Local Representations of Rotations on Discrete Configuration Spaces

Abstract

A spin-half system may be characterised as having a set of two-valued observables which generate infinitesimal rotations in three dimensions. This abstract formulation can be given a concrete realization using ensembles on configuration space. We derive very general probabilistic models for ensembles that consist of one and two spin-half systems. In the case of a single spin-half system, there are two main requirements that need to be satisfied: the configuration space must be a discrete set, labelling the outcomes of two-valued spin observables, and these observables must provide a representation of so(3). These two requirements are sufficient to lead to a model which is equivalent to the quantum theory of a single qubit. The case of a pair of spin-half systems is more complicated, in that additional physical requirements concerning locality and subsystem independence must also be taken into account, and now the observables must provide a representation of $so(3) \oplus so(3)$. We show in this case that, in addition to a model equivalent to the quantum theory of a pair of qubits, it may also be possible to have non-quantum local models.

Appendix: Invariance of Subsystem Independence Constraints Under Rotations

We show that the constraints $C_1 = 0$ and $C_2=0$ given by Eq. (7.51) and Eq. (7.56) are preserved under local rotations.

We first consider the invariance of $C_1$. A simple computation shows that we have $\{C_1,M_3\} = \{(q_3-2q_1q_2),q_1\}=0$ and $\{C_1,N_3\} = \{(q_3-2q_1q_2),q_2\}=0$, as required. Next we compute the Poisson bracket with $M_1$,

$$\begin{aligned} \{C_1,M_1\} =&\{(q_3-2q_1q_2),M_1\}\nonumber \\ =&\frac{\partial M_1}{\partial p_3} - 2 q_2 \frac{\partial M_1}{\partial p_1}\nonumber \\ =&-(1/2 + q_2)\sqrt{\left( \frac{1}{2}\right) ^2-\left( q^{(1)}_+\right) ^2} \; \sin \left( p_1+p_3+b\left( q^{(1)}_+\right) \right) \nonumber \\&+(1/2 - q_2)\sqrt{\left( \frac{1}{2}\right) ^2-\left( q^{(1)}_-\right) ^2} \; \sin \left( p_1-p_3+b\left( q^{(1)}_-\right) \right) \nonumber \\&+2 q_2(1/2 + q_2)\sqrt{\left( \frac{1}{2}\right) ^2-\left( q^{(1)}_+\right) ^2} \; \sin \left( p_1+p_3+b\left( q^{(1)}_+\right) \right) \nonumber \\&+2 q_2(1/2 - q_2)\sqrt{\left( \frac{1}{2}\right) ^2-\left( q^{(1)}_-\right) ^2} \; \sin \left( p_1-p_3+b\left( q^{(1)}_-\right) \right) \nonumber \\ \rightarrow&-q_2 L_2(q_1,p_1) + q_2 L_2(q_1,p_1)~~~\mathrm{when}~~~ C_1=C_2=0\nonumber \\ =&0, \end{aligned}$$

(7.70)

as required. Under the canonical transformation $q_1 \rightarrow q_1$ and $p_1 \rightarrow p_1 + \pi /2$, we have $C_1 \rightarrow C_1$ and $M_1 \rightarrow M_2$, which leads to $\{C_1,M_2\}=0$. Finally, under $q_1 \leftrightarrow q_2$ and $p_1 \leftrightarrow p_2$, we have $C_1 \rightarrow C_1$ and $M_j \rightarrow N_j$, which leads to $\{C_1,N_1\}=\{C_1,N_2\}=0$. Thus we conclude that $\{C_1,M_j\}=\{C_1,N_k\}=0$ when $C_1=C_2=0$.

Now we look at the invariance of $C_2$. A simple computations shows that we have $\{C_2,M_3\} = \{(p_3-n\pi ),q_1\}=0$ and $\{C_2,N_3\} = \{(p_3-n\pi ),q_2\}=0$, as required. Next we compute the Poisson bracket with $M_1$,

$$\begin{aligned} \{C_2,M_1\} =&\{(p_3-n\pi ),M_1\}\nonumber \\ =&-\frac{\partial M_1}{\partial q_3}\nonumber \\ =&~\frac{q^{(1)}_+}{2\sqrt{(1/2)^2-\left( q^{(1)}_+\right) ^2}}\;\cos \left( p_1+p_3+b\left( q^{(1)}_+\right) \right) \nonumber \\&- \frac{q^{(1)}_-}{2\sqrt{(1/2)^2-\left( q^{(1)}_-\right) ^2}}\;\cos \left( p_1-p_3+b\left( q^{(1)}_-\right) \right) \nonumber \\&+(1/2+q_2)\sqrt{(1/2)^2-\left( q^{(1)}_+\right) ^2}\;\sin \left( p_1+p_3+b\left( q^{(1)}_+\right) \right) \frac{b'}{1+2q_2}\nonumber \\&+(1/2-q_2)\sqrt{(1/2)^2-\left( q^{(1)}_-\right) ^2}\;\sin \left( p_1-p_3+b\left( q^{(1)}_-\right) \right) \frac{-b'}{1-2q_2}\nonumber \\ \rightarrow&\frac{q_1-q_1}{2\sqrt{(1/2)^2-q^2_1}}\;\cos \left( p_1+b\left( q_1\right) \right) \nonumber \\&+ \sqrt{(1/2)^2-q^2_1}\;\sin \left( p_1+b\left( q_1\right) \right) \frac{b'-b'}{2}~~~\mathrm{when}~~~ C_1=C_2=0\nonumber \\ =&0, \end{aligned}$$

(7.71)

as required. Using the same type of canonical transformations considered in the case of $C_1$, we can show that the Poisson bracket of $C_2$ with the remaining generators also vanishes. Thus we conclude that $\{C_2,M_j\}=\{C_2,N_k\}=0$ when $C_1=C_2=0$.

This shows that the generators satisfy all of the constraints imposed by subsystem independence.