Past exam of the theoretical physics I course of the University of Cambridge / 2014 / Paper 1 / 5

For the supplied Lagrangian, variation with respect to ${\varphi }^{\ast }$ usesThus $\square \varphi +2ie{A}^{\mu }{\partial }_{\mu }\varphi +ie({\partial }_{\mu }{A}^{\mu })\varphi +m^2\varphi =f$. In the Lorenz gauge,For real external fields and driving, the conjugate equation has $-2ie{A}^{\mu }{\partial }_{\mu }$ acting on ${\varphi }^{\ast }$. The supplied model has no $e^2{A}^2|\varphi {|}^2$ term; one must not insert such a term from a different minimally coupled Lagrangian.

First fix the source at the spatial origin. Fourier transform replaces ${\partial }_t$ by $-i\omega$ and $\nabla$ by $i\mathbf{k}$ in the free term. Multiplication by a static background becomes a momentum convolution; the derivative in $A\cdot \partial G$ acts on the Green function, so it carries the internal momentum ${\mathbf{k}}^{\prime }$. ConsequentlyThe transformed point source supplies the right-hand side one. The signs follow from $A^{\mu }{\partial }_{\mu }=A^0{\partial }_t+\mathbf{A}\cdot \nabla$ in these conventions.

Source-position qualification. For a nonuniform $A(\mathbf{r})$, the general Green function is not translation invariant. In the printed basis $e^{i\mathbf{k}\cdot (\mathbf{r}-{\mathbf{r}}^{\prime })}$, its coefficient must also depend on ${\mathbf{r}}^{\prime }$, and the convolution contains an extra factor $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot {\mathbf{r}}^{\prime }}$. The displayed source equation is valid for ${\mathbf{r}}^{\prime }=0$, or with the background itself expressed in coordinates centred on that source. Equivalently, a general spatial Fourier Green function has two momentum arguments and a momentum-delta-function source. Staticity alone permits the time-difference transform, not a general position-difference-only transform.

The delta function in the artificial constant potential collapses the convolution. Put $b=e\gamma$ and $K^2=k^2+m^2$. ThenInverse Fourier transform gives the stated real-frequency integral with exponential $e^{-i\omega (t-t^{\prime })}$. For $K^2>b^2$, the poles have opposite real parts and common imaginary part $b$. For $0<K^2<b^2$, both are imaginary, at $i[b\pm \sqrt{b^2-K^2}]$. They remain in the upper half-plane if $b>0$, or the lower half-plane if $b<0$. At equality they merge into a second-order pole at $ib$. For $K^2=0$ one pole reaches the real axis, requiring a prescription.

Let $\tau =t-t^{\prime }$ and $\Omega =\sqrt{K^2-b^2}>0$. Because $|e^{-i(u+iv)\tau }|=e^{v\tau }$, close below for $\tau >0$ and above for $\tau <0$. The lower closure is clockwise. The two residues sum to $i{e}^{b\tau }\sin (\Omega \tau )/\Omega$. Hence, for nonzero $b$,In particular, the printed sign with $e\gamma >0$ gives zero for $t>t^{\prime }$ and an advanced Green function for $t<t^{\prime }$. Reversing that sign puts the poles below and gives a retarded Green function. Both solutions have the unit derivative jump required by $({\partial }_{\tau }^2-2b{\partial }_{\tau }+K^2)G=\delta (\tau )$. If $b=0$, the bare integral has real-axis poles and requires an additional contour prescription; a unique retarded or advanced answer cannot be inferred without one.