Computing the spin degrees of freedom for a massless particle in $D$ dimensions

Question

According to the paper A Lagrangian formulation of the classical and quantum dynamics of spinning particles, a relativistic spinless particle in $D$ spacetime dimensions can be described by the Lagrangian $$L = \frac12 \dot{X}_\mu \dot{X}^\mu.$$ This is well-known, but the paper goes on to claim that spin can be modeled by adding the Grassmann variables $\psi^\mu$, with Lagrangian $$L = \frac12 \dot{X}_\mu \dot{X}^\mu + \frac i2 \psi_\mu \dot{\psi}^\mu.$$

I'm confused about this because it doesn't seem to have the right number of degrees of freedom. Following Wigner's classification, in $D$ spacetime dimensions, the little group of the massless particle is $E_{D-2}$. The finite dimensional representations of $E_{D-2}$ have the translations acting trivially, so the little group is effectively $SO(D-2)$. Now, I think that the particles we call "spinors" should correspond to the spinor representation of $SO(D-2)$, but the dimension of that representation grows exponentially in $D$, while here we only have $D$ Grassmann degrees of freedom.

What's going on here? How does the number of degrees of freedom line up?

Answer 1

1. Ref. 2 argues that OP's Lagrangian (extended with the necessary SUSY world-line (WL) reparametrization invariance, cf. e.g. this Phys.SE post) describes a Dirac-spinor, which has $2^{[D/2]}$ complex off-shell DOF and $2^{[D/2]-1}$ complex on-shell DOF.

2. The main point is that the Clifford algebra$^1$ $$\{\psi^{\mu}, \psi^{\nu}\}_{PB}~=~-i\eta^{\mu\nu} \qquad\Leftrightarrow\qquad \{\hat{\psi}^{\mu}, \hat{\psi}^{\nu}\}_+~=~\hbar\eta^{\mu\nu}\mathbb{1}$$ with real generators $$\psi^{\mu}, \qquad \mu~\in~\{0,1, \ldots, D\!-\!1\},$$ has only $[D/2]$ anticommuting variables $$\theta^1~=~(\psi^1-\psi^0)/\sqrt{2} ,\qquad \theta^a~=~(\psi^{2a-2}+i\psi^{2a-1})/\sqrt{2} , $$ $$ a~\in~\{2,3, \ldots, [D/2]\},$$ constructed from appropriate linear combinations of the $\psi$s.

3. Phrased differently, the Poisson supermanifold is $M=\mathbb{R}^{2D|D}$ with global coordinates $(x^{\mu},p_{\mu},\psi^{\mu})$. In geometric quantization we next pick an isotropic subspace, say, $N=\mathbb{R}^{D|1}\times \mathbb{C}^{[D/2]-1}$ with global coordinates $(x^{\mu},\theta^a)$, corresponding to a choice of polarization.

4. The first-quantized wavefunction $\Psi(x,\theta)\in{\cal L}^2(N)$ in a Schrödinger-like representation can be expanded in these Grassmann-odd anticommuting $\theta$s into $2^{[D/2]}$ component fields, which precisely form a Dirac spinor in $x$-space$^2$. In particular, one may show that these component fields transform as a Dirac spinor under Lorentz rotations.

5. The vanishing Hamiltonian constraint $H=p^2/2\approx 0$ imposes the (massless) mass-shell condition, while the vanishing supercharge constraint $Q=p_{\mu}\psi^{\mu}\approx 0$ imposes the (massless) Dirac equation.

References:

1. L. Brink, P. Di Vecchia & P. Howe, Nucl. Phys. B118 (1977) 76; Chapter 4.

2. C.M. Hull & J.-L. Vazquez-Bello, arXiv:hep-th/9308022; Chapter 2, p. 7-8.

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$^1$ Conventions: We use the Minkowski sign convention $(-,+,\ldots,+)$ and we work in units where $c=1$.

$^2$ It seems Ref. 1 effectively fails to go to an isotropic subspace in eq. (4.12).

Answer 2

The gamma matrices are related to their transpose, which are again related to their negatives, both by a similarity transform [1]. The representations corresponding to these are then equivalent. I believe the second fact takes you from $2^{d+1}$ to $2^d$, while the first gets you down to $2^{d/2}$ dimensions. I think there are also some 1d representations, but those aren't the ones you're looking for, probably because they can't anti-commute unless they themselves are Grassmann valued (I don't know why that can't be the case though if you really wanted it to be though...).

[1] [Higher-dimensional gamma matrices]: https://en.wikipedia.org/wiki/Higher-dimensional_gamma_matrices