# Particles as Force Carriers - How Does It Work?

1. Jun 17, 2012

### RomanL

I have long been mystified by how particles act as force carriers. Apparently, the exchange of force-carrier particles (or virtual particles?) causes force action. OK, but how does this work?

1) Some forces cause attraction and some repulsion (and others decay). How can an exchange of the same type of particle cause both repulsion (e.g. for particles with a negative charge) and attraction (e.g. for a particle with a negative charge and a particle with a positive charge)?

2) How come photons carry both electric and magnetic forces?

3) If photons are electromagnetic force carriers, why do sources of photons (e.g. a domestic lamp or an X-ray machine) not cause magnetic/electric attraction or repulsion? I mean, shining a torch on an iron bar will not move it closer nor push it further away...

4) What's the deal with virtual particles and do they really exist if they are ' virtual'? How do they relate to real particles?

2. Jun 17, 2012

### tiny-tim

Hi RomanL!
they don't

virtual particles are just a mathematical device used in perturbation expansions of the S-operator (transition matrix) of an interaction in quantum field theory

3. Jun 17, 2012

### Bob S

Quote from Romani-
Does the off-the-mass-shell electron in the Feynman diagram for Compton scattering that connects the two electron-photon-electron scattering vertices carry anything besides charge?

4. Jun 18, 2012

### tiny-tim

"the Feynman diagram"?

there are infinitely many Feynman diagrams for a compton scattering of one electron and one photon

only the two lowest-order diagrams (the "H" diagram and the "stick-man" diagram, see eg http://upload.wikimedia.org/wikipedia/commons/8/8a/Compton_Scattering.svg) have only one virtual electron

4-momentum and charge are conserved at each vertex (i don't know about spin ), so yes, you could say that the electron carries 4-momentum, if that helps you to understand, or to remember, the maths

5. Jun 19, 2012

### RomanL

Hmm, that's interesting. Yet, I still feel that I am missing something. For example, the scientists keep on searching for the Higgs Boson/graviton (I am not sure if they are the same particle or not) that is supposed to be a force carrier for the gravitational force. If it is just a mathematical device, why search for this particle experimentally? I know you said that virtual particles (not actual particles) are a mathematical device, but then the real particles are not actually force carriers?

6. Jun 19, 2012

### cosmik debris

I think that you are thinking of particles as little solid chunks. In Quantum Field theory they are excitations of an underlying field, not little bullets. You can measure these excitations. You should not expect these ideas to correspond to your ideas of the "normal" world.

7. Jun 19, 2012

### tiny-tim

Hi RomanL! smile:
the virtual carrier boson has a fixed mass m associated with it

although it is "off-mass-shell", so that its 4-momentum q can be anything, it has a propagator q2 - m2 + iε, for fixed m

from this we deduce that there must be a real particle of mass m

8. Jun 19, 2012

### K^2

So are real particles. Particle representation of fields is just a convenient way to represent properties of linear fields in an intuitive way. So while the answer, "The particles aren't real" is a correct answer, it's not a useful one.

@OP.

1) Because a virtual particle can in principle carry any amount of momentum. It doesn't have to propagate in the same direction as momentum it carries. So simply because it moves to the right doesn't mean it can't transfer quantity of motion to the left.

2) Electric and magnetic fields are really parts of the same field. It just happens to be convenient to describe it as two separate components.

3) They do! If you shine a light on something, you do push it away. It just takes a LOT of photons to provide any significant push. Oh, and since these are real photons, they can only push, not pull.

4) Well, that's the whole thing. Neither real nor virtual particles are more real than the other. They are just quantizations of appropriate fields. There are a whole bunch of reasons why we treat the two differently, but it gets a little complicated. I was going to recommend tiny-tim's link, but it seems a little iffy. I'll have to take a closer look at it later. Try Wikipedia to get you started.

9. Jun 20, 2012

### tiny-tim

surely the big distinction between real and virtual particles is that (while virtual particles are only in the maths) we can make things out of real particles … the earth, the stars, experimental apparatus, etc?

(and if you say, we can't make an electric field out of real particles, i reply that the energy of an electric field isn't the sum of the energies of virtual particles, unlike the energy of matter, which is the sum of the energy of the individual particles (ok, i admit, ± some "binding energy" … but at least adding the energies is possible, and gets us close ))

(and if we claim that fields are made out of virtual particles, then we're making the claim not only about virtual bosons, but also virtual fermions … there are roughly twice as many virtual electrons as virtual photons in the expression of the electromagnetic interaction)

but how does it get from one (real) particle to the other?

there are infinitely many virtual particles being "exchanged", and most of them have velocities that take them nowhere near the "target" particle (either in space or time)

(or do the off-mass-shell virtual particles "target" each other, as in a fluid? with about half of them going backwards in time?)

in the maths, real particles have creation and annihilation operators, but off-mass-shell virtual particles don't

10. Jun 20, 2012

### Staff: Mentor

If you look closer (up to the level where you apply quantum field theory to the particles), you can describe all stuff as a big collection of virtual particles in some way. The transition to macroscopic collections then usually depends on the interpretation of QM.

Virtual particles are objects which are produced and annihilated within the interaction, and it is usually implied that the interaction is negligible "before" and "after" the process. If that is not true, you cannot simply stop your evaluation at some point.
Real particles are just particles which live longer than your calculation.

11. Jun 20, 2012

### tiny-tim

are you talking about off-mass-shell virtual particles?

i don't see any off-mass-shell virtual particles being created or annihilated in the maths

i do see creation and annihilation operators for real particles

where in the maths are the creation or annihilation operators for off-mass-shell virtual particles?

12. Jun 20, 2012

### K^2

More than half of the mass of protons and neutrons are virtual particles. That means more than half of the "stuff" you are building "things" with is virtual particles.

You cannot emit a virtual particle if it isn't absorbed somewhere. It would violate momentum conservation. It's the same reason you can't excite an electron in an orbital with the wrong energy photon. So you do only have virtual particles whose propagator takes them from one particle to another. Neither of these has to be real, of course. Virtual particles can emit and absorb more virtual particles no different from real particles.

Of course they do. They are the same operators. The difference is when you have real particles created/destroyed the momentum is conserved. Virtual particles do not conserve energy or momentum, and so they have to be integrated out. You cannot have the creator/annihilator operators of these particle acting on final states, in other words. But you can (and do) have virtual particle creator/annihilator operators sandwiched in you interaction.

Simple example from QED. Take the QED Lagrangian, sandwich it between <x| and |y> states for a fermion, and take a look at what you can get in between. That's your basic propagator, of course. At zero order, you'll get just S(y-x), which is a creator-annihilator pair. At second, you'll get a one-photon exchange. At fourth, you can finally have a connected fermion loop. Take a look at the propagators for it. They are the same creator-annihilator pairs as the real fermion propagator. But for these, you integrate over all possible momenta. So the fermions in the fermion loop are entirely off shell. Yet, they have the same creator/annihilator operators, and they absorb/emit the same virtual photon that interacts with your real electron. And oh, by the way, between virtual photon emission/absorption, you don't have to have a real electron at all, since the photon momentum is also integrated over. That means, the only real electron propagators are the ones that take you to and from the <x| and |y> states. So there will be at most two pairs of creator/annihilator operators actually acting on your real particles. All the rest will be acting on something that's virtual.

13. Jun 22, 2012

### tiny-tim

are we talking about the same virtual particles, ie off-mass-shell?

off-mass-shell virtual particles do conserve energy and momentum … the hamiltonian contains delta functions which do precisely that … those delta functions ensure that no off-mass-shell virtual particle is used unless it is in a term in which they are conserved

(and even on-mass shell virtual particles conserve (3-)momentum, for the same reason)

and aren't the creation and annihilation operators always written as functions of p, the 3-momentum of the on-mass shell virtual particles (whose energy is √(m2 + p2)), not q, the 4-momentum (energy and momentum) of the off-mass shell virtual particles?

ie no creation or annihilation operator for any off-mass shell virtual particle appears in the maths?​

14. Jun 22, 2012

### K^2

You are right about momentum conservation. You do pick up extra delta functions for each vertex when you do Fourier transform to momentum space. And yeah, it makes sense in retrospect. Not sure what I was thinking.

An off shell particle still can't propagate freely. So it still has to be absorbed somewhere. But I'll have to think up a more convincing argument for this now.

As far as creators/annihilators, I think you're getting your QFT and RQFT mixed up.

15. Jun 22, 2012

### tiny-tim

oooh, i didn't realise there was a difference

which one am i using?

16. Jun 22, 2012

### K^2

In non-relativistic QFT, time and spacial coordinates are often treated as entirely different. Metric doesn't even come up. In RQFT, a partial derivative will always be a 4 derivative. So if you have a derivative of momentum, it's a derivative of 4-momentum. Otherwise, what you are writing doesn't really have a chance of being Lorentz invariant, and Lagrangian ought to be.

17. Jun 22, 2012

### tiny-tim

i'm confused

the QFT i'm used to is from weinberg, "the quantum theory of fields" volume I, particularly chapter 6, viewable free at http://books.google.co.uk/books?id=h9kR4bmCPIUC&pg=RA1-PR19

i call it QFT, but it's lorentz invariant throughout, so i suppose it could be RQFT

from the start of Chapter 6 (p.259):
In previous chapters, the use of covariant free fields in the construction of the Hamiltonian density has been motivated by the requirement that the S-matrix satisfy Lorentz invariance and cluster decomposition conditions. With the Hamiltonian density constructed in this way, it makes no difference which form of perturbation theory we use to calculate the S-matrix; the results will automatically these invariance and clustering conditions in each order in the interaction density.

Nevertheless, there are obvious practical advantages in using a version of perturbation theory in which the Lorentz invariance and cluster decomposition properties of the S-matrix are kept manifest at every stage in the calculation. This was not true for the perturbation theory used in the 1930s, now known as 'old-fashioned perturbation theory', described at the beginning of section 3.5.

The great achievement of Feynman, Schwinger, and Tomonaga in the late 1940s was to develop perturbative techniques for calculating the S-matrix, in which Lorentz invariance and cluster decomposition properties are transparent throughout.​

18. Jun 22, 2012

### K^2

Dirac is RQFT, of course. And it's all 4-momentum. Where did you see any 3-momentum operators in that theory?

Where non-relativistic QFT shows up is condensed matter. There, you would often just use Shroedinger equation, but with second-quantized fields. I don't deal with these, though. So I'm a bad person to ask about any details.

19. Jun 22, 2012

### tiny-tim

ah, weinberg just calls it QFT

(i suspect it's generally assumed that everything is "R", so there's no need to say it)
i don't

i see creation and annihilation operators which are functions of the 3-momentum p

i see them, for example, as a(p,σ,n) throughout §6.1 of weinberg (pp.259-274)

i do not see any creation and annihilation operators in the form a(q,σ,n), as functions of the 4-momentum q

nor do i see any creation and annihilation operators at all in §6.3

(which is why i say there are no creation or annihilation operators for off-mass-shell virtual particles in the maths)

§6.1 deals with the coordinate-space representation of the S-matrix, in which only 3-momentum is conserved at each vertex …

by the end of §6.1, the creation and annihilation operators have been eliminated, resulting in a non-operator formula which, however, is a function of ps, and is not lorentz invariant​

§6.3 deals with the momentum-space representation of the S-matrix, in which 4-momentum is conserved at each vertex …

by defining an extra variable, the non-operator formula from §6.1 is made lorentz invariant, resulting in a non-operator formula which is a function of qs, and is lorentz invariant​

so i say that coordinate-space (ie feynman diagrams of the 1st type) has on-mass-shell 4-momentums, and creation and annihilation operators for (on-mass-shell virtual) particles with those 4-momentums,

but momentum-space (ie feynman diagrams of the 2nd type) has off-mass-shell 4-momentums, but no creation or annihilation operators at all

ie off-mass-shell virtual particles don't even exist in the maths, because the maths doesn't include any creation or annihilation operators for them