Spin-Orbit Splitting in the Hydrogen Atom

Introduction

Here I will compute the spin orbit calculation the way Niels Bohr would have done starting with his very successful results for the simple energy levels of the hydrogen atom before he knew about the spin magnetic moment of the electron.  Once he realized that there was a spin magnetic moment as well as the obvious magnetic field due to the orbital motion of the electrons he would have been able to compute the Spin-Orbit splitting on completely classical grounds.  For accuracy I have also included the correct gyromagnetic factor, g=2, in this calculation.  If he had known that the electron’s spin angular momentum was h/(4p) Bohr would have guessed that g=1 on classical grounds.

Ratio of Magnetic Moment, m, to Angular Momentum L for a Classical Charged Rotating Body

To start just compute the ratio for very thin toroid of radius r, charge q, mass m and rotational frequency w:

                                                                                                                           (1)

For a similar charged toroid of radius r and total charge q:

                                                                                                      (2)

The ratio is then:

                                                                                                                             (3)

It turns out that the ratio in equation 3 is the same regardless of the radial distributions of mass and charge in the body that contains the charge or mass elements.  The following equations are the general expressions for the momentum and magnetic moment of bodies that have volumetric charge distributions q(r) and mass distributions r(r):

                   

 where h is the height of the mass and charge distributions and the sub-integrals:

                                                      

 

 

Electron Spin Magnetic Moment

            First we need the magnetic moment of the electron.  We will obtain this by analogy with the current loop result in the previous section.  For a quantum object, L in equations 1 and 3 is renamed to S so that:

                                                                                                             (4)

where g is the unitless g factor, 2.002, e is the electronic charge, s is the spin angular momentum quantum number (either ½ or -½ for an electron), and  m is the mass of the electron.  The factor g stems from quantum electrodynamics which is beyond the scope of this document.

Magnetic Field in which the m Finds Itself

            Since there is no relative movement of the orbiting electron with respect to its spin, there is no magnetic field due to that motion.  However, as seen by the orbiting magnetic moment, the nucleus is rotating at the same radius and the same angular speed but in the opposite sense.  This provides a magnetic field that can interact with the electron’s magnetic moment.  Effectively the apparent orbiting nucleus is a current loop.  The magnetic field at the center of a current loop is computed by the Biot-Savart Law as:

                                                                                                                             (5)

where m0 is the magnetic permeability of vacuum, I is the effective current, and r is the radius of the loop.  In the case of the hydrogen atom in the n=1 state, the expression for r is

                                                              

and for the n=2 state the radius is

                                                                 

 

Since we know the charge of the nucleus, Ze, the current, I, can be computed if we know the rate of rotation, w, of the electron then

                                                                

However that rate can be computed from the electron’s angular momentum:

                                                                       

                                                                                                                   (6)

where n is the radial state quantum number and can be any integer greater than of equal to 1.

Solving equation 6 for w:

                                                                                                                             (7)

and then using the result in equation 5 we have:

                                                                                             (8)

For the n=2 state of the hydrogen atom (where Z=1) it turns out that B is about 0.4 Tesla.

 

Spin-Orbit Energy Splitting

            Since the energy splitting is the difference between m.B and –m.B, our result is obtained by doubling the product of equations 4 and 8:

                            (9)

For the n=2 state, r=2a0 and with s=1/2 we have the numerical result

                                                

where I have divided the result in equation 9 by e in order to obtain electron Volts. This agrees with the full quantum mechanical result which is much more complicated to calculate.

 

Summary of Part I

            A couple of comments are in order here.  First, Bohr would have believed that the n=1 level should be split because he didn’t know that the n=1 level has no average angular momentum.  Second, he would have believed that the splittings of the levels with n>2 would be simple two-fold splittings like the n=2 levels, but of much lower energy difference.  These beliefs all changed when a full quantum mechanical model of hydrogen, where j=l+s was worked out.