Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Schwinger has on his grave (it seems) the relation between the g-factor of the electron and the fine structure constant:

$$g~=~2+\frac{\alpha}{\pi}+{\cal O}(\alpha^2)$$

Did Schwinger or somebody else ever give a simple explanation for the second term of the right hand side? The 2 appears from Dirac's equation. The second term is due to the emission and absorption of a photon. Is there a simple way to see that this process leads to the expression $\frac{\alpha}{\pi}$?

share|improve this question
    
This is a loop integral which is done elegantly in Weinberg, Schwinger's way. This is as simple as it gets. –  Ron Maimon Jul 18 '12 at 18:08
    
Also see chapter 6 of Peskin. –  DJBunk Jul 18 '12 at 18:34
add comment

1 Answer 1

The additional correction to the magnetic moment of the electron, aptly called the 'anomalous magnetic moment,' arises from a one loop Feynman diagram calculation in quantum electrodynamics. To be specific, the Landé $g$ factor is given by,

$$g=2[1+F_2(0)]$$

where $F_2$ is a 'form factor.' The electron vertex scattering amplitude is given by,

$$\require{cancel} \int \frac{\mathrm{d}^4k}{(2\pi)^4} \, \frac{-i\eta_{\nu \rho}}{(k-p)^2}\bar{u}(p')(-ie\gamma^\nu)\frac{i(\cancel{k'}+m)}{k'^2-m^2}\gamma^\mu \frac{i(\cancel{k}+m)}{k^2-m^2}(-ie\gamma^\rho) u(p)$$

$$=2ie^2 \int \frac{\mathrm{d}^4 k}{(2\pi)^4} \, \frac{\bar{u}(p')[\cancel{k}\gamma^\mu \cancel{k}' +m^2 \gamma^\mu -2m(k+k')^\mu]}{(k-p)^2(k'^2 -m^2)(k^2-m^2)}$$

Several methods are required to compute the amplitude, including the introduction of Feynman parameters to combine the propagators, Wick rotation and Pauli-Villars regularization, which roughly corresponds to the change,

$$\frac{1}{(k-p)^2 +i\epsilon} \to \frac{1}{(k-p)^2 +i\epsilon} - \frac{1}{(k-p)^2 -\Lambda^2 +i\epsilon}$$

introducing a fictitious 'photon mass.' The final result is given by,

$$\frac{\alpha}{2\pi} \int_0^1 \mathrm{d}x \, \mathrm{d}y \, \mathrm{d}z \, \delta(x+y+z-1) \times \bar{u}(p')\left(\gamma^\mu \left[ \log \frac{z\Lambda^2}{\triangle} +\frac{1}{\triangle} ((1-x)(1-y)q^2 +(1-4z+z^2)m^2)\right]+\frac{i\sigma^{\mu \nu}q_\nu}{2m}\left[ \frac{1}{\triangle} 2m^2 z(1-z)\right] \right)$$

where new variables have been introduced for convenience. Precisely how the result is derived is available in Chapter 6 of Peskin and Schroeder. One can extract the form factor from the amplitude, and one obtains (after many manipulations),

$$F_2(q^2 =0)=\frac{\alpha}{\pi}\int_0^1 \mathrm{d}z \, \int^{1-z}_0 \mathrm{d}y \, \frac{z}{1-z} = \frac{\alpha}{2\pi}$$

The coupling constant of quantum electrodynamics is $e$, and appears in the Feynman rules due to an interaction term $e\bar{\psi}\cancel{A}\psi$, and hence its appearance in the original amplitude. The fine structure constant, $\alpha = e^2 /4\pi$ and the famous result by Schwinger is usually stated in terms of it. The correction to the $g$ factor, numerically, is

$$a_e = \frac{g-2}{2}=\frac{\alpha}{2\pi} \approx 0.0011614$$

Although a minute correction, the result was highly significant and monumental. Loop calculations are by no means trivial. For Schwinger's original paper, see Physical Review, 73, 416L (1948).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.