Feynman Diagram of a Coulombic Attraction?

[mk17]

Whenever I see examples of the diagram for the Coulomb interaction it always seems to be two electrons interacting via an electron and being repulsed. The diagram looks intuitive in terms of momentum conservation.

I was wondering how (for example) a proton - electron interaction, and subsequent attraction, would be drawn? Does the electron still emit a photon in the direction of the proton?

Thanks.

 

[tiny-tim]

Welcome to PF!

Hi mk17! Welcome to PF!

Whenever I see examples of the diagram for the Coulomb interaction it always seems to be two electrons interacting via an electron and being repulsed.

(you meant "via a photon" )

"the diagram" ?

There's no such thing as "the diagram" for an interaction … there are infinitely many diagrams for the same interaction, of which that one is simply the simplest.

I was wondering how (for example) a proton - electron interaction, and subsequent attraction, would be drawn? Does the electron still emit a photon in the direction of the proton?

The Feynman diagrams would be the same.

But the virtual photons and virtual electrons in the middle of the diagrams (twice as many electrons as photons in most of the diagrams) don't repel or attract, beacuse they don't exist …

virtual particles don't exist (the clue's in the name ), they're just maths.

 

[Dmitry67]

There difference between virtual and real particles is frame-dependent. Those who deny the existence of virtual particles should also deny the existence of 'real' ones.

 

[tiny-tim]

Hi Dmitry67!

There difference between virtual and real particles is frame-dependent. Those who deny the existence of virtual particles should also deny the existence of 'real' ones.

uhh?

Are you saying that there is a frame in which the virtual photons and virtual electrons in a Feynman diagram for electron-electron repulsion are real?

 

[nrqed]

There difference between virtual and real particles is frame-dependent. Those who deny the existence of virtual particles should also deny the existence of 'real' ones.

What do you mean? A virtual particle is one for which

[tex] P_\mu P^\mu \neq m^2 [/tex]

and this is a frame independent statement.

 

[Dmitry67]

In some accelerated frames they are real. Different accelerated frames don't agree on the number of real and virtual particles of the same macroscopic events. Check Unruh effect for example. Hawking radiation is another one (free falling body does not see that radiation).

My personal opinion: Feynman Diagrams describe what is actually happening. Sum on infinite number of histories. However, as there is no collapse and measurement, there are no 'real' particles, no incoming particles and no particles which go away to be 'registered'. Our world is just one infinite Feynman diagram without borders.

 

[tiny-tim]

In some accelerated frames they are real …

Are you saying that there is an accelerated frame in which the virtual photons and virtual electrons in a Feynman diagram for electron-electron repulsion are real?

 

[Dmitry67]

No, because if they participate in the repulsion, then they go from one particle to another without any absorption -> they could not leave any tracks (there are cases however when virtual particles create some macroscopic effects)

But, on the other hand, claim that 'they are just pure mathematics' is too strong.

 

[tiny-tim]

No, because if they participate in the repulsion, then they go from one particle to another without any absorption -> they could not leave any tracks

Sorry, but that doesn't make any sense … what does "without any absorption" mean, and what does an accelerated frame have to do with it?

 

[Dmitry67]

Ok, I was trying to say that these virtual photons can not be registered directly, but their effects are registered indirectly. But it is not enough to say that they are 'not real' and 'pure mathematics'.

For example, wavefunction itself can't be registered directly. However, in many interpretations wavefunction is real. So the claim 'virtual particles are pure mathematics' for me is equivalent to Copenhagen Interpretation (where it was valid, and as CI was dominating for a long time, people got used to that point of view).

 

[Dickfore]

Feynman diagrams do not sketch a real world situation where particles collide. They are a pictorial way of representing complicated mathematical expressions in terms of a few primitive objects (propagator, vertex, external source).

 

[tiny-tim]

Ok, I was trying to say that these virtual photons can not be registered directly, but their effects are registered indirectly …

hmm … I notice that you're now ignoring the virtual electrons in the same interaction …

there are twice as many virtual electrons as virtual photons in most of the diagrams …

I suspect that you've heard that the electromagnetic field is "mediated by" photons, and so the virtual electrons in the diagrams somehow don't "mediate" it.

Well, as Dickfore says, Feynman diagrams are a pictorial representation of mathematical expressions. Virtual photons and virtual electrons participate in the same way, and since an infinite number of diagrams is needed, so an infinite number of virtual particles is needed … so how many virtual photons and electrons do you think there are when two real electrons repel, and when do they appear?

 

[Dmitry67]

Feynman diagrams do not sketch a real world situation where particles collide. They are a pictorial way of representing complicated mathematical expressions in terms of a few primitive objects (propagator, vertex, external source).

Whats about the repulsion between, say, 2 electrons with not only 1 virtual photon, but including the loop of virtual e+/e-? and (on very short distances) even t+/t- (top quarks)? Or even ?+/?-, where ? is not-yet-discovered supersymmetric particle? Why that 'pure mathematics' mimics the 'real' world so literally?

 

[xepma]

If we had a way of determining quantum amplitudes directly (scattering processes, the ground state of a system or just the propagation of a particle), we would never need the concept of a virtual particle. It's just a mathematical side of perturbation theory. For example, in some integrable quantum field theories (e.g. certain conformal field theories) you do not need perturbation theory so to whole concept of Feynman diagrams and 'intermediate' states is completely circumvented! The concept of a virtual particle (= 'intermediate' state) never pops up!

Perturbation theory, in the end, is nothing but trying to solve an interacting theory using a non-interacting formulation -- completely similar to using a Taylor series to approximate an analytic function. The different terms in the Taylor expansion is similar to the different Feynman diagrams (which contain the virtual particles) in the perturbation series ...

 

[tom.stoer]

The concept of Coulomb force is gauge dependent!

In the Lorentz gauge you will not see a Coulomb potential at all. In Coulomb gauge (that's where it's name is coming from) there is a "photon term" plus a static Coulomb potential term" in the Hamiltonian. So the static Coulomb potential is not due to virtual photons but looks exactly as in electrostatics.

The choice of the gauge is arbitrary, but it should fit to the problem you want to study. For (relativistic) scattering experiments the Lorentz gauge is nice, but e.g. for Lamb shift calculations it's awful.

 

[xepma]

That gauge choice actually named after Ludwig Lorenz (no t!), and not Hendrik Lorentz ;-)

 

[tom.stoer]

correct; sorry for that

 

[mk17]

(you meant "via a photon" )

I did indeed, oops.


Anyway, thanks for the replies everybody, obviously I was thinking Feynman diagrams have a physical significance that they don't.

Eventually, hopefully, I'll be competent enough to understand some of the mathematics behind them and actually be able to use them!

 

[alxm]

The idea of using diagrams for perturbation theory spread to different areas either, so quantum chemistry has e.g. Hugenholtz diagrams and Goldstone diagrams.

 

[Dmitry67]

If we had a way of determining quantum amplitudes directly (scattering processes, the ground state of a system or just the propagation of a particle), we would never need the concept of a virtual particle.

In SM - yes.
But it is just one interpretation.

 

[Dmitry67]

Example:
Kepler law can be used to calculate positions of planets. And it works. However, you can't claim that 'gravity is just pure math used to calculate the orbits' just because the results match the calculations.

Another example: using the same logic you use, you can deny the existence of Gravitational waves, saying that 'they are just pure math used to calculate the effects of distant rotating masses on the macroscopic objects'

 

[tom.stoer]

One could even argue that in reality only virtual particles exist, no real particles! Why? Because if we detect a particle (and this is the only way to see if it's here) the particle is absorbed by the detector and therefore it is an internal line, a propagator, ending at a vertex in the detector.

=> I agree with Dmitry67 that ontology based on mathematical concepts in physics is not always straightforward.

I do not deny that virtual particles may exist - whatever this means - I only want to warn people not to take everything for granted which is written in popular books (we have some threads here discussing the violation of energy conservation due to vacuum fluctuations; a misleading concept).

 

[tiny-tim]

Example:
Kepler law can be used to calculate positions of planets. And it works. However, you can't claim that 'gravity is just pure math used to calculate the orbits' just because the results match the calculations.

Gravity is a physical effect, not a particle.

This effect definitely exists, because it shows itself in the results (to use your word).

The analogy in electromagnetism is that the electromagnetic field is a physical effect which shows itself in the results, and definitely exists, but the virtual particles are only in the maths.

(and the continuation of the gravity example would be that, although gravitons may well exist, virtual gravitons don't exist and aren't "exchanged" in gravitational attraction any more than electrons or photons are in electromagnetic interaction )

Another example: using the same logic you use, you can deny the existence of Gravitational waves, saying that 'they are just pure math used to calculate the effects of distant rotating masses on the macroscopic objects'

No, because gravitational wave theory says that the wave has a definite (ok, slightly fuzzy) position at any time … the theory itself (of gravitation) says that the wave exists: quantum field theory does not say that virtual electrons and photons exist!

A more interesting example is, do vortices (in flowing water) exist? They have definite position (you can even literally pick them up and transfer them to a different stream!), but there's no "vortex particle", nor even a separate state of matter … so we could argue that they're just a mathematical effect (a singuarity) with no physical existence.

However, virtual particles aren't even analogous to vortices …

they don't have a definite position, not even a really fuzzy one.

Nothing in quantum field theory even purports to describe where or when they appear, or in what numbers.

 

[DrDu]

Back to the original question. I think that at the level of drawing a diagramm for an exchange of only one photon really makes no difference between attraction and repulsion. It would correspond to a scattering in the first Born approximation and the angular distribution of the scattered particles would be identical, whether they are of like or unlike charge.
To see a difference between attractive and repulsive interactions, one has to consider also processes where more than one photon is exchanged.

 

[Dmitry67]

This effect definitely exists, because it shows itself in the results (to use your word).

There is absolutely no difference between, say, G-wave affecting the detector, and 2 Casimir plates attracting. In both cases 2 macroscopic objects change their position because of something, not directly detectable.

However, virtual particles aren't even analogous to vortices …
they don't have a definite position, not even a really fuzzy one.

As well as real ones - they don't have definite position too. (You can argue that you can pinpoint their position exactly in the detector, but it is so Copenhagen and obsolete! That process is an interaction of a particle (wave) and the 'particles' of the detector, leading to some macroscopic events, which we interpret as 'measuring the position')

Nothing in quantum field theory even purports to describe where or when they appear, or in what numbers.

Yes, you know that there are 3 quarks in a proton, but you can't say how many gluons are there. But not being able to count something can't deny the physical existence of that something. Do you deny the physical existence of Higgs condensate?

 

[tom.stoer]

Yes, you know that there are 3 quarks in a proton, but you can't say how many gluons are there.

No, not even three quarks is correct according to deep inelastic scattering results for nucleon structure functions.

The question is about the ontologically "reality" of mathematical concepts like
wave functions, state vectors, operators (gauge invariant - not gauge invariant, hermitian = observables, unitary ones), matrix elements, propagators, path integrals, currents and charges, energy-momentum density, energy (as an integral over three-space it is not defined in GR ...), ...

 

[tiny-tim]

There is absolutely no difference between, say, G-wave affecting the detector, and 2 Casimir plates attracting. In both cases 2 macroscopic objects change their position because of something, not directly detectable.

That's not true … the detector sits there until a "G-wave" goes past, it wiggles a little, then sits there for a long time again, then wiggles when another G-wave goes past, and so on. The G-wave has a definite position, and follows a definite course, at a definite time (and would be detected by other detectors at appropriate positions and times).

But in the 2 Casimir plates case, what is being detected? Only a field! It doesn't go anywhere, there isn't a time when it appears or disappears … yes, the field exists, but if you're claiming that virtual particles exist between the plates (you don't actually say), then when do they appear and disappear? what velocity do they have, and where do they go to and from? how many are there? … none of these questions is answered (or even asked ) by the theory!

As well as real ones - they don't have definite position too. (You can argue that you can pinpoint their position exactly in the detector, but it is so Copenhagen and obsolete! That process is an interaction of a particle (wave) and the 'particles' of the detector, leading to some macroscopic events, which we interpret as 'measuring the position')

Oh, come off it! … a real electron is (for example) ejected from one specific atom, follows a specific course, and is absorbed at another specific atom, all at specific (though slightly "fuzzy") times, and the number of electrons is also specific.

The disappearance of the electron from one atom, its travel, and its reappearance at another atom, are not "just maths".

ok, you can construct a theory in which a philosopher could argue that the electron doesn't have a position etc, but there are perfectly good theories in which the electron obviously does have a position, and there is a definite number of electrons … but there are no theories in which the virtual particles in an interaction can be said to have a position or any other property, including number …

the only theories that there are introduce virtual particles as a mathematical trick, and do not even purport to say anything about their position, or when or where they are created and destroyed.

Yes, you know that there are 3 quarks in a proton, but you can't say how many gluons are there. But not being able to count something can't deny the physical existence of that something. Do you deny the physical existence of Higgs condensate?

There are no real gluons in a proton. And virtual gluons have the same status in the strong interaction as virtual photons and electrons have in the electromagnetic interaction … just maths.

(And the http://en.wikipedia.org/wiki/Higgs_mechanism" is only a field, isn't it? )

 

[Dmitry67]

"not even 3 quarks" - could you provide links for further reading?
I am very curious.

 

[tom.stoer]

Of course.

It's rather simple: you have exactly three quarks provided that you treat them as "frozen". In lattice gauge theory this is called "quenched approximation".

But in QFT you must take both "virtual" gluons and "virtual" quarks into accounts. The particle number operator fpr quarks does not commute with the Hamiltonian and therefore the number of quarks in an energy eigenstate is no longer defined (only its expectation value).

In deep inelastic scattering one probes the scaling limit of QCD, that means one looks at "asymptotically-free" quarks interacting with a single electron. But the quarks are only treated as asymptotically-free during the interaction with the electron; they have some "residual knowledge" that they were bound in a nucleon. This is described by so-called structure functions telling you which fraction of the nucleon momentum was carried by the single quark interacting with the electron. This fraction is not equal to 1/3!

Try http://www.phy.uct.ac.za/courses/phy400w/particle/dis.pdf which has an excellent drawing on page 35 showing the difference between "3 quarks inside the proton" and "? quarks inside the proton".,

 

[Dmitry67]

That's not true … the detector sits there until a "G-wave" goes past, it wiggles a little, then sits there for a long time again, then wiggles when another G-wave goes past, and so on. The G-wave has a definite position, and follows a definite course, at a definite time (and would be detected by other detectors at appropriate positions and times).

Is it your definition or reality?
To have definite position and momentum?
It is a definition of macroscopic reality, not the reality in general

Oh, come off it! … a real electron is (for example) ejected from one specific atom, follows a specific course, and is absorbed at another specific atom, all at specific (though slightly "fuzzy") times, and the number of electrons is also specific.

At first, even 'real' electron does not have an exact trajectory. We can see its track in the camera just because we are macroscopically very huge.

Then, electron are not 'absorbed by specific atom'. Electron wave hits all atoms on an absorber, putting it into superposition of billions of states. These states decohere into YOU-observing-spot-on-some-specific-place.

ok, you can construct a theory in which a philosopher could argue that the electron doesn't have a position etc, but there are perfectly good theories in which the electron obviously does have a position, and there is a definite number of electrons … but there are no theories in which the virtual particles in an interaction can be said to have a position or any other property, including number …

In non-collapse theories electron does not have a position because it is a wave, and particle behavior is an illusion created by quantum decoherence. There are other interpretations where it is not true. That is what I was saying: claim 'it is just a math' is interpretation-dependent.

(And the http://en.wikipedia.org/wiki/Higgs_mechanism" is only a field, isn't it? )

It is not only a field. On high temperature condensate decays into 'real' Higgs particles. The same as Bose-Eistein condensate. Do you agrue that when Rubidium atoms form BEC, they stop being 'real', because they no longer have well-defined position, momentum and even number?

 

[Dmitry67]

tom, thank you, interesting. The number of additional quark-antiquark pairs participating in very short pocesses is not a surprise at all. I hope tiny-tim can tell 'real' quarks from 'virtual' ones inside the proton :)

 

[tom.stoer]

... but there are perfectly good theories in which the electron obviously does have a position, and there is a definite number of electrons …

Can you tell me which theory you are talking about?

Can you tell me how to proof that the number of electrons in QED is fixed? It will not work, simply because the operator counting the number of electrons does not commute with the Hamiltonian of QED.

 

[reilly]

The concept of Coulomb force is gauge dependent!

In the Lorentz gauge you will not see a Coulomb potential at all. In Coulomb gauge (that's where it's name is coming from) there is a "photon term" plus a static Coulomb potential term" in the Hamiltonian. So the static Coulomb potential is not due to virtual photons but looks exactly as in electrostatics.

The choice of the gauge is arbitrary, but it should fit to the problem you want to study. For (relativistic) scattering experiments the Lorentz gauge is nice, but e.g. for Lamb shift calculations it's awful.

>>>>>>>>>>>>>>>>>>>>>>>>>>

Not true at all. E&M allows various gauges for potentials, but not for forces. Now, experiment demonstrates that charged particles, moving slowly, exert Coulomb forces on each other. Thus, no matter what gauge is involved, there must be a Coulomb potential to generate this force -- as is demonstrated in potential theory, and, for that matter, in freshman physics. Any relativistic theory of EM regardless of gauge, classical or quantum, must have a NR limit of Coulomb interactions. And, it thus becomes evident, a single photon exchange generates this interaction -- see Yukawa's explanation of meson exchange as a generator of nuclear forces.

Note also, that a charged particle exerts only a Coulomb interaction in its rest frame. However, the exact solutions, Coulomb wave functions for an NR electron scattering in a Coulomb field, don't indicate anything like photon exchange. So, ...

Virtual particles real? Well, consider the photoelectric effect. Clearly, the ejected electron is not on its mass-shell when inside the metallic target. Yet, who would say that this electron does not exist prior to ejection?

Regards,
Reilly Atkinson