# Gauge Bosons

Forces are mediated by gauge bosons.
If, for instance, you look at a collision between 2 electrons, is there a definite/calculable number of photons exchanged? Do they have a particular frequency? Do we expect the same thing of gravitons? Does that mean the action of forces is quantized?

Or are these not valid questions?

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Fredrik
Staff Emeritus
Gold Member
Given the state of the two electrons before the interaction, you can calculate the probability of a particular state after the interaction. The probability is calculated by expanding the mathematical expression that represents the probability amplitude into a series, and then the terms of the series are calculated one at a time. The first term will give give you a rough idea what the amplitude is. The second will be a correction to the first, and so on. The first term can be represented by a Feynman diagram with two vertices (i.e. describing the exchange of one photon). The second term can be represented by a Feynman diagram with four vertices (describing the exchange of two photons). The third has six vertices (three photons), and so on.

So to get the correct amplitude you have to add up contributions from the exchange of every positive integer number of photons. Fortunately, it's usually sufficient to only include a few terms in the calculations.

They don't have a particular frequency. Actually, inside each term, you integrate over all photon momenta (which means that you include all frequencies).

A probability amplitude is a complex number such that when you take the square of its magnitude you get a probability.

This method doesn't work for gravitons, because the result when you calculate any of the terms is infinity. (Hm, maybe the first one is finite, but only tells us exactly the same thing as classical general relativity. I don't remember how it is). This is why people are working so hard to find a quantum theory of gravity that isn't a quantum field theory.

Last edited:
Given the state of the two electrons before the interaction, you can calculate the probability of a particular state after the interaction. The probability is calculated by expanding the mathematical expression that represents the probability amplitude into a series, and then the terms of the series are calculated one at a time. The first term will give give you a rough idea what the amplitude is. The second will be a correction to the first, and so on. The first term can be represented by a Feynman diagram with two vertices (i.e. describing the exchange of one photon). The second term can be represented by a Feynman diagram with four vertices (describing the exchange of two photons). The third has six vertices (three photons), and so on.

So to get the correct amplitude you have to add up contributions from the exchange of every positive integer number of photons. Fortunately, it's usually sufficient to only include a few terms in the calculations.

They don't have a particular frequency. Actually, inside each term, you integrate over all photon momenta (which means that you include all frequencies).

A probability amplitude is a complex number such that when you take the square of its magnitude you get a probability.

This method doesn't work for gravitons, because the result when you calculate any of the terms is infinity. (Hm, maybe the first one is finite, but only tells us exactly the same thing as classical general relativity. I don't remember how it is). This is why people are working so hard to find a quantum theory of gravity that isn't a quantum field theory.

This is a clear and detail explanation, but there seems to be a minor problem. (Please correct me if I were wrong, thx)

The momentum of photon or photons in the tree level diagram are fixed.
Only those diagram contain loops have independent momentum, hence we
have to integrate over all possible momentum. This is also the origin of the divergence in field theory.

Fredrik
Staff Emeritus
Gold Member
I think you're right. It's been a while since I actually did one of these calculations, so I forgot about that.

nrqed
Homework Helper
Gold Member
This is a clear and detail explanation, but there seems to be a minor problem. (Please correct me if I were wrong, thx)

The momentum of photon or photons in the tree level diagram are fixed.
Only those diagram contain loops have independent momentum, hence we
have to integrate over all possible momentum. This is also the origin of the divergence in field theory.
yes, this is correct. To the OP: the energy
and the three-momentum of the exchanged photon in a
tree-level diagram is fixed by conservation of four-momentum at the vertices.
Note that the relation E = c p that we usually have for photons is not valid here (we say that the exchnaged photon is off-shell when this relation is not valid).

CarlB
Homework Helper
Only those diagram contain loops have independent momentum, hence we have to integrate over all possible momentum. This is also the origin of the divergence in field theory.
It's interesting that in 't Hooft's introduction to QFT, he says that the difference between diagrams for classical and quantum physics is that classical does not have loops:

THE CONCEPTUAL BASIS OF QUANTUM FIELD THEORY
http://www.phys.uu.nl/~thooft/lectures/basisqft.pdf

Fredrik, awesome explanation thank you.
Can you give any qualitative explanation of why its valid to include all photon momenta (for each photon)? and why ever number of mediating photons?

For a particular interaction, can measurements be made to determine the exact number/frequency of photons exchanged?

Fredrik, awesome explanation thank you.
Can you give any qualitative explanation of why its valid to include all photon momenta (for each photon)? and why ever number of mediating photons?

For a particular interaction, can measurements be made to determine the exact number/frequency of photons exchanged?
I try to give an explanation. (Everyone is welcome to correct me or discuss with me~)

First of all, we cannot determine the intermediate momenta only for those diagrams contain loops, i.e. the momenta flow in loops can be arbitrary. Moreover, one can show that loop expansion for a perturbation theory is equivalent to the \hbar expansion.

This tells us that the quantum effect allows the intermediate particles to be off-shell. From the view point of uncertainty principle, since we include all possible momenta which flowing in loops, we pretend that we know all the physics up to arbitrary high energy actually. However, QFT is just an effective theory of some unknown fundamental theory, hence the calculation of loop diagrams always gives you infinity. However, if the theory is not sensitive to high energy physics, we can cure this divergence by some tricky process called reonormalisation process, and this kind of theories are called renormalizable.

For your second question, I guess this is a no-can-do. Experimentally, what we measured are usually the cross sections for some scattering processes. A cross section can be calculated perturbatively by summing over diagrams. In each loop diagram, all possible momentum of intermediate particles contributes to the cross section.