I Pion+/- decay - what *Exactly* is happening?

[Asgrrr]

I have seen vague explanations and (to me) misleading diagrams, and I am still not completely sure what is supposed to be happening in pion+/- decay. This Feynman diagram is a starting point.
https://en.wikipedia.org/wiki/Pion#/media/File:PiPlus_muon_decay.svg
What is happening? What happens to the quarks? I'm not asking for the unknowable, just a reasonable description of events and the fate of particles.

Thank you.

 

[mfb]

What is happening?

A pion decays to a muon plus a neutrino. The coupling is mediated by the W field.
The quarks stop existing.

Don't take Feynman diagrams too literally.

 

[Orodruin]

To build on that, Feynman diagrams are really mathematical objects that represent different terms in a series expansion. It is easy to be misled into thinking that there are particles running around like that and physicists will often use a language that suggests that.

 

[Asgrrr]

I'm grateful for the replies I'm sure. However, I'm working on teaching materials for high school/college. I am hoping for some way to explain pion decay in the terms of already established weak interactions involving the W particle. Help with that would be much appreciated.

 

[Orodruin]

I am not sure you can explain Feynman diagrams to high school students without a significant amount of "lying to children". You need a large amount of background to just understand what a particle physicist means when (s)he says "particle", let alone what the lines in a Feynman diagram represent.

 

[ChrisVer]

Since it's for teaching matterial:
The level of the class should be considered. As a result if you, as a a teacher, can't interpret well the lines of a Feynman Diagram, you should rethink of introducing the decay through it... If Ws are already introduced (wow), then you can just tell that the udbar quarks annihilate via a virtual W boson which couples to a lepton+neutrino.
How would you answer the simpler question of "what happens to the muons" when a muon and antimuon annihilate to electron+positron?
In fact, outside the context of a quantum theory, the only way you can introduce Feynman diagrams is by giving the students a list of conservation laws (eg charge conservation) and rules (eg flavor changing happens only via Ws), and making them play with the possible Feynman Diagrams (since they are made such that all conservations laws are satisfied in every vertex). They can't be thought as what "really" happens as others mentioned, or as you can understand by the fact that you integrate over all the spacetime the vertices of the internal lines.

 

[Asgrrr]

To clarify, it was never my intention to use Feynman diagrams as the basis of the treatment. I might show one or two for curiosity. What I am after in this thread is connecting pion decay to specific, established weak interaction(s) involving quarks, that's all.

 

[ChrisVer]

as an up-quark can be transformed to a down-quark by emitting a W (eg in beta processes), an up-quark and an anti-down-quark can "decay" into a W (virtual or real depending on the available energy, for the pion decay that's virtual)... The W decays to quarks (giving birth to mesons/hadrons) or leptons (e/mu/tau+neutrino), however because the pion is the lightest meson, the only energy-conserving decays are those to leptons (and in particular to electron or muon).

 

[Asgrrr]

PS: for some reason a quote from "Fear and Loathing in Las Vegas" keeps coming to mind when looking at this thread:

"Finish the f-ing story! What happened? What about the glands?"

For "glands" insert "quarks".

Don't take it the wrong way please.

 

[mfb]

See above: They stop existing.

 

[ChrisVer]

what about them? they're are no more (before you had a pion, then you have the leptons)... if they didn't disappear we wouldn't call it a decay but radiation (or something).

How would you answer the simpler question of "what happens to the muons" when a muon and antimuon annihilate to electron+positron?

 

[Asgrrr]

Sorry I didn't see your reply earlier, ChrisVer, my silly quote was not a response to you. I feel we are getting somewhere now.

I'll express antiquarks as ~x.

We start with a let's say u~d (Pion). That decays into W, disappearing both u and ~d.

That doesn't feel right to me as the end of the story. I have a problem with two particles decaying at once in this manner.

Can we say that the established decay of u -> d + W leaves us with d and ~d which immediately annihilate each other in the established way? Photons are not produced because they are not needed. The W carries away the energy and momentum.

That sounds much better to me.

 

[ChrisVer]

That doesn't feel right to me as the end of the story. I have a problem with two particles decaying at once in this manner.

what's that problem? They don't decay at once (it's not like a double beta decay), they annihilate each other.

Can we say that the established decay of u -> d + W leaves us with d and ~d which immediately annihilate each other in the established way?

I don't understand what you mean here, is ~d a \bar{d} ? And why would you need a ddbar (or where did you find it)? I also don't understand the "photons are not needed"...

 

[Asgrrr]

Yes, a ~d is as you print it, anti-d. I'm not that good with symbols here.

My problem is, what is it that gets u and ~d to annihilate each other? What allows that? They are not each other's antiparticle, and it doesn't happen inside protons for example (or does it?). This doesn't feel like a fundamental interaction, which leads me to think there's more to the story.

What I was trying to say, is that if we start out with a pion, we have a u~d. That's a pion. Now, if the u decays to d via

u → d +W

which is an established, basic decay mode, then what we are left with is d~d, or a particle and its antiparticle. THAT can reasonably decay in an established fashion (pair annihilation). Usually, pair annihilation would produce photons to carry away the energy and momentum of the destroyed particles, but when the W is forming at just about the same time, the W can carry away the energy/momentum, so there is no need for photons to be emitted. This is a much better explanation of pion decay because it only appeals to already established decay modes.

Thank you for your prompt response and your patience, ChrisVer.

PS: scratch the proton example, I was mistaken there.

 

[ChrisVer]

My problem is, what is it that gets u and ~d to annihilate each other? What allows that?

Weak interactions do (via the charged W bosons). W bosons allow transitions/couplings between up-type quarks (u,c,t) and down-type quarks (d,s,b) and generally violate the flavour conservation (it's the only Standard Model interaction which does that)...
The vertices will then have any combination depending on what is incoming/outgoing (so for the ud, you can have any vertex with ud, \bar{d}\bar{u}, \bar{u}d or u\bar{d} )

then what we are left with is d~d

in your u\rightarrow d W where is the \bar{d} ? This process has the following outcomes:
u \rightarrow d W \rightarrow d \ell \nu
u \rightarrow d W \rightarrow d q q'

Usually, pair annihilation would produce photons to carry away the energy and momentum of the destroyed particles

Pair annihilation between quark-antiquarks will preferably happen via gluons, rather than photons, because of the stronger strong interactions. The only exception would be the neutral pion because again of energy conservation (at least for its decay).
The photons can't transform an up-quark to a down-quark (this would even violate charge conservation), so there can be no ud, \bar{u}d or u\bar{d} vertices with a photon coming out of them.

 

[Asgrrr]

Thank you ChrisVer.

I think you have perhaps misunderstood some things I've said, but I'm satisfied with your explanations.

 

[Orodruin]

What I was trying to say, is that if we start out with a pion, we have a u~d. That's a pion. Now, if the u decays to d via

u → d +W

which is an established, basic decay mode, then what we are left with is d~d, or a particle and its antiparticle.

This is incorrect. What is established is the vertex between the u, d, and W with a particular fermion and boson charge flow. It certainly is no "basic decay mode" as the W is much heavier than both quarks. What is established from the theory are the Feynman rules and later you have to figure out what this means for the interactions among the asymptotic states.

 

[ChrisVer]

OK I think I see what you mean now: you are saying that a charged pion coul decay to a neutral pion and a lepton(and neutrino) like a semileptonic decay? however that's very unlikely and it's a very rare way for the charged pion to decay (because a lot of energy will be kept by the neutral pion, and the W would have only energy to transfer equal to the charged minus neutral pions' masses difference (which makes it even more off-shell W).
If that's what you meant, then the neutral pion will indeed decay to 2 photons (main decay channel for the neutral pions)...

 

[mfb]

I have a problem with two particles decaying at once in this manner.

That is the process that happens. Nature doesn't care if you have a problem with its processes.

 

[Orodruin]

That is the process that happens. Nature doesn't care if you have a problem with its processes.

There is only one decaying particle, the pion. If you look at the quark level it is a binary interaction between the bond state constituents.

 

[Asgrrr]

That is almost what I mean but not quite. A neutral pion is (u~u - d~d)/root2 or something like that. But after u decays into d in the charged pion we would have a pure d~d. I realize this is a state that would probably be impossible to detect, I'm looking for a rationalization.

 

[vanhees71]

http://pdglive-temp.lbl.gov/Particle.action?init=0&node=S009

 

[ChrisVer]

uubar or ddbar can be neutral pions... the expression you show doesn't make sense in terms of quark content (what does MINUS stand for?)...
Nevertheless, the \pi^\pm \rightarrow \pi^0 e^\pm \nu exists, however the dominant decay is the fully leptonic one to muons (mostly) and electrons (less)

 

[Orodruin]

uubar or ddbar can be neutral pions... the expression you show doesn't make sense in terms of quark content (what does MINUS stand for?)...

Uhmmmm, it is the linear combination of states that makes a neutral pion. The other linear combination is not a pion.

 

[Asgrrr]

uubar or ddbar can be neutral pions... the expression you show doesn't make sense in terms of quark content (what does MINUS stand for?)...
Nevertheless, the \pi^\pm \rightarrow \pi^0 e^\pm \nu[\itex] exists, however the dominant decay is the fully leptonic one to muons (mostly) and electrons (less)

<br /> <br /> <a href="https://en.wikipedia.org/wiki/Pion" target="_blank" class="link link--external" rel="nofollow ugc noopener">https://en.wikipedia.org/wiki/Pion</a><br /> <br /> See table at bottom of article.

 

[ChrisVer]

Uhmmmm, it is the linear combination of states that makes a neutral pion. The other linear combination is not a pion.

However a neutral pion is either uubar or ddbar...

 

[mfb]

However a neutral pion is either uubar or ddbar...

No, it is the linear combination, as Orodruin said already.

 

[ChrisVer]

No, it is the linear combination, as Orodruin said already.

OK, draw the Feynman Diagram of the linear combination for me and I would really like to see how you'd represent the (-) and where the /sqrt(2) appears, try the decay \pi \rightarrow \pi^0 e \nu... or diagrams like in Fig.2(a,b) and the blue neutral pion (for I don't see a uubar-ddbar), neither do I know a Z decaying to both a uubar and ddbar to give a neutral pion (d) http://www2.kek.jp/en/press/2008/BellePress12e.html
The combination appears only when you want to write the neutral pion's wavefunction in flavor space... afterall superpositions aren't natural.

 

[vanhees71]

uubar or ddbar can be neutral pions... the expression you show doesn't make sense in terms of quark content (what does MINUS stand for?)...
Nevertheless, the \pi^\pm \rightarrow \pi^0 e^\pm \nu exists, however the dominant decay is the fully leptonic one to muons (mostly) and electrons (less)

The mesons are grouped according to isospin (or its SO(3) generalization if you wish). Take $\psi=\binom{u}{d}$ for the first-generation quarks, forming an iso-doublet. Out of those you can build the pions by
$$\vec{\pi} = \bar{\psi} \vec{\tau} \gamma_5 \psi,$$
making them an isovector (isospin 1) triplet, the charged pions are given by $\pi^{\pm}=\pi_1 \pm \mathrm{i} \pi_2$ and the neutral pion is $\pi^0=\pi_3$. The $\vec{\tau}$ are the isospin-Pauli matrices and thus indeed $\pi^0=\bar{u} \gamma_5 u-\bar{d} \gamma_5 d$ (without caring about normalizations here). The iso-singlet is the scalar $\sigma$ meson (or $f^0$): $\sigma=\bar{\psi} \psi=\bar{u} u + \bar{d} d$, forming the other neutral combination of light quarks.

Whether or not you put factore $1/\sqrt{2}$ or $1/2$ there is a matter of convention, and there are at least three conventions for the pion-decay constant in the literature all different by factors $\sqrt{2}$ or $2$ from each other. For more details with all the factors (hopefully ;-)) in place, see theory lecture I under

http://th.physik.uni-frankfurt.de/~hees/hgs-hire-lectweek17/

 

[ChrisVer]

The mesons are grouped according to isospin (or its SO(3) generalization if you wish). Take $\psi=\binom{u}{d}$ for the first-generation quarks, forming an iso-doublet.

Whether or not you put factore $1/\sqrt{2}$ or $1/2$ there is a matter of convention, and there are at least three conventions for the pion-decay constant in the literature all different by factors $\sqrt{2}$ or $2$ from each other. For more details with all the factors (hopefully ;-)) in place, see theory lecture I under

Still I think this works only for effective field theories, since you mention the f_\pi (pion's decay constant). In that case you wouldn't represent the pion as a combination of quarks but as a single line (scalar field) in a Feynman Diagram (see eg some neutral pion decay diagrams), which is exactly what you do when you try to define it as a "field".

 

[vanhees71]

Sure, and the effective models are governed by chiral symmetry and thus take the form they take. Pions are a bound state with the specific quark content. Already these quarks are constituent quarks of course!

 

[mfb]

OK, draw the Feynman Diagram of the linear combination for me

You would draw both diagrams, and you have to calculate both diagrams for the decay and add their amplitudes.

afterall superpositions aren't natural.

Everything is a superposition - in most bases. Saying "this is not a superposition" without specifying a basis is meaningless. Superpositions are much more natural than the special case of pure states.

 

[ChrisVer]

You would draw both diagrams, and you have to calculate both diagrams for the decay and add their amplitudes.

in any case in the 1st diagram you would end up with a ddbar or with a uubar... so according to the OP, this wouldn't be a neutral pion.

 

[Orodruin]

in any case in the 1st diagram you would end up with a ddbar or with a uubar... so according to the OP, this wouldn't be a neutral pion.

The point is that you have to draw all the other diagrams as well and take the appropriate linear combination when you actually compute the decay amplitude. Indeed the uubar state is not a neutral pion, but the neutral pion projection onto the uubar direction (or vice versa) is non-zero so you have to include it when you compute the amplitude.

 

[Orodruin]

afterall superpositions aren't natural.

Somebody should tell the $K-\bar K$ oscillations ...

 

[ChrisVer]

Everything is a superposition

I disagree...
The fact that you can have an electron with spin up or spin down, doesn't mean that a single electron would have both... A collection of electrons would of course have 50% up and 50% down (given that the orientation symmetry is preserved), the single electron would have either up or down (you'd need to measure it).

 

[Orodruin]

I disagree...
The fact that you can have an electron with spin up or spin down, doesn't mean that a single electron would have both... A collection of electrons would of course have 50% up and 50% down (given that the orientation symmetry is preserved), the single electron would have either up or down (you'd need to measure it).

Before the measurement an electron may very well be in a superposition of spin up and down. It is your measurement process that forces it to select either by singling out a basis. After your measurement each individual electron is in an eigenstate - in your chosen direction. If you change from the z to the x direction each electron is in an equal parts linear conbination so whether it is a linear combination of seceral states really is a question of your choice of basis.

 

[vanhees71]

A single electron has the spin it prepared in. Usually it will be in a mixed state (aka unpolarized electron).

 

[mfb]

I disagree...
The fact that you can have an electron with spin up or spin down, doesn't mean that a single electron would have both... A collection of electrons would of course have 50% up and 50% down (given that the orientation symmetry is preserved), the single electron would have either up or down (you'd need to measure it).

Sorry Chris, but you are wrong, and repeating wrong statements won't improve them.
Up and down in which direction? Your perfectly polarized beam of electrons "up" in the z-direction will be in a superposition of up and down in all other directions. And yes, the electrons do have both up and down in these directions. It is important to consider both directions and their relative phase to describe the electron correctly in such a basis. If you want to assign them a specific orientation in a different basis, then you don't get your perfectly polarized beam in z-direction any more.

A pure (edit) measurement eigenstate is a very special case, in general you have superpositions.

 

[Orodruin]

A pure state is a very special case, in general you have superpositions.

A superposition can be a pure state as well. I believe you mean "a measurement eigenstate".

 

[ChrisVer]

OK people, I will agree with you and say that the neutral pion is only the u\bar{u}-d\bar{d} state... In that manner the decay \pi \rightarrow \pi^0 e \nu can't happen since there is no meson with the d\bar{d} (or u\bar{u}) alone quark content...
Also what's the meson that has the state: u\bar{d} \pm u\bar{s} \pm u \bar{b} \pm c \bar{s}\pm c \bar{d} \pm c \bar{b} (that would come from the W hadronic decay addition of Feynman Diagrams)?

 

[vanhees71]

How do you come to that conclusion? According to the particle data booklet the decay indeed happens (although quite rarely):

http://pdglive-temp.lbl.gov/BranchingRatio.action?desig=4&parCode=S008

 

[ChrisVer]

How do you come to that conclusion? According to the particle data booklet the decay indeed happens (although quite rarely):

http://pdglive-temp.lbl.gov/BranchingRatio.action?desig=4&parCode=S008

because I couldn't produce a neutral pion according to the definition I am getting: I either end up with processes resulting to u\bar{u} or to d\bar{d} (so projections of the neutral pion, as well as the sigma mesons)... I originally said that the decay was possible (like a semileptonic meson decay), since those were actually neutral pions but I was wrong.

 

[vanhees71]

because I couldn't produce a neutral pion according to the definition I am getting: I either end up with processes resulting to u\bar{u} or to d\bar{d} (so projections of the neutral pion, as well as the sigma mesons)... I originally said that the decay was possible (like a semileptonic meson decay), since those were actually neutral pions but I was wrong.

But it's not wrong, as you can see from the particle data booklet (the branching ratio is only about $10^{-8}$, but it's a measured decay of the charged pions). I still don't understand, why you come to the conclusion that this decay cannot exist, only because the pion is the given superposition?

 

[mfb]

They have a neutral pion component.

Also what's the meson that has the state: u\bar{d} \pm u\bar{s} \pm u \bar{b} \pm c \bar{s}\pm c \bar{d} \pm c \bar{b} (that would come from the W hadronic decay addition of Feynman Diagrams)?

There is no such thing, and a W cannot decay to a single meson anyway.

We had a great question about partial widths a while ago, where you can see the effect of the superpositions of quark states directly in the branching fractions.

 

[ChrisVer]

I still don't understand, why you come to the conclusion that this decay cannot exist, only because the pion is the given superposition?

Because that decay ends up with u\bar{u} or d\bar{d} and not in a superposition of the two... and that is not a neutral pion (as I was repeatedly told in this discussion)...
On the other hand, if we are to "add" every possible outcome (and so say we can have both the u\bar{u} and d\bar{d} in a combination, al), then I would ask on this counter-example (a possible W decay to a meson):
https://arxiv.org/pdf/1001.3317.pdf (Fig2)
D^0 \rightarrow \rho^+ \pi^- : c \bar{u} \rightarrow W (d\bar{u}) \rightarrow (u \bar{d}) (d \bar{u})
my question would then be why do we get a \rho^+ (with the u\bar{d} decay of a W ) and not a \alpha u\bar{d} + \beta u \bar{s} (forgetting the bottoms due to energy). Instead we call the 1st a rho, and the second a possible Kaon (2 different outcomes)... Is it related to a flavour symmetry or something?

 

[Orodruin]

Because that decay ends up with u¯uuu¯u\bar{u} or d¯ddd¯d\bar{d} and not in a superposition of the two...

This is self contradictory. If it can end up in uubar and ddbar, clearly it can end up in a linear combination of those states. It is a question of what the amplitude is.

my question would then be why do we get a ρ+ρ+\rho^+ (with the u¯dud¯u\bar{d} decay of a W ) and not a αu¯d+βu¯sαud¯+βus¯ \alpha u\bar{d} + \beta u \bar{s} (forgetting the bottoms due to energy).

What makes you think it doesn't decay to both linear combinations? The only difference is that the s quark is heavy enough for the mass eigenstates to be essentially equivalent to the flavour states, which does not happen in the pion system.