I read in a book that for $\beta$-decay the electrons have always been found to have an expectation value for their helicity of $h=-v/c$.

Then ist is said in the book, that it follows from this fact that such electrons are in a left-handed chiral state which is characteristic for the weak interaction.

In another article I read that the chiral state of an electron is not conserved in time. The electron will soon evolve a component with a right-handed chiral state and it will be a mixture of right- and lef-handed chiral states.

Suppose after the decay one electron moves like a free particle.
When it evolves a right-handed chiral component in addition to the left-handed component it starts off with, how can its helicity be conserved?


2 Answers 2


General explanation

Chirality and helicity are in general different quantities. Chirality is connected with the representation of the Lorentz group (left or right) while helicity is connected with projection of spin on momentum direction and becomes to characterize representation only in massless case.

I have meaned following. Particularly (!), representation with spin $s$ of the Lorentz group can be given (in terms of the irreducible representations of the lorentz group) as $\left( s, 0\right) $ (left chiral) or as $\left(0 , s \right)$ (right chiral). These representations are equal only for massive case (there exist an operator which connects them), and the second one is transformed as complex conjugated first one. You may get the second one by acting of spatial inversion or time inversion operators on the first one, and vice versa.

Helicity is the spin projection on the momentum direction. In general, it means that there are $2s + 1$ values of helicity for massive case. But massless case is radically different from massive one. Massless representations are characterized by helicity; There is only one lorentz-invariant (under continuous transformations) helicity value for massless particle (if theory isn't invariant under spatial inversion), and helicity $\lambda $ representation in particular case may be given as $\left( \lambda , 0\right)$, while helicity $-\lambda$ representation may be given as $\left( 0, \lambda \right)$. So only in massless case we may "equalize" helicity and chirality. You may understand this as one of the demonstration of relativistic aberration effect: when the speed of inertial frame is near $c$ the spin vertor "lays down" on the momentum direction independently of it's projections distribution at rest frame. So there is only one value of projection. If you inverse spatial coordinates, you will get the value with minus sign.

So this is a kind of accident that for massless case helicity "coincides" with chirality. Formally they are totally different quantities. For example, gravitons have only two helicities, $2$ (right graviton) and $-2$ (left graviton), similarly to photons, $1$ (right), $-1$ (left). But theory of left photon doesn't coincide with left graviton theory, because they have different helicities.

Let's talk about Dirac spinor case.

Particular case

Let's discuss particular case - Dirac spinor.

Helicity operator is $\hat {h} = \frac{(\hat {s} \cdot \hat{\mathbf p})}{|\mathbf p|}$, while chirality operator is $\hat {c} = \gamma_{5}$. You can show that for massless case $$ \hat{c}u(p) = \hat{h}u(p), \quad p_{0} > 0, $$ and $$ \hat{c}u(p) = -\hat{h}u(p), \quad p_{0} < 0. $$ Here $u(p)$ is the spinor wave: $$ \hat{\Psi} = \sum_{\sigma}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3}2E_{\mathbf p}}}\left( u_{\sigma}(\mathbf p)\hat{a}_{\sigma}(\mathbf p) e^{-ipx} + v_{\sigma}(\mathbf p)\hat{b}_{\sigma}^{\dagger}(\mathbf p)e^{ipx}\right). $$

But in massive case these operators are different. This statement is evident from the fact that helicity is conserved in time but it isn't Lorentz invariant $$ \tag 1 \dot{\hat{h}} = [\hat{h}, \hat{H}] = 0, \quad [e^{\sigma^{\mu \nu}\omega_{\mu \nu}}, \hat{h}] \neq 0, $$ while chirality isn't conserved in time but it is lorentz-invariant: $$ \tag 2 \dot{\hat{c}} = [\hat{c}, \hat{H}]= -2m\gamma_{0}\hat{c}, \quad [e^{\sigma^{\mu \nu}\omega_{\mu \nu}}, \hat{c}] = 0, $$ (so an axial current in electroweak lagrangian isn't conserved) and only in massless case they have identical behaviors under time evolution and Lorentz group transformations.

In terms of previous "paragraph" ("General explanation") these all can be summarized into following statements.

Dirac representation is $\left( \frac{1}{2}, 0\right) \oplus \left(0 , \frac{1}{2} \right)$ (so it is sum of left and right chiral states, they are mixed by mass term). Dirac particle has two possible values of helicity. In massless case we may separate this representation to $\left( \frac{1}{2}, 0\right)$ (left chirality) and $\left( 0 , \frac{1}{2} \right)$ (right chirality) with constant and lorentz-invariant helicities. So we may "equalize" left chirality and helicity $-1$ and right chirality and helicity $+1$.

Axial currents in Standard model

In Standard model's electroweak interactions part you have the expression for charged current in electroweak theory: $$ L_{Ch} = \bar{e}\gamma^{\mu}(A + B\gamma_{5})\nu_{e} W_{\mu} + h.c., $$ or $$ L_{Ch} = J^{\mu}W_{\mu} + h.c., \quad J^{\mu} = \bar{e}\gamma^{\mu}(1 + \gamma_{5})\nu_{e}. $$ Let's get the 4-derivative of axial part $J_{Ax}^{\mu}$ of this current, i.e., part proportional to $\gamma_{5}$: $$ \partial_{\mu}J_{Ax}^{\mu} = (\partial_{\mu}\bar{e}\gamma^{\mu})\gamma_{5}\nu_{e} + \bar{e}\partial_{\mu}\gamma^{\mu}\gamma_{5}\nu_{e} \approx $$ $$ = \left|(\partial_{\mu}\bar{e}\gamma^{\mu}) \approx -m_{e}\bar{e}, \quad [\gamma_{\mu}, \gamma_{5}]_{+} = 0, \quad \gamma^{\mu}\partial_{\mu}\nu_{e} \approx 0\right| = $$ $$ \approx -m_{e}\bar{e}\gamma_{5}\nu_{e} $$ (here $\approx$ means that I have neglected hermitean conhugated summand in $J_{Ax}$ which contracts nonmass terms and doubles mass term).

So, as you can see, an axial component of charged current isn't conserved in time in electroweak processes because electron has mass (or, equivanently, it can "move" from one chiral state to another, as the previous part of answer claims). This conclusion is equal to $(2)$ (that's why I have written it).

But helicity's current is always conserved in time (this fact doesn't depend on mass; this is expressed in $(1)$). You also may build the expression for helicity's weak current and see that it is different from axial current. It can be easily understood because (see first two parts of the answer) in massive case helicity isn't equal to chirality.

So there is nothing strange in the fact that helicity is conserved while chirality doesn't.

  • $\begingroup$ You put a lot of effort and a lot of theory in this answer, still it doesn't help me to understand how the chirality can change in time and helicity not. Perhaps you can elaborate on the sub-question: what kind of process is it, that evolves a r-h-ch. component? At time t=0 there is 100% l-h-ch. Where does the r-h-ch. component come from? $\endgroup$
    – Gerard
    Aug 18, 2014 at 21:34
  • $\begingroup$ @Gerard : I have added some info into the answer. Maybe this makes the answer better. $\endgroup$ Aug 18, 2014 at 21:55
  • $\begingroup$ So, am I right in seeing that (2) implies chirality goes to zero as t goes to infinity for massive particles? If so, does that mean any massive particle tends to equalize its handedness (i.e. in the infinite time limit it will be just as right handed as it is left handed)? $\endgroup$
    – adhanlon
    Sep 13, 2016 at 5:14

Helicity (the correlation of spin and momentum) and chirality (whether a particle couples to the left-handed or right-handed part of the weak interaction) are strongly correlated in the relativistic limit $v\to c$, but not strongly coupled at low speeds.

The electron, following the decay, is happy to believe that it's in its own rest frame and to evolve from a purely left-handed state into a mixture of left- and right-handed components. However, the evolution of its helicity is governed by conservation of angular momentum.

This is, for instance, why $\pi^+\to\mu^+\nu_\mu$ is strongly favored over $\pi^+\to e^+\nu_e$ in charged pion decays, even though the latter would release much more energy. The pion has spin zero, and in the rest frame the decay leptons must have opposite momentum. The weak interaction wants to make a (chirally) left-handed lepton and right-handed antilepton. But to conserve angular momentum you must have (in helicity) to right-handed or two left-handed decay products. The $\mu^+$ is born going lots slower than the $e^+$, so the correlation between spin and helicity is weaker and the muon can be polarized the "wrong way."


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