On Feynman diagrams, why do antiparticles have their arrows pointing backwards

[jeebs]

...compared to normal particles?

I was told this was something about, for a particle moving forward in spacetime, its antiparticle can be considered as moving backwards in space time. but that really doesn't mean anything to me.

what's wrong with putting a forward arrow on an antiparticle, as surely they can in reality move in one direction through space and time just as easily as normal particles?

 

[tiny-tim]

hi jeebs!

it's so that the vertices in https://www.physicsforums.com/library.php?do=view_item&itemid=235" preserve their charge (and it's more a question of "running the film backwards" than of anything actually "living backwards") …

if a particular vertex has an arrow in and an arrow out (and a photon), then that represents an electron in and an electron out, which preserves the charge

but if one of them was a positron, and you had a positron in and an electron out, you'd be losing 2 units of charge …

you're only allowed a positron and an electron both being destroyed or both being created …

that preserves the charge, but means that an arrow in has to represent a positron out (and vice versa)

ie creating an electron is the same as destroying a positron, which would look like creating a positron if you ran the film backwards (and vice versa)

 

[bcrowell]

Baez has a good FAQ about virtual particles here: http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html

I don't know the answer to your question, but it seems to me that it is clearly related to the fact that virtual particles can travel faster than c. Since they can travel faster than c, there is no way to time-order the emission and absorption, and therefore it wouldn't make sense to try to make a distinction between the ones that were propagating forward in time and the ones that were propagating backward. Such a distinction would be frame-dependent.

But this isn't really an answer to your question. It just shows that answering your question is equivalent to answering the question of why it's helpful to think of virtual particles as propagating at speeds higher than c.

Here's a *possible* answer to that question, but I don't know if it's right. You want to be able to say that all real particles propagate along a least-action path, because only along such a path do the partial waves add constructively. This whole argument requires that you be able to discuss whether such a path has an extremum of the action relative to all other paths that differ from it by a small amount. If you want to apply that to a photon, then you need to compare with paths that don't travel at the "right" speed, because the "right" speed is supposed to be an output of the calculation, not an input.

 

[bcrowell]

Tiny-tim, I could be wrong, but it seems to me that your #2 does not really answer the OP's question. It seems to me that you're making an argument like this: Suppose you start with a vertex with two charged particles and one neutral particle. Then by continuously deforming the vertex you can make either charged particle change its role from that of an absorbed particle to that of an emitted particle. So in order to preserve charge conservation, you need to adopt the convention about forward and backward being interpreted as matter and antimatter.

I don't have any objection to your argument, but it starts from the assumption that it should be legal to continuously deform a vertex in any way. It seems to me that the OP is asking why we can't forbid deformations that turn a forward-going particle into a backward-going particle.

 

[GreyBadger]

It's purely a matter of convention, but the idea is that fermion lines are continuous (see comments above). Now, if you want to think of it in terms of 'backwards in time' terms, then the reason is that applying the two operators C and P (Charge and Parity, CP) yields an antiparticle from a particle. These are mathematically equivalent (see textbooks ad infinitum) to the T (Time) operator. So, a particle reversed in time is equivalent to an antiparticle moving forwards in time.

 

[tiny-tim]

hi bcrowell!

It seems to me that you're making an argument like this: Suppose you start with a vertex with two charged particles and one neutral particle. Then by continuously deforming the vertex you can make either charged particle change its role from that of an absorbed particle to that of an emitted particle. So in order to preserve charge conservation, you need to adopt the convention about forward and backward being interpreted as matter and antimatter.

I don't have any objection to your argument, but it starts from the assumption that it should be legal to continuously deform a vertex in any way. It seems to me that the OP is asking why we can't forbid deformations that turn a forward-going particle into a backward-going particle.

(i don't understand what you mean by "deforming a vertex" )

no, I'm saying that quantum field theory represents any creation operator by a "field" composed, on an equal basis, of creation operators for particles and annihilation operators for anti-particles …

the justification for this is basically "that it works!" (ok, i know there are more sophisticated arguments ) …

but when we ask why something is the way it is in a Feynman diagram, surely we're talking in Feynman diagram language, ie as if the diagram actually represents a physical process? …

so instead of giving the mathematical reason for defining "field" that way, and continuing in terms of momentum-preserving delta functions etc, we talk as if the diagram represents a series of collisions, with each vertex representing a single collision …

in those terms, the correct mathematical explanation translates into one about electrons etc being destroyed or created, and conservation of momentum charge etc

(though i like GreyBadger's CP = T explanation also )

 

[Vanadium 50]

I have a very different view. Feynman diagrams are mnemonic devices for calculations. The arrows are there to remind you to use fermionic or antifermionic fields in the right place. Nothing more, nothing less.

 

[tiny-tim]

isn't that what i said?