## #010: A classical analogue for relativistic spin

Category: Classical mechanics - Tags: Quantum mechanics, spin, Hamiltonian mechanics, symplectic geometry

Provide a classical analogue for spin that works relativistically.

Mathematical problem. In Hamiltonian mechanics, the phase space of a free particle is given by the conjugate quantities $$\{q^i, p_i\}$$ and the Hamiltonian $$H=\frac{1}{2m} p_i g^{ij}p_j$$. In the extended phase space, the variables become $$\{q^\alpha, p_\alpha\}=\{t, q^i, -E, p_i\}$$ and the extended Hamiltonian is $$\mathcal{H}=\frac{1}{2m} p_\alpha g^{\alpha\beta}p_\beta - \frac{1}{2}mc^2$$. Note that the extended Hamiltonian functions both as a constraint and as the generator of the affine parameter (proper time for particles and minus proper time for antiparticles).

If we take a spatial direction as a degree of freedom, phase space is the two-sphere (the only sphere that is also a symplectic manifold) identified by all unitary vectors of components $$S_i$$. The conjugate quantities are given by the angle over a plane and the component perpendicular to the plane. For example, we have $$\{\theta^{xy}, S_z\}$$. The Hamiltonian is given by $$H=\mu B^{k}S_k$$. See here for details.

The problem is to find the “correct” space with the “correct” conjugate variables for the relativistic case.

Physical significance. The idea is that spin in quantum mechanics is the quantization of a directional degree of freedom. It is not clear whether the correct generalization is a direction in space-time, or a plane of rotation in space-time. In the first case, we would expect spin to generalize to a four-vector while in the second case to a two-form (i.e. an anti-symmetric rank two tensor).

The other issue is to fully understand how the units work. In the same way that conjugate momentum is covariant under changes of position, $$S$$ should be covariant under changes of angles. Somehow, radian units for angles are “better”. For one, if we take units of radians for the angle, the range of the angle is $$2\pi$$. If we take the range of $$S_z$$ to be $$[-\hbar/2, \hbar/2]$$, then the area is $$2\pi \hbar = h$$. Therefore, if we have a sphere with the surface area equal to the Planck constant, by using radians for angle we find the range of the spin’s vertical component to match that of a spin $$1/2$$ system.