This is given in Greiner, Relativistic Quantum Mechanics

For a free particle solution and antiparticle solution with momentum $\vec{p}$ the current is given by $e\frac{c^2\vec{p}}{E_p}$. The current is same for both particles and antiparticles. But why are the currents for particle and antiparticle with energy $E_p$ and momentum $\vec{p}$ equal although the charges are different? The reason is that for antiparticles with momentum $\vec{p}$ the velocity operator is $-c^2\vec{p}/E_p$, i.e. velocity and momentum have opposite directions. One says that the antiparticles move "backwards in time".

My question is: How can velocity and momentum of antiparticles be opposite in direction as given in the paragraph? $\vec{p}=\gamma m_0\vec{v}$ where the symbols have their usual meanings. So velocity and momentum should be in the same direction. It also states that the antiparticles move "backward in time". What does it mean? I thought it might be related to Time Reversal but couldn't find anything.

  • $\begingroup$ Hello Asit! Could you please take a moment to read this post about the usage of images in the questions here at the PSE? And if you have the time, also this post about the typesetting of equations in the site? By following these guidelines you will enable other people with the same question to find the answer through a search engine. $\endgroup$ – S V Aug 19 '19 at 21:44
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    $\begingroup$ Yeah, I'll edit it. $\endgroup$ – Asit Srivastava Aug 19 '19 at 21:52
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    $\begingroup$ Thank you, and in case you need it, here is a basic tutorial for MathJax, I believe it is helpful as a reference. $\endgroup$ – S V Aug 19 '19 at 21:54
  • $\begingroup$ Thank you, it'll definitely help. $\endgroup$ – Asit Srivastava Aug 19 '19 at 21:59
  • $\begingroup$ This does not make much sense. Is this really what Greiner wrote? $\endgroup$ – my2cts Aug 19 '19 at 22:09

In Klein-Gordon a free particle, for either positive or negative energies, has probability current density $\vec j = \vec p / E_p$ ($\hbar = c = 1$, natural units). With a normalization that imposes positive and negative charges on positive- and negative-energy solutions, respectively, this is quite surprising. The charge current density is the same regardless of the sign of the charge and energy.

Another difficulty is that while the positive-energy solution is written as $\Psi_+ \sim exp (-i E_p t + i \vec p \cdot \vec x)$ for $E = +E_p$, the negative-energy solution is written as $\Psi_- \sim exp (i E_p t + i \vec p \cdot \vec x)$ for $E = -E_p$. The latter expression is not Lorentz invariant.

A way to fix these inconsistencies is to say that the negative-energy particles move backward in time. This not only reverses the sign of $\vec p$, but also lets the energy be positive in the exponent of the negative-energy solution.

We can associate the positive-energy solution to a particle and the negative-energy solution to an antiparticle with charges of opposite sign.

Note: If you assume that the antiparticle moves backward in time, velocity and momentum have the same sign.

  • $\begingroup$ The energy is positive for positive and negative frequency alike. $\endgroup$ – my2cts Aug 21 '19 at 23:09
  • $\begingroup$ @my2cts. You misread my post. I stated clearly that the assumption of the move backward in time for the negative-energy solution would cause the energy to be positive also for that solution. I ask you to remove your negative mark. $\endgroup$ – Michele Grosso Aug 22 '19 at 14:39
  • $\begingroup$ The energy is positive, independent of any assumptions. Apart from the normalisation factor N, which should be positive, energy is proportional to $+\omega^2 $ and momentum to $+\omega k$, for a plane wave $e^{i\omega t - kx} /N$. For negative $\omega $ the forward moving wave also has momentum proportional to $+\omega k$. Velocity and momentum have the same direction. $\endgroup$ – my2cts Aug 27 '19 at 19:08
  • $\begingroup$ @my2cts. What you say is not how historically the Klein-Gordon was developed. The $E$ which solves the equation is $E = \pm \sqrt{p^2 + m^2}$ in natural units. From that you get an apparently negative energy. Then with some reasoning (refer to my post) you succeed to recover a positive energy. You can find this in any text of relativistic quantum mechanics. $\endgroup$ – Michele Grosso Aug 28 '19 at 7:38
  • $\begingroup$ The negative root must be discarded since antimatter experimentally has positive energy and the Noether energy of a free particle is positive. The sign of the frequency does determine the sign of the charge. Which historical paper are you referring too? I hope not some Dirac sea like reasoning. @MicheleGrosso $\endgroup$ – my2cts Aug 28 '19 at 15:27

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