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My question is related to the interpretation of antiparticles. According to the so called Feynman-Stueckelberg interpretation a negative energy solution of the Dirac equation corresponds to a positron which then runs apparently backwards in time. But if I consider the vertex of a electron-positron annihilation I have at this vertex an incoming electron with positive energy which under "emission" of a virtual photon turns into a (the vertex) leaving positron which runs backwards in time. But this way several conservation laws seem to be violated.

I would in fact expect the positron not to leave the vertex but entering the vertex. In that case the conservation laws would be fulfilled.

However, for me a solution of ~exp(iEt) is either a negative energy electron running forwards in time or a positive energy positron running backwards in time. But I cannot fit that with the electron-positron annihilation (see above). I would be grateful if somebody could dissipate by confusion.

bobie
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Frederic Thomas
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  • For one thing, the Dirac equation is not a good physical model for the vacuum, for another, how does one go backward in time? I think one should leave some "interpretations" of the past where they are: in their historic perspective. They were tools to establish a path to better descriptions of reality but they have outlived their usefulness once better ones were discovered. – CuriousOne Dec 15 '14 at 14:58
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    Which conservation laws do you think to be violated? – ACuriousMind Dec 15 '14 at 15:05
  • @CuriousOne: see http://physics.stackexchange.com/a/17781/6389 – Christoph Dec 15 '14 at 17:45
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    @Christoph: So you are saying that the positrons in my accelerator can actually meet my dead grandmother some time in the past? Cool! Of course that is not what is meant by that in the Feynman formalism. It was, by the way, Feynman, who has warned people against interpreting the terms in any particular choice of perturbation series as actual physical processes/entities. They are mathematical representations for terms on a sheet of paper, of which ONLY the final result has any meaningful reality. – CuriousOne Dec 15 '14 at 17:50
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    @CuriousOne: So you are saying that the positrons in my accelerator can actually meet my dead grandmother some time in the past? Yes, but the process is indistinguishable from (actually identical to) your grandmother emitting an electron. – Christoph Dec 15 '14 at 18:00
  • @Christoph: Except that you are discounting that my dead grandmother hasn't been emitting electrons for some 30 years... so there is a bit of a problem there for the interpretation. :-) – CuriousOne Dec 15 '14 at 18:10
  • @CuriousOne: once you have a Cauchy surface, past and future are fixed; sure, you only can send a positron into the past if your grandmother emitted an electron; but in the Wheeler-Feynman absorber formulation of EM, you also only can send a light ray into the future it it gets absorbed eventually by, say, your unborn granddaughter; hence the transactional interpretation of QM, which is philosophically interesting, but sadly worthless from the point of view of theoretical physics – Christoph Dec 15 '14 at 18:16
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    @CuriousOne: or formulated another way: just because humans only remember the past doesn't necessarily mean that the universe has to operate that way – Christoph Dec 15 '14 at 18:17
  • @Chrsitoph: Curiously, my dead grandmother has left a causal artifact in this universe (me). I can turn off my accelerator and stop the production of positrons. Now you tell me how my positrons are going to reach back into the past and hit my dead grandmother? Look, I get it that you are not comfortable with the difference between a local symmetry and its globally broken consequences... but there it is. A symmetry on paper is a symmetry on paper, but that means absolutely nothing for the dynamic structure that develops as its consequence. – CuriousOne Dec 15 '14 at 18:22
  • Let us continue this discussion in chat. – Christoph Dec 15 '14 at 18:22
  • #Christoph: I already said everything there is to say about the subject, and so did Mr. Feynman. If you don't take my word in the matter, maybe you should consider taking his. He never pretended that all the parts of his formalism were physically real. All that counts is that one can calculate final results that are. – CuriousOne Dec 15 '14 at 18:25

2 Answers2

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A negative energy particle running backwards in time is mathematically equivalent to a positive energy antiparticle running forwards in time.

Since the time dependence of the wavefunction is of the form $exp[-iEt]$ for particles that run forward in time, then the simultaneous transformation $t \to -t$ and $E \to -E$ will give you:

$$ exp[-i(-E)(-t) \equiv exp[-iEt]$$

i.e the time dependence is unchanged. Hence the two pictures are mathematically indistinguishable.

Lets illustrate this with an example. Consider the two vertices below:

enter image description here

I claimed before that these 2 vertices are mathematically equivalent. How so? Well, in the left diagram, a positive energy electron emits a photon of energy 2E, and in order to conserve energy, a negative energy electron is emitted propagating backwards in time (from the solutions of the Dirac equation).

This is equivalent to the second diagram because you can interpret it like this: The positive energy electron is annihilating with the positive energy positron (both here propagating forwards in time) and in order to conserve energy they produce a photon of energy 2E.

For further reading I suggest the book Modern Particle Physics by Mark Thompson

PhotonBoom
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  • @FredericThomas I'm not sure what you are disagreeing with. :) – PhotonBoom Dec 22 '14 at 16:31
  • It is just a detail. I find a negative energy solution and at the moment looking at it this "strange" solution is still running forward in time. Then trying making sense of it I change the interpretation, I get a positron with positive energy (So I got rid of my "strange solution" that's nice), however this positron then runs backward in time. It seems to be that right from the beginning of the "raisonnement", and well before the switch from the electron to the positron takes place the time arrow is changed "ad hoc" while the energy of the particle is maintained negative. – Frederic Thomas Dec 22 '14 at 17:24
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    @FredericThomas I think you are confusing what you have initially. Initially you have a negative energy particle propagating *backwards* in time, not forwards. Then I showed above that this is equivalent in having a positive energy positron running *forwards* in time. Its only when you draw the Feynman diagram afterwards that you switch the arrow of the positron, without really any real reason behind it, other than to confuse people :P – PhotonBoom Dec 22 '14 at 17:34
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According to the so called Feynman-Stueckelberg interpretation a negative energy solution of the Dirac equation corresponds to a positron which then runs apparently backwards in time.

A negative energy solution corresponds to an electron running backwards in time, which is the same thing as a positron running forwards in time.

A positron runnning backwards in time would be an electron.

To be a bit more explicit:

Say we have an equation that describes particles with negative charge. While analyzing its implications, we find that it requires solutions with negative energy $p_0 < 0$, ie the particle's momentum points in the direction opposite to coordinate time!

Because time travel is a Bad Thing as far as physicists are concerned, we have to discard our equation or try to 'argue away' its problems.

So let's assume such particles did indeed exists. How would they manifest, eg what would their trajectories look like in a magnetic field? Well, they would essentially look like an ordinary particle with opposite (thus positive) charge!

So we reinterpret a particle with negative energy (ie having momentum pointing towards the past) as an anti-particle with positive energy.

Christoph
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  • @Christopf, okay, but I would also like to know about the energies of these particles. I guess "your" electron running backwards in time has positive energy, which energy has the positron, also positive energy ? I thought that a particle with 4-momentum -p running backward in time would be equivalent to an antiparticle with 4-momentum p running forward. According to this my positron would have again negative energy, so the negative energy "problem" would not be solved. So how can the positron have positive energy after all ? – Frederic Thomas Dec 22 '14 at 13:51
  • @FredericThomas: see edit – Christoph Dec 22 '14 at 14:11