# Elementary processes in Feynman Diagrams

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## Main Question or Discussion Point

Hello there.

I'm attending an introductory course in particle physics. We're supposed to know how to draw first-order tree level Feynman diagrams for simple reactions.
I've been struggling to understand the method I should follow in order to correctly draw them.
As I understand it now, we can combine the elementary vertexes of each interaction to get any possible diagram.

I attached my pdf file with my notes, with the questions clearly identified in boxes - everything inside a box is a question.

If you could please answer some of those and perhaps see if I got the concepts right, I'd be very grateful.

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mfb
Mentor
There are things that cannot be calculated with perturbation theory, and Feynman diagrams are a graphical way to display perturbative calculations. There are things that do not have a proper Feynman diagram.

Mesons are not fermions.

> Are there variants other than rotations?
Depends on what exactly you count as rotation, but as long as that vertex is present it is fine.

> In this case, the gluon will be blue and anti-red
If you want to assign colors like that, the gluon should also have a clear time-direction in the diagram (here: outgoing). This also applies to the W discussed later.

You are missing the gluon self-interaction and the vector boson couplings.

Diagram (7) is not the same as (6), but if you mirror/rotate it in time and then replace particles by antiparticles you get 7 from 6.

Why is there no tree-level neutral flavor changing process? That is a very good question, we just do not observe it.

Energy and momentum are conserved at every vertex.

> Is isospin, parity, charge conjugation and avor conserved in QCD vertexes??
Yes.

• Xico Sim
• By rotation I meant only proper rotations... Reflections are also fine? If they are, do the (proper+improper) rotations of the drawn vertex generate all possible vertexes? And what do you mean by "as long as the vertex is present"?
• About the arrows in the gluon and the W bosons: I know, but I couldn't get Lyx to render the arrows, I don't know why. I'll fix that later.
• I thought the self-interactions were unimportant for 1st order diagrams...
• Energy and momentum are conserved at every vertex? But if we consider the elementary QED vertex and consider the CM reference frame, we have $E_i = mc^2$ but $E_f=mc^2+E_\gamma$, meaning that $E_i \neq E_f$ Besides, there are problems with conservation of momentum too if we consider the vertex corresponding to the electron-positron annihilation, obtained by rotating the elementary QED vertex. Griffiths says something about this too, and so I understood that energy and momentum are not conserved when virtual particles are involved (in single vertexes), but only when we talk about whole processes... am I wrong
Thank you for your answer, by the way. I'm learning a lot!

mfb
Mentor
By rotation I meant only proper rotations... Reflections are also fine?
Up and down do not have a meaning, so a reflection is the same as a rotation by 180 degrees.
And what do you mean by "as long as the vertex is present"?
2 fermions and a photon coupling to each other, with conserved quantum numbers. It does not matter in which direction the individual legs point.
I thought the self-interactions were unimportant for 1st order diagrams...
1st order of what in what? For fermion interactions, yes. For boson interactions, no.

• Energy and momentum are conserved at every vertex? But if we consider the elementary QED vertex and consider the CM reference frame, we have $E_i = mc^2$ but $E_f=mc^2+E_\gamma$, meaning that $E_i \neq E_f$ Besides, there are problems with conservation of momentum too if we consider the vertex corresponding to the electron-positron annihilation, obtained by rotating the elementary QED vertex. Griffiths says something about this too, and so I understood that energy and momentum are not conserved when virtual particles are involved (in single vertexes), but only when we talk about whole processes... am I wrong
Well, it tells you that this process on its own is not physical, you need a second vertex to produce an actual interaction. At least one of the particles won't satisfy the usual energy/momentum relation. Electron-positron annihilation cannot go to a single photon for the same reason. Add another photon at a second vertex and you can conserve energy and momentum at both vertices.

Up and down do not have a meaning, so a reflection is the same as a rotation by 180 degrees.
Fair enough.

2 fermions and a photon coupling to each other, with conserved quantum numbers. It does not matter in which direction the individual legs point.
Ok.

1st order of what in what? For fermion interactions, yes. For boson interactions, no.
I'm not sure: in my course we don't really get to understand what the Feynman Diagrams are, so it's all very superficial. Sadly, I must (for now) just find a recipe to follow.

Add another photon at a second vertex and you can conserve energy and momentum at both vertices.
As I understand, the total process does conserve energy and momentum, but not each vertex. I did not understand how what you said goes against that.

I will post an update of my pdf now. I think I got all EM vertexes right.

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mfb
Mentor
Energy and momentum are conserved exactly at each vertex.
I think I got all EM vertexes right.
Don't think of them as 6 separate things, they are the same interaction.

ChrisVer
Gold Member
As I understand, the total process does conserve energy and momentum, but not each vertex. I did not understand how what you said goes against that.
The energy and momentum have to be conserved exactly at each vertex....
At the vertex of the lower right Fig9 of your plot for example, you can say that $\bar{a}$ has energy and momentum $E_1,\vec{p}_1$ and the $a$ has $E_2, \vec{p}_2$.... then the photon you write should have energy and momentum $E_\gamma = E_1+E_2$ and momentum $\vec{p}_\gamma = \vec{p}_1 + \vec{p}_2$.
This is energy/momentum conservation at each vertex...
What may be true though is that the photon is off-shell, meaning that it doesn't obey the $E^2_\gamma - | \vec{p}_\gamma|^2 = m^2_\gamma = 0$ relation... In that case it can't appear as a physical particle in the final state but needs to be connected to another vertex.

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• Xico Sim
The energy and momentum have to be conserved exactly at each vertex....
At the vertex of the lower right Fig9 of your plot for example, you can say that $\bar{a}$ has energy and momentum $E_1,\vec{p}_1$ and the $a$ has $E_2, \vec{p}_2$.... then the photon you write should have energy and momentum $E_\gamma = E_1+E_2$ and momentum $\vec{p}_\gamma = \vec{p}_1 + \vec{p}_2$.
This is energy/momentum conservation at each vertex...
What may be true though is that the photon is off-shell, meaning that it doesn't obey the $E^2_\gamma - | \vec{p}_\gamma|^2 = m^2_\gamma = 0$ relation... In that case it can't appear as a physical particle in the final state but needs to be connected to another vertex.
But how can momentum be conserved in the process I mentioned above? (Electron goes to electron plus photon)

ChrisVer
Gold Member
Point 1 to your above mentioned process: the $E_\gamma =0$ allows the conservation of energy and momentum. Meaning that the photon carries no energy but momentum...
Point 2 to your above process: the photon in that case is off-shell, as mfb said it doesn't satisfy the known energy/momentum relation $E_\gamma^2 - | \vec{p}_\gamma|^2 =m^2_\gamma =0$.. This means that you don't observe it but it has to be connected to some other vertex to have the complete interaction...If you do so then of course it can happen, an electron scattering off some other charged particle is done with such a vertex.

• Xico Sim
Point 1 to your above mentioned process: the $E_\gamma =0$ allows the conservation of energy and momentum. Meaning that the photon carries no energy but momentum...
Point 2 to your above process: the photon in that case is off-shell, as mfb said it doesn't satisfy the known energy/momentum relation $E_\gamma^2 - | \vec{p}_\gamma|^2 =m^2_\gamma =0$.. This means that you don't observe it but it has to be connected to some other vertex to have the complete interaction...If you do so then of course it can happen, an electron scattering off some other charged particle is done with such a vertex.
So we can think of virtual partiicles as having no energy? Always?
Even so, the conservation of momentum in my example is still impossible, right?

ChrisVer
Gold Member
So we can think of virtual partiicles as having no energy? Always?
No. Virtual particles are particles that donnot obey the energy/momentum relation... they can have 0 energy and momentum, they can have energy and 0 momentum, and of course they can have both energy and momentum but their combination/4momentum squared is not equal to the physical particle's mass. The particle cannot be observed and has to be connected to another vertex... by the way, a vertex itself hardly ever shows an "interaction".

Even so, the conservation of momentum in my example is still impossible, right?
Why? In fact it's not right, since by construction vertices conserve both energy and momentum.
May I also ask which phrase you misinterpret as stating that momentum or energy are not conserved at a vertex?

mfb
Mentor
But how can momentum be conserved in the process I mentioned above? (Electron goes to electron plus photon)
That process on its own is not physical - it does not happen.

Let me try to me clear, here.

I know that the process "Electron goes to electron plus photon" does not happen. In fact, it couldn't happen, it does not conserve momentum or energy.
This is precisely the reason why I'm saying that vertexes don't conserve energy or momentum: this example corresponds to a vertex but violates conservation of energy and conservation of momentum!

May I also ask which phrase you misinterpret as stating that momentum or energy are not conserved at a vertex?
Not exactly one phrase, but this whole argument: all I was saying was:
Look! Here is a vertex which represents a process which does not conserve energy or momentum! It doesn't matter if this process is physical or not: I'm only asking if p and E are conserved, and the answer is "no".
Let me put this differently: In the afore-mentioned process (and in the CM frame), the total initial energy is mec². The final energy is $m_{e}c^2+E_\gamma$. Therefore, E is conserved is conserved iff $E_\gamma=0$. Momentum conservation raises no problems.
About the process $e^- + e^+ \rightarrow \gamma$: Energy conservation raises no problems. Momentum is conserved iff $p_e=0$ and $p_\gamma=0$.

That said, I know realised that the first process conserves energy only if the photon has no energy, whereas the second one conserves momentum only if the photon has no momentum. Are you saying that, for some reason, this is allowed?

mfb
Mentor
This is precisely the reason why I'm saying that vertexes don't conserve energy or momentum: this example corresponds to a vertex but violates conservation of energy and conservation of momentum!
It would have to violate it, if that vertex would be a physical process. This vertex is not a physical process - your argument about outgoing particles does not apply, because at least one of those lines does not correspond to an actual particle, and the energy/momentum relation for that line will not follow the usual relation from special relativity.

ChrisVer
Gold Member
That said, I know realised that the first process conserves energy only if the photon has no energy, whereas the second one conserves momentum only if the photon has no momentum. Are you saying that, for some reason, this is allowed?
In the center of mass frame of the electron/positron you don't have to impose $p_e=0$.. just for the photon $p_\gamma=0$ is enough.... In any other reference frame, the photon can still have other momenta $\vec{p}_\gamma= \vec{p}_{e^-} + \vec{p}_{e^+}$...
I am not also saying it is allowed, I am saying that by construction [and what mathematically vertices represent] energy/momentum are conserved at each vertex.... you would have a non-conservation if you couldn't deduce the energy/momentum of the photon by the energy/momenta of the electron and positron... Whether then the process is possible or not, depends on whether your "final state particle" is on-shell (real) or not (virtual).... A virtual particle cannot be a final state because they don't get observed.... Especially depending on how off-shell the particle is, the "faster" (if that would make sense) it decays...If you get pretty close to the particle's mass ($E^2_x - p^2_x \approx m_x^2$, then it can also act as a resonance...
For example the $e^- e^+ \rightarrow \gamma$ is not observable... the $e^- e^+ \rightarrow \gamma \gamma$ is.
The interaction $e^- e^- \rightarrow \gamma \rightarrow \mu^- \mu^+$ is allowed (given that the electrons have enough energy to produce the heavier muons) and in both vertices ($e^-e^+\gamma$ and $\mu^- \mu^+ \gamma$) separately the energy/momentum is conserved.

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vanhees71
Gold Member
2019 Award
Let me try to me clear, here.

I know that the process "Electron goes to electron plus photon" does not happen. In fact, it couldn't happen, it does not conserve momentum or energy.
This is precisely the reason why I'm saying that vertexes don't conserve energy or momentum: this example corresponds to a vertex but violates conservation of energy and conservation of momentum!
I think a lot of the problems people have with QFT and Feynman diagrams is that they got from some popular-science sources the wrong idea that Feynman diagrams are pictures for processes in the real world. In a certain sense it's true, but only in a certain sense!

Feynman diagrams are not pictures of how physicists think that scattering processes really happen but they are symbols for mathematical expressions to calculate S-matrix elements. E.g., the QED vertex (electron-positron-photon vertex) stands for $\mathrm{i} e \gamma^{\mu}$ (the $\mu$ is the Lorentz index carried by the photon line).

Then there are rules to define the meaning of the Feynman diagrams when you evaluate an S-matrix element, i.e., transition-rate amplitudes for the real processes observable in nature. Let's now just take the above mentioned vertex and consider it as a real processes. Drawing the photon line pointing down and the electron and positron lines pointing up, this would be the S-matrix element for the decay of a photon to an electron-positron pair (time conventionally running from bottom to top). Now the Feynman rules tell you that external lines stand for wave functions of asymptotic free particles, i.e., the momenta of external lines must fulfill the on-shell condition $p \cdot p=m^2$ for each of the lines with $m$ being the invariant mass of the corresponding particle. Further on each vertex energy-momentum conservation must be fulfilled. Analysing this with the "photon decay" we consider here, you'll see that you cannot fulfill the on-shell conditions together with energy-momentum conservation. That's understandable, because for a particle to decay into two other particles it must have a mass bigger than the sum of the masses of the decay products ("threshold" for decay). Since the photon mass is 0, it cannot decay to anything. So also interpreting the elementary vertex as a real process leads to the correct result that a photon cannot decay into an electron-positron pair. Of course the same holds if you arrange the diagram to describe the decay of an electron to a positron and a photon or a positron decaying into an electron and a photon. For none of these processes you can fulfill the kinematical constraints (on-shell conditions and energy-momentum conservation), and thus all these processes do not happen.