Apparently there are 2 electron self-energy graphs possible. The first, the more "familiar", where the incoming electron at time $t_1$ splits up in a photon and an virtual electron. At $t_2>t_1$ the virtual photon joins the electron again. But the Feynman-propagator also allows $t_1>t_2$, where apparently the incoming electron hits a positron at $t_1$ which was created at time $t_2<t_1$. But for the creation of the electron-positron pair at $t_2$ the photon has to provide the positive energy at $t_2$ for the creation process, so this photon seems to be created at $t_1>t_2$, so the photon is apparently moving backward in time. The other possibility is of course apply Feynman's saying: Backward in time running particles with energy $E$ can be interpreted as forward in time running particles with $-E$ (assuming $E$ can have positive or negative sign in general). Therefore, it would be equivalent to say that for the electron self-energy diagram where $t_1>t_2$, at $t_2$ a photon is created with energy $-E$ ($E$ being the energy needed for the creation of the electron-positron pair) and then moving in time forward to $t_1$ to deliver this negative energy to destroy the (incoming) electron-(virtual) positron pair. Therefore I come to the conclusion that virtual photons can have negative energy in running in time forward or positive energy running in time backward.

What about real photons? I would be astonished about real negative energy photons. Or would the dispersion relation $\omega^2=k^2$ allow for negative frequencies (keeping in mind that these negative frequency solutions would again correspond to photons as anti-photons and photons are the same)?


3 Answers 3


Note that you have to choose between momentum representation and space-time representation, for Feynmann diagrams, but you cannot use the $2$ together, this is quantum mechanics. So you cannot speak, at the same time, of precise times $t_1,t_2$, and precise energy $E$

"Virtual" particles are not particles at all. The Feynman propagator $D(x)$, in space-time representation, just express amplitudes to go from, say, the origin $0$ to a point $x$. It is better to see this as field correlations between the points $0$ and $x$

However, your general idea in the first paragraph of your question is quite correct, in the expression of the simple (massless scalar field) Feynman propagator :

$D(x) = -i\int \frac{d^3k}{(2\pi)^3 2\omega_k}[\theta(x^0)e^{-i(\omega_k x^0- \vec k.\vec x)}+\theta(-x^0)e^{+i(\omega_k x^0- \vec k.\vec x)}] \tag{1}$

with $\omega_k = |\vec k|$

we see that positive $x^0$ are associated with a collection of positive energies $\omega_k$, while negative $x^0$ are associated with a collection of negative energies $-\omega_k$

But once more, it is better to speak of field perturbations, or field correlations, but not "particles"

Real particles, on the other hand, have always, in momentum representation, a positive energy. So they belong to some representation of the Poincaré group, with, for massless particles as photons, $p^0>0$ and $p^2=0$. These real particles have creation and anihilation operators, etc..

  • $\begingroup$ virtual particles are "not particles at all" is a boring, meaningless mantra. They become real thermal photons when the observer is accelerated, or whenever there is some sort of event horizon $\endgroup$
    – lurscher
    May 4, 2018 at 15:51

A negative energy photon means that an excited atom state, upon absorption of this negative energy photon (with frequency matching the energy level), would, instead of stimulated emission, collapse back to ground level. The energy of the photon is absorbed in the past emitter. This is just another way of saying that advanced photons (with positive energy) are the same thing as a retarded photon with negative energy.

So far, advanced photons have never been observed experimentally, and no one really understands why. Fundamental equations are time-symmetric, but this time-symmetry seems to be broken in the actual universe

  • $\begingroup$ Would photons under a potential have negative energy? $\endgroup$
    – jinawee
    Jun 10, 2014 at 21:25
  • $\begingroup$ "... and no one really understands why." Because the theory is wrong? Noooo..., can't be. $\endgroup$ Jun 11, 2014 at 10:50

By construction, the real photons have positive energies.

Concerning virtual photons, especially from the self-energy diagram, let us remember that this diagram is not left intact in calculations, but modified with subtractions. In addition, it is better to speak of the exact renormalized solution rather than of a (perturbative) piece of it.


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