Hermitian conjugation of part d gives the corresponding charge insertion on the bra. The <Ward identity> $i\langle[Q,\mathcal O]\rangle=\langle\delta_b\mathcal O\rangle$ therefore becomes
$$
\boxed{
\lim_{\mathbf q\to0}a^3b^i\left[
\frac{\dot f_q}{f_q}\langle A_i^V(-\mathbf q)\mathcal O\rangle
-\frac{\dot f_q^*}{f_q^*}\langle\mathcal O A_i^V(\mathbf q)\rangle
\right]
=-i\langle\delta_b\mathcal O\rangle}.
$$
For a neutral operator, $\delta_b\mathcal O=0$, so the two soft limits, multiplied by their respective wavefunctional coefficients, are equal. If $\mathcal O$ is charged, the right side is nonzero. For a product of fields of charges $e_a$ at positions $\mathbf x_a$, it is proportional to $\sum_ae_a b_i x_a^i\langle\mathcal O\rangle$; in momentum space this becomes the corresponding momentum derivative. This is a soft-vector <Ward-Takahashi identity>.

Solution

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

 Solution

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