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A quantum spin $1/2$ particle does not return to itself upon a rotation by $360^\circ$, but rather itself up to a sign. This is acceptable, because this extra phase is unobservable. In general the fact that phases don't matter is what allows projective representations to be used in quantum mechanics.

But now consider the spinor field, as introduced in a standard quantum field theory text. Usually, we begin by defining the Dirac spinor field classically, i.e. as a collection of four ordinary numbers, that transform under the Lorentz group in a prescribed way. The field does not return to itself upon a rotation by $360^\circ$, but rather itself up to a sign. But classical fields do not have the same phase ambiguity of quantum states, so why is this acceptable?

Physically, this particular case doesn't present a problem because the components of a spinor field are not observable; only spinor bilinears are. Is that what is always going on when we consider classical projective representations, i.e. are we always postulating that the elements of these representations cannot be directly measured?

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  • $\begingroup$ Do you know any classical theory -- with phenomenological relevance -- that uses projective representations? $\endgroup$ – AccidentalFourierTransform May 21 '18 at 18:47
  • $\begingroup$ Minor comment to the first paragraph: phases being unobservable requires projective representations rather than allows them. It is not that you are free to include them, but that you must. $\endgroup$ – AccidentalFourierTransform May 21 '18 at 18:48
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Quantum mechanics is linear. It can be described by means of Hilbert spaces, thus it is natural to use linear representations to describe its symmetries. Classical mechanics, on the other hand, is non-linear in general. Apart from translational motion of system of particles on a flat manifold, the spaces of states of classical systems are non-linear Poisson manifolds, for example a particles moving on a curved manifold, or the two-sphere $S^2$ which is the phase space of spin etc. Poisson spaces do not possess natural linear realization of symmetries (I'll get to the example given in the question in the sequel). It is possible to define nonlinear representations of Poisson algebras which are the classical analogs of the linear representations in quantum mechanics. This has been done in this beautiful paper by N.P. Landsman.

I would even say that if one really wants to understand the quantum-classical transition, it is almost illegitimate to talk about linear representations of classical systems (except the linear case mentioned above).

As very well known, projective representations are representations of a universal cover or a central extension of the group. So instead of talking about projective representations in classical mechanics, I'll answer the question: can universal covering or central extensions act as classical symmetries.

The answer, which might be somewhat surprising is that not only that they act, there are cases that they must act, i.e., a classical symmetry can only be realized by means of a universal cover or a central extension of a group.

The first example is the Galilean group which can act on the phase space of a free particle, only by means of a central extension where the center is the mass of the particle. Since, in this case the phase space is linear; this fact can be re-expressed in terms of the linear representation terminology, namely: the Galilean group can act on the phase space of a free particle only by means of a projective representation. Please, see the following answer of mine for elaboration and references.

The second example is related to the rotation group $SO(3)$ vs $SU(2)$. It is customary to think that in classical system every $SU(2)$ action can be reduced to an $SO(3)$ action and the SU(2) center acts only trivially on classical systems. Here is a counter example. Consider, a three dimensional isotropic harmonic oscillator. It is phase is $\mathbb{C}^3$, parametrized by the coordinates $z_i = p_i + i q_i$, $i=1,2,3$. ($p$ and $q$ are the positions and momenta). Consider now, this oscillator constrained to a constant energy hypersurface:

$$\sum_i |z_i|^2 = \mathrm{const}.$$

The reduced phase space can be parametrized by: $\zeta_1 = \frac{z_2}{z_1}, \zeta_2 = \frac{z_3}{z_1}$. In fact, the reduced phase space is $\mathbb{C}P^2$ (please see the following answer of mine).

The group $SU(3)$ of automorphisms of the $3-D$ isotropic harmonic oscillator phase space acts also on the reduced phase space of the energy hypersurface, its $SU(2)$ subgroup fixing the point $(0, 0)$ acts on a general point as:

$$\zeta_1' = \alpha \zeta_1 + \beta \zeta_2$$ $$\zeta_2' = -\bar{\beta }\zeta_1 + \bar{\alpha} \zeta_2$$ For the group element $\begin{pmatrix} \alpha & \beta \\-\bar{\beta }& \bar{\alpha} \end{pmatrix}$.

As you can see, the center elements $\begin{pmatrix}1 & 0 \\0 & 1 \end{pmatrix}$ and $\begin{pmatrix}-1 & 0 \\0 & -1 \end{pmatrix}$ act differently . Thus this is an example where an $SU(2)$ acts on a classical system and the action cannot be reduced to an $SO(3)$ action.

Returning to your example, the use of spinors in classical field theory has many advantages. May be its greatest advantage comes from their use in the path integrals of field theoretical models which resulted in many marvelous results in perturbation theory. The use of Grassmann variable fields effortlessly generates the constraints of spin-statistics theorem. However, this realization, per se, is hybrid. The spin degrees of freedom are already quantized (the spin acts by a matrix representation on a vector space of spinors), while the translation degrees of freedom are not quantized.

There exist models where both the spin and the translational degrees of freedom are not quantized, for example the Bargmann Michel Telegdi model. (However the use of this model seems to be harder to in applications for example in cross section computation). Thus it is not surprising that the $SU(2)$ action in this case reduces to an $SO(3)$ action, as the classical observables are only bilinears and the use of the spinor components is only an auxiliary tool.

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