- #1
- 683
- 412
- TL;DR
- 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.
$$\bar{u}(p')\left(\gamma^i F_1(q^2)+\frac{i \sigma^{i\nu}q_\nu}{2m}F_2(q^2)\right)u(p)$$
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.