If you have two fermions (with spin $\pm \frac{1}{2}$) that form a weak-isospin doublet and their respective right-handed fermions which are weak-isospin singlets, what would it imply if the doublet had the weak hyper charge $0$? How would the particles couple to the photon and the Z boson?

For this to happen the fermions must obviously have the charge $\pm \frac{1}{2}$ according to : $ Q = T_3 + \frac{1}{2} Y_W$ And the righthanded parts then have weak hyper charge $\pm 1$


It is a direct plugin of the standard model formula you find in all texts and the PDG.

You have posited four fermions with $(Q,T_3,Y_W)$ quantum numbers; call them $$ \psi_L \qquad (1/2,1/2,0) \\ \chi_L \qquad (-1/2,-1/2,0) \\ \psi_R \qquad (1/2, 0,1) \\ \chi_R \qquad (-1/2, 0,-1). $$

Their EM coupling is vector, so there is no point in separating L from R in it, $$ -\frac{e}{2} A_\mu (\bar{\psi} \gamma^\mu \psi -\bar{\chi} \gamma^\mu \chi ). $$

By contrast, the neutral current coupling unfolds its ventriloquist weirdness, $$ -\frac{g}{2\cos \theta_W} Z_\mu \Bigl (-\sin^2 \theta_W ~\bar{\psi}\gamma^\mu \psi_R +\sin^2 \theta_W ~\bar{\chi}\gamma^\mu \chi_R \\ + \cos^ 2 \theta_W ~\bar{\psi}\gamma^\mu \psi_L - \cos^ 2 \theta_W ~\bar{\chi}\gamma^\mu \chi_L \Bigr ). $$ I have not bothered with chiral projectors on the bar sides, since the implicit projectors on the right commute to the left to enforce selection of the same fermions. (You might choose to squint and reorganize the neutral current by resolving $\cos^ 2 \theta_W = -\sin^ 2 \theta_W +1$ leading to a vector piece with the R term, and, lopsidedly, a left-handed piece, to better grasp the weirdness.)

Note in the limit $g'\to 0, \Longrightarrow \theta_W\to 0$, the right handers decouple from the Z, while the left handers couple to it the way they do to the Ws, as required by (physical) law.

Further note the right handers couple to just $(\tan\theta_W Z_\mu - A_\mu)$, so, then to just B and are completely decoupled from $W^3$, as required.

Symmetrically, the left handers only couple to $W^3$, but not B, as required by the definition of the problem, e.g., for the $\psi_L$, $$ -\frac{g}{2} (\sin\theta_W ~ A_\mu + \cos\theta_W ~ Z_\mu)\bar{\psi}\gamma^\mu \psi_L= -\frac{g}{2} W_\mu^3 ~ \bar{\psi}\gamma^\mu \psi_L, $$ since $Z= \cos \theta_W W_3 - \sin \theta_W B$, $\qquad A = \sin \theta_W W_3 + \cos \theta_W B$, and $e=g\sin \theta_W$.

  • $\begingroup$ Okey, thank you ! $\endgroup$ – DrShellyCooper Feb 5 at 23:20

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