# Dirac equation plane wave solutions for antiparticles (Peskin QFT book overlooked?)

I am puzzled by the derivation given here about the Dirac equation plane wave solutions of Peskin QFT book (shown in (3.59) and (3.62) on the scanned image below): $$(i \gamma^\mu \partial_\mu -m) \psi=0 \tag{eq.0}$$

1. For $$\psi(x) =u(p) e^{-i p\cdot x}$$, we get the equation of motion (eq.0) becomes $$(\gamma^0 E -\gamma^j p^j -m) u(p) =0 \tag{eq.1}$$ with $$j=1,2,3.$$ Here we choose $$p^0=E>0$$.

Say I agree with the solution $$u(p)=\begin{pmatrix} \sqrt{ p \cdot \sigma} \zeta^s \\ \sqrt{ p \cdot \bar\sigma} \zeta^s \end{pmatrix} \tag{eq.1-sol}$$ 2. For $$\psi(x) =v(p) e^{+i p\cdot x}$$, we get the equation of motion (eq.0) becomes $$(-\gamma^0 E +\gamma^j p^j -m) v(p) =0 \tag{eq.2}$$ with $$j=1,2,3$$, here we still choose $$p^0=E>0$$

Naively, I can derive $$v(p)$$ by a mapping from the known solutions (eq.1-sol)? Naively, I thought that we just either map $$(E,p^j) \mapsto (-E,-p^j)$$ then we plug in (eq.1-sol) to get a rewriting of eq.2 to $$(+\gamma^0 E -\gamma^j p^j -m) v(-p) =0 \tag{eq.2}$$ so we have also a change of $$\zeta$$ to $$\eta$$ $$v(-p)=\begin{pmatrix} \sqrt{ p \cdot \sigma} \eta^s \\ \sqrt{ p \cdot \bar\sigma} \eta^s \end{pmatrix} \tag{eq.2-sol-a}$$ or $$v( p)=\begin{pmatrix} \sqrt{ -p \cdot \sigma} \eta^s \\ \sqrt{ -p \cdot \bar\sigma} \eta^s \end{pmatrix} \tag{eq.2-sol-b}$$ But these are all different from Peskin's $$v( p)=\begin{pmatrix} \sqrt{ p \cdot \sigma} \eta^s \\ -\sqrt{ p \cdot \bar\sigma} \eta^s \end{pmatrix} \tag{eq.2-sol}$$

Why do I get (eq.2-sol-a) or (eq.2-sol-b) instead of (eq.2-sol)? Can you correct my mistakes?

The way that Peskin and Schroeder (P&S) solves the Dirac equation is by going to the rest frame first. The first solution is then given by $$u(p_0)=\sqrt{m}\left(\begin{array}{c} \xi \\ \xi \end{array} \right) ,$$ as given in (3.47) in P&S. The full solution then comes from a Lorentz boost of this solution.
When you do the same for the antiparticle solution, you'll get $$v(p_0)=\sqrt{m}\left(\begin{array}{c} \xi \\ -\xi \end{array} \right) .$$ So the minus sign in the lower entry does not have anything to do with the momentum vector or its direction. Note that these are positive energy solutions, so $$E>0$$ for both the particle and the antiparticle.
• then in that case, how do you tell which component is a particle or anti-particle just by staring at a component of $\xi$ in your notation? Sep 22, 2021 at 2:25