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It is known that the value of 2 of the electron g-factor arises from the Dirac equation. As far as I can see from the various sources, this value is obtained in non-relativistic limit, in particular by reducing Dirac equation to Pauli equation.

The anomalous magnetic moment of electron is further explained to arise from the electron's interaction with the surrounding electromagnetic field.

Is it possible that the anomalous magnetic moment is due to additional terms (such as an electric moment) omitted in the non-relativistic limit?

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up vote 2 down vote accepted

g=2 comes from the Dirac equation, and is a relativistic effect. In the Pauli equation, it would have to be added by hand.

Radiation corrections give additional terms to the Dirac equation and modify the value by a little bit.

The corrections are in fact corrections of the form factors. There are an electric and a magnetic form factor. The magnetic form factor (and only that) determines g, while the electric form factor determines the Lamb shift.

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Thanks @Arnold. Does it mean that Dirac theory does NOT give a complete picture? Aren't radiation corrections the consequences of Dirac theory? – Murod Abdukhakimov Mar 5 '12 at 13:51
The Dirac theory gives the bulk of g; but the complete picture is given by QED, which modifies the Dirac equation a little, to approx. 2.0023. See… - Radiative corrections come from interaction with the electromagnetic field, hence needs QED, which couples the Dirac equation and the Maxwell equations, and is a multiparticle theory, whereas the Dirac equation describes only a single particle. – Arnold Neumaier Mar 5 '12 at 16:06
"g=2 comes from the Dirac equation, and is a relativistic effect. In the Pauli equation, it would have to be added by hand." This is folklore. Feynman pointed out (as Sakurai notes in Advanced QM, p.79, footnote) g=2 follows from the non-relativistic Schrodinger equation by writing the $H_0 = \mathbf{p}\cdot\mathbf{p}/(2m) = (\boldsymbol{\sigma}\cdot\mathbf{p})(\boldsymbol{\sigma}\cdot\mathbf{p})/(2m)$ and then making the minimal substitution $\mathbf{p}\to\mathbf{p}-e\mathbf{A}$. – MarkWayne Jun 17 '15 at 19:56
@MarkWayne: interesting. By writing H_0 in other quadratic patterns before doing the minimal substitution one can probably get an arbitrary value of g. – Arnold Neumaier Jun 17 '15 at 21:07
@ArnoldNeumaier: example? – MarkWayne Jun 18 '15 at 4:45

No, the correction is a strictly quantum effect. If you were to keep track of $\hbar$ while doing this computation, you'd find the correction is $O(\hbar)$. Generically, this corresponds to one-loop corrections in quantum field theory.

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