Why is G-2 of Lande Factor Positive?

[Ancient_Nomad]

Hello Everyone,

In a course on field theory I was asked to calculate the first order correction for Lande g factor.
I noticed that this comes out to be positive. ie g-2 > 0

I am wondering if there is a physical reason why this must be positive. Or is it just a matter of chance, that this comes out as it does.

Thanks in advance.

 

[maverick280857]

Hello Everyone,

In a course on field theory I was asked to calculate the first order correction for Lande g factor.
I noticed that this comes out to be positive. ie g-2 > 0

For a proton, it turns out that g-2 > 0, whereas for a neutron, instead of g-2 < 0 (c.f. http://arxiv.org/abs/physics/0405126 for exact values).

I am wondering if there is a physical reason why this must be positive. Or is it just a matter of chance, that this comes out as it does.

This is a good question..I am going to try and offer an explanation which may not be physically satisfying, but maybe you can think of it as a first order corrective explanation ;-)

The value '2' is due to the contribution of the charge form factor (recall that the vertex function for a general electromagnetic vertex is written in terms of the electric and magnetic form factors $F_1$ and $F_2$, among other things), whereas the 'correction' to 2 is due to the magnetic form factor.

For a particle with nonzero normalized charge Q (so the physical charge = eQ where Q = -1 in HEP units for an electron), one has

$$g = 2 - \frac{4m}{Q}F_{2}(0)$$

I'm of course skipping several steps here...assuming you have derived such an expression before. Now, after a tedious calculation involving expressing $F_{2}(0)$ in terms of Feynman parameters (let me look for a reference in a book, and I'll refer to it in a subsequent post), one gets

$$F_{2}(0) = \frac{\alpha}{4\pi m}$$

Substituting it back into the above equation you get

$$g = 2 + \frac{\alpha}{\pi}$$

So, your question about the underlying basis for the correction being positive definite can be equivalently framed in terms of the positive contribution of the magnetic form factor (since it is Q = -1 sticking outside which makes the overall contribution of the anomalous term positive definite). So, why is the magnetic form factor positive definite?

Rather, why should the magnetic form factor be positive definite? For that one needs to look at the lowest order correction to the electromagnetic vertex function. To be honest, offhand I cannot think of a reason why it is part of the design that the vertex function matrix element be positive definite...

 

[Ancient_Nomad]

For a proton, it turns out that g-2 > 0, whereas for a neutron, instead of g-2 < 0 (c.f. http://arxiv.org/abs/physics/0405126 for exact values).

I am sorry. I should have specified that I was talking about the electron not bound states like proton or neutron.

But yes, in principle we should try to extend any answer we get (if we do) to them.

Rather, why should the magnetic form factor be positive definite?

That is true. I guess my question can be rephrased as 'why should the magnetic form factor is positive definite' as you suggested.

Although, I would also like to know 'why should the electric form factor be exactly 2'.
I know that the corrections to $F_1(0)$ get canceled due to corrections to the residue of the electron propagator ($Z_2$) But is there a physical reason behind this exact cancellation?