# Writing lepton sector kinematic term in terms of N_L, E_L projections of Dirac fields

1. Feb 10, 2012

### LAHLH

Hi,

In srednicki (ch88) he starts off considering the electron and associated neutrino, by introducing the left handed Weyl fields $l, \bar{e}$ in the representations (2,-1/2), (1,+1) of SU(2)XU(1).

The covariant derivaties are thus

$(D_{\mu}l)_i=\partial_{\mu}l_i-ig_2A_{\mu}^a(T^a)_i^jl_j-ig_1(-1/2)B_{\mu}l_i$ and $D_{\mu}\bar{e}=\partial_{\mu}\bar{e}-ig_1(+1)B_{\mu}\bar{e}$ where the T's are SU(2) gens and Y=-1/2I for l, and Y=+1 for the $\bar{e}$. He then relabels the SU(2) components of $l=\left( \begin{array}{c} \nu\\ e\end{array} \right)$ and defining the Dirac field $\varepsilon=\left( \begin{array}{c} e\\ \bar{e}^{\dagger}\end{array} \right)$ and finally the Majorana feld for the neutrino:$N=\left( \begin{array}{c} \nu\\ \nu^{\dagger}\end{array} \right)$.

Now the kinematic term in the Lagrangian starts life in terms of the Weyl fields looking like:
$L_{\text{kin}}=i\nu^{\dagger}\bar{\sigma}^{\mu}(D_{\mu}\nu)+ie^{\dagger}\bar{\sigma}^{\mu}(D_{\mu}e)+i\bar{e}^{\dagger}\bar{\sigma}^{\mu}(D_{\mu}\bar{e})$

My question is how exactly do I rewrite this in terms of the Dirac fields?

I believe if you define $N_L:=P_L N=\left( \begin{array}{c} \nu\\ 0\end{array} \right)$ then the first term can be expressed as $i\bar{N_L}\gamma^{\mu}D_{\mu}N_L$. This is the case because $\bar{N_L}=(0, \nu^{\dagger})$ so we get:

$$i\bar{N_L}\gamma^{\mu}D_{\mu}N_L =i(0, \nu^{\dagger})\left( \begin{array}{cc} 0 & \sigma^{\mu}\\ \bar{\sigma}^{\mu}&0\end{array} \right)\left( \begin{array}{c} D_{\mu}\nu\\ 0\end{array} \right) =i\nu^{\dagger}\bar{\sigma}^{\mu}(D_{\mu}\nu)$$

The second term of the kinematic Lagrangian is similarly $i\bar{\varepsilon_L}\gamma^{\mu}D_{\mu}\varepsilon_L$ as far as I can tell, but then the third term does not seem to fall into such an expression?