D.84 Classical spin-orbit derivation

This note derives the spin-orbit Hamiltonian from a more intuitive, classical point of view than the Dirac equation mathematics.

Picture the magnetic electron as containing a pair of positive and negative magnetic monopoles of a large strength $q_m$. The very small distance from negative to positive pole is denoted by $\vec{d}$ and the product $\vec\mu=q_m\vec{d}$ is the magnetic dipole strength, which is finite.

Next imagine this electron smeared out in some orbit encircling the nucleus with a speed $\vec{v}$. The two poles will then be smeared out into two parallel “magnetic currents” that are very close together. The two currents have opposite directions because the velocity $\vec{v}$ of the poles is the same while their charges are opposite. These magnetic currents will be encircled by electric field lines just like the electric currents in figure 13.15 were encircled by magnetic field lines.

Now assume that seen from up very close, a segment of these currents will seem almost straight and two-dimensional, so that two-dimensional analysis can be used. Take a local coordinate system such that the $z$-axis is aligned with the negative magnetic current and in the direction of positive velocity. Rotate the $x,y$-plane around the $z$-axis so that the positive current is to the right of the negative one. The picture is then just like figure 13.15, except that the currents are magnetic and the field lines electric. In this coordinate system, the vector from negative to positive pole takes the form $\vec{d}=d_x{\hat\imath}+d_z{\hat k}$.

The magnetic current strength is defined as $q_m'v$, where $q'_m$ is the moving magnetic charge per unit length of the current. So, according to table 13.2 the negative current along the $z$-axis generates a two-dimensional electric field whose potential is

$$\varphi_\ominus = -\frac{q_m'v}{2\pi\epsilon_0c^2} \thet... ...rac{q_m'v}{2\pi\epsilon_0c^2} \arctan\left(\frac{y}{x}\right)$$

To get the field of the positive current a distance $d_x$ to the right of it, shift $x$ and change sign:

$$\varphi_\oplus = \frac{q_m'v}{2\pi\epsilon_0c^2} \arctan\left(\frac{y}{x-d_x}\right)$$

If these two potentials are added, the difference between the two arctan functions can be approximated as $-d_x$ times the $x$ derivative of the unshifted arctan. That can be seen from either recalling the very definition of the partial derivative, or from expanding the second arctan in a Taylor series in $x$. The bottom line is that the monopoles of the moving electron generate a net electric field with a potential

$$\varphi = \frac{q_m' d_x v}{2\pi\epsilon_0c^2} \frac{y}{x^2+y^2}$$

Now compare that with the electric field generated by a couple of opposite electric line charges like in figure 13.12, a negative one along the $z$-axis and a positive one above it at a position $y=d_{\rm {c}}$. The electric dipole moment per unit length of such a pair of line charges is by definition $\vec\wp{\,'}=q'd_{\rm {c}}\,{\hat\jmath}$, where $q'$ is the electric charge per unit length. According to table 13.1, a single electric charge along the $z$-axis creates an electric field whose potential is

$$\varphi = \frac{q'}{2\pi\epsilon_0}\ln \frac{1}{r} = - \frac{q'}{4\pi\epsilon_0}\ln\left(x^2+y^2\right)$$

For an electric dipole consisting of a negative line charge along the $z$ axis and a positive one above it at $y=d_{\rm {c}}$, the field is then

$$\varphi = - \frac{q'}{4\pi\epsilon_0}\ln\left(x^2+(y-d)^2\right) + \frac{q'}{4\pi\epsilon_0}\ln\left(x^2+y^2\right)$$

and the difference between the two logarithms can be approximated as $-d_{\rm {c}}$ times the $y$-derivative of the unshifted one. That gives

$$\varphi = \frac{q'd_{\rm {c}}}{2\pi\epsilon_0} \frac{y}{x^2+y^2}$$

Comparing this with the potential of the monopoles, it is seen that the magnetic currents create an electric dipole in the $y$-direction whose strength $\vec\wp{\,'}$ is $q_m'd_xv/c^2\,{\hat\jmath}$. And since in this coordinate system the magnetic dipole moment is $\vec\mu{\,'}=q_m'(d_x{\hat\imath}+d_z{\hat k})$ and the velocity $v{\hat k}$, it follows that the generated electric dipole strength is

$$\vec\wp{\,'} = - \vec\mu{\,'} \times \vec v/c^2$$

Since both dipole moments are per unit length, the same relation applies between the actual magnetic dipole strength of the electron and the electric dipole strength generated by its motion. The primes can be omitted.

Now the energy of the electric dipole is $-\vec\wp\cdot\skew3\vec{\cal E}$ where $\skew3\vec{\cal E}$ is the electric field of the nucleus, $e{\skew0\vec r}/4\pi\epsilon_0r^3$ according to table 13.1. So the energy is:

$$\frac{e}{4\pi\epsilon_0c^2}\frac{1}{r^3}{\skew0\vec r}\cdot(\vec\mu\times\vec v)$$

and the order of the triple product of vectors can be changed and then the angular momentum can be substituted:

$$-\frac{e}{4\pi\epsilon_0c^2}\frac{1}{r^3}\vec\mu\cdot({\sk... ...e}{4\pi\epsilon_0c^2m_{\rm e}}\frac{1}{r^3}\vec\mu\cdot\vec L$$

To get the correct spin-orbit interaction, the magnetic dipole moment $\vec\mu$ used in this expression must be the classical one, $-e\vec{S}/2m_{\rm e}$. The additional factor $g_e=2$ for the energy of the electron in a magnetic field does not apply here. There does not seem to be a really good reason to give for that, except for saying that the same Dirac equation that says that the additional $g$-factor is there in the magnetic interaction also says it is not in the spin-orbit interaction. The expression for the energy becomes

$$\frac{e^2}{8\pi\epsilon_0m_{\rm e}^2c^2}\frac{1}{r^3}\vec S\cdot\vec L$$

Getting rid of $c^2$ using $\vert E_1\vert=\frac12\alpha^2{m_{\rm e}}c^2$, of $e^2/\epsilon_0$ using $e^2/4\pi\epsilon_0=2\vert E_1\vert a_0$, and of $m_{\rm e}$ using $\vert E_1\vert=\hbar^2/2{m_{\rm e}}a_0^2$, the claimed expression for the spin-orbit energy is found.