Re: Relation of g^uv = (1/2) {gamma^u,gamma^v} to gravitational fields

Jay R. Yablon wrote:

I have focused on the latter part of this discussion in some replies,
but I want to make sure I also understand exactly what you are saying
when you say "why do you think that the anticommutators of the corrected
vertex factors Gamma^u will have anticommutators proportional to the
Clifford identity?"

I have been studying the Schwinger calculation carefully and closely.
If sigma_uv denotes the object constructed from antisymmetric [gamma,
gamma] combinations, and p, p' designate the ingoing and outgoing
electron four-momenta, and alpha is the EM coupling, then it looks to me
like the form of the Schwinger "correction" Lambda is:

Lambda = (alpha/2pi)(i sigma (p'-p)/2m) (1)

or, alternatively, decomposed:

Lambda = (alpha/2pi)gamma - (alpha/2pi)((p'-p)/m) (2)

In both, the full vertex factor is:

Gamma ^u = gamma^u + Lambda^u (3)

Looking specifically at (2), when you say "anticommutators proportional
to the Clifford identity?" do you mean that although the term
(alpha/2pi)gamma in (2) anticommutes proportionally to the Dirac
Clifford algebra generators gamma, the second term (alpha/2pi)((p'-p)/m)
does NOT? And, is this the source of the issue you have identified
here?

First, I don't understand how you got your equations (1) and (2). Using
Weinberg's QFT v.1 book as a reference, I'm looking at equation
(11.3.29):

Gamma^u = gamma^u F_1 + i/2 sigma^uv (p'-p)_v F_2,

where F_1 and F_2 are scalar structure factors that depend only on the
square of the momentum transfer. Also F_1 ~ 1 + O(alpha) and F_2 ~
O(alpha). Some combination of these O(alpha) terms gives the correction
to the magnetic moment. However, this equation looks neither like your
(1) nor (2).

And instead of wondering whether the anticommutators of Gamma^u are
proportional to the identity or not, why don't you just check. Have you
tried to compute the anticommutators? The Clifford algebra is
16-dimensional, with a basis given by the identity, gamma^u, products
of two, three, and four gamma matrices. So, every algebra element has a
canonical form when expressed in this basis. Calculate the
anticommutator and express it in this basis, then you'll know whether
it's proportional to the identity or not.

Igor

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