Photons traveling backwards in time? Imagine that two widely separated charged particles $A$ and $B$ exchange a photon.
Because they are far apart one can imagine that there is a major contribution to the photon propagator that travels at the speed of light from $A$ at a time $T_0$ to $B$ at a time $T_1$ where $T_1 > T_0$.
But in that case is there also a major contribution to the photon propagator that travels backwards in time at the speed of light from $B$ at time $T_1$ to $A$ at time $T_0$?
The forwards-in-time photon imparts momentum to particle $B$ whereas the backwards-in-time photon imparts a reaction momentum back to particle $A$.

 A: Suppose $A$ is at the space-time origin $0$, and $B$ is at space-time event $x$. You suppose that a real photon could go from $A$ to $B$, so this means that $A$ and $B$ are separated by a light-like interval, that is $x^2 = (x^0)^2- \vec x^2=0$. This means that $x^0>0$, too. 
Now, the propagator $D_{\mu\nu}(x)$ represents the amplitude for a photonic field perturbation to go from $A$ to $B$ (implicitely you have electronic sources $J(A)$ and $J(B)$)
The (Feynman) propagator may be written (skipping polarizations indices for simplicity):
$D(x) = -i\int \frac{d^3k}{(2\pi)^3 2\omega_k}[\theta(x^0)e^{-i(\omega_k x^0- \vec k.\vec x)}+\theta(-x^0)e^{+i(\omega_k x^0- \vec k.\vec x)}] \tag{1}$ 
where $\omega_k = |\vec k|$, is a positive value.
Now, with your hypothesis ($x^0>0, x^2=0$), equivalent to $x^0=|\vec x|$, the propagator may be written :
$D(\vec x, |\vec x|) = -i\int \frac{d^3k}{(2\pi)^3 2\omega_k} e^{-i(\omega_k |\vec x|- \vec k.\vec x)} \tag{2}$
However, even with this expression, the propagator is still a field perturbation which "propagates" from $0$ to $x$, and you cannot consider it as a particle. A possibility, in this very special case, would be to consider the propagator as a "kind-of" sum of contributions (with a weight) of pseudo-classical-real-particles, with momentum $|\vec k|$ a and positive energy $\omega_k =|\vec k|$, and , "supposed" going from $0$ to $x$ (it would be "possible" because $x^2=0$). But I don't think this is a good idea, because this pseudo-pattern is no more applicable for $x^2>0$ and $x^2<0$, so it is better to think at the propagator as representing a field perturbation which may take different representations following the sign of $x^0$ and/or the values of $x^2$, and clearly this field perturbation cannot be considered as a particle. 
In fact, the term "propagator" is  not the best one, one should better think of $D(x)$ as a correlation amplitude between the sources $J(0)$ and $J(x)$.
For instance, a analogy is to think about entanglement, you may have spatially separated sub-systems which could be however correlated.
A: Its impossible for a photon to travel backwards in time sense it keeps disappearing because it keeps giving up its energy to other particles like an electron, either part of it or all of it which means it wont have enough energy to warp space time or even have enough energy to create closed time like curve and travel backwards in time. Here's how it works when a photon is absorbed by an electron, it is completely destroyed. All its energy is imparted to the electron, which instantly jumps to a new energy level. The photon itself ceases to be.
