# Does Bell's inequality hold for different particles?

Having read Bell's theorem and proofs. It seems it builds upon the assumption that we measure identical particles. Usually these thought experiments involve giving A and B an identical scratchcard. If they scratch the same box they will find the same symbol. If they scratch a different, then we can calculate the minimum coincidence rate thus set up the Bell's inequality.

But does this inequality hold if you give A and B a different scratchcard (aka measuring different particles)?

EDIT:

It seems the question is a bit misunderstood.

Okay let me clarify. Let's imagine an extreme case. I'm an evil scratchcard dealer and deal cards to A and B. A and B don't know anything about the cards. They may think they are identical or entangled but they won't find out anything till they measure it anyway. And I don't know which boxes they will scratch on which card. I have full control of the variables I give to A and B and there are no constraints other than I must conserve the quantities at least statistically. While reserving the right that each card in a particular pair can be completely different and there is no computable dependency between them.

Can I deal cards such way that I can fool A and B, so they find that their measurements violate the inequality, just becuase I dealed the cards to them in a clever pattern without entanglement or other? Or does Bell's theorem explicitly proves the such shenanigans are not possible at all?

• Are you asking if Bell's theorem would say non-identical particles should obey Bell's inequality if the laws of physics are local realistic? If so, see WillO's answer. Or are you asking if non-identical particles can, like identical particles, be put into entangled states where they will be measured to violate Bell's inequality, thus demonstrating that local realism cannot explain their behavior? If so, see lionelbrits' answer. – Hypnosifl Jan 1 '15 at 14:53
• @Hypnosifl See edit. – Calmarius Jan 1 '15 at 15:31

Bell's inequality is a theorem about classical random variables. It can be applied to observations of identical particles, or to observations of different particles, or to observations of the stock market and the weather. (The point is that because the theorem never mentions particles in the first place, the types of the particles to which it's applied cannot be relevant.)

You could imagine a neutral, spinless particle at rest decaying into an electron and positron, in which case the final state is of the form $$\left|\Psi\right\rangle = \int\!dx_1\,dx_2\,\; \left( \psi_{\uparrow\downarrow}(x_1,x_2) \left|x_1\right\rangle \left|x_2\right\rangle \frac{1}{\sqrt{2}}\left|\uparrow\downarrow\right\rangle + \psi_{\downarrow\uparrow}(x_1,x_2) \left|x_1\right\rangle \left|x_2\right\rangle \frac{1}{\sqrt{2}}\left|\downarrow\uparrow\right\rangle \right).$$ This is just a tensor product of the position and spin degrees of freedom, where I have absorbed all "coefficients" into $\psi_{\uparrow\downarrow}(x_1,x_2)$, etc. I have set the coefficients $\psi_{\uparrow\uparrow}$ and $\psi_{\downarrow\downarrow}$ to zero, because by conservation of angular momentum and spin, one can show that the state can be written as follows: $$\left|\Psi\right\rangle = \int\!dx_1\,dx_2\,\; \psi(x_1,x_2) \left|x_1\right\rangle \left|x_2\right\rangle \frac{1}{\sqrt{2}}\left(\left|\uparrow\downarrow\right\rangle -\left|\downarrow\uparrow\right\rangle \right)$$ This is not the same as the EPR or Bell state, but my point is to illustrate the fact that whether the particles are distinguishable or indistinguishable is not important for entanglement. To make it a Bell state, imagine Alice and Bob trying to measure one or the other particles using a Stern-Gerlach apparatus. Alice's particle detector will register $0$ if particle 1 is in state $\uparrow$, while Bob's particle detector will register $0$ if particle 2 is in state $\downarrow$, etc. This is just a matter of notation; a mental unitary transformation, if you will. Then we can write the above state as $$\left|\Psi\right\rangle = \int\!dx_1\,dx_2\,\; \psi(x_1,x_2) \left|x_1\right\rangle \left|x_2\right\rangle \frac{1}{\sqrt{2}}\left(\left|00\right\rangle -\left|11\right\rangle \right).$$ Note that Alice can only measure the first component of spin, while Bob can only measure the second. This is because the particles are assumed to be spatially separated. But there is only one state, and therefore only one "scratch card", as you put it.

Edit: Since you modified your question to clarify, you won't be able to deal the cards in any way and violate the inequalities, unless your cards can somehow talk to one another.

• Hi Lionel did you know "the Tangent Bundle" seems not to be working (at least I saw errors that seem to be coming from the server side). – Selene Routley Jan 3 '15 at 12:35
• @WetSavannaAnimalakaRodVance, thanks, between family and work I haven't had time to fix it. I think an automatic upgrade broke something. – lionelbrits Jan 3 '15 at 16:13

I assume you're referring to the scratch lotto card example I posted earlier--I posted it in two different answers, but this one goes into more detail about various possible loopholes (if you haven't read that answer you might find it helpful). The short answer to your question is that indeed, it will be impossible for the "evil scratchcard dealer" to assign hidden fruits to the cards in such a way that the two experimenters (Alice and Bob) will always find the same fruit when they scratch the same box on their cards, but will have a probability of less than 1/3 of finding the same fruit when they scratch different boxes, assuming that they have an equal probability of picking either of the three boxes and that their choices are uncorrelated with each other, and that the dealer has no way of predicting what choice they will make on each trial when assigning the hidden fruits.

There are a few loopholes though, like if the dealer can predict their choices in advance and correlate the hidden fruits to them, or if the fruits aren't determined in advance, and the experimenters scratch the boxes at different times in such a way that the choice made by the first experimenter of which box to scratch can be communicated as a signal to the second card before the other experimenter scratches it (either implying faster-than-light communication, or just that the distance/time between the two events of the experimenters scratching their cards was less than or equal to the speed of light). This paper has what seems to be a nice rigorous derivation of a Bell inequality, which makes explicit a series of assumptions 1-9 that must be met in order to be certain that a local realist theory must predict the inequality is obeyed. These assumptions are stated in a somewhat abstract way, but I can translate what they would mean for the scratch lotto example:

1. "Perfect correlation": In the lotto example, this means that whenever the experimenters both choose to scratch the same box on their respective cards, they are guaranteed to get the same fruit.

2. "Separability": When the two experimenters scratch their respective cards, these are distinct events (so we can talk about the probability of one event given the other, or the probability of either one given some other event).

3. "Locality 1": Neither of these two events is "causally relevant" for the other--so this is ruling out the loophole I mentioned where the fruits aren't just predetermined, and the first experimenter's choice of what box to scratch can affect the probability the other experimenter will see a given fruit (see the example with the touchscreen devices in my other answer).

4. "Principle of Common Cause": If two events are correlated, and this correlation can't be explained by a direct causal influence from one to the other (see assumption 3) or by "event identity" (see assumption 2), there must be some "common cause" that explains the correlation. So in the lotto card example, if the experimenters always see the same fruit when they choose the same box, then given 2 and 3 the fruits they each get must have been predetermined by something in their common past, in this case the dealer choosing what hidden fruits to print under each box on a given pair of cards.

5. "Exactly one of exactly two possible outcomes": An assumption about probabilities that in the example would boil down to the fact that whenever the experimenter scratches a given box, the probability they will see a cherry plus the probability they will see a lemon must add up to 1, and the probability the experimenter will see a cherry and a lemon at once is 0.

6. "Locality 2": An assumption that says that if there's a common cause as in assumption 4, and if knowledge of which boxes both experimenters scratched on a trial and knowledge of the common cause on that trial (i.e. what hidden fruits the dealer put behind each box) is sufficient to determine that a cherry will be seen by one of the experimenters, say Alice, then knowledge of what box was scratched by Bob should actually be irrelevant--just knowing the common cause and what box Alice scratched should be sufficient to determine that Alice will see a cherry.

7. "No outcome without measurement": This one is basically just saying that the only way to learn what fruit is behind a given box on a card is for an experimenter to scratch that box.

8. "Locality 3": An assumption almost identical to #6, except saying that if knowledge of which boxes both experimenters scratched on a trial and knowledge of the common cause on that trial (i.e. what hidden fruits the dealer put behind each box) is sufficient to determine that one of the experimenters (say Alice) will not see a cherry, then knowledge of what box was scratched by Bob should be irrelevant--just knowing the common cause and what box Alice scratched should be sufficient to determine that Alice will see a not see a cherry.

9. "No conspiracy": This assumption says the common cause events--the dealer's choice of what hidden fruits are behind each box on a given trial--are not statistically correlated with the experimenters' choice of which box to scratch at a later time in the same trial. In other words, the dealer can't have foreknowledge of what Alice and Bob will do and choose what series of hidden fruits to assign based on that knowledge. See the paragraph with the bolded title "Violation of freedom" in my previous answer that I linked to at the top for a bit more on this.

So if all these assumptions hold, there's no possible way the probabilities of different combinations of measurement outcomes could fail to satisfy Bell's inequality, as proved formally in the paper. Thus if we set up an experiment where the Bell inequality is consistently violated, at least one of these premises must be violated too.

• Do actual experiments show perfect correlation? – Calmarius Jan 1 '15 at 18:45
• The equations of QM would predict perfect correlation for identical measurements on entangled pairs, but in practice I think sources of entangled pairs will also produce some non-entangled individual particles of the same type so experimenters have to rely on methods like timing of detection to decide which pairs of detection-events were measurements of an entangled pair, and these methods probably aren't 100% error-free. There are however more complicated Bell inequalities that don't depend on perfect correlation, like the CHSH inequality. – Hypnosifl Jan 1 '15 at 19:25