3
$\begingroup$

I'm reading Climbing the Mountain by Mehra, in which he explains Schwinger's source theory by an example. He appears to end up giving an explicit form for the effective action of QED and I don't understand the reasoning.

He starts by giving the vacuum persistence amplitude in the presence of photon source $J$ and electron source $\eta$ for the interactionless case $$ \langle 0_{+} | 0_{-} \rangle^{\eta J} = e^{iW} , $$ in which the effective action $W$ is $$ W = \frac{1}{2} \int (dx) (dx') \eta(x) \gamma^0 G_+ (x-x') \eta(x') + \frac{1}{2} \int (dx) (dx') J^\mu(x) D_+ (x-x') J_\mu(x') $$ with $G_+$ and $D_+$ the propagators for the electron and photon fields.

He then claims that we could write instead $$ W = \int (dx) \left[ \psi(x) \gamma^0 \eta(x) + A^\mu J\mu(x) + \mathcal{L}(\psi,A) \right] $$ with electron field $\psi$ and photon field $\psi$ and free-field Dirac+Maxwell Lagrangian density $\mathcal{L}(\psi,A)$. I assume this is fine since we have no interactions and this is just an equivalent statement.

However, he then goes on to apparently argue that we simply need to make the substitution $$ \partial_\mu \rightarrow \partial_\mu - i e q A_\mu , $$ i.e. the usual minimal substitution, to get the effective action with interactions. Then does some algebra and writes $$ W = \frac{1}{2} \int (dx) \left[ J_\mu A^\mu + \eta \gamma^0 \psi\right] + \frac{1}{2} \int (dx) \psi \gamma^0 e q \gamma A \psi + \\\frac{1}{2} \int (dx) (dx') \frac{1}{2} (\psi \gamma^0 e q \gamma^\mu \psi) (x) D_+ (x-x') \frac{1}{2} (\psi \gamma^0 e q \gamma^\mu \psi) (x') + \\ \frac{1}{2} \int (dx) (dx') \psi(x) e q \gamma A(x) G_+ (x-x') e q \gamma A(x') \psi(x')+\cdots $$ He says that "we can infer a sequence of increasingly elaborate ``interaction skeletons"" as if to suggest this is the first-order approximation. But it's not clear what approximation scheme he is using or how one would get an improved approximation.

What is Mehra actually trying to say here?

Climbing the Mountain is a biography of Schwinger and this calculation apppears in the Nobel Prize chapter.

$\endgroup$
0

0