Summation of an exponential operator on quantum amplitude

For a quantum Dirac field interacting with a classical EM field, one can (through the Quantum Dynamical Principle) write the vacuum transition amplitude as $$\langle0_+|0_-\rangle=\exp\left[ie_0\int dx\ \frac{\delta}{\delta\eta}\gamma^\mu A_\mu\frac{\delta}{\delta\bar{\eta}}\right]\exp\left[i\int dxdx'\ \bar{\eta}(x)S(x-x')\eta(x')\right]\tag{1}$$ with $$S(x)=\int \dfrac{d^4p}{(2\pi)^4}\dfrac{e^{ipx}}{\gamma^\nu p_\nu-m}.\tag{2}$$

On the other hand by taking the Fourier transform of the corresponding Dirac equation with source term the above is $$=\exp\left[i\int dxdx'\ \bar{\eta}(x)S_A(x-x')\eta(x')\right]\tag{3}$$ with $$S_A=\int\dfrac{d^4p}{(2\pi)^4}\dfrac{e^{ipx}}{\gamma^\nu p_\nu-ie_0\gamma^\nu A_\nu-m}.\tag{4}$$

I would think that the two expressions can be seen to be equal by expanding the exponential containing the operators, but the series is an expression with sums of products of integrals, whereas the Taylor series of the latter expression would involve products of $$\gamma^\mu A_\mu$$ and the free propagator $$S$$ in one integral.

Perhaps the most straightforward way would be to expand the fermion propagators for the interacting and noninteracting cases in a "Fourier space": $$S_A=\exp\left[ie_0\int dx\ \bar{\psi}(x)\gamma^\mu A_\mu\psi(x)\right]S$$ $$=\sum_{n=0}^\infty ie_0\left(\gamma^\mu A_\mu\int \bar{\eta}(x)S(x-x')\int S(x-x')\eta(x)\right)^n S$$ $$\stackrel{?}{=}\sum_{n=0}^\infty \left(ie_0\gamma^\mu A_\mu S\right)^nS$$ $$=S_A\tag{5}$$ where I have written $$\psi(x)$$ to mean $$\int dx'\ S(x-x')\bar{\eta}(x')$$. The second to last equation is the Taylor expansion of $$S_A$$ in terms of $$A$$

Is the approach feasible in summing the former operator expression to the latter? How would one show equality between the two (with any approach)?

• @Qmechanic I am reading Manoukian's Quantum Field Theory I. It doesn't really explain anything about Feynman diagrams... – Quantumness Feb 28 at 3:34
• Which page in Manoukian? – Qmechanic Feb 28 at 9:46
• @Qmechanic The author uses the Quantum Dynamical Principle somewhat throughout but goes more in depth in section 4.6, starting page 168. The series expressions are mostly my own doing, with the exception of an integral equation derived on page 80. – Quantumness Feb 28 at 22:11

1. There are functional determinants missing from the right-hand sides of OP's eqs. (1) and (3) from performing a Gaussian Grassmann integration over the fermion fields $$\psi$$ and $$\bar{\psi}$$. This will lead to 1-loop corrections.

2. The two exponential factors on the right-hand side of OP (1) generate all Feynman diagrams built out of 2-vertices $$ie_0\gamma^{\mu}A_{\mu}$$, bare propagators $$iS$$, and sources $$\eta$$, $$\bar{\eta}$$. Since we only have 2-vertices, there are not so many possible types of Feynman diagrams.

3. If we take the logarithm, we get all connected Feynman diagrams $$iW_c$$, cf. the linked cluster theorem. There are only two types $$iW_c = iW_{c,0} +iS_A$$, namely

• 1-loop bubbles $$iW_{c,0}$$ (closed necklaces). These will generate the above-mentioned 1-loop corrections to the right-hand side of eq. (3).

• and tree 2-pt function/dressed propagator $$iS_A$$ (open necklaces).

4. The dressed propagator $$iS_A$$ is a geometric series, as OP already seems to have observed.

• I appreciate this answer, but am not very familiar with many aspects of Feynman diagrams: could you elaborate on why (1) represents the 2-vertex diagrams (as opposed to others) and explain your later analysis? – Quantumness Feb 26 at 21:57
• Is this saying that the log of the expression in (1) (and therefore the first series equality in (5)) can be separated into the bubbles and dressed propagator diagram, and therefore the two expressions are not exactly equal? – Quantumness Feb 27 at 22:40
• I updated the answer. – Qmechanic Feb 28 at 11:38