0
$\begingroup$

enter image description here

In Zee's QFT book, Section II.6, he gives the amplitude for this pictured process of an electron scattering from a structureless nucleon as

$$ \mathcal{M}(P,P_N) = \cfrac{-ie^2}{k^2-m_\gamma^2} \,\bar u(P)\gamma^\mu u(p)\,\bar u(P_N)\gamma_\mu u(p_N)~~. \tag{II.6.3}$$

Here the subscript $N$ refers to the nucleon, $m_\gamma$ is a hypothetical photon mass that we will set to zero later and $k = P - p$. I use notation such that $\gamma_\mu=\eta_{\mu\nu}\gamma^\nu$. Zee says that since the mass of the nucleon is much greater than the electron mass, we can approximate the nucleon spinors are being in the rest frame. The Dirac equation in momentum space is

$$ \big(\gamma^\mu p_\mu -m\big)u=\big(\gamma^0 p_0+\gamma^i p_i -m\big)u=0 ~~, $$

so in the rest frame of $p_i=0$, we have

$$ \big(\gamma^0 p_0 -m\big)u=0 ~~. $$

In units of $c=1$, we have $p_0=m$ so we find $\gamma^0u=u$. Using the identity $\bar uu=1$ and the rest frame condition $P_N=p_N$, we have

$$ \bar u(P_N)\gamma^0 u(p_N)= \bar u(P_N) u(p_N)=1~~. $$

However, Zee says we also find that the rest frame condition gives

$$ \bar u(P_N)\gamma^i u(p_N)=0~~, $$

such that the stationary nucleon approximation of $\mathcal{M}$ is

$$ \mathcal{M}(P,P_N) \approx \cfrac{-ie^2}{k^2-m_\gamma^2} \,\bar u(P)\gamma^0 u(p) ~~.$$

I don't see where Zee gets $\bar u(P_N)\gamma^i u(p_N)=0$. It seems like the $\gamma^i u(p_N)$ terms disappear in the Dirac equation due to multiplication with $p_i=0$ and not because they have to vanish on their own. Where does it come from that

$$ \bar u(P_N)\gamma^i u(p_N)=0~~? $$

$\endgroup$
2
  • 1
    $\begingroup$ Is it possible to do this using that a spin basis for a rest-frame Dirac spinor is $u^{(1)} = [1 \ 0\ 0\ 0]$ and $u^{(2)} = [0 \ 1 \ 0 \ 0]$? $\endgroup$ Dec 2 '20 at 6:42
  • $\begingroup$ Yes, thank you. I cranked out the matrix algebra and I see that it is true. $\endgroup$ Dec 2 '20 at 13:27
1
$\begingroup$

Given that $u^{(1)}(0) = \begin{bmatrix} 1\\0\\0\\0 \end{bmatrix}$ and $u^{(2)}(0) =\begin{bmatrix} 0\\1\\0\\0 \end{bmatrix}$ form a (plane wave) basis for a rest-frame spinor, we can calculate $\bar{u}(0)\gamma^i u(0)$ by direct matrix multiplication. However, you can also note that the $\gamma^1$, $\gamma^2$ and $\gamma^3$ in the Dirac basis only have off-diagonal elements, so the inner product between $u^\dagger\gamma^0$ and $\gamma^i u$ will be zero since at least one of the two has its $j$-th component equal to zero.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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