I need to show that helicity is Lorentz invariant (under the proper Lorentz transformation) for the massless particles. I heard about most frequently used argument which contains an idea of impossibility to "outrun" the massless particle, so the sign of helicity is Lorentz invariant. But how about the absolute value of the helicity (when we don't divide this operator on the spin operator norm)?

I want to ask about the method of proof of Lorentz invariance of helicity value $h$, which is determined for the massless particles as $$ W_{\mu} = hp_{\mu}, \qquad (1) $$ or in the spinor language $$ W_{c \dot {c}}\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}} = hp_{c \dot {c}}\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{k}}, \qquad (2) $$ where $h = \frac{n - k}{2}$ and $\psi_{a_{1}...a_{n}\dot {b}_{1}...\dot {b}_{n}}$ has only one independent component.

Do expressions $(1), (2)$ give us the authomatical proof of the invariance of the helicity value? For example, the left part of $(2)$ transforms under spinor representation of the Lorentz group as the product of $n + 1$ undotted spinors and $k + 1$ dotted spinors, so the right side must transforms by the same way, so it means that $h$ is Lorentz scalar? Analogical thinking may be passed for $(1)$.

  • $\begingroup$ Are you sure about those definitions? You can show that helicity is equal to chirality for massless particles, would that suffice? $\endgroup$
    – innisfree
    Feb 19, 2014 at 19:32
  • $\begingroup$ @innisfree : yes, I'm sure. "...You can show that helicity is equal to chirality for massless particles, would that suffice?..", - can I do it in general for an arbitrary spin particle? $\endgroup$ Feb 19, 2014 at 19:41
  • $\begingroup$ @AndrewMcAddams: What do you mean by "But how about the absolute value of the helicity"? It also doesn't change under boosts, just by the definition of helicity, $h={\bf J} \cdot {\mathbf{p }} / \left| {\mathbf{p}} \right| $, as boosts don't change the direction of ${\mathbf{p}}$ for massless particles. $\endgroup$
    – JeffDror
    Feb 20, 2014 at 9:36
  • $\begingroup$ @JeffDror : what is a helicity? It is a scalar product of a 3-vectors. Yes, boosts don't change the direction of $\mathbf p $, but they may change the absolute value $h$ (sometimes one uses operator $h = \frac{(\mathbf J \cdot \mathbf p )}{|\mathbf J||\mathbf p |}$, so it formally may be only $\pm 1$. $\endgroup$ Feb 20, 2014 at 16:52
  • $\begingroup$ @AndrewMcAddams: If I understand correctly, what you then want to show is that the momentum of a massless particle can't change direction under a boost. If you take momentum to be $p^\mu=(E,0,0,E) $ and then boost it in the $z$ direction, you find that if $\gamma >\gamma \beta $ ($\gamma$ is the Lorentz factor, $\beta $ is the velocity), then $p^\mu$ won't change direction. Since $\beta<1 $ this is always true. $\endgroup$
    – JeffDror
    Feb 20, 2014 at 17:27

1 Answer 1


Rotational invariance follows easily from the Poincare algebra. The non-trivial part is the invariance under boosts. Using the Poincare algebra one finds:

$[\frac{J\cdot P}{H},K_i]=i\bigg(\frac{\epsilon_{ijk}K_jP_k}{H}+J_i-\frac{P_i}{H^2}J\cdot P\bigg)\qquad(1)$

So it does not follow from the Poincare algebra that $\frac{J\cdot P}{H}$ is boost invariant. Instead only for those states for which $(\sigma P^{\mu}-W^{\mu})|\sigma\rangle=0$ do we have that helicity is boost invariant. One can show this by using that this condition requires

$\sigma\vec{P}=H\vec{J}-\vec{P}\times\vec{K}\qquad (2)$

Plugging (2) into (1) and acting on such a state gives:

$[\frac{J\cdot P}{H},K_i]|\sigma\rangle=0$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.