Feynman diagram for weak interaction.

[rkrsnan]

For a leptonic weak interaction diagram, we have an election converted into a neutino and emitting a w- particle. But only left handed part should undergo weak interaction. So basically you should have the left handed part of electron converted into a neutrino(left handed) emitting a w- particle. My question is what happens to the right handed part of the election during this interaction? In diagrams that I see in books etc, I don't see the weak interaction diagrams with only left handed part.
Same problem happens for quarks. You have an up quark going to down quark emitting a w+. But shouldn't it be that it is true only for the left handed parts. But from the diagrams it seems to me that the whole up quark is converted to down quark. How can the right handed part of the up quark get converted to the right handed part of the down quark by the weak interaction?

 

[chrispb]

The physical states of the theory are a left-handed electron and a right-handed electron. You don't simultaneously get both from a creation operator. Left-handed electrons can interact with Ws and Zs, and right-handed ones can't. Chirality isn't conserved, so photons can turn right-handed electrons into left-handed ones.

I think.

 

[rkrsnan]

I don't understand. Electron being a four component dirac particle, any electron wave function must have both left handed and right handed parts. So when you create an electron you create both left handed and right handed parts of its wave function. May be you are referring to electrons with right and left helicity? I am confused. In weak interactions we say only left handed part of the wavefunction takes part. But does that have anything to do with the helicity of the electron?(I mean even for a right polarised electron beam, its four component wavefunction has both left and right handed parts)

 

[Hawkwind]

For a leptonic weak interaction diagram, we have an election converted into a neutino and emitting a w- particle. But only left handed part should undergo weak interaction. So basically you should have the left handed part of electron converted into a neutrino(left handed) emitting a w- particle. My question is what happens to the right handed part of the election during this interaction? In diagrams that I see in books etc, I don't see the weak interaction diagrams with only left handed part.
Same problem happens for quarks. You have an up quark going to down quark emitting a w+. But shouldn't it be that it is true only for the left handed parts. But from the diagrams it seems to me that the whole up quark is converted to down quark. How can the right handed part of the up quark get converted to the right handed part of the down quark by the weak interaction?

I would put it this way: the strength of the weak interaction (the probability of the process to happen) goes with the left handed projection of the lepton or quark wave function. So, a purely right handed fermion has zero probability to decay via weak interaction. However, of course the fermion as a whole will be subject to change.

 

[rkrsnan]

... So, a purely right handed fermion has zero probability to decay via weak interaction...

But this confuses me. I guess by "a purely right handed fermion" you mean the right handed part of the fermion wave function. Right handed projection itself is not a particle. For a particle you need both the right and left handed parts, right?
I did some reading and in Griffiths book he says when a negative pion decays into a muon and an antineutrino, always the muons are found to be right handed experimentally. But he is talking about the helicity of the muon. He is not saying that the muon wave function has only right handed part, muon wavefuction obviously should have both right and left handed parts. What he says is that muon has right helicity.
So my question boils down to this. In weak interaction lagrangian only "left handed projection of the wavefuction" is there. How does this correspond to "particles with left and right helicities" interacting differently in weak interaction ? (and in a boosted frames we can even invert helicities)

 

[Hawkwind]

But this confuses me. I guess by "a purely right handed fermion" you mean the right handed part of the fermion wave function. Right handed projection itself is not a particle. For a particle you need both the right and left handed parts, right?

We are talking about helicity, i.e. the spin projection in the direction of the particle's momentum. Particles having spin parallel to momentum are called right-handed and in opposite direction left-handed. Under a "purely right handed fermion" I understand a particle having a sharp spin value +1/2 in direction of its momentum. Such particle may exist and it will not interact weakly.

I did some reading and in Griffiths book he says when a negative pion decays into a muon and an antineutrino, always the muons are found to be right handed experimentally. But he is talking about the helicity of the muon. He is not saying that the muon wave function has only right handed part, muon wavefuction obviously should have both right and left handed parts.

Hmm, it's my understanding that helicity is a property of a particle's wave function, isn't it ?
In general, a fermion will be in a state of superposition of right- and lefthanded helicity, i.e. its wave function is a linear combination

|Psi> = crh * |+1/2> + clh * |-1/2>

where |+1/2> and |-1/2> denote the helicity eigenstates.

The probability for such a particle to interact weakly goes with the coefficient clh (or its square); a particle in the eigenstate |+1/2> (clh=0,"right-handed") will not take part in weak interactions. This is what I was trying to say.

 

[Hawkwind]

We are talking about helicity, i.e. the spin projection in the direction of the particle's momentum. Particles having spin parallel to momentum are called right-handed and in opposite direction left-handed. Under a "purely right handed fermion" I understand a particle having a sharp spin value +1/2 in direction of its momentum. Such particle may exist and it will not interact weakly.

A correction: this is only exactly true for massless particles.
Strictly speaking, it's the chirality (and not the helicity) of a fermion which determines its coupling to the weak interactions. With increasing mass of a particle, this makes an increasing difference.
This is the reason why you observe muons with wrong (r.h.) "helicity" in the beta decay of a negative pion. If they were massless, this decay would not happen.
Unfortunately, helicity is a more abstract concept than helicity, based upon the Lorentz transformation of spinors.
Sorry in case I have created confusion instead of clarification. :(