastrophe]

Expand ## e^{ i \alpha_f \gamma_5} = 1 + i \alpha_f \gamma_5 - \frac{\alpha_f^2}{2!} \gamma_5 \ldots = \cos{\alpha_f} 1 + i \sin{\alpha_f} \gamma_5 ,## since ## \gamma_5^2 = 1##.

The first term in the lagrangian goes to ##\rightarrow \bar{\psi} e^{i \alpha_f \gamma_5} ( 1 + \gamma_5 ) e^{i \alpha_f \gamma_5} \psi ##

Using the expansion above, we find that
[tex] e^{2 i \alpha_f \gamma_5} = \cos{2 \alpha_f} 1 + i \sin{2 \alpha_f} \gamma_5 [/tex]
and
[tex] e^{i \alpha_f \gamma_5} \gamma_5 e^{i \alpha_f \gamma_5} = \cos{2\alpha_f} \gamma_5 + i \sin{2\alpha_f} 1 [/tex]
so
[tex] e^{i \alpha_f \gamma_5} ( 1 + \gamma_5 ) e^{i \alpha_f \gamma_5} = e^{2 i \alpha_f} (1 + \gamma_5) [/tex]

 

[ChrisVer]

Hi, thanks for clarifying... so I have to expland the whole exponential... [that is somewhat confusing, other times people expland it up to 1st order, other times they write the whole expansion].

 

[ChrisVer]

Could someone help me with the metric he obtains at 23.4.10?
I have tried all possible ways but I haven't been able to reproduce his result... Here is my last try for solution...Obviously my results are not the same even for the easies ase of (44) component..

Used:

[itex]Tr[\tau_a \tau_b]= \frac{1}{2} \delta_{ab}[/itex]
[itex]Tr[\tau_a \tau_b \tau_c]= \frac{i}{8} \epsilon_{bca}[/itex]
[itex]Tr[\tau_a \tau_b \tau_c \tau_d]= \frac{1}{8} \delta_{ab} \delta_{cd} + \frac{1}{32} ( \delta_{cb} \delta_{ad} - \delta_{ca} \delta_{bd})[/itex]

\begin{multline}

\\

g=\theta_4 1 +2 i \theta_i \tau_i \\

g^{-1}= \theta_4 1 -2 i \theta_i \tau_i \\

A_{ij}=g^{-1} (\partial_i g) g^{-1} (\partial_j g)\\

\end{multline}
\begin{equation}

\gamma_{ij}(\theta) = -\frac{1}{2} Tr A_{ij}

\end{equation}
So:
\begin{multline}

A_{44}=g^{-1} (\partial_4 g) g^{-1} (\partial_4 g)=g^{-1} g^{-1} 1= \theta_4^2 1- 4 \theta_a \theta_b \tau_a \tau_b-4i \theta_4 \theta_a \tau_a \\

\gamma_{44}= - \theta_4^2 + \theta^2 = 2 \theta^2 -1 \\

A_{4i}= g^{-1} (\partial_4 g) g^{-1} (\partial_i g)= g^{-1} g^{-1} 2i \tau_i=2i \theta_4^2 \tau_i - 8i \theta_a \theta_b \tau_a \tau_b \tau_i +8 \theta_4 \theta_a \tau_a \tau_i\\

\gamma_{4i}=-\frac{1}{2} \theta_a \theta_b \epsilon_{bia}-2 \theta_4 \theta_i = - 2 \theta_4 \theta_i = \gamma_{i4} \Big|_{checked}\\

A_{ij}=-4g^{-1} \tau_i g^{-1} \tau_j= -4 (\theta_4 1 -2 i \theta_a \tau_a)\tau_i (\theta_4 1 -2 i \theta_b \tau_b)\tau_j\\

\gamma_{ij}=-\frac{1}{4} (3 \theta^2 \delta_{ij}- 4\delta_{ij}+ 5 \theta_{i} \theta_{j})\\

\end{multline}