# Can Bell's inequality be tested on a single particle?

In general, we apply Bell's inequality in the context of two entangled particles that travel away from each other, usually making some measurements in a non-separable state that can be describes as:
$| \psi \rangle=|\uparrow\rangle|\downarrow\rangle-|\downarrow\rangle|\uparrow\rangle$ where the first bracket on each term corresponds to one particle, and the second to the other.

But I could not find any assumption in the derivation that stops us from applying the inequality to a single particle state, also described by a state like the one above, but in which the first and second brackets represent two different observables on the same particle, let us say spin and angular momentum (assuming they are not coupled).

Is this correct? will these measurements on a single particle also violate Bell's inequality? If this is so, can we still argue that a possible reason for the violation of the inequality is non-locality, due to the fact that the two measurements are local?

UPDATE: After the original post I found this, which seems to answer my question.

• I'm not sure what you mean but Bell's inequalities are inequalities of correlations between particles so by definition cannot be applicable to one particle. – ZeroTheHero Mar 15 '17 at 17:36
• @ZeroTheHero I might be wrong, but I do not think that is the case. Could you indicate me where, in your favorite demonstration, it is assumed that the two "systems" need to be different particles as opposed to different properties of a single particle? – user126422 Mar 15 '17 at 17:54
• ZeroTheHero is right, the point of Bell's inequality is to show that there are limits to correlations between two entangled particles (photons). Spooky action at a distance (between TWO particles) is part of the controversy of entanglement. – Bill Alsept Mar 15 '17 at 18:19
• @HughMungus The particles need not be entangled of course (if they are not the inequalities will certainly not be violated) but you need at least two since the inequalities tell you about the probable outcomes of one particle given the outcomes of the other. – ZeroTheHero Mar 15 '17 at 18:26
• I guess I have to change my question. It seems to me that nothing in the demonstration requires two different particles. It is all about the correlations between results on different measurements. What equation would be violated if we forget that a and b are measurements on different particles and interpret it as different measurements of different properties in the same particle?cds.cern.ch/record/111654/files/vol1p195-200_001.pdf – user126422 Mar 15 '17 at 18:51

Entanglement between two different degrees of freedom of the same particle is known as "classical entanglement," as first laid out in Reference 1 (this is also called "classical non-separability" or "classically non-separable states" by those in the community that don't like calling anything quantum unless it is essentially quantum/has no classical analogue). As described in that paper [Ref 1], anything that you can do with two entangled particles you can also do with classical entanglement (including violating Bell-like inequalities as shown in Ref 2) with two caveats. The first and most obvious caveat is there is no non-locality. Second, you lose the exponential speedup in any quantum computation (so you don't gain anything using classical entanglement for computing).

The reason why this type of entanglement is called "classical," is because it is incredibly easy to make such states in an optics lab using classical beams of light and simple, linear optical components. For instance you can prepare the polarization of a (normal) laser beam in a way that the state of polarization depends on the location within the beam that you measure. Such beams are then "classically entangled" between the polarization and transverse degrees of freedom. An image of some different beams taken from Reference 3 is shown below. Figure a) is known as a radially polarized beam.

## References

1. RJC Spreeuw, "A Classical Analogy of Entanglement," Foundations of Physics 28, 361–374 (1998). doi:10.1023/A:1018703709245. Online pdf here.

2. XF Qian, B Little, JC Howell, and JH Eberly, "Shifting the quantum-classical boundary: theory and experiment for statistically classical optical fields," Optica, OPTICA 2, 611–615 (2015) doi:10.1364/OPTICA.2.000611.

3. F. Töppel, A. Aiello, C. Marquardt, E. Giacobino, and G. Leuchs, "Classical entanglement in polarization metrology," New J. Phys. 16, 73019 (2014) doi:10.1088/1367-2630/16/7/073019.

• This is very interesting! Do you have any intuition for why quantum computation fails for classical entanglement? – Rococo Mar 16 '17 at 23:19
• @Rococo Sure. The problem is you can't physically separate the different degrees of freedom. So if you have 2 qbits, you can first entangle the 2 qbits and then send them on to be independently operated on at the same time in a single step, but this would take 2 sequential steps with classical entanglement b/c the degrees of freedom can't be separated. So (for instance) if you wanted to entangle N qbits with 2-gate CNOT operations, it would take order ~log(N) steps with nonclassical entanglement, but ~N steps with classical entanglement. – Punk_Physicist Mar 17 '17 at 18:58

It is possible to entanglement two degrees of freedom of a single particle. For example, in Ion traps researchers entangle the electronic state (spin) of an ion with the motional state of the ion. If you prepared the system in such an entangled state then you would see a degree of correlation consistent with the predictions of quantum mechanics, i.e. similar degree of correlation that you see in a Bell test. However, the interesting part about the Bell inequalities is that the degree of correlation you see between the distantly (space-like) separated particles is larger than that you would ever see between two similarly separated classical particles. This is the non-local part of entanglement and the striking evidence that we must reject classical theories in favor of quantum theories.

If instead the classical particles were only time-like separated then it is possible the two classical particles could conspire via some classical causal, but as yet unknown, channel to achieve the higher degrees of correlation seen in experiment typically attributed to quantum entanglement. This is what we would call a 'local hidden variables' theory

So the point is that yes, you could entangle two degrees of freedom of the same particle and observe a degree of correlation consistent with quantum entanglement, but the degree of correlation would also be consistent with a hidden variables theory which doesn't require quantum weirdness.

The reason particles in a Bell test are separated very far from each other is that the goal of Bell inequality experiments is typically to rule out local hidden variables theories, the experiment you have proposed would not rule out such a theory.

• Violation of the inequality by two entangled particles rule out "local hidden variables", so QM is compatible either with non-local hidden variables, or either QM is local but there is no counterfactual definiteness. If you can entangle two degrees of freedom then a violation of the inequality must be linked to a lack of counterfactual definiteness, not to non-locality – user126422 Mar 16 '17 at 18:30

Bell's inequality is related to entanglement. Same property gets entangled. So, you can not entangle spin of an electron with itself.

As a minimum, you need one pair. You entangle the pair and do the measurement. Suppose you are able to collect/capture the pair and entangle it again, and measure and you are able to repeat this process, then you can achieve the test of Bell's inequality with a single pair.

I do not think it would be possible with a single particle because 1) I think same property is entangled, and 2) even if it was possible to entangle two different properties of same particle, as soon as you measure one property, you can not say spooky because it is same particle and spooky action at a distance is not even true, whatever the outcomes are. Light speed is sufficient for communication within the single particle.

It seems to me that on a reasonable interpretation, the answer to your question is yes.

Let $E_\theta$ be a spin measurement at angle $\theta$. Consider a particle initially in an eigenstate of $E_0$.

Then if you measure $E_A$ followed by $E_B$, the probability they will disagree is

$$P(A,B)=\cos^2(A)\sin^2(B-A)+\sin^2(A)\cos^2(B-A-\pi/2)= \sin^2(B-A)$$

Now if we contemplate measurements in four different directions, $A$, $B$, $C$, $D$ and expect these to behave like classical random variables, then we should certainly have $$P(A,D)\le P(A,B)+P(B,C)+P(C,D)$$ which you can think of as a Bell Inequality.

But for $A=0$, $B=\pi/8$, $C=\pi/4$, $D=3\pi/8$, this inequality is violated.

• I was unaware you could do that with a single degree of freedom. Is this violation trivial? do you think this imply that the violation of the inequality must be linked to a lack of counterfactual definiteness, and not to non-locality ? (see comment below @jgerber answer) – user126422 Mar 16 '17 at 18:38
• Is this the Leggett-Garg inequality? – Rococo Mar 16 '17 at 23:27