Every proof of Bell's Theorem I can find relies on there being three different "traits" of particles that we can choose to measure. Usually this is the spin of the particles along three different axes. However, trying the same logic and math with only two traits does not seem to work. Does this suggest that non-locality only becomes an issue when there are three possible traits a particle can have, or is it conceivable for particles to only have two measurable traits and have Bell's Theorem still hold?


2 Answers 2


No, classical local models can reproduce any correlations between two measurable traits, that is, Alice choses at random to measure either the $\theta_1$ or $\theta_2$ direction of the spin, and the same with Bob. For QM to predict something that classical correlations cannot reproduce, you need at least 3 different angles. Such as Alice to chose between $\theta_1$ and $\theta_2$, and Bob between $\theta_1$ and $\theta_3$.


There is no way to have quantum-mechanical particles with only two traits. Bell's Inequalities are usually discussed in the context of spin-$\frac{1}{2}$ systems, but they apply to any system with a two-dimensional Hilbert space.

The simplest possible quantum Hilbert space would be only one-dimensional, but in that case the only observable is represented by an operator proportional to the $1\times 1$ identity matrix. With only one state, everything is always completely known.

So the simplest system with nontrivial physics is one in which the state space is two-dimensional. The state vectors have the forms $$|\psi\rangle=\left[\begin{array}{c} a_{1} \\ a_{2} \end{array}\right]$$ $a_{1}$ and $a_{2}$ are complex numbers, and they can be normalized so that $|a_{1}|^{2}+|a_{2}|^{2}=1$. Moreover, the relative phase of $a_{1}$ and $a_{2}$ is also unphysical; multiplying the state vector by an overall phase factor $e^{i\alpha}$ does not change the value of any observable $\langle\psi|M|\psi\rangle$, because $\langle\psi|$ is the transpose and complex conjugate of $|\psi\rangle$.

However, if we have a two-dimensional state space, we can always construct observables that are implemented by the operators $$\sigma_{1}=\left[\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right],\quad \sigma_{2}=\left[\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right],\quad \sigma_{3}=\left[\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right].$$ These are the three "traits" you are asking about. What they correspond to physically depends on the specific two-level system under consideration. For a spin-$\frac{1}{2}$ particle, they correspond to spins along three perpendicular axes. For the polarization states of a photon, they correspond to measurements of the polarization in three different bases:

  • $\sigma_{1}$: vertical versus horizonal polarization (the $+$ polarization basis),
  • $\sigma_{2}$: diagonal polarizations rotated $45^{\circ}$ relative to the vertical/horizontal basis (the $\times$ basis),
  • $\sigma_{3}$: right versus left circular polarization.

For other two-state systems, the three observables may be even more different, and they may be difficult to realize in practice. Classically, they might seem like very peculiar things to measure. However, they always in principle exist. There are always three traits; for a given system you can figure out what they would be, so that it is always possible to set up Bell's Inequalities.


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