# Spin non-conservation in Coulomb interaction

In non-relativistic limit, in the lowest order of perturbation, QFT reproduces the classical Coulomb potential. A nice result is that in coulomb interaction the spin of the particles remain conserved separately. However, if we consider next order relativistic correction, I find, this condition is not recoverable.

The Feynman diagram for the process,

The amplitude for the process,

\begin{align} i\mathcal{M}&=\bar{u}^{s^\prime}(\mathbf{p^\prime})\left(-ie\gamma^\mu\right)u^s(\mathbf{p})\frac{-i\eta_{\mu\nu}}{\left(p-p^\prime\right)^2}\bar{u}^{r^\prime}(\mathbf{k^\prime})\left(-ie\gamma^\nu\right)u^r(\mathbf{k})\\ &\approx\bar{u}^{s^\prime}(\mathbf{p^\prime})\left(-ie\gamma^0\right)u^s(\mathbf{p})\frac{-i}{\left(p-p^\prime\right)^2}\bar{u}^{r^\prime}(\mathbf{k^\prime})\left(-ie\gamma^0\right)u^r(\mathbf{k}) \end{align}

In the last step we have taken non-relativistic limit. Now we shall focus on the term,

\begin{align} &\bar{u}^{s^\prime}(\mathbf{p^\prime})\gamma^0 u^{s}(\mathbf{p})\nonumber\\ &=u^{s^\prime\dagger}(\mathbf{p^\prime}) u^{s}(\mathbf{p})\nonumber\\ &=\begin{pmatrix} \xi^{s^{\prime}\dagger}\sqrt{p^\prime\cdot\sigma}&& \xi^{s^{\prime}\dagger}\sqrt{p^\prime\cdot\bar{\sigma}} \end{pmatrix} \begin{pmatrix} \sqrt{p\cdot\sigma}\xi^s\\ \sqrt{p\cdot\bar{\sigma}}\xi^s \end{pmatrix}\nonumber\\ &=\xi^{s^{\prime}\dagger}\left(\sqrt{\left(p^\prime\cdot\sigma\right)\left(p\cdot{\sigma}\right)}+\sqrt{\left(p^\prime\cdot\bar{\sigma}\right)\left(p\cdot\bar{\sigma}\right)}\right)\xi^s \end{align} Now, \begin{aligned} &\left(p^{\prime} \cdot \sigma\right)(p \cdot \sigma) \\ =& p_{\mu}^{\prime} \sigma^{\mu} p_{\nu} \sigma^{\nu} \\ =& p_{0}^{\prime} p_{0}+\left(p_{0}^{\prime} p_{i}+p_{i}^{\prime} p_{0}\right) \sigma^{i}+p_{i}^{\prime} p_{j} \sigma^{i} \sigma^{j} \\ =& p_{0}^{\prime} p_{0}+\left(p_{0}^{\prime} p_{i}+p_{i}^{\prime} p_{0}\right) \sigma^{i}+p_{i}^{\prime} p_{j}\left(\delta^{ij}+i \epsilon^{i j k} \sigma^{k}\right) \\ =& p_{0}^{\prime} p_{0}+\left(p_{0}^{\prime} p_{i}+p_{i}^{\prime} p_{0}\right) \sigma^{i}+p_{i}^{\prime} p_{j} {\delta}^{i j}+i \vec{\sigma} \cdot\left(\vec{p}^{\prime} \times \vec{p}\right) \\ =& p_{0}^{\prime} p_{0}-\left(p_0^{\prime} \vec{p}+p_{0} \vec{p}^{\prime}\right) \cdot \vec{\sigma}+\vec{p}^{\prime} \cdot \vec{p}+i \vec{\sigma} \cdot\left(\vec{p}^{\prime} \times \vec{p}\right) \end{aligned} Also, $$\left(p^{\prime} \cdot \bar{\sigma}\right)(p\cdot \bar{\sigma})=p_{0}^{\prime} p_{0}+\left(p_0^{\prime} \vec{p}+p_{0} \vec{p}^{\prime}\right) \cdot \vec{\sigma}+\vec{p}^{\prime} \cdot \vec{p}+i \vec{\sigma} \cdot\left(\vec{p}^{\prime} \times \vec{p}\right)$$ In the non-relativistic limit, considering the next order relativistic correction, which is not generally done in textbooks (or rather I have not seen it being done), $$\sqrt{\left(p^\prime\cdot\sigma\right)\left(p\cdot{\sigma}\right)}+\sqrt{\left(p^\prime\cdot\bar{\sigma}\right)\left(p\cdot\bar{\sigma}\right)}\approx2\sqrt{p^\prime_0p_0}+\frac{1}{\sqrt{p^\prime_0p_0}}\left(\vec{p}^{\prime} \cdot \vec{p}+i \vec{\sigma} \cdot\left(\vec{p}^{\prime} \times \vec{p}\right)\right)$$ With this, $$\xi^{s^{\prime}\dagger}\left(\sqrt{\left(p^\prime\cdot\sigma\right)\left(p\cdot{\sigma}\right)}+\sqrt{\left(p^\prime\cdot\bar{\sigma}\right)\left(p\cdot\bar{\sigma}\right)}\right)\xi^s=2\sqrt{p^\prime_0p_0}\delta^{s^{\prime}s}+\frac{1}{\sqrt{p^\prime_0p_0}}\left(\vec{p}^{\prime} \cdot \vec{p}\delta^{s^{\prime}s}+i \xi^{s^{\prime}\dagger}\vec{\sigma}\xi^s \cdot\left(\vec{p}^{\prime} \times \vec{p}\right)\right)$$ The only interesting term to us is $$\xi^{s^{\prime}\dagger}\vec{\sigma}\xi^s$$, which is not proportional to $$\delta^{s^{\prime}s}$$. In fact in general I find it to be,

$$\xi^{s^{\prime}\dagger}\vec{\sigma}\xi^s\cdot\left(\vec{p}^{\prime} \times \vec{p}\right)=\left(1-\delta^{s^{\prime}s}\right)\left(p^\prime_2p_3-p^\prime_3p_2\right)+(-1)^{s^{\prime}}i\left(1-\delta^{s^{\prime}s}\right)\left(p^\prime_3p_1-p^\prime_1p_3\right)+(-1)^{s^{\prime}+1}\delta^{s^{\prime}s}\left(p^\prime_1p_2-p^\prime_2p_1\right)$$

Also this term is not canceled by other term in the amplitude when taken the next order relativistic correction. Therefore, the conclusion is, the spin of a particle engaging in Coulomb interaction is not conserved! Is this surprising? Does it make any sense? Is my calculations somehow wrong? Or is it perfectly normal and expected to happen? Please shed some light on this.

Also as this extra piece is not proportional to $$\delta^{s^{\prime}s}$$, how does one applies Born approximation to extract out the corrected potential?

Few mathematical conventions and results:

$$\gamma^\mu=\begin{pmatrix} 0 & \sigma^\mu\\ \bar{\sigma}^\mu & 0 \end{pmatrix}$$ where, $$\sigma^\mu=(1,\sigma^i)$$, and $$\bar{\sigma}^\mu=(1,-\sigma^i)$$.

$$\sigma^1= \left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right),\quad \sigma^2= \left(\begin{array}{ll} 0 & -i \\ i & 0 \end{array}\right),\quad \sigma^3= \left(\begin{array}{ll} 1 & 0 \\ 0 & -1 \end{array}\right)$$ $$\xi^1=\begin{pmatrix} 1\\ 0 \end{pmatrix},\quad \xi^2=\begin{pmatrix} 0\\ 1 \end{pmatrix}$$ $$\xi^{r \dagger} \sigma^{1} \xi^{s}=\left(1-\delta^{r s}\right)$$ $$\xi^{r \dagger} \sigma^{2} \xi^{s}=(-1)^ri\left(1-\delta^{r s}\right)$$ $$\xi^{r \dagger} \sigma^{3} \xi^{s}=(-1)^{r+1}\delta^{r s}$$

• An answer was added and later deleted by user @VacuuM . In which the user suggested the extra piece may not be a part of Coulomb interaction but of a magnetic-like interaction. If anyone answers the question, please take this into consideration as well. If this is correct mention why and if this is wrong also mention why may that be. Apr 5, 2020 at 10:31
• Particles that carry spin often carry a magnetic moment along the spin axis. For elementary particles the moment is predicted by the possible relevant interactions (ie QED) and is non-zero for charged particles, while for composite particles you should allow higher-dimensional operators (think EFT with cutoff the order of the size of the particle) which can modify the magnetic moment (if I recall correctly). E.g., neutron has a non-zero magnetic moment. These moments interact in a fashion that does not preserve individual spins (decaying as $r^{-3}$) in classical ED, so there is no surprise. Apr 7, 2020 at 6:21
• Moreover, it is useful to think about the classical situation in Gaussian units, which make factors of $1/c$ more explicit in classical ED formulas than SI units. I bet that if you assume gyromagnetic ratio of order 1 and compute the torque between two particles in classical ED, it will have the same number of $1/c$ factors as your next-order-correction. Apr 7, 2020 at 6:27
• Here is a dimensional analysis approach. We know that the interaction should be $1/r^3$ and be a torque. Torque is evolution of angular moments, which are order $\hbar$ in question and also involve particle mass. This means that if we want to write a classical formula for torque and compare it with your equations, we should allow ourselves to use $\hbar$ and $m$. Then we want to write torque = $e^2 r^{-3} c^a \hbar^b m^d$ for some $a,b,d$ (Gaussian units). A simple calculation shows that $c^a\hbar^b m^d$ should be length${}^2$, which gives $a=-b=d=-2$. So the classical effect is $1/c^2$. Apr 7, 2020 at 6:49
• @PeterKravchuk Thank you for your comments. There are difficulties I see at the moment. First, if we go to the rest frame of one charge then the torque on the other is $\vec{r}\times\vec{F}$ with the Coulomb force. But the problem is the expression I have derived is in Fourier space. So it directly does not convert to something like $\vec{r}\times\vec{F}$. Moreover, from amplitude via Born's approximation we can get the potential and then I have to do an integration to get the force. So I do not see any direct connection. Perhaps, you can write an answer with all the calculations. Apr 7, 2020 at 7:04

The spin of the particles involved in Coulomb interaction is always conserved. Though your calculation is right, it does not contradict this fact about Coulomb interaction. The Feynman diagram you have drawn corresponds to $$A_\mu\bar{\psi}\gamma^\mu\psi$$. As you can see the involvement of $$A_\mu$$ makes the interaction an 'electromagnetic' interaction and thus is not limited only to 'electric' interaction (read Coulomb interaction). In higher order relativistic correction you should see not only electric effects but magnetic effects as well and the later effects exists independent of the Coulomb interaction. And let's ask why there should be any magnetic effects at all. These exist because of the fact that unlike in the case of Coulomb potential where the charges are at rest, we have moving charges in this particular scattering. Also recall that electric and magnetic fields are frame dependent concepts and thus if in one frame you see magnetic field it does not guarantee the existence of a magnetic field in all inertial frames. With these facts in hand let's examine the extra term that is bothering you. One of the extra terms is the following,
$$e^2\xi^{s^\prime\dagger}\frac{1}{\sqrt{p_0^\prime p_0}}i\boldsymbol{\sigma}\cdot(\mathbf{p^\prime}\times\mathbf{p})\xi^s\frac{i}{|\mathbf{p^\prime}-\mathbf{p}|^2}2\sqrt{p_0^\prime p_0}(\xi^{r^\prime\dagger}\xi^{r})_{\mathbf{k}}$$ The physical meaning of this term is not quite visible in the momentum space, so let's transit back to the position space by Fourier transforming, $$-e^2\int\frac{d^3\mathbf{q}}{(2\pi)^3}\xi^{s^\prime\dagger}\frac{1}{4m^3}i\boldsymbol{\sigma}\cdot(\mathbf{p^\prime}\times\mathbf{p})\xi^s\frac{1}{|\mathbf{q}|^2}2m(\xi^{r^\prime\dagger}\xi^{r})_{\mathbf{k}}e^{i\mathbf{q}\cdot(\mathbf{x_1}-\mathbf{x_2})}$$ where, I have set $$p_0=p^\prime_0\approx m$$, and $$\mathbf{q}$$ is the difference between the momenta of the particles, $$\mathbf{q}=\mathbf{p^\prime}-\mathbf{p}=\mathbf{k}-\mathbf{k^\prime}$$ and $$\mathbf{x_1}$$ and $$\mathbf{x_2}$$ are the position vectors of the two particles. Moreover, I have removed one $$-i$$ and divided by $$4m^2$$ in order to acquire potential from the amplitude due to the Born approximation. To perform the integration assume $$\mathbf{p}$$ to be a fixed vector and vary the vector $$\mathbf{p^\prime}$$ to generate all the possible $$\mathbf{q}$$ vectors. Now, write $$\mathbf{p^\prime}=\mathbf{q}+\mathbf{p}$$. Thus, the cross-product simplifies to $$\mathbf{p^\prime}\times\mathbf{p}=\mathbf{q}\times\mathbf{p}=-\mathbf{p}\times\mathbf{q}$$. Having done all these, let's do the integration, \begin{align} &-e^2\frac{1}{4m^2}\int\frac{d^3\mathbf{q}}{(2\pi)^3}\xi^{s^\prime\dagger}i\boldsymbol{\sigma}\cdot(\mathbf{p^\prime}\times\mathbf{p})\xi^s\frac{1}{|\mathbf{q}|^2}2(\xi^{r^\prime\dagger}\xi^{r})_{\mathbf{k}}e^{i\mathbf{q}\cdot(\mathbf{x_1}-\mathbf{x_2})}\\ &=e^2\frac{1}{4m^2}\int\frac{d^3\mathbf{q}}{(2\pi)^3}\xi^{s^\prime\dagger}\boldsymbol{\sigma}\xi^s\cdot i(\mathbf{p}\times\mathbf{q})\frac{1}{|\mathbf{q}|^2}2\delta^{r^\prime r}e^{i\mathbf{q}\cdot(\mathbf{x_1}-\mathbf{x_2})}\\ &=2e^2\frac{1}{4m^2}\delta^{r^\prime r}\xi^{s^\prime\dagger}\boldsymbol{\sigma}\xi^s\cdot(\mathbf{p}\times\boldsymbol{\nabla}_{\mathbf{x}})\int\frac{d^3\mathbf{q}}{(2\pi)^3}\frac{1}{|\mathbf{q}|^2}e^{i\mathbf{q}\cdot\mathbf{x}} \end{align} where we have called $$\mathbf{x_1}-\mathbf{x_2}=\mathbf{x}$$. Continuing, \begin{align} &-e^2\frac{1}{4m^2}\int\frac{d^3\mathbf{q}}{(2\pi)^3}\xi^{s^\prime\dagger}i\boldsymbol{\sigma}\cdot(\mathbf{p^\prime}\times\mathbf{p})\xi^s\frac{1}{|\mathbf{q}|^2}2(\xi^{r^\prime\dagger}\xi^{r})_{\mathbf{k}}e^{i\mathbf{q}\cdot(\mathbf{x_1}-\mathbf{x_2})}\\ &=2e^2\frac{1}{4m^2}\delta^{r^\prime r}\xi^{s^\prime\dagger}\boldsymbol{\sigma}\xi^s\cdot(\mathbf{p}\times\boldsymbol{\nabla}_{\mathbf{x}})\frac{1}{4\pi|\mathbf{x}|}\\ &=-2e^2\frac{1}{4m^2}\delta^{r^\prime r}\xi^{s^\prime\dagger}\boldsymbol{\sigma}\xi^s\cdot\frac{\mathbf{p}\times \mathbf{x}}{4\pi|\mathbf{x}|^3}\\ &=-2\frac{e}{2m}\delta^{r^\prime r}\xi^{s^\prime\dagger}\frac{1}{2}\boldsymbol{\sigma}\xi^s\cdot\frac{e\frac{\mathbf{p}}{m}\times \mathbf{x}}{4\pi|\mathbf{x}|^3} \end{align} Now, recall the definition of magnetic moment, $$\boldsymbol{\mu}=2\left(\frac{e}{2m}\right)\mathbf{S}$$ with setting the Lande $$g$$-factor to the value $$2$$ at the lowest order of QED perturbation. The spin operator is defined as, $$\mathbf{S}=\frac{1}{2}\boldsymbol{\sigma}$$ and $$\xi^{s^\prime\dagger}\frac{1}{2}\boldsymbol{\sigma}\xi^s$$ can be written as $$\langle\mathbf{S}\rangle$$. Now, the magnetic moment is in dot product with, $$\frac{e\frac{\mathbf{p}}{m}\times \mathbf{x}}{4\pi|\mathbf{x}|^3}$$ which is the magnetic field created due to the motion of one of the particles. This is the standard Biot-Savart law for a magnetic field due to a moving charge. Thus, the extra piece corresponds to a potential $$-\langle\boldsymbol{\mu}\rangle\cdot\mathbf{B}$$ due to one particle, via its magnetic moment, coupling to the magnetic field generated by the other particle's motion.