In my relativistic quantum mechanics course, we found plane wave solution to the Dirac equation by first studying it the reference frame of the particle. Using a plane wave solution for both positive and negative energy we got

$$ (i\not \partial -m)u(m,0)e^{-ip_\mu x^\mu} = 0 \implies (\not P - m)u(m,0) = (\gamma^0-1)u(m,0)=0 $$

for the positive energy solution. Then we evaluated the negative energy solution in the same way to get, in the end

$$ u(m,0) = \alpha \left( \begin{matrix} 1\\0\\0\\0 \end{matrix} \right) + \beta \left( \begin{matrix} 0\\1\\0\\0 \end{matrix} \right) = \alpha u^{(1)}(m,0)+\beta u^{(2)}(m,0) \\ v(m,0) = \alpha^\prime \left( \begin{matrix} 0\\0\\1\\0 \end{matrix} \right) + \beta^\prime \left( \begin{matrix} 0\\0\\0\\1 \end{matrix} \right) = \alpha v^{(1)}(m,0)+\beta v^{(2)}(m,0) $$

Then to get the general solution we had to do a Lorentz transformation: at this point the professor said something that it's not so clear to me. I quote

Since $(\not P -m)(\not P +m)=0$ we can easily do a reference frame transformation such that $$ \left\{ \begin{matrix} u^{(\alpha)} = c_{\alpha}(\not P +m)u^{(\alpha)}(m,0)\\ v^{(\alpha)} = d_{\alpha}(\not P -m)v^{(\alpha)}(m,0) \end{matrix} \right. $$

so, leaving the normalisation factors behind for a moment, he's implying that the representation of a general Lorentz transformation for the spinor components of the Dirac spinor (note that I'm working in standard, non chiral, representation) are given by $(\not P-m)$ and $(\not P +m)$. It is clear why the first statement is useful: so that when we plug in the solutions made in this way they are in fact solution to the Dirac equation. My question then is

How can this be proven?

Probably it would be easy to prove it in chiral representation, but even still I'm not able to do it.


1 Answer 1


Consider the Lorentz transformation of the spinor parametrizing by the matrix $S$: $$ u(\mathbf p) \to S u(\mathbf{0}) $$ Here $S$ corresponds to the Lorentz transformation $\Lambda_{\mu}^{\nu}p_{\nu,\text{rest}} = p_{\mu}$, where $p_{\nu,\text{rest}} = (m,\mathbf{0})$ and $p_{\mu} = (E,\mathbf{p})$. $\gamma$ matrices have a property $$ S^{-1}\gamma_{\mu}S = \Lambda_{\mu}^{\ \nu}\gamma_{\nu}, $$ using which we obtain $$ \tag 1 \color{red}{\gamma_{\mu}p^{\mu}S =} S\Lambda_{\mu}^{\ \nu}p^{\mu}\gamma_{\nu} = S(\Lambda^{-1} p)^{\nu}\gamma_{\nu} = \color{red}{mS\gamma_{0}} $$ This property is failed to satisfy if $S \propto \gamma_{\mu}p^{\mu}+m$ as far as $\gamma_{0} \neq \mathbb{1}$. The latter is impossible.


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