In the Dirac-Pauli Representation $\gamma^{0}= \left( \begin{smallmatrix} I&\ \ 0\\ 0&-I \end{smallmatrix} \right)$ , $\gamma^{i}= \left( \begin{smallmatrix} 0&\sigma_i\\ -\sigma_i&0 \end{smallmatrix} \right)$, the solutions to the Dirac Equation $(\gamma^{\mu} \partial_{\mu}-m)\psi=0$ are, acording to Griffith's Elementary Particle book:

\begin{equation} \psi_1= \left( \begin{matrix} 1 \\ 0 \\ \frac{\sigma_i p_i}{E+m} \left( \begin{matrix} 1 \\ 0 \\ \end{matrix} \right) \end{matrix} \right) e^{-ip_{\mu}x^{\mu}} \end{equation}

\begin{equation} \psi_2= \left( \begin{matrix} 0 \\ 1 \\ \frac{\sigma_i p_i}{E+m} \left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right) \end{matrix} \right) e^{-ip_{\mu}x^{\mu}} \end{equation}

\begin{equation} \psi_3= \left( \begin{matrix} \frac{\sigma_i p_i}{E-m} \left( \begin{matrix} 1 \\ 0 \\ \end{matrix} \right) \\ 1 \\ 0 \end{matrix} \right) e^{ip_{\mu}x^{\mu}} \end{equation} \begin{equation} \psi_4= \left( \begin{matrix} \frac{\sigma_i p_i}{E-m} \left( \begin{matrix} 0 \\ 1 \\ \end{matrix} \right) \\ 0 \\ 1 \end{matrix} \right) e^{ip_{\mu}x^{\mu}} \end{equation}

Where it can be seen that $\psi_1$ and $\psi_2$ evolve as $e^{-ip_{\mu}x^{\mu}}$ while $\psi_3$ and $\psi_4$ evolve as $e^{ip_{\mu}x^{\mu}}$. In this case, $p_0$ is a positive number and the minus sign is explicitly put in the exponential.

However, in my course (and in Wikipedia and other books) every solution to the Dirac Equation evolves with $e^{-ip_{\mu}x^{\mu}}$ and the only thing that changes is that $p_0$ is positive for the first two and negative for the other. Altough this may seem as different conventions, notice that with the first way time and space differ always in sign while with the last one antiparticles evolve like $e^{-i(Et-\vec{p}.\vec{x})}$ with $E<0$.

Are this two equivalent? If not, Which one is correct? If it is of any help, The difference in the two solutions happens when one realices that $p_{\mu}p^{\mu}=m^2$ and Griffiths says "obviusly the two solutions are $p^{{\mu}} = \pm (E,\vec{p})$ while the other says "obviusly the two solutions are $E=\pm \sqrt{m^2+p^2}$

  • $\begingroup$ That's just different views of one problem. $\endgroup$
    – Turgon
    Commented Apr 5, 2017 at 6:06
  • $\begingroup$ Hello! Could you say a little more? It seems to me that the two aproaches are not equivalent. $\endgroup$ Commented Apr 5, 2017 at 6:11
  • $\begingroup$ I'm sure you understand the standard negative energy form, so I'll explain how Griffiths thinks. His solution gives negative $p^{\mu } $, which is an obvious solution as negative energy. So he chooses a reversed $x^{\mu } $, that's anti particle explanation. $\endgroup$
    – Turgon
    Commented Apr 5, 2017 at 6:21

1 Answer 1


Actually, the description considering solutions $p^{\mu}=(\pm E,\pm \vec{p})$ prevails in most books about relativistic quantum mechanics and QFT like the classic Bjorken, Drell up to Peskin, Schroeder including many others. The solution with the positive sign on all components of the 4-momentum is called $u(p)\exp(-ipx)$ and the solution with a negative sign on all components of the 4-momentum is called $v(p)exp(ipx)$ (always $px = Et-\vec{p}\vec{x}$ and with $E=+\sqrt{\vec{p}^2+m^2}$ ) The principal justification for it is that they fulfill the Dirac equation. (The algebraic matrix equations $(\gamma_\mu p^{\mu}-m)u(p)=0$ and $(\gamma_\mu p^{\mu}+m)v(p)=0$ can be easily fulfilled as prove the equations you give in your post (these namely are the solutions)).

The more interesting point of course is why this "convention" was adopted as a standard. It makes the interpretation of the solutions $v(p)exp(ipx)$ easier. So what does $v(p)exp(ipx)$ describe ? If $u(p)\exp(-ipx)$ describe electrons with 4-momentum $p^{\mu}$, then $v(p)exp(ipx)$ should describe an electron with negative energy $-E$ and negative momentum $-\vec{p}$. Now comes the Feyman-Stueckelberg interpretation comes in. Principally it says that the emission of a charged particle of $-p^{\mu}$ corresponds to the absorption of an opposite-charged particle of momentum $p^{\mu}$. On the other hand the absorption of a charged particle of $-p^{\mu}$ corresponds to the emission of an opposite-charged particle of momentum $p^{\mu}$. Therefore we can interprete the solutions $v(p)exp(ipx)$ as opposite-charged electrons, i.e. positrons with momentum $+p^{\mu}$ and thereby give the awkward negative energy solutions a proper sense.

Looking at the both solutions (1) $u(p)\exp(-ipx)$ and (2) $v(p)exp(ipx)$ the swap of emission and absorption can be very well observed: The first solution describes an in-going wave (absorption) whereas the second describes and out-going wave (emission). This would be not so obvious if in the establishment of the negative energy solutions the sign of the 3-momentum were not changed too. Moreover, describing propagating particles with positive 4-momentum a phase advance $px = Et-\vec{p}\vec{x}$ with $E>0$ is necessary (changing the relative sign between the 2 parts would be awkward).

Well, the price for the reinterpretation to pay is that emission and absorption process of particles has to be swapped, the consequence of this is that the negative energy electrons are supposed to run backwards in time so that the positrons move forward in time. (Another interpretation is the one of the Dirac sea which is still rather popular but actually a bit old-fashioned as it has some drawbacks. It reinterprets the negative $-p^{\mu}$ solutions in a similar way as positive $p^{\mu}$ solutions.)

If solutions $p^{\mu}=(\pm E, \vec{p})$ had be taken instead, the Feyman-Stueckelberg interpretation can of course also be applied, but the management of the signs would be more complicated, the positrons would have certainly positive energy and negative momentum what is a priori no problem, but the whole formalism would be more complicated if not rather awkward.

Anyway, the solutions of your course and can be transformed in the Griffiths solutions and vice versa by a change of sign in the 3-momentum, and this is no problem as for free particles if there is a solution with momentum $\vec{p}$, there exists also the solution of $-\vec{p}$. It's just rellabeling solutions with different indices.


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