Is it possible to derive the Coulomb's law using the principles of quantum electrodynamics? How?


1 Answer 1


There are multiple ways to interpret Coulomb's law in quantum electrodynamics (QED). Interestingly, they don't lead to quite the same conclusion (but there is no inconsistency because they are not defined in the same way).

The most commonly used way (that ACuriousMind refers to in his comment) consists in relating the notion of classical potential with that of the scattering cross section between two charges. The reason why this is the most commonly used method is purely field-related in the sense that (high-energy) particle physicists are the ones who use the most QED and what they measure in fact is cross sections; so that makes sense.

In this scattering interpretation of the Coulomb's law, the interaction is related to the S matrix expression in terms of Feynman diagrams. The first diagram gives the classical Coulomb's law which is independent of the Planck constant $h$ while higher order terms contain quantum corrections that vanish as $h \rightarrow 0$. An example for scalar QED can be found here (page 12).

There is however another way to interpret the Coulomb's law that is linked to the field of low energy physics (out of which we get, atoms, molecules and so forth).

In this interpretation (more adapted to low energy physics), the Coulomb potential is the QED ground state energy of a system of two point charges $q_1$ and $q_2$ pinned at two locations separated by a distance $r_{12}$.

If we write first the hamiltonian of the free Electro Magnetic (EM) field, we have:

\begin{equation} \hat{H}_{R} = \sum_{i = 1,2} \sum_{\mathbf{k}} \hbar \omega(\mathbf{k}) \:a^{\dagger}_{i,\mathbf{k}} a_{i,\mathbf{k}} \end{equation} where $i$ stands for the two possible transverse polarizations and $\mathbf{k}$ is the wave vector of the photon.

With this definition of the free EM energy, we have that the vacuum energy for the free EM field is

\begin{equation} \langle 0| \: \hat{H}_{R} \: |0 \rangle = 0 \end{equation}

Now, when we introduce the two pinned charges, we need to account for the coupling between them and the EM field. The hamiltonian of the whole system is then:

\begin{equation} \hat{H}_{tot} = \hat{H}_{R} + \hat{H}_{coupling} \end{equation}


\begin{equation} \hat{H}_{coupling} = \int d^3r \: A_{\mu}\: j^{\mu} \end{equation}

where $\{A_{\mu}\}$ is the 4-potential of the EM field and $\{j^{\mu}\} \equiv (c\rho, \mathbf{j})$ is the 4-current density. In our case of pinned charges we have that $\{j^{\mu} \} = (c[q_1\delta(\mathbf{r}-\mathbf{r}_1)+q_2\delta(\mathbf{r}-\mathbf{r}_2)], \mathbf{0})$.

In a covariant formulation of QED, the field $A_s = A_0$ can be written in term of scalar photons (which have some unusual properties):

\begin{equation} A_s(\mathbf{r}) = \int d^3k \sqrt{\frac{\hbar}{2\varepsilon_0 \omega (2\pi)^3}}(a_s(\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{r}} + a^{\dagger}_s(\mathbf{k})e^{-i\mathbf{k}\cdot \mathbf{r}}) \end{equation} which leads to \begin{equation} \hat{H}_{coupling} = \int d^3k c\sqrt{\frac{\hbar}{2\varepsilon_0 \omega (2 \pi)^3}}\left[a_s(\mathbf{k})(q_1 e^{i\mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{i\mathbf{k}\cdot \mathbf{r}_2}) + a^{\dagger}_s(\mathbf{k})(q_1 e^{-i\mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{-i\mathbf{k}\cdot \mathbf{r}_2}) \right] \end{equation}

which are not observable as they do not contribute to the radiation field energy $\hat{H}_R$.

Because there are charges present in the system, the ground state energy is now different from zero. We can try to estimate the shift $\Delta E$ by which the vacuum energy has been changed by using perturbation theory up to the first non zero term:

\begin{equation} \Delta E = \langle 0| \hat{H}_{coupling} | 0 \rangle + \sum_{n \neq 0} \frac{|\langle n|\hat{H}_{coupling}| 0 \rangle |^2 }{- E_n} + .... \end{equation}

as it is quite obvious, the first term is zero by definition of the vacuum of the free field. The second term is only non zero for Fock states where there is a scalar photon produced. We thus get:

\begin{equation} \Delta E \approx \int d^3k \frac{|\langle \mathbf{k};s|\hat{H}_{coupling} | 0 \rangle |^2}{-\hbar \omega} \end{equation}

We now have that:

\begin{equation} \langle \mathbf{k};s|\hat{H}_{coupling} | 0 \rangle = c\sqrt{\frac{\hbar}{2 \varepsilon_0 \omega(\mathbf{k})}}(q_1 e^{i \mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{i \mathbf{k}\cdot \mathbf{r}_2}) \end{equation}

This is now where the weirdness of the scalar photons shows up (in this very sketchy calculations). When we compute the square norm of the matrix element, it turns out that it is negative for scalar photons (this trick enables to make more troublesome thing disappear and long story short that's how it works). We thus have that:

\begin{equation} |\langle \mathbf{k};s|\hat{H}_{coupling} | 0 \rangle |^2 = - \frac{c^2 \hbar}{2 \varepsilon_0 \omega(\mathbf{k})} |(q_1 e^{i \mathbf{k}\cdot \mathbf{r}_1} + q_2 e^{i \mathbf{k}\cdot \mathbf{r}_2})|^2 \end{equation}

Upon replacing in the expression for $\Delta E$, one finds that:

\begin{equation} \Delta E \approx \epsilon_1 + \epsilon_2 + V_{coul} \end{equation}


\begin{equation} \epsilon = \int d^3k \frac{q^2}{2 \varepsilon_0 k^2} \end{equation}is the self interaction energy of the charges with themselves (can be interpreted as the emission and absorption of a scalar photon by the same charge)


\begin{equation} V_{coul} = \int d^3k \frac{q_1 q_2 e^{i\mathbf{k}\cdot(\mathbf{r}_1-\mathbf{r}_2)}}{2\varepsilon_0 k^2} = \frac{q_1q_2}{4\pi \varepsilon_0 |\mathbf{r}_1-\mathbf{r}_2|} \end{equation}is the Coulomb interaction potential.

Note that although for simplicity I have only used the perturbation expansion up to second order, it turns out that in fact the exact EM ground state with pinned charges at fixed locations is exactly equal to the $\Delta E$ we have calculated. To make the point clear, there is no such thing as quantum correction to the Coulomb interaction in this setup as it is an exact result from QED. In this view, quantum corrections arise when one introduces a dynamic field theory for the charged particles as well; hence the differences with the scattering picture.

For more on this picture, I recommend greatly this book from which the above summary is heavily inspired (although written in a simpler fashion).


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