Assume that a stationary electron $A$ emits a virtual photon with $4$-momentum $k$ and a stationary electron $B$ absorbs it.

Let us assume a description in which time is moving forwards.

At the instant that electron $A$ recoils with $4$-momentum $-k$ can we say that the photon has $4$-momentum $k$?

I assume not as this would imply that the photon is real with a definite $4$-momentum.

It seems to me that, under the assumption that time is going forwards, we can only assert that electron $B$ has $4$-momentum $k$ once the virtual photon has been absorbed.

So how can we understand electron $A$'s recoil?

Let us assume a description in which time is moving backwards.

In that case electron $B$ emits a virtual photon with $4$-momentum $-k$ and electron $A$ absorbs it.

Now we can describe electron $A$'s recoil with $4$-momentum $-k$, without implying a real photon, again once the time-reversed virtual photon has been absorbed.

So to fully describe the exchange of a photon between electron $A$ and electron $B$ it seems to me that we need to assume time both goes forwards and backwards.

Is this right?

  • $\begingroup$ You have to choose between position space and momentum space, but you cannot use the two together. So, you may consider Feynman diagram in position space, with transition amplitudes depending on positions, or Feynman diagram in momentum space, with transition amplitudes depending on momenta, but not all together. $\endgroup$ – Trimok Jan 22 '14 at 19:10
  • $\begingroup$ I would have thought that energy and momentum is always conserved at vertices whether you are in position space or momentum space. $\endgroup$ – John Eastmond Jan 23 '14 at 9:34
  • $\begingroup$ QFT is a quantum theory. By definition, momentum/energy (as a real quantity, that is, as an eigenvalue) is a notion only available in momentum space. $\endgroup$ – Trimok Jan 23 '14 at 10:53

Feynman diagrams are the generic mathematical model algortithms that describe particle interactions and define "virtual", as in "virtual photon".

photon exchange

How is this algorithm changed into a mathematical formula which will give finally the probability of electron positron elastic scattering? There exists a one to one correspondence with each vertex and each line to a mathematical form. The symbolic graph allows us to visualize the interaction as an exchange of a virtual photon, it has all the quantum numbers of a photon except it is off mass shell.

Off mass shell means that its four energy/momentum vector does not retain its "length" , the "length is the rest mass, it means that it is off mass shell, i.e. for the mass of the photon is different than zero. This means that energy and momentum conservation for the process can only be imposed to the incoming and outgoing lines ( and the vertices on them), and are irrelevant on a virtual line because the four momentum exchanged between the electron and positron is carried pictorially by the virtual photon whose mass is unphysical. This definition of virtual holds for all particle internal exchanges between vertices.

It has no meaning to speculate about the distances and times, which will be unphysical, only the result of the final integration of the whole formula has physical meaning.

Talking of "virtual" particles helps us in writing the propagator because the rest mass of the real particle enters there, and it helps us keep track of the quantum numbers. It is a mathematical representation that works beautifully but it is a mistake to extend it to a realm where it does not apply: one cannot verify or experiment in the virtual space and validation comes from experiments agreeing with the total computation.

So a thought experiment within the virtual regime has no meaning in momentum conservation or time direction, it cannot be carried out since the particles are "virtual", as defined above.

  • $\begingroup$ "This means that energy and momentum conservation for the process can only be imposed to the incoming and outgoing lines, and are irrelevant on the virtual ones". Sorry, no. If you draw a Feynman diagram in momentum space, you have conservation of energy/momentum for all lines, including internal lines (conservation of momentum/energy works at vertex). $\endgroup$ – Trimok Jan 22 '14 at 19:03
  • $\begingroup$ I am trying to answer the "recoil" . The internal line does not obey it, though, no? $\endgroup$ – anna v Jan 22 '14 at 19:22
  • $\begingroup$ Well, I added a comment under the question: OP must choose between position space and momentum space for Feynman diagrams. So, choosing momentum space, conservation of momentum/energy works for all vertex. $\endgroup$ – Trimok Jan 22 '14 at 19:28

No, time does not "go backwards". This expression is sometimes used to say that the real parameter $t$ describing time instant can be run backwards, but this happens only in the mind of the theorist. The mathematical concept of time is used in such a way that it always increases along the progression of every physical process, that is past events have lower $t$ and future events have higher $t$.

In relativistic theory, if body(electron) sends a packet of energy in one direction, it has to send momentum as well and if we believe in conservation of momentum, the body has to either recoil back as soon as it sends the packet or it has to send more than one packet, in such a way as to conserve the momentum. In relativistic theory, as soon as this happens, the mass of the electron has to decrease, which was never observed - all electrons have the same mass. So my view on this is that electron does not send packets of energy in definite direction. Electrons send EM waves that can together carry energy around, but most of the time, this does not come from the masses (internal energies) of the electrons, but is just present in the space around the particles.


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