# How does the amplitude for photon exchange reproduce magnetic forces?

One of the most beautiful things you learn in an introductory QFT course is the derivation of the $$1/r$$ Coulomb potential, from the $$1/k^2$$ amplitude for single-photon exchange. This follows from simply applying the Born approximation at leading order, in the nonrelativistic limit.

However, I'm confused about the case where the charges are moving relativistically. Because the single-photon exchange amplitude is derived in relativistic quantum field theory, it should still apply. But we know that when relativistic charges scatter, the physics is much more complicated than the Coulomb potential. For example, charges produce magnetic fields a factor $$v/c$$ smaller than their electric fields, which implies that magnetic forces are $$v^2/c^2$$ smaller than electric forces. So in the relativistic case the two are comparable.

In ordinary quantum mechanics, we could treat this effect using the Darwin Lagrangian. However, this same information must somehow be implicit in the single-photon exchange amplitude $$1/k^2$$. (It can't be in a higher-order diagram, because the magnetic force is the same order in $$e$$ as the electric force, and there are no other leading-order diagrams.) How can this be seen explicitly?

• Considering QED gives you Maxwell's equations, and considering you can derive with them the relativistic Lorentz force (and subsequent $v^2/c^2$ expansions), doesn't this yield your desired description? Or, are you asking something else? Sep 1, 2021 at 8:25
• @Angry Refrigerator I’m asking how to derive it directly from the scattering amplitude, which should contain the relevant information. Sep 1, 2021 at 17:04
• Not exactly what you are looking for but Weinberg (vol1) explains how to get the point-particle Dirac equation from QED. This equation already contains the full EM interactions and can be non-relativistically expanded to the Schrodinger eq, plus EM terms. Sep 6, 2021 at 18:43

## How to separate electric and magnetic effects

The QED lagrangian has the form $$\newcommand{\opsi}{\overline\psi} L = -\frac{1}{4e^2}F^{ab}F_{ab} + \opsi\gamma^a A_a\psi + L_\psi \tag{1}$$ where $$L_\psi$$ is the kinetic term for the matter field $$\psi$$. The coupling constant $$e$$ has been moved from the interaction term to the gauge-field kinetic term by rescaling the gauge field. The gauge-field kinetic term has the form $$-\frac{1}{4e^2}F^{ab}F_{ab} =\frac{E^2-B^2}{2e^2}. \tag{2}$$ To separate electric and magnetic effects from each other, we can temporarily generalize (2) to $$\frac{E^2}{2e_E^2}-\frac{B^2}{2e_B^2} \tag{3}$$ where $$e_E^2$$ and $$e_B^2$$ are independent parameters. This ruins Lorentz symmetry, but only temporarily. After finishing the calculation and seeing how the electric/magnetic distinction manifests itself in the result, we can restore Lorentz symmetry by setting $$e_E^2=e_B^2=e^2$$.

## The scattering amplitude

Since we've temporarily compromised Lorentz symmetry anyway, we might as well use the temporal gauge $$A_0=0$$. Then the action is \begin{align} L &= \frac{1}{2}\sum_{jk}\int_{\omega,p} A_j \left(\frac{\omega^2\delta_{jk}}{e_E^2} -\frac{p^2\delta_{jk}-p_j p_k}{e_B^2} \right)A_k \\ &+ \int_x \big(\opsi\gamma^k A_k\psi + L_\psi\big). \tag{4} \end{align} with $$j,k\in\{1,2,3\}$$. The four-momentum is $$(\omega,p_1,p_2,p_3)$$. From (4), we can read off the form of the propagator for the gauge field: \begin{align} D_{jk}(\omega,p) &= \left(\frac{\omega^2}{e_E^2}-\frac{p^2}{e_B^2}\right)^{-1} \left(\delta_{ij} - \frac{p_j p_k/e_B^2}{\omega^2/e_E^2}\right) \\ &= e_E^2\frac{e_B^2\delta_{jk}-e_E^2 p_j p_k/\omega^2}{ e_B^2\omega^2-e_E^2p^2}. \tag{5} \end{align} The scattering amplitude is given by the usual expression, but with this propagator in place of the usual one.

## Interpretation in the pure electric case ($$e_B=0$$)

When $$e_B=0$$ in the lagrangian, a nonzero magnetic field is infinitely costly, so the magnetic field is effectively constrained to be zero. Therefore, when $$e_B=0$$, we would expect (5) to describe the pure Coulomb interaction. This is not obvious the way (5) is written, because setting $$e_B=0$$ gives $$D_{jk}(\omega,p)= e_E^2\frac{ p_j p_k}{p^2\omega^2}. \tag{6}$$ Despite appearances, this is consistent with a pure Coulomb interaction. To see why, recall that the interaction term in the original lagrangian is $$J^a A_a$$, where $$J^a=\opsi\gamma^a\psi$$ is the current. Gauge invariance implies current conservation $$\partial_a J^a=0$$, whose momentum-space version is $$\omega J^0+p_k J^k=0$$. Use this with (6) to get $$J^j D_{jk} J^k = e_E^2 \frac{J^0 J^0}{p^2} \tag{7}$$ in momentum space. This shows that the right-hand side of (7) is the expected pure Coulomb interaction. Throughout this answer, the current $$J$$ is an operator. The scattering amplitude (with $$e_B=0$$) is obtained by sandwiching (7) between the initial and final two-particle states.

## Interpretation in the pure magnetic case ($$e_E=0$$)

When $$e_E=0$$ in the lagrangian, a nonzero electric field is infinitely costly, so the electric field is effectively constrained to be zero. Therefore, when $$e_E=0$$, we might have expected (5) to describe a purely magnetic interaction... but setting $$e_E=0$$ in (5) makes the propagator zero, which means no interaction at all! What's going on here?

In hindsight, setting $$e_E=0$$ in (5) gives exactly what we should have expected. In the temporal gauge, constraining the electric field to be zero is the same as constraining the gauge field to be independent of time. With this constraint, the current is unable to influence the gauge field, so the gauge field is unable to mediate any interaction between currents.

Another perspective: The electric and magnetic fields are not entirely independent of each other, because they're both written in terms of the gauge field. Constraining $$B=0$$ only constrains the transverse components of the gauge field, but constraining $$E=0$$ constrains the time-dependence of all components of the gauge field (in the temporal gauge).

## How the amplitude reproduces magnetic forces

From now on, let's focus on the part of (5) that is not included in the purely electric part (6). I'll avoid calling this the purely magnetic part, for the reason given in the preceding paragraphs, but we can at least say that it includes the magnetic effects that the question is asking about. The difference between (5) and (6) is \begin{align} \delta D_{jk}(\omega,p) &= e_E^2\frac{e_B^2\delta_{jk}-e_E^2 p_j p_k/\omega^2}{ e_B^2\omega^2-e_E^2p^2} - e_E^2\frac{ p_j p_k}{p^2\omega^2}. \tag{8} \end{align} Now that we've isolated the part of interest, we can set $$e_E^2=e_B^2=e^2$$ to get \begin{align} \delta D_{jk}(\omega,p) &= e^2\frac{\delta_{jk}-p_j p_k/\omega^2}{ \omega^2-p^2} - e^2\frac{ p_j p_k}{p^2\omega^2} \\ &= e^2\frac{\delta_{jk}-p_j p_k/p^2}{\omega^2-p^2}. \tag{9} \end{align} The corresponding contribution to the electron-electron interaction is $$J^j \delta D_{jk} J^k. \tag{10}$$ The $$p$$ in (9) is the momentum transferred from one current to the other, and the currents $$J$$ here are operators. The scattering amplitude is obtained by sandwiching (10) between the initial and final two-particle states.

Now consider the Fourier transform of (10) in the center-of-mass frame, where the energy-transfer is $$\omega = 0$$. The result (shown in this Wikipedia page) is proportional to $$e^2 J^j(x)\frac{\delta_{jk}+r_j r_k/r^2}{r}J^k(y) \tag{11}$$ with $$\vec r=\vec x-\vec y$$. Insofar as the (quantum) particle has any well-defined location and velocity, the current $$J(x)$$ is proportional to the velocity and $$x$$ is the location, so (11) reproduces the Darwin lagrangian that was mentioned in the question.

The last paragraph was a bit loose with its interpretation of the currents $$J$$. Again, the currents in (10) are operators. By treating the currents as though they were classical currents, the last paragraph glossed over some important features of the scattering amplitude, including information about the particles' spins.

• Does introducing $e_E\neq e_B$ have any effect on $E=\nabla \phi+\dot A$, $B=\nabla\times A$? I guess $dF$ is not quite 0, so $F\neq dA$? Sep 6, 2021 at 18:15
• @AccidentalFourierTransform (This is a refinement of my original pair of comments.) We can define $E$ and $B$ to be the time-space and space-space parts of $F_{ab}\equiv \partial_a A_b-\partial_b A_a$ ($F=dA$), which satisfies $dF=0$ identically. These def'ns don't rely on the form of the action, so they're not affected by $e_E\neq e_B$. The action says something about the tensor nature of the fields based on what kinds of transforms leave the action invariant, but we can still define $F$ in terms of $A$ in a particular coordinate system without assuming anything about how they transform. Sep 7, 2021 at 13:13
• Yes, that sounds convincing. Thanks! Sep 7, 2021 at 13:21
• A brilliant answer, as always! Sep 10, 2021 at 6:15