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)?
Best Answer
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..