Electron's magnetic moment

[Gaussian97]

TL;DR Summary

Show that $\bar{u}(p')\left(\gamma^i F_1(q^2)+\frac{i \sigma^{i\nu}q_\nu}{2m}F_2(q^2)\right)u(p)=0$ for $\vec{q}=0$

Is Peskin and Schroeder book, page 187 when they try to connect the electron form factors to its magnetic moment they get the expression
$$\bar{u}(p')\left(\gamma^i F_1(q^2)+\frac{i \sigma^{i\nu}q_\nu}{2m}F_2(q^2)\right)u(p)$$
Where $p$, $p'$ are the momenta on on-shell electrons and $q=p'-p$.
And they say that this vanishes at $\vec{q}=0$.

I see that, because $p,p'$ are on-shell $\vec{q}=0 \Longrightarrow q^0=0$ so in this limit the expression is
$$\bar{u}(p')\gamma^i u(p)$$
Because $F_1(0)=1$. But I can't see why this must be zero.

 

[AndyW]

The reason why this expression must be zero at $\vec{q}=0$ is because of the Dirac equation, which is given by:
$$\left(\gamma^\mu p_\mu - m\right)u(p)=0,$$
where $\gamma^\mu$ are the Dirac matrices, $p_\mu$ is the four-momentum, and $m$ is the mass of the electron. When we take the limit of $\vec{q}=0$, we are essentially taking the limit where the momentum transfer $q$ goes to zero, which means that the final state electron and the initial state electron have the same four-momentum. In this case, the Dirac equation becomes:
$$\left(\gamma^\mu p_\mu - m\right)u(p)=\left(\gamma^\mu p'_\mu - m\right)u(p'),$$
which simplifies to:
$$\left(\gamma^\mu p_\mu - m\right)u(p)=\left(\gamma^\mu p_\mu - m\right)u(p'),$$
since $p'=p$ when $\vec{q}=0$. Therefore, we can see that in this limit, the expression becomes:
$$\bar{u}(p')\gamma^i u(p)=\bar{u}(p)\gamma^i u(p)=0,$$
since the Dirac equation vanishes for both the initial and final state electron. This is why the expression vanishes at $\vec{q}=0$.