Is the Charge Conservation Correct in This Meson Decay Feynman Diagram?

[songoku]

Homework Statement

Please see below

Relevant Equations

Conservation of charge
Conservation of baryon number
Conservation of lepton number

I have trouble understanding the diagram and the answer key.

1) I am thinking time is the horizontal axis. Since A is π⁺, the quark composition will be $u$ and $\bar{d}$. There are 2 vertices, let say P (on the left) and Q (on the right). I don't understand the conservation of charge at vertex P. $u$ is going in the vertex and $\bar{d}$ is going out the vertex. I think W⁺ is going to the right so out of the vertex. In terms of charge, the charge going in is +2/3 but the charge going out is 1/3 + 1 = 4/3. Is the diagram wrong or is something wrong with my understanding?

2) Now about vertex Q. μ⁺ is the antiparticle of muon so why the arrow is to the right, not left? I also don't understand about the conservation of charge at vertex Q and what B is. Based on the answer key, B is antimuon neutrino. I imagine the equation for the decay is like this:
$$\pi^+ \to \mu^+ + \bar{v_\mu}$$

The lepton number on the left side is 0 but on the right side is -2. For the conservation of charge, if I reverse the arrow of μ⁺ (since it is an antiparticle) then both μ⁺ and W⁺ is into the vertex and B is out so the charge going in is +2 but going out is 0. I think there is something wrong with my understanding.

Thanks

 

[Orodruin]

1) If you consider time going to the right, the dbar is going into the vertex, not out of it.

2) The arrow is in the wrong direction if it is supposed to display fermion flow. The decay is $\pi^+ \to \mu^+ + \nu_\mu$.

 

[songoku]

1) If you consider time going to the right, the dbar is going into the vertex, not out of it.

Why? $\bar{d}$ is an antiparticle so if time is going to the right, shouldn't the arrow for antiparticle is to the left, so out from the vertex?

2) The arrow is in the wrong direction if it is supposed to display fermion flow. The decay is $\pi^+ \to \mu^+ + \nu_\mu$.

This makes much more sense, so B is not antimoun neutrino but muon neutrino so the lepton number and family lepton number are conserved. But sorry I still don't understand the conservation of charge at point Q. Both μ⁺ and W⁺ is into the vertex and $\nu_\mu$ is out so the charge going in is +2 but going out is 0.

If I consider time to be going upwards, then the arrow for $\bar{d}$ should be going down, not up, which does not fit the diagram from the question. That's why I assume time is to the right, but unfortunately it contradicts to the direction of μ⁺ given by the question.

Thanks

 

[Orodruin]

Why? $\bar{d}$ is an antiparticle so if time is going to the right, shouldn't the arrow for antiparticle is to the left, so out from the vertex?

Arrows typically indicate fermion flow. If you have a d-line with an arrow pointing out of it, you can consider it a d going out of the vertex or a dbar going into the vertex. These are equivalent. A d-line going out of the diagram with an arrow against the time direction is a dbar, one with it pointing in the time direction is a d.

This makes much more sense, so B is not antimoun neutrino but muon neutrino so the lepton number and family lepton number are conserved.

Indeed, as it must be in the perturbative SM.

But sorry I still don't understand the conservation of charge at point Q. Both μ⁺ and W⁺ is into the vertex and $\nu_\mu$ is out so the charge going in is +2 but going out is 0.

Same issue as above. The easiest way to think about it is using the arrow indicating the particle (in this case $\mu^-$) going into the vertex or the antiparticle ($\mu^+$) going out. Antiparticles go against the fermion flow.

If I consider time to be going upwards, then the arrow for $\bar{d}$ should be going down, not up, which does not fit the diagram from the question.

No, that would violate lepton number. If you let time go up instead of right, the d line is a d, not a dbar.

That's why I assume time is to the right, but unfortunately it contradicts to the direction of μ⁺ given by the question.

See above. The muon artow js arguably drawn in the wrong direction.

 

[songoku]

See above. The muon artow js arguably drawn in the wrong direction.

So I think the correct arrow for the vertex on the right is B to the right and $\mu^+$ to the left

Arrows typically indicate fermion flow. If you have a d-line with an arrow pointing out of it, you can consider it a d going out of the vertex or a dbar going into the vertex. These are equivalent. A d-line going out of the diagram with an arrow against the time direction is a dbar, one with it pointing in the time direction is a d.

I am sorry I still do not fully understand about the conservation of charge.

Based on the question, the dbar line is going out of the vertex so it is the same as d-line going into the vertex.
a) If I consider it as dbar line going out from the vertex, it means there is charge +1/3 going out, +1 going out from the $W^+$ and +2/3 going in so charge is not conserved?

b) If I consider it as d-line going in the vertex, it means there is charge -1/3 going in, 2/3 going in and +1 going out so the charge is still not conserved?

Thanks

 

[Orodruin]

Based on the question, the dbar line is going out of the vertex so it is the same as d-line going into the vertex.

It is not going out of the vertex. The arrow is pointing away from the vertex.

So I think the correct arrow for the vertex on the right is B to the right and μ+ to the left

Yes

b) If I consider it as d-line going in the vertex

It is a d-line going out of the vertex. The arrow points away from the vertex.

 

[songoku]

It is not going out of the vertex. The arrow is pointing away from the vertex.

You mean although the dbar arrow is pointing away from the vertex, dbar is still considered going into the vertex? The arrow just shows that it is antiparticle, not actually showing that the antiparticle is going out of the vertex?

Thanks

 

[Orodruin]

You mean although the dbar arrow is pointing away from the vertex, dbar is still considered going into the vertex? The arrow just shows that it is antiparticle, not actually showing that the antiparticle is going out of the vertex?

Thanks

The arrow is the fermion flow. If you consider the line a dbar line, the fermion flow goes against the direction of the dbar. If you consider it a d line, it goes with the arrow. In reality it is not a question of d or dbar, it is simply a d-line, which has a fermion flow direction.

 

[songoku]

I think I get it. Thank you very much for the help and explanation Orodruin