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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?


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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.

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  • $\begingroup$ thanks. Then I have a question: suppose I boost my solution to such that among the 4 components, only 1 of them is nonzero (this is possible see Peskin book, depending on whether the you have spin up or down, left or right chirality), $\endgroup$ Sep 22, 2021 at 2:24
  • $\begingroup$ 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? $\endgroup$ Sep 22, 2021 at 2:25
  • $\begingroup$ Are your referring to the massless limit? In that case you have Weyl spinors, which are not the same thing as the particle and antiparticle solutions. $\endgroup$ Sep 22, 2021 at 8:03

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