Instantaneous Coulomb interaction in QED

Question

It seems I am stuck with a (at a first sight) trivial problem.

It's from the "Quarks and Leptons" (Halzen, Martin) book page $141$, where one considers the following integral:

$$\tag{1} T_{fi} = -i\int \!d^4x \, J_0^A(t_A,\vec{x}_A)\,J_0^B(t_A,\vec{x}_A)\frac{1}{|\vec{q}|^2}. $$ In equation $(1)$, $J_0^A$ and $J_0^B$ are the zeroth component of two electron currents: $$J_\mu(x) = j_\mu\mathrm{exp}[(p_f-p_i)\cdot x].$$

Now, according to the authors, one can rewrite $(1)$ by making use of the Fourier transform

$$\tag{2} \frac{1}{|q|^2} = \int\! d^3x\, e^{i\vec{q}\cdot\vec{x}}\frac{1}{4\pi|\vec{x}|}, $$ to the following $$ \tag{3} T_{fi}^{Coul} = -i\int \!dt_A\int d^3x_A\int d^3x_B \, \frac{J_0^A(t,\vec{x}_B)\,J_0^B(t,\vec{x}_B)}{4\pi|\vec{x}_B-\vec{x}_A|}. $$

Equation $(3)$ is then interpreted as the instantaneous$^1$ Coulomb interaction between the charges of the particles, $J_0^A$ and $J_0^B$.

The derivation of this is given in the answer below.

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$^1$I.e. interaction without retardation at time $t_A$.

Answer 1

It seems to me that there is a typo in the book. Your starting equation should be the following: \begin{equation} T_{fi} = -i\int \frac{d\omega \,d^{3}\textbf{q}}{(2\pi)^{4}} \tilde{A}(\textbf{q},\omega)\tilde{B}(-\textbf{q},-\omega) \frac{1}{|\textbf{q}|^{2}}, \end{equation} where $\tilde{A}$ denotes the Fourier transform of $A$. Then, using \begin{equation} \tilde{f}(\textbf{q},\omega) = \int dt\, d^{3}\textbf{x} \,f(\textbf{x},t)\, e^{-i(\textbf{q}\cdot \textbf{x}-\omega t)} \end{equation} will lead to the desired result.

Answer 2

I suspected that one needed to go back to the definition of the currents and indeed, in doing so one can derive the result. Here's a short version.

The electron current is defined as [see equation (6.6) in 1] $$\tag{1}J_\mu(x) = -e\bar{u}_f\gamma_\mu u_i \times\mathrm{exp}[(p_f-p_i)\cdot x]$$ which we write as $$\tag{2}J_\mu(x) = j_\mu\mathrm{exp}[(p_f-p_i)\cdot x]. $$

We will also need to use $$\tag{3}q = p_i^A-p_f^A = p_f^B-p_i^B$$

Then the integral $(1)$ in the original post can be written $$ \tag{4} T_{fi} = -i\int \!dt_Ad^3x_A d^3x \,\, j^Aj^B e^{i(p_f^{A0}-p_i^{A0})t_A}e^{i(p_f^{B0}-p_i^{B0})t_A}\frac{1}{|\vec{x}|}e^{i\vec{q}\cdot\vec{x}}. $$

Now shifting $\vec{x}=\vec{x}_B-\vec{x}_A$ with $d^3x=d^3x_B$ and using $(3)$ and $$(\vec{p}_f^A-\vec{p}_i^A)\cdot(\vec{x}_B-\vec{x}_A) = -(\vec{p}_f^B-\vec{p}_i^B)\cdot\vec{x}_B-(\vec{p}_f^A-\vec{p}_i^A)\cdot\vec{x}_A, $$

equation $(4)$ becomes

$$\tag{3} T_{fi} = -i\int \!dt_A\int d^3x_A\int d^3x_B \, \frac{J_0^A(t_A,\vec{x}_A)\,J_0^B(t_A,\vec{x}_B)}{4\pi|\vec{x}_B-\vec{x}_A|}, $$ where $J_0^A(t_A,\vec{x}_A)$ corresponds to the OP's $A(t_A,\vec{x}_A)$ and so on.

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References: See appendix of J. H. Field, Classical electromagnetism as a consequence of Coulomb's law, special relativity and Hamilton's principle and its relationship to quantum electrodynamics