D.82 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 $\vec\mu$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q_m\vec{d}$ 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 $xy$-​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 $\vec{d}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $d_x{\hat\imath}+d_z{\hat k}$.

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

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

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 $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $d_{\rm {c}}$. The elec­tric di­pole mo­ment per unit length of such a pair of line charges is by de­f­i­n­i­tion $\vec\wp{ '}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q'd_{\rm {c}} {\hat\jmath}$, 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 $y$ $\vphantom0\raisebox{1.5pt}{$=$}$ $d_{\rm {c}}$, 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 $\vec\mu{ '}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $q_m'(d_x{\hat\imath}+d_z{\hat k})$ 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}$$\raisebox{.5pt}{$/$}$​$4\pi\epsilon_0r^3$ 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:

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

To get the cor­rect spin-or­bit in­ter­ac­tion, the mag­netic di­pole mo­ment $\vec\mu$ used in this ex­pres­sion must be the clas­si­cal one, $\vphantom{0}\raisebox{1.5pt}{$-$}$$e\vec{S}$$\raisebox{.5pt}{$/$}$​$2m_{\rm e}$. The ad­di­tional fac­tor $g_e$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\frac12\alpha^2{m_{\rm e}}c^2$, of $e^2$$\raisebox{.5pt}{$/$}$​$\epsilon_0$ us­ing $e^2$$\raisebox{.5pt}{$/$}$​$4\pi\epsilon_0$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\vert E_1\vert a_0$, and of $m_{\rm e}$ us­ing $\vert E_1\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar^2$$\raisebox{.5pt}{$/$}$​$2{m_{\rm e}}a_0^2$, the claimed ex­pres­sion for the spin-or­bit en­ergy is found.