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In the 'Standard Model' book by Y. Grossman and Y. Nir, in chapter 7 (the leptonic standard model) on page 93 after defining the charge of the broken symmetry generator, i.e $Q=T_3+Y$

they say that the $W_{\mu}^{\pm}$ are eigenstates of $T_{3}$ with eigenvalues $\pm1$, and have Y = 0. Hence, their electromagnetic charges are: $q(W_{\mu}^+ ) = +1; q(W_{\mu}^{-} ) = -1$

Both $W_{3\mu}$ and $B_{\mu}$ are eigenstates of $T_3$ with eigenvalue 0, and have Y = 0. Hence, they are neutral under $U(1)_{EM}$ and so is any linear combination of them: $q(Z^0_{\mu}) = 0; q(A^0_{\mu}) = 0$

My question is how they found those eigenvalues. What are the matrices represent $W_{\mu}^{\pm}$,$W_{3\mu}$ and $B_{\mu}$ I need to diagonalise and find those values?

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    $\begingroup$ Write down the weak isotriplet of the W s, the three T generators of SU(2) in the triplet representation, and see how these three matrices act on your triplet. As an isosinglet, $B_\mu $ is defined to be a number not acted upon by these matrices. You may represent Y as the 3x3 identity matrix. $\endgroup$ – Cosmas Zachos Jun 15 at 14:49

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