# What is the speed of light doing in the spin half hamiltonian?

## Main Question or Discussion Point

I'm currently reading Sakurai's 'Modern Quantum Mechanics' (Revised Edition) and at page 76 he introduces a spin half hamiltonian

$$H = - (\frac{e}{mc}) \vec S \cdot \vec B.$$

But what is c doing in this hamiltonian? Clasically the energy of a magnetic moment in a magnetic field is

$$E = - \vec \mu \cdot \vec B = - \frac{e}{2m} \vec S \cdot \vec B$$

and as far as i know what one does to go to QM is to introduce the g-factor. What am I missing here?

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dextercioby
Homework Helper
Standard question, standard answer: the spin-orbit coupling is a specially relativistic effect, even though the spin notion itself can be derived without special relativity. The <c> comes from the Dirac equation.

Bill_K
Say what? If this term is spin-orbit coupling, where is L??

The c comes about because according to Dirac the magnetic moment μ of the electron is -eħ/2mc, or -e/mc times the spin ħ/2.

ah, thanks. I guess I have to be reading up on that then.

dextercioby
Homework Helper
Sorry, not paying attention. Disregard my post above.

If the <c> is there or not must be in the end a matter of convention, just like the as the case of CGS vs MKS in electromagnetism.

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