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Helicity and Weyl spinors 
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#1
Jun1409, 01:20 PM

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Weyl spinors are not eigenstates of the helicity operator when the mass is not zero.
But they have welldefined chiralities no matter what the mass is. Yet, it seems to me that references keep talking of Weyl spinors as if they have welldefined helicities, regardless of the mass. First, people keep talking of the Weyl spinors as lefthanded and righthanded, even when there is a mass present. I guess that they are really talking about the chirality here, even though they use the language of handedness. Am I correct? Second, consider CPT. I have read that the CPT conjugate of a lefthanded spinor is the righthanded antiparticle state. It seems to me here again that we are trully talking about chirality, not helicity. And yet I have read some authors saying that CPT conjugates have opposite helicities. Third, consider Lorentz boosts. Now it definitely seem that under a Lorentz boost, we may change a positive helicity state into a negative helicity state. But that does not imply that we are transforming a lefthanded Weyl spinor into a righthanded Weyl spinor, since these do not have welldefined helicities. And yet, that seems to be what is implied by some authors. Anyoen can shed some light on these points? Thanks! Patrick 


#2
Jun1409, 02:14 PM

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I'm not sure if this helps, but if the fermion has mass, then there is really no such thing as a left or right chiral field. The free Dirac equation says you have to have both when your particle has mass. So one speaks about fermions with helicities and not chiralities since the free Dirac equation allows the existence of fermions with left and right helicities. For something that's fast moving like neutrinos, helicity is almost equal to chirality, so you can mix up the two terms and it doesn't really change the outcome of the discussion.
As for CPT, I think helicity does change. Spin transforms like angular momentum, rxp, so under CPT this changes sign as p remains the same and r changes sign. Helicity is spin dotted with momentum, and since momentum doesn't change, helicity overall changes sign. As for Lorentz boosts, just take the case of fast moving particles. So say you have a righthelicity particle and you boost so that it becomes a lefthelicity one. Well, since the particle is fast, chirality is almost equal to helicity, so you've also almost changed a rightchiral particle to a leftchiral one. 


#3
Jun1409, 02:36 PM

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Thansk for your input 


#4
Jun1409, 02:47 PM

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Helicity and Weyl spinors
The term Weyl spinor has different meanings. Some call only massless particles Weyl spinors. Others call Weyl spinors the (1/2,0) and (0,1/2) representations that make up SO(4).
I'm not sure handedness is exclusive to helicity. I think I've read handedness refer to chirality before too. It could be that people are just confusing the terms helicity and chirality. You have left and right helicity, and left and right chirality, so sometimes maybe people just call them both handedness. I agree the nomenclature is horrible and misleading. The reason I excluded chirality is because for experimentalists there's no such thing. They prepare an initial free state and a final free state, and since the states are free, it's not a state of definite chirality, but they can prepare states of definite helicity. Also a massive particle state can have welldefined helicity. As for my last statement on Lorentz boosts, that was a mistake. If something is moving really fast in one direction, close to the speed of light, then you cannot overtake it and still have it approaching the speed of light. I knew something was fishy about my last statement since you can't change chirality by Lorentz boosts. 


#5
Jun1409, 03:53 PM

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That's right. Some use "lefthanded" and "righthanded" for the eigenvalues of chirality and others for the eigenvalues of helicity or, often, for both. Unfortunately, when it is used for both, then any statement made is ambigous. My questions are these: a) One often reads that the charge conjugated state of a lefthanded particle is a rigthanded antiparticle. If I understand correctly, here ``lefthanded" and "righthanded" refers to chirality. So CPT changes the chirality. It changes also the helicity but again, for a massive particle, this is not an invariant property so it is not really relevant b) A Lorentz boost changes helicity but, as far as I can tell, it does not change the chirality. Now the fact that the notation is so confusing becomes extremely tricky. Because if we use lefthanded and righthanded to denote chirality, then we would have to say that a Lorentz boost changes the helicity even though a lefthanded particle remains a leftanded particle (i.e. [itex] \psi_L [/itex] remains [itex] \psi_L [/itex]). Am I getting this right? 


#6
Jun1409, 04:48 PM

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Let me ask a slightly different questions:
1) Do CPT conjugates have opposite chiralities? 2) Does a Lorentz boost change the chirality? It seems to me that the answers are respectively yes and no, but I must be mistaken. If we consider a Majorana particle and take its CPT conjugate, we should get back something equivalent to the state we started from, since a Majorana particle is its own antiparticle. If taking the CPT changes the chirality but a Lorentz boost doesn't, then we can never recover the initial state and the particle is distinct from its antiparticle. Therefore, a Lorentz boost must change the chirality!? This seems wrong since chiral eigenstates are irreducible representations of the Lorentz group.... Anybody can shed some light on this issue? Thanks 


#7
Jun1409, 08:35 PM

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operators explicitly as [tex] P = e^{i\phi}\gamma_0 [/tex] [tex] C = i \gamma_2 \gamma_0 [/tex] [tex] T = \gamma_1 \gamma_3 K [/tex] where K is the operator of complex conjugation. (I.e., [itex]K i =  i K[/itex] and [itex]K \gamma_2 =  \gamma_2 K[/itex]) The chirality operator is [tex] \gamma_5 = i \gamma_0 \gamma_1 \gamma_2 \gamma_3 [/tex] The Lorentz transf operators are [tex] \gamma_{\mu\nu} ~:=~ i/4 [\gamma_\mu , \gamma_\nu] [/tex] So [itex]P \gamma_5 =  \gamma_5 P[/itex] , showing that a parity inversion also inverts chirality. Also [itex]\gamma_{\mu\nu} \gamma_5 = \gamma_5 \gamma_{\mu\nu}[/itex] , showing that a Lorentz transformation leaves chirality unchanged. Similarly, it looks to me that C and T both commute with [itex]\gamma_5[/itex] but you'd better doublecheck that. :) This would mean that CPT changes the sign of the chirality. But this is errorprone since you've gotta make sure all factors of i are correct everywhere, since the K operator is involved. Cf. Peskin & Schroeder p71 where they summarize the action of the various discrete operators on Dirac bilinears: their pseudoscalar is written as [itex]i\bar{\psi}\gamma_5\psi[/itex] and CPT leaves it unchanged (see their table). BTW, in the thing about "handedness" for helicity and chirality.... helicity handedness is visualized by pointing your thumb in some direction and wrapping your fingers around it to indicate spin. However, chirality handedness is visualized by holding your thumb, 1st finger and 2nd finger mutually at right angles. Doing this with both your left and right hands shows that no amount of rotation or boosting can make the right hand look like the left. HTH. 


#8
Jun1409, 08:52 PM

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We say that a Majorana spinor is its own antiparticle. Let's say I start with a rightchiral, righthanded state. Under CPT, we then get a leftchiral, lefthanded antiparticle state. But this must somehow be equivalent to the state we started with. A Lorentz boost flips the helicity but it does not change the chirality, so how can we say that a Majorana particle is its own antiparticle? Thanks!! 


#9
Jun1509, 09:27 PM

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and "antiparticle" transform according to the same Poincare irrep, but belong to different inequivalent representations (which can arise when complex conjugation is in play). E.g., for a 2complexdimensional Lorentz rep there are in fact two inequivalent reps, obtainable from each other by complex conjugation. These reps are "inequivalent" because you can't use a (constant) matrix similarity transformation to achieve complex conjugation. However, a Majorana spinor belongs by definition to a real representation (not a complex rep as the more familiar spinors do). Hence complex conjugation has no effect  it doesn't give you an inequivalent rep in this case. Wiki has a bit more info... http://en.wikipedia.org/wiki/Majorana_equation http://en.wikipedia.org/wiki/Antiparticle What matters is the particular representation the particle transforms according to. The set of distinguishable reps determines the set of particle "types". 


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