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I'm studying the Higgs mechanism of the spontaneus symmetry breaking in the SM.

The expression for the $\rho$-parameter is $$ \rho=\frac{M^2_W}{{\cos^2\theta_w}M^2_Z}, $$ that, in the case of a $SU(2)$ doublet with hypercharge $Y/2=1/2$: $\phi=\frac{1}{\sqrt{2}}(0\;\;\;v+h(x))^T$, is $\rho=1$.

I did a straightforward calculation: evaluating $|(D_\mu\phi)|^2$, where $$ D_\mu=\partial_\mu+igT^aW^a_\mu+i\frac{g'}{2}B_\mu, $$

in which $T^a$ are the generators of $SU(2)$ in doublet representation.

However, I found that in general, for several Higgs multiplets, with $Y_i$, $T_i$ and VEV $v_i$, the parameter becomes

$$ \rho=\frac{\sum_iv_i^2[T_i(T_i+1)-Y_i^2]}{2\sum_iv_i^2Y_i^2}. $$

Searching the web I found this, in which I understand the new way to write the $W^a$, but I don't really understand the way to write the multiplets and how to get to the masses of $W$ and $Z$.

Any hints?

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Considering several Higgs multiplets with $T_i$ and $Y_i$, whose neutral ($Q_i=0$) states acquire $v_i$, we can write the kinetic part of the Lagrangian as $$ (D_\mu\phi_i)(D^\mu\phi_i)=\left|i\partial_\mu-\frac{g}{\sqrt{2}}(T_i^+W_\mu^++T_i^-W_\mu^-)\phi_i-\left(gT^3_iW_\mu^3+\frac{g}{2}Y_iB_\mu\right)\phi_i\right|^{\;2}, $$ where $T^{\pm}=T^1\pm iT^2$ and $\phi_i$ is a multiplet.

Using the $W^3,B\rightarrow Z, A$ transformation: $$ \sum_i |D_\mu\phi_i|^2=\\\sum_i\left|i\partial_\mu\phi_i-\frac{g}{\sqrt{2}}(T_i^+W_\mu^++T_i^-W_\mu^-)\phi_i-\frac{g}{\cos\theta_W}(T_i^3-\sin^2\theta_WQ_i)Z_\mu\phi_i-ig\sin\theta_WQ_iA_\mu\phi_i\right|^{\,\,2} $$

($g'\cos\theta_W=g\sin\theta_W=e$).

Then, recalling that $Q_i=0$ and $T_i^3=-Y_i$, we can find $M_W$ and $M_Z$:

$$ M_W^2=\sum_i\frac{g^2}{2}\phi^+_i(T_i^+T_i^-+T_i^-T_i^+)\phi_i=\sum_i\frac{g^2v_i^2}{2}[T_i(T_i+1)-(T_i^3)^2]=\sum_i\frac{g^2v_i^2}{2}[T_i(T_i+1)-Y_i^2] $$ and $$ M_Z^2=\sum_i\frac{g^2}{\cos^2\theta_W}(T_i^3)^2v_i^2=\sum_i\frac{g^2}{\cos^2\theta_W}v_i^2Y_i^2. $$

Finally: $$ \rho=\frac{\sum_iv_i^2[T_i(T_i+1)-Y_i^2]}{2\sum_iv_i^2Y_i^2}. $$

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