Polarisation, spin, helicity, chirality and parity keep confusing me. They seem to be related, but exactly how they are related is unclear to me. Can someone maybe give a short overview about what these quantities mean and how they are related? What are the values for e.g. basic particles like electron or photon?


2 Answers 2

  • Spin is determined from the representation of the Lorentz group the quantum field transforms in. The projective finite-dimensional representations of the Lorentz group are labeled by two half-integers $(s_1,s_2)$. The spin of a field is the sum $s = s_1+s_2$. For example, a scalar transforms in $(0,0)$, a vector field in $(\frac{1}{2},\frac{1}{2})$, a Dirac spinor in $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$, and those have spin $0$, $1$ and $\frac{1}{2}$, respectively.

  • Helicity is the projection of the spin vector upon momentum. Formally, if $S^{\mu\nu}$ are the generators of the Lorentz group in a given representations, then helicity is given by the operator $$ h = \epsilon_{ijk} p^i S^{jk}$$ where $p$ is the momentum operator. For massive particles, it is not Lorentz-invariant, but for massless particles, which lack a rest frame, it is a Lorentz invariant notion.

  • Chirality only makes sense for Dirac spinors and similar objects whose representation decomposes into smaller representations, or where exactly one of the two $s_i$ labelling their spin is zero. The Dirac spinor representation $(\frac{1}{2},0)\oplus(0,\frac{1}{2})$ decomposes into the left-handed $(\frac{1}{2},0)$ and the right-handed $(0,\frac{1}{2})$, which are also called the (left/right)-handed Weyl spinors. For massless fermions, the evolution equations that couple the right-handed part of a massive Dirac spinor to its left-handed part decouple, meaning a massless Dirac spinor is equivalently a theory to two uncoupled Weyl spinors. The projection operator into the chiral subspaces of the Dirac spinor representation is $$ \mathbb{P}_\pm = \frac{1}{2}(1\pm\gamma^5)$$ for $\gamma^5$ the usual product of 4D gamma matrices. It is to be noted that, for massless fermions, the chiral subspaces are precisely the eigenspaces of helicity.

  • Parity is the unitary operator $P$ upon the quantum space of states that is associated to the classical symmetry $\vec x\mapsto -\vec x$. It is equivalently one of the generators of the $\mathbb{Z}/2\mathbb{Z}$ when presenting the Lorentz group as the semi-direct product $$ \mathrm{SO}_0(1,3)\ltimes \left(\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\right)$$ of its component $\mathrm{SO}_0(1,3)$ connected to the identity with the group $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z} = \{1,P,T,PT\}$

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    $\begingroup$ What about polarization? $\endgroup$
    – asmaier
    Feb 7, 2016 at 15:55
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    $\begingroup$ @asmaier: I'm not sure polarizations has a well-defined meaning that's distinct from these. E.g. the two polarization states of a photon arise from its two helicity states, and I haven't really heard polarization for anything else $\endgroup$
    – ACuriousMind
    Feb 7, 2016 at 22:10
  • $\begingroup$ Late follow up: how can the eigenspaces of Chirality and Helicity be the same in the massless case, when one operator is purely a matrix and one is a derivative? Or am I misunderstanding that momentum operator? $\endgroup$
    – Craig
    Jan 7, 2022 at 5:18
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    $\begingroup$ @Craig Ah, this is for the solutions of the (massless) Dirac equation in momentum space: The solution is a Dirac-spinor valued function $u(p)$, and solutions with $hu(p) = \pm\frac{1}{2}u(p)$ are precisely those that take values only in one of the chiral eigenspaces. $\endgroup$
    – ACuriousMind
    Jan 7, 2022 at 8:39

Polarization is, generically, a normalized vector in spin-space, i.e. the spin-state of a particle (When I say spin, I actually mean total angular momentum, spin+orbital angular momentum). Usually we take "fundamental" polarizations to be an orthonormal basis in spin-space.

  • For scalars (spin-0), there is only a trivial polarization because the spin-space is 1-dimensional.
  • For spin-1/2 particle you can have a normalized linear-combination of spin-up or spin-down, $c_1|\uparrow\rangle+c_2|\downarrow\rangle$. Actually this isn't Lorentz covariant notation, but never mind that.
  • For spin-1 massive you have $3$ spin degrees-of-freedom. The usual choice of basis polarizations here are $\epsilon^\mu_{\pm}=(0,1,\pm i,0)$ and $\epsilon^\mu_{3}=k^{\mu}/m$.
  • For spin-1 massless (which means helicity-1, since the concept of 'spin' only strictly applies to massive particles) you have $2$ spin degrees-of-freedom, which can be taken to be right/left handed polarizations $\epsilon^\mu_{\pm}=(0,1,\pm i,0)$, or any linear combination of them.
  • etc.

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