Given the definition of helicity as $\lambda = \vec{p} \cdot \vec{J}$ up to normalization, does it even make sense to define helicity for a particle at rest (i.e. $\vec{p} = 0$)?

If it doesn't make sense, then what happens to the helicity of a state when it is boosted to its rest frame?


The correct definition of helicity is $\lambda = \frac{\vec{p} \cdot \vec{J}}{|\vec{p}|}$.

To clarify my question, I have added more details here.

My question was due to some confusion in reading up on the helicity formalism (first introduced by Jacob and Wick) from the book 'Angular Momentum Techniques in Quantum Mechanics' by Devanathan. This is discussed in chapter 13 of that book (in case you have it handy). The notation in that book for a helicity state is $\psi_{p,λ}$, where $p$ is the momentum and $\lambda$ is the helicity. Everything in that chapter seems fine up until Eq. (13.5) given by

$\mathcal{P} \psi_{0, \lambda} = \eta \psi_{0, \lambda}$,

where $\mathcal{P}$ is the parity operator, and $\eta$ is a constant. This seemed strange to me, because I believed the parity operator should flip the sign of the helicity. But, further up in that chapter, they write "Under space reflection about the origin (i.e. parity operation), the helicity $\lambda$ of a moving particle changes sign." So, of course, they say the parity only flips the helicity for a moving particle. And, in the equation above, the particle is at rest. This then made me ask if helicity is even well defined for a particle at rest. Now, in order for it to be well-defined, I can now see how the normalization is absolutely relevant, because taking the limit as $p$ goes to zero would then give a finite value for the helicity, but it seems it would depend on from what direction you take this limit...(i.e. not it actually is not well-defined). If anyone could shed some light on this, that would be great. Is Devanathan simply doing something they probably shouldn't be?

  • $\begingroup$ Helicity is a quantity that is defined only for massless particles. Hence, there is no rest frame. $\endgroup$
    – Prahar
    Jan 22, 2018 at 16:51
  • $\begingroup$ @Prahar Why do you think that? Nothing I read limits helicity to massless particles. $\endgroup$
    – adhanlon
    Jan 22, 2018 at 20:40
  • $\begingroup$ You can define whatever you want of course. So you can certainly consider the quantity $h = {\hat p} \cdot \vec{J}$. However, this quantity is not Lorentz invariant for massive states and is therefore not a useful quantity to talk about. For massive states, the correct object is the spin $s$ which is defined as $J^2 = s(s+1)$. $\endgroup$
    – Prahar
    Jan 22, 2018 at 21:22
  • $\begingroup$ That doesn't mean it isn't defined for massive particles. $\endgroup$
    – adhanlon
    Jan 22, 2018 at 21:33
  • $\begingroup$ True, but unless its a useful quantity there is no reason to expect any sort of nice properties whatsoever. I could also define the quantity $Q = \frac{1}{|\vec{p}|}$ for massive states, but then wondering why $Q$ is ill-defined in the $\vec{p} \to 0$ limit is not good. $\endgroup$
    – Prahar
    Jan 22, 2018 at 21:36

1 Answer 1


First, read up. The normalization is not irrelevant.

Helicity is defined as $\lambda =\hat{p} \cdot \vec{J}= \vec{p}\cdot \vec{J}/|\vec{p}| ~,$ so for a massless particle, which has no rest frame, it is a relativistic invariant.

For a massive particle, it is only meaningful to use it for high speeds/momenta, when the mass of the particle can virtually be overlooked. For massive fermions, for instance, distinct chirality states (e.g., as occur in the weak interaction charges) have both positive and negative helicity components, in ratios proportional to the mass of the particle. See, e.g., here. For low momenta, however, you reach a point of reckoning: Boost the particle to a frame where its momentum is reversed! Its helicity flips to the oposite of the original one, so it is not a relativistic invariant.

Putting it all together, first boost your particle to an infinitesimal momentum in its original direction. Then boost it to the opposite infinitesimal momentum so it has the opposite helicity. The helicity is then a step function of the momentum. What is the value at the midpoint of a step function, your 0/0, in your rest frame? 0? Note the spin in that frame is identical to the helicity in the upper, positive step of the step function.

  • Edits in response to comments: Helicity is the projection of spin on the direction of the momentum, not the momentum itself. No problem depends on the projection on the momentum itself. Yes, "left handed fermions interact weakly" means they are left-chiral states, so they contain both helicities if they are massive; for example, in charged $\pi^-$ decay, the coupling is left-chiral, but conservation of angular-momentum/helicity dictates the charged lepton have positive helicity. Because this is rare (see 2nd link above) the high-mass decay product ($\mu^-$) is chosen over the lower-mass one (e-)!

  • As for parity, it leaves spin alone and flips momentum, and therefore its direction. For the origin of momenta, there is nothing to flip--the same step-function argument as above. Sorry I don't have the Devanathan text you mention, but any good review of the helicity formalism will do the trick: In the rest frame, if you approached it from above, 0+, you can take the helicity to be the spin. Then the parity operation does not affect the (spin) helicity, and helicity flip is achieved by this Y, (13.10) of this reference. For moving states, use (13.60). It's actually easier to look at an applied particle-decay example, Section 13.4.

  • Here is a sleazy informal mnemonic: For vanishing mass, helicity and chirality coincide. As you turn on a mass, the two are still close to each other, with a "contamination of the wrong kind" proportional to the mass.

  • $\begingroup$ Thank you for the answer! I have a few follow up questions: 1) As for your comment about normalization not being irrelevant. Do you just mean that helicity would no longer be a relativistic invariant without this normalization, or are there other consequences? 2) Your answer got to the heart of some of my confusion about the difference between chirality and helicity. When they say that only 'left-handed fermions' interact weakly, they mean chirality, not helicity, right? A massive 'left-handed fermion' (i.e. chirally left-handed) can have both helicities, correct? 3) Edits in original question $\endgroup$
    – adhanlon
    Jan 22, 2018 at 19:02
  • $\begingroup$ @probably_someone That is not true. For a massless particle helicity is a relativistic invariant. I understand that there is an important distinction between chirality and helicity. Your claim that chirality is a relativistic invariant implies helicity is a relativistic invariant for massless particles, because chirality and helicity are equivalent for massless particles. $\endgroup$
    – adhanlon
    Jan 22, 2018 at 20:42
  • $\begingroup$ @Silmaril89 Sorry, wasn't thinking. Response deleted. $\endgroup$ Jan 22, 2018 at 20:43
  • $\begingroup$ @Cosmas Zachos Thanks again for your update. Your example of the charged $\pi^-$ is fascinating. I'm still confused by the book I was reading, but I don't think we will reach an answer here. I will take a look at the review you provided. Also, I'm curious what caused you to say "Helicity is the projection of spin on the direction of the momentum, not the momentum itself." I ask, because as far as I can tell, I never made the claim that helicity is momentum itself. If I made a mistake somewhere, I would appreciate it being pointed out! $\endgroup$
    – adhanlon
    Jan 22, 2018 at 20:47
  • $\begingroup$ @Cosmas Zachos The link to the review on the helicity formalism you shared is exactly what I have been reading! It is chapter 13 from Devanathan. It is eq. (13.5) that had me confused when they act parity on an at rest state and the helicity does not change. $\endgroup$
    – adhanlon
    Jan 22, 2018 at 20:50

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