In Appendix B of QFT in a nutshell by Zee, a review of group theory is given. In the last paragraph of the appendix on page 533, he briefly discusses the helicity quantization of massless particles.

Firstly, he takes a rotation of $4\pi$ and considers possible helicities by using $exp(i4 \pi h) = 1$, where h is the helicity. Why he considers a rotation by $4\pi$ is not clear to me, and also why the path traced by this rotation can be shrunk to a point. Secondly, why is he not quantizing the helicity algebraically as conventionally done as in the case of angular momentum, but instead applying topological arguments. Is there a relation between the two?


1 Answer 1

  1. The path traced out by a rotation by $4\pi$ can be shrunk to a point in $\mathrm{SO}(3)$ because such a rotation corresponds to a closed curve in the double cover $\mathrm{SU}(2)\cong S^3$, which is simply connected and therefore any closed curve in it can be shrunk to a point. Curves which can be shrunk in the cover can also be shrunk in the base since covering maps are injective on the fundamental groups. Contrast this with a rotation by $2\pi$, which in $S^3$ is a path between two antipodal points, hence not closed and cannot be shrunk. This works the same way for the Lorentz group $\mathrm{SO}(1,3)$ and its universal (double) cover $\mathrm{SL}(2,\mathbb{C})$.

  2. Helicity is different from angular momentum because the angular momentum algebra is the Lie algebra of the massive little group $\mathrm{SO}(3)$, while helicity is for massless particles and hence must be related to the massless little group, which is a group variously denoted as $\mathrm{ISO}(2),\mathrm{SE}(2)$ or similar - it is the symmetry group of two-dimensional affine Euclidean space, hence consists essentially of $\mathrm{SO}(2)$ together with two-dimensional translations. Since the representations of translations are trivial, the question of the representations of the massless little group is reduced essentially to the representation theory of $\mathrm{SO}(2)\cong\mathrm{U}(1)$.

    This representation theory can be determined algebraically and one arrives at quantized values for helicity , but this cannot be done by the usual physicist's approach of representing the Lie algebra - the Lie algebra of $\mathrm{SO}(2)$ is simply the trivial Lie algebra $\mathbb{R}$, and there appear no quantization conditions in its representation theory - the quantization instead comes from the global structure of $\mathrm{SO}(2)\cong\mathrm{U}(1)\cong S^1$ and the fact that the possible unitary irreducible representations must be one-dimensional, hence completely classified by representation maps $\mathrm{U}(1)\to\mathrm{U}(1)$, since $\mathrm{U}(1)$ is the unitary group of a one-dimensional Hilbert space.

    It is straightforward to see that such maps $\rho_n : \mathrm{U}(1)\to\mathrm{U}(1), \mathrm{e}^{\mathrm{i}\phi}\mapsto\mathrm{e}^{\mathrm{i}n\phi}$ are simply classified by how often they wind the $S^1$ around itself, i.e. by the integers $n\in\mathbb{Z}$.

  3. So how does this helicity relate to $4\pi$ rotations? The generator of this $\mathrm{U}(1)\subset\mathbb{R}^4\rtimes\mathrm{SL}(2,\mathbb{C})$ (the r.h.s. is the universal cover of the Poincaré group) is physically the projection of spin onto momentum, $S\cdot p$ (how exactly you work this out depends on how exactly you determined that the little group of the massless particle is $\mathrm{SE}(2)$). So a full $4\pi$ rotation around $p$ acts as $\rho_n(1) = \rho_n(\mathrm{e}^{2\pi\mathrm{i}})$ on our massless particle: $$ \mathrm{e}^{4\pi\mathrm{i}h} = \mathrm{e}^{\mathrm{2\pi\mathrm{i}n}},$$ hence $h=\frac{n}{2}$.

  • $\begingroup$ A minor doubt, for massless particles where you consider the massless little group, why is Zee considering $4\pi$ rotations, instead of $2\pi$ rotations? Is it because $4\pi$ rotations can be shrunk to a point on $S_1$ while $2\pi$ won't? $\endgroup$
    – Bruce Lee
    Nov 19, 2017 at 4:33
  • $\begingroup$ @BruceLee The $4\pi$ rotation lives not in the little group, but still in the ordinary rotation group. From the physical definition of helicity, there is an axis such that a rotation by $\alpha$ acts on the state as $\mathrm{e}^{\mathrm{i}\alpha h}$, and we know that $\alpha = 4\pi$ needs to act as the identity. $\endgroup$
    – ACuriousMind
    Nov 19, 2017 at 11:41
  • $\begingroup$ That's still a little bit confusing. For massless particles, the little group is SO(2). Why do we still use the stuff (the 4pi rotation) from SO(3) which is the little group for massive particles? Shouldn't the discrete labels are completely determined by the little group? That is what Wigner tells us, right? $\endgroup$
    – Wein Eld
    Apr 9, 2018 at 9:38
  • $\begingroup$ @ACuriousMind could you please further explain how the 1/2 helicity corresponds to winding number 1? $\endgroup$
    – Andrea
    Dec 16, 2021 at 15:32
  • $\begingroup$ @Andrea Perhaps it's not clear what I meant by "winding number" here - I mean the winding number of the $S^1$ around itself via the representation map $\rho_n:\mathrm{U}(1)\to\mathrm{U}(1). \mathrm{e}^{\mathrm{i}\phi}\mapsto\mathrm{e}^{\mathrm{i}n\phi}$. $n=1$ corresponds to $h=1/2$ with $h$ as in the question. (See my first comment above for why) $\endgroup$
    – ACuriousMind
    Dec 16, 2021 at 15:52

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