# If helicity can suppress weak decay of $\pi^{\pm}$, doesn't that make chiralty observer dependent?

I have been under the impression that chirality is at least in part a way to sneak in a Lorentz invariant version of helicity for particles that mass. Flip Tanedo seems to espouse this view in a somewhat cutesy fashion, for example, though he seems to be using the math rightly. (Mea culpa BTW, with cute little bee-like particles $e$lly $p$eter $p$aul and $\mu$ary in fact.) Tanedo makes this observation:

... is there some property which is equivalent to helicity in the massless limit [and] all observers in valid reference frames would measure to be the same for a given particle[?] The good news is that such a property exists; it is called chirality.

So, having convinced myself that chirality must somehow manage to remain experimentally independent of helicity -- since helicity is of course frame dependent -- imagine my surprise when I noticed Michael Brown's comment about helicity suppressing weak decays of the $\pi^{\pm}$:

Indeed, you see "helicity" suppressed weak decays of the $\pi^{\pm}$. You could just as well call it chirality suppression since what is being measured is the rotation between chirality and helicity eigenstates which is the result of the mass term.

Eh?

Someone help me with this: Even if the various constituent components of such decays are by themselves Lorentz independent, what good does that do if their relative ratios are dependent on helicity, which would then as best I can understand it would again make the decay rate of $\pi^{\pm}$ decay frame dependent?

I'm sure I'm missing something here, but I honestly don't know what. I've been looking for the papers on the original experiments, but I was hoping someone could point out where my understanding took an, um, left-hand turn...

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I think a good reference for this is Halzen and Martin, section 12.6. The main point is that neutrinos (of the Standard Model variety) always come as either left handed neutrinos or right handed antineutrinos. Since they travel at the speed of light (SM massless neutrinos), you cannot boost to a frame where they would have different helicity. (And even with beyond SM neutrinos, I think this holds as a very good approximation). A relevant quote, since there's no way I could do a better job:

The pion is spinless, and so, by the conservation of angular momentum, the outgoing lepton pair $(e^- \, \bar \nu_e)$ must have $J = 0$. As the $\bar \nu_e$ has positive helicity, the $e^-$ is also forced into a positive helicity state, see Fig. 12.7. But recall that this is the "wrong" helicity state for the electron. In the limit $m_e = 0$, the weak current only couples negative helicity electrons, and hence the positive helicity coupling is highly suppressed. Thus, in the $\pi^-$ decay, the $e^-$ (or $\mu^-$) is forced by angular momentum conservation into its "wrong" helicity state. This is much more likely to happen for the $\mu^-$ than for the relatively light $e^-$, in fact, $10^4$ times more likely.

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Jack Wimberley, thanks! Your quick description from Halzen and Martin about pion decay really helps. So: There's no "applied" helicity involved (my mental stumble point on this), only tracking of decay outcomes that after-the-fact express the suppression. Now that is an idea I can wrap my head around much more easily! –  Terry Bollinger Mar 22 '13 at 3:18

I liked Flip Tanedo's discussion. It is accurate as far as I can see. The key point is that mass mixes the chirality states. An electron might be produced as a pure left-chiral state in a weak interaction, but because it has mass it rapidly becomes a mixture of left and right chiral components. Another way of saying it is that the chirality operator (I've never seen it, but you could surely construct it - at least on the single particle sector of Fock space) does not commute with the mass term in the Hamiltonian.

All of this is independent of the helicity question. In pion decay the electron must be ejected with a certain helicity to conserve angular momentum, but it is the "wrong" helicity for the chirality state produced in the weak decay. Remember that for a massless Dirac particle helicity = chirality. The process is only allowed by the fact that the electron has a nonzero mass, so the helicity and chirality are mixed up slightly. I probably wasn't being particularly clear when I wrote that comment - it was very late, this this is all I really meant.

This section is a more mathematical way of describing the situation.

Everything comes from the representations of the Lorentz group, so let's examine that. A massive review of two component spinor techniques in four dimensions is

Dreiner, H. K., Haber, H. E., & Martin, S. P. (2010). Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry. Physics Reports, 494(1-2), 1–196. doi:10.1016/j.physrep.2010.05.002 (http://arxiv.org/abs/0812.1594)

I recommend following that to make sure you understand all the conventions etc. It also includes a useful appendix connecting all of this to the usual 4 component spinor notation.

In four dimensions the (double cover of the) Lorentz group is the product of two copies of SU(2): a left handed and right handed group. The generators of the two SU(2) are

$$\begin{array}{lcl} \left[N_{i}^{\pm},N_{j}^{\pm}\right] &=& i\epsilon_{ijk}N_{k}^{\pm},\\ \left[N_{i}^{\pm},N_{j}^{\mp}\right] &=& 0, \end{array}$$

and the generators of Lorentz rotations and boosts can be written

$$\begin{array}{lcl} J_i &=& N_i^+ + N_i^-, \\ K_i &=& -i (N_i^+ - N_i^-). \end{array}$$

Using the $N_i^\pm$ makes the representation theory simple, though I don't know any nice physical interpretation for them. Note that parity and conjugation both change $N_i^+ \leftrightarrow N_i^-$. By convention the $+$ generators correspond to the left hand group and the $-$ ones correspond to the right hand group. The irreducible representations are $(l,r)$ where $l,r$ are the spins under the left/right-handed SU(2)s respectively. You can see that chirality is not an observer-dependent thing - the separation of the Lorentz group into two pieces is invariant and the two pieces don't mix with each other.

Now you can write down left handed spinor fields which transform in the (1/2, 0) representation and right handed fields which transform in (0, 1/2), though you can also write these as the conjugates of left handed fields. From now on I'll use left handed fields exclusively, and following DHM will denote a right handed field by conjugation.

Now, given two left chiral spinor fields $\psi,\xi$ there are two types of mass terms you can construct:

• Majorana masses $\frac{1}{2} m_\psi (\psi \psi + \psi^\dagger \psi^\dagger) + (\psi\rightarrow\xi)$
• Dirac mass $m_D (\psi \xi + \xi^\dagger \psi^\dagger )$

There is some matrix structure hidden in the notation for field products. Check DHM if you want more detail. The upshot is that combinations like $\psi \psi$ etc. are Lorentz invariant.

Now the Majorana mass terms are ruled out for an electron because they violate charge conservation (there is no way to assign charges $Q_\psi,Q_\xi$ such that the terms are invariant under $U(1)_{EM}$ rotation). But, if we give $\psi$ and $\xi$ opposite charges the Dirac mass term is invariant. We interpret the fields as

• $\psi$ annihilates left handed electrons and creates right handed positrons
• $\xi$ annihilates left handed positrons and creates right handed electrons
• $\psi^\dagger$ annihilates right handed positrons and creates left handed electrons
• $\xi^\dagger$ annihilates right handed electrons and creates left handed positrons

We call the $\psi$ the field for left-chiral electrons and $\xi^\dagger$ the field for right-chiral electrons, though we have written it in terms of the left-chiral field $\xi$ which annihilates positrons. Confusing conventions. Oh well.

The weak interaction is chiral because only the $\psi$ component is charged under $SU(2)_L$. You have the operator $\psi^\dagger \bar{\sigma}^\mu \nu W^-_\mu$ (among others) which annihilates a $W^-$ boson and creates an electron and corresponding anti-neutrino. But this state has zero overlap with the right-handed electron that has to appear in the final state. If this was the end of the story there would be no matrix element and the decay would be impossible.

But the Dirac mass is there. There is another field $\xi$ which is mixed up with the $\psi$ due to the mass term, and a propagating electron eventually becomes a mixture of these two fields. The mass term annihilates a left-handed electron and produces a right-handed electron. It is the $\xi^\dagger$ which eventually annihilates the state to give the matrix element.

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Michael Brown, this is an excellent and interesting mini-tutorial expansion on the topic, thanks! I gave you a plus up, but alas, the answer nod goes to Jack Wimberley for the specificity with which he nailed my point of confusion and provided a very apt reference quote. Both are excellent answers, a case where I wish SE allowed dual answer nods. Thanks again! –  Terry Bollinger Mar 22 '13 at 3:23
@TerryBollinger No worries, I'm not offended. Jack gave a good answer. –  Michael Brown Mar 22 '13 at 5:13