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Дата изменения: Wed Aug 9 20:40:47 2000
Дата индексирования: Mon Oct 1 22:52:17 2012
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Interaction of vector bosons with fermions

Experiments in particle physics show that the $W$-bosons interact with left-handed components of fermions

$$\Psi^L = \frac{1 - \gamma^5}{2} \Psi\;.$$

Thus, the $SU(2)$ group must transform only the left-handed fermion components. The initial Lagrangian of fermion field has the corresponding global symmetry only if all fermions are massless. Indeed, the Lagrangian of free massless fermion field splits into two independent parts:

$$\begin{eqnarray*} L &=& \frac{i}{2}(\bar{\Psi} \gamma_\mu \partial^\mu \Psi - (... ...tial^\mu \Psi - (\partial^\mu \bar{\Psi}^R)\gamma_\mu \Psi^R)\;, \end{eqnarray*}$$

which allows to apply the $SU(2)$ gauge transformations to the left-handed components of fermion doublets. In the same time the mass term contains a product of left-handed and right-handed fermion components:

$$m \bar{\Psi} \Psi = m ( \bar{\Psi}^L \Psi^R + \bar{\Psi}^R \Psi^L)\;,$$

what forbids an appearance of such terms in the invariant Lagrangian. Later on we shall show how such fermion particles will acquire masses in result of the gauge invariant interaction of formerly massless fermion fields with the Higgs doublet.

Whereas $SU(2)$ transforms only the left-handed components of doublets, $U(1)$ interacts with both. Left-handed and right-handed components of fermions must have the same electric charge. It allows to find the hypercharge of right-handed components if a hypercharge of left-handed doublet is known. The $U(1)$ gauge field $B_\mu$ is equal to $\cos{\Theta_w} A_\mu - \sin{\Theta_w} Z_\mu$. So, the electromagnetic coupling constant for right-handed fermion with a hypercharge $Y^R$ is $(g_1 Y^R \cos{\Theta_w})/2$. Comparing it with the expression (11) we see that the hypercharges of right-handed components of doublet are

$$\begin{eqnarray*} Y^R_1&=&1+Y\;, \\ Y^R_2&=&Y-1\;. \end{eqnarray*}$$

Thus, we may unambiguously write down the vertices of interactions associated with the covariant derivative:

$$\begin{eqnarray*} &&\frac{g_2}{2} \left( \begin{array}{cc} \bar{\Psi}_1 & \bar{\... ...amma^\mu \Psi^R_1 \\ \gamma^\mu \Psi^R_2 \end{array} \right)\;. \end{eqnarray*}$$

After matrix multiplication we obtain:

$$\frac{g_2}{2\sqrt{2}} \left( W^-_\mu \bar{\Psi}_2 \gamma^\mu (1 - \gamma^5)\Psi_1 + W^+_\mu \bar{\Psi}_1 \gamma^\mu (1 - \gamma^5)\Psi_2 \right)$$

$$+ \frac{g_1}{4 \sin{\Theta_w}} Z_\mu \left(\; \bar{\Psi}_1 \gamma^\mu ( 1 - \gamma^5 -2(Y+1) \sin^2{\Theta_w} ) \Psi_1 \right.$$

$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left. - \bar{\Psi}_2 \gamma^\mu (1-\gamma^5 +2(Y-1)\sin^2{\Theta_w} ) \Psi_2 \;\right)$$

$$+ \frac{g_1}{2}\cos{\Theta_w} A_\mu \left( (Y+1) \bar{\Psi}_1 \gamma^\mu \Psi_1 + (Y-1) \bar{\Psi}_2 \gamma^\mu \Psi_2 \right)\;.$$ (19)