User:100Sapiens/Non-classicality of Spin-1 Systems

Spin 1 systems, or three dimensional quantum systems, are the smallest quantum systems that can be shown to be non-classical. It was first shown by Kochen and Specker that any hidden variable theory that can describe a spin 1 system must necessarily be contextual, that is, the outcomes of a group of compatible measurements must depend on the other measurements in that group and cannot be described by a joint probability distribution. Klyachko et al. derive an Bell-like inequality that is the simplest necessary and sufficient condition for a non-contextual hidden variable theory for quantum mechanics. This is referred to as the Klyachko-Can-Binicioglu-Shumovsky (KCBS) inequality, the pentagram inequality, or pentagon inequality. This is because the inequality is maximally violated for an equally-spaced quintuplet of measurements on the spin 0 state. Since the system is not separable into a bipartite system, the KCBS inequality shows that entanglement is not necessary for the non-classicality of quantum states. The violation of this inequality was experimentally realized by Lapkiewiz et al. in a photonic qutrit.

Derivation of the Inequality
Consider a set of five measurements, $ A_1, A_2, A_3, A_4, A_5 $. Let $ a_1, a_2, a_3, a_4,$ and $ a_5 $  be the variables corresponding to the outcomes of their respective measurements. Assume that these variables can only take values $$ a_i = \pm 1 $$ and remained fixed. This is the assumption of non-contextual realism. The observables have preassigned values the outcomes of a measurement are independent of the other measurements and their outcomes. Therefore, the outcomes of the measurements should be described by a joint probability distribution of all of the observables. With these assumptions we can now derive the Bell-like inequality:

$$ a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_5 + a_5 a_1 \geq -3 $$

Note that the product of the summands on the left-hand side is 1 $ (a_1 a_2) (a_2 a_3) ( a_3 a_4) ( a_4 a_5 ) ( a_5 a_1 ) = (a_1 a_2 a_3 a_4 a_5)^2 = 1$. So at most, 4 of the monomial terms on the left-hand side of the inequality can be -1. The result follows. Let p be probability function for all of the outcomes. By definition, for all values of the $$ a_i $$, $ p(a_1, a_2, a_3, a_4, a_5)\geq 0 $ and $ \sum p(a_1, a_2, a_3, a_4, a_5) = 1 $. Let $ f(a_1, a_2, a_3, a_4, a_5) = a_1 a_2 + a_2 a_3 + a_3 a_4 + a_4 a_5 + a_5 a_1 $. Taking the expectation value gives us:

$$ \langle f(A_1, A_2, A_3, A_4, A_5) \rangle = \sum_{a_1, a_2, a_3, a_4, a_5} p(a_1, a_2, a_3, a_4, a_5) f(a_1, a_2, a_3, a_4, a_5) $$

By the previous inequality, since the expectation value is a convex combination of $f$ s, the lower bound of the expectation value of f is the same as the f. Combining this with the linearity of the expectation value, we derive the inequality:

$$ \langle A_1 A_2 \rangle + \langle A_2 A_3 \rangle + \langle A_3 A_4 \rangle + \langle A_4 A_5 \rangle +\langle A_5 A_1 \rangle \geq -3 $$

For a spin 1 system, we set $A_i = 2 S_{\ell_i}^{2} - 1 $, where $S_{\ell_i} $ is the spin projection along $$ \ell_i $$, a unit vector in Hilbert space. These unit vectors are chosen such that, $ \ell_{i} \perp \ell_{i+1} $, where the indices are taken modulo 5 (so $$ \ell_{5} \perp \ell_{1} $$). This orthogonality implies $ [S_{\ell_i}, S_{\ell_{i+1}}] = 0 $, which ensures that the measurements for each expectation value is compatible, an essential assumption for a test of this inequality. To neglect this compatibility requirement could lead to a "compatibility loophole" in the experiment. A convenient choice for maximal violation of the inequality is when the $$ \ell_i $$, represented in 3D euclidean space, form a regular pentagram, with the input state, $$ |\psi \rangle $$, having a magnitude of spin equal to 0 along the symmetry axis of the pentagram. For this reason we call this inequality the pentagram inequality. This representation allows for a geometric calculation of the expectation values. Ultimately, quantum mechanics predicts that the left-hand side of the inequality is $ 5 - 4 \sqrt{5} \approx -3.944 $, showing violation of the inequality.

Experimental Test
The violation of the pentagram inequality was experimentally verified by Lapkiewicz et al. by applying the formalism derived above to a photonic qutrit. The experimental setup consists of a single photon source, an initialization stage that spreads the photons amongst 3 modes, and the measurement stage. The measurement stage is configured with 5 stages of two mode transformations that correspond to the five summands $ \langle A_i A_{i+1} \rangle $ in the inequality. Two detectors each measure a mode, corresponding to the measurements $ A_i $ and $ A_{i + 1} $. A detector click corresponds to $ A_i = -1 $ and no click corresponds to $ A_i = 1 $.

In the experiment, the single photon source is achieved by creating an entangled pair of oppositely polarized photons via spontaneous parametric down conversion. The pair is then split using a polarizing beam splitter, sending one to a detector and the other to the measurement stage of the experiment. The detection of the former heralds the detection of the latter in the measurement stage. An additional detector is required to monitor the mode not represented by $ A_i $ or $ A_{i + 1} $. If a heralded photon does not result in a detection, it is registered as a photon loss and that data point is disregarded. Two-mode transformations are implemented by first spatially combining two of the modes using calcite crystal as a polarizing beam splitter, and then applying a waveplate. The left-hand side of the pentagram inequality calculated by Lapkiewiz et al. was -3.8936, which is in close agreement to the value predicted by quantum mechanics