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

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?

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).

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

• So I looked it up more and this answer physics.stackexchange.com/q/409011 takes you to Appendix E2, where it is explained much better than I have could Jan 15, 2019 at 21:21