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Weinberg QFT Vol2 pg457

  1. Nov 23, 2014 #1

    ChrisVer

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    Gold Member

    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?
     
  2. jcsd
  3. Nov 24, 2014 #2
    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]
     
  4. Nov 27, 2014 #3

    ChrisVer

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    Gold Member

    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].
     
  5. Nov 27, 2014 #4

    ChrisVer

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    Gold Member

    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}
     
    Last edited: Nov 27, 2014
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