Define
$$
W_\mu^\pm=\frac{W_\mu^1\mp iW_\mu^2}{\sqrt2},
\qquad
\tan\theta_W=\frac{g'}g,
$$
and rotate the neutral fields by
$$
\begin{pmatrix}Z_\mu\\A_\mu\end{pmatrix}
=\begin{pmatrix}\cos\theta_W&-\sin\theta_W\\
\sin\theta_W&\cos\theta_W\end{pmatrix}
\begin{pmatrix}W_\mu^3\\B_\mu\end{pmatrix}.
$$
The quadratic mass terms from $|D_\mu\langle H\rangle|^2$ are diagonal in this basis and give
$$
m_W=\frac{gv}{2},
\qquad
m_Z=\frac v2\sqrt{g^2+g'^2},
\qquad
m_A=0.
$$
The scalar mass is
$$
m_h^2=2\lambda v^2.
$$
The unbroken generator is $Q=T^3+Y$, and $A_\mu$ is its gauge field. Its exact masslessness and coupling $e=g\sin\theta_W=g'\cos\theta_W$ identify it as the photon.


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 305 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-305/2/c/solution