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

[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.