Chapter 15 - Electroweak Unification

$Y = 2(Q - I_W^{(3)}$)

Summarizing weak interaction

W transforms neutrino $\leftrightarrow$ leptons within the same flavor

W transforms up-type quarks (u, c, t) $\leftrightarrow$ down-type quarks (d', s', b') within the same generation in weak basis (not flavor basis, which is (d, s, b))

Therefore, construct a weak doublet: $ \begin{pmatrix}\nu_e\\ e \end{pmatrix}_L$, $\begin{pmatrix}\nu_\mu\\ \mu \end{pmatrix}_L$, $\begin{pmatrix}\nu_\tau\\ \tau \end{pmatrix}_L$, $\begin{pmatrix}u\\ d' \end{pmatrix}_L$, $\begin{pmatrix}c\\ s' \end{pmatrix}_L$, $\begin{pmatrix}t\\ b' \end{pmatrix}_L$

Weak doublet is described by weak isospin: $I_w = \frac{1}{2}, I_w^{(3)}(\nu_e) = \frac{1}{2}, I_w^{(3)}(e^-) = -\frac{1}{2}$

Upper and lower components differ in electric charge by $1 e$; Only LH particle chiral states interact weakly; RH particles do not interact weakly, e.g. $e^-_R, d_R$ have weak charge $I_w = I_w^{(3)}=0$

$SU(2)_L$

Local gauge symmetry: $\phi(x) \rightarrow \phi'(x) = e^{-ig_w \mathbf{\alpha(x)}\cdot \mathbf{T}} \phi(x)$

$\mathbf{T} = \frac{1}{2}\mathbf{\sigma}$ $\rightarrow$ 3 generators $\rightarrow$ new interaction $ -ig_w \frac{1}{2}\sigma_k\gamma^\mu W^k_\mu\phi_L$ $\rightarrow$ 3 weak currents $\rightarrow$ 3 fields $ \mathbf{W^k_\mu}$, k={1,2,3}, two charged, one neutral.

Two of the charged fields give rise to physical W bosons: $\mathbf{W^{\pm}_\mu} = \frac{1}{\sqrt{2}}(W^{(1)}_\mu \mp W^{(2)}_\mu)$

Unification

replace $U(1)$ $\rightarrow$ $U(1)_Y$ local sym with weak hypercharge: $\phi(x) \rightarrow \phi'(x) = e^{-i g' \frac{Y}{2} \zeta(x)} \phi(x)$

Interaction term: $g' \frac{Y}{2}\gamma^\mu B_\mu \psi$

analogy: $Y g'/2 \leftrightarrow Qe$

Together with $SU(2)_L$, physical photon and Z boson are: $A_\mu = \cos(\theta_W) B_\mu + \sin(\theta_W) W^{(3)}_\mu$, $Z_\mu = -\sin(\theta_W) B_\mu + \cos(\theta_W) W^{(3)}_\mu$

$W^{(3)}_\mu$ is left over; this is not a physical state!

Since we know QED and CC weak interaction, should predict everything about $Z_\mu$ once we know $\theta_W$!

Y from $U(1)_L$ should be a combination of Q (from $A_\mu$) and $I_W^{(3)}$ (from $Z_\mu$).

$Y = 2(Q - I_W^{(3)}$)

$e = g' \cos(\theta_W) = g_W \sin(\theta_W)$, $\tan(\theta_W) = \frac{g'}{g_W} $, and $g_Z = \frac{g_W}{\cos(\theta_W)}$

$\frac{\alpha}{\alpha_W} = \sin^2(\theta_W) = 0.23$

Screen Shot 2022-03-03 at 11.12.32 PM

Z (neutral current) couples to both LH and RH chiral states, but not equally! $j^\mu_Z = g_Z(c_L \bar{u}_L\gamma^\mu u_L + c_R \bar{u_R}\gamma^\mu u_R) $

$c_L = I_w^{(3)} - Q_f \sin^2(\theta_W) $, $c_R = -Q_f \sin^2(\theta_W)$

Z boson interaction vertex: $-ig_Z\frac{1}{2}\gamma^\mu(c_V - c_A \gamma^5)$

$c_V = c_L + c_R = I_W^{(3)} - 2Q\sin^2(\theta_W)$, $c_A = c_L - c_R = I_W^{(3)}$

Z decay

$\Gamma(Z \rightarrow f\bar{f}) = \frac{g_Z^2 m_Z}{48 \pi} \Big((c^f_{V})^2 + (c^f_{A})^2\Big)$

Total width to $f\bar{f}$: $\Gamma_{Z} = 3 \Gamma_{\nu_e\bar{\nu}_e} + 3 \Gamma_{e\bar{e}} + 3 \times 2 \Gamma_{u\bar{u}} + 3\times 3 \Gamma_{d\bar{d}} =$ 2.5 GeV.

Note: $W^{\pm}$ is $I = 1, I_W^{(3)} = \pm 1$. For RH anti-particle, $I_W^{(3)}(\bar{f}) = - I_W^{(3)}(f)$

$BR(Z\rightarrow \nu\bar{\nu})=21\%$, $BR(Z\rightarrow l^+l^-)=10\%$, $BR(Z\rightarrow hadron)=69\%$ (jets)

Could not decay to top quark (173 GeV > 91 GeV); 3 comes from color d.o.f; applied radiative correction to hadronic decay.