At least as normally formulated, the law of transformation of a wave function solution of the Dirac equation to another inertial frame seems to indicate that if observer 1 is certain the particle is an electron, observer 2 (in a frame moving relative to the first) will judge that it might be an electron or it might be a positron. But shouldn't charge be invariant under Lorentz transformation?

To express this is more detail, the Lorentz transformation of the 4-component wave function as given for example in (4.22) of http://www.damtp.cam.ac.uk/user/tong/qft/four.pdf , generally changes a wave function of the form (1,0,0,0) in frame O to (a,b,c,d) in frame Oprime. I took this to mean O is sure the particle is an electron spin up, while Oprime assigns a non-zero probability to the particle being a positron (since c and d are not zero). Thanks to Sanya for requesting this addition.

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    $\begingroup$ Could you add a source or the transformation law you are referring to?The transformation laws I could find (see e.g. damtp.cam.ac.uk/user/tong/qft/four.pdf ) did not seem to mix electron- and positron-components if I understand them correctly. $\endgroup$
    – Sanya
    Jun 16, 2016 at 20:02

1 Answer 1


Even if you interpret the Dirac equation the original way (not as a quantum field theory), the Lorentz transformation doesn't mix electron and positron. Consider two types of plane wave with positive frequency and negative frequency $$u(p)e^{-i (E t-p x)},\quad v(p)e^{+i (E t-p x)}$$ The idea is the energy $E$ is always positive, but the positive frequency describes an electron and negative frequency a positron.

Under a Lorentz transformation the 4-vector (E,p) will change to (E',p') but a proper Lorentz transformation does not change the sign of the time component, i.e. the energy, so it does not mix the two types of plane wave.

If you want to describe charge in a Lorentz invariant way, the quantity $\bar{\psi}\gamma^\mu \psi$ is a 4-vector that describes the current.

  • $\begingroup$ Don't you also have to apply a Lorentz transformation matrix $\Lambda$ to the spinor components when boosting into another frame? I think it's this aspect that the question is most concerned with. $\endgroup$ Jun 16, 2016 at 20:33
  • $\begingroup$ You can write the spinor transformation as $U(\Lambda)u(p)=u(p')$ for some matrix $U$ or you can just figure out what the form of $u(p)$ has to be in general and update $p$ to $p'$. I think the original question is confusing the lower two components of the spinor in the Dirac basis as being positron coponents, which is not the case. $\endgroup$
    – octonion
    Jun 16, 2016 at 20:36
  • $\begingroup$ Yes I was thinking the two lower components where positron components (whatever frame one was in). If this is not the case the problem disappears. But in that case what do the components mean? Can you refer me to any account of the Dirac equation that expressly says the components are generally mixed up in this way and how they should be interpreted? $\endgroup$
    – John Hemp
    Jun 18, 2016 at 16:45
  • $\begingroup$ Maybe try reading Peskin and Schroeder's quantum field theory book, chapters 3.3 and 3.4. The point is what distinguishes positron and electron is the time dependent frequency part of the solution. The properties of the spinors u and v follow from substituting this into the Dirac equation, and yes u and v will have different nonzero components in the rest frame. But there is no step where you identify certain components of u, which are just coefficients of the positive frequency solution, with an antiparticle. $\endgroup$
    – octonion
    Jun 18, 2016 at 18:44
  • $\begingroup$ Thanks for your help octonion and Michael. I think I understand now. One constructs eigenfunction solutions - some have +ve energy time factor some -ve energy time factor. Then any wave function can be expressed as a linear combination of these, and the squared moduli of the coefficients will give probabilities for the particle being an electron or positron with some other properties (like total angular momentum) as well. $\endgroup$
    – John Hemp
    Jun 21, 2016 at 18:24

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