For constant $b_i$, the transformation is $\delta A_i^V=b_i$. Its <Noether charge> is therefore
$$
\boxed{Q=\int d^3x\,b_i\Pi^i
=\lim_{\mathbf q\to0}b_i\Pi^i(\mathbf q)}
$$
up to the Fourier-sign convention. Expand a transverse polarization as
$$
A_i^V(\mathbf q)=f_q a_i(\mathbf q)+f_q^*a_i^\dagger(-\mathbf q),
\qquad
\Pi_i(\mathbf q)=a^3\bigl(\dot f_q a_i(\mathbf q)
+\dot f_q^*a_i^\dagger(-\mathbf q)\bigr).
$$
Only the creation term survives on the vacuum, while $A_i^V(\mathbf q)|0\rangle=f_q^*a_i^\dagger(-\mathbf q)|0\rangle$. Hence
$$
\boxed{Q|0\rangle=\lim_{\mathbf q\to0}
\left[\frac{a^3\dot f_q^*}{f_q^*}
b^iA_i^V(\mathbf q)|0\rangle\right]}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 312 / 2 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-312/2/d/solution