# Is $v(p)\exp(ipx)$ really the positron wave function?

In many textbooks the negative energy solution of the Dirac equation is quoted as describing the positron. Actually I don't understand this. For me $v(p)\exp(ipx)$ is the wave function of an electron of negative energy, therefore part of the famous Dirac sea (sorry for sticking to this old model). But a positron is not an electron of negative energy, but the absence of such an electron. Furthermore, if I apply the Hamilton operator on $v(p)\exp(ipx)$ I get a negative energy as a solution. Furthermore, why should a particle which has exactly the same properties as an electron (a part from the charge which does not appear neither in $u(p)$ nor in $v(p)$) have a different wave function ?

Another argument: In Feynman diagrams $u(p)\exp(-ipx)$ describes an in-going electron, but $v(p)\exp(ipx)$ is used for an out-going positron (or if you like an in time backwards in-going negative energy electron). This is my problem from the viewpoint of the Feynman-Stueckelberg interpretation: If $v(p)\exp(ipx)$ were a positron wave function should it should describe an in-going particle, but it describes an out-going particle with positive energy. Assuming that positrons should in time forward running with positive energy and only the negative energy solutions in time backward running, then $v(p)\exp(ipx)$ describes an out-going particle with positive energy, then charge conjugation transforms an in-going electron $u(p)$ in an out-going positron $v(p)$. But for me the charge conjugation should transform an in-going electron into an in-going positron. I don't know how to resolve the contradictions. At last if it were said $v(p)\exp(ipx)$ is NOT the positron function, how shall I understand the charge conjugation operation which apparently makes of an electron function $u$ a positron function $v=Cu(=Uc \bar{u})$ ? Thank you for your help.

## 1 Answer

The Dirac sea is a nice concept, but it does not hold up too much unfortunately. The simplest argument against it is just bosons. The Dirac sea works okay for fermions, but without the exclusion principle, there's no reason a boson cannot drop energy levels forever, radiating an infinite quantity of energy.

If you go with actual QFT, this is not a problem :

Furthermore, if I apply the Hamilton operator on $v(p)\exp(ipx)$ I get a negative energy as a solution.

$v(p) e^{ixp}$ is not a wavefunction, it is an operator (well more accurately $b^\dagger v(p) e^{ixp}$ is). The wavefunction is something else, usually expressed purely as a Hilbert vector (I saw it expressed as a function here and there but it is very rare). Using this, the energy of a positron remains positive.

Furthermore, why should a particle which has exactly the same properties as an electron (a part from the charge which does not appear neither in $u(p)$ nor in $v(p)$ have a different wave function ?

As a free theory, it does not matter too much I guess, but then again that argument could be made of any other fermion. But once you add interactions in, it becomes pretty important. For a start, electrons and positrons annihilate. But if you look at the weak interaction, results can be even more drastic : if you assume the neutrino massless (which was done until rather recently, and is still done for ease of calculations), the two chiral parts of the wavefunction are uncorrelated, and due to the parity violation of the weak interaction, this means that neutrinos always have left-handed chirality, and anti-neutrinos always right-handed.

If $v(p)\exp(ipx)$ were a positron wave function should it should describe an in-going particle, but it describes an out-going particle with positve energy.

That is more of a mathematical trick due to various symmetries of the equations. By applying appropriate symmetries and transformations, you can flip Feynman diagram or rotate them