# Questions on the EPR Paradox

• honzik
Bob gets a lemon, Alice gets a lemon)So, on average, Alice and Bob should find the same fruit on 1/3 of the trials, but in fact they always find a different fruit.In summary, the classical assumption that each card's "+'s and -'s are the same as the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must also have been created with the hidden fruits A+,B+,C---forces you to the conclusion that on those trials where Alice and Bob pick different boxes to scratch, they should find the same fruit on at least 1/3 of the trials. For example, if we imagine Bob and Alicef

#### honzik

Hi,
I am little bit lost in what is so called EPR paradox (see also http://en.wikipedia.org/wiki/EPR_paradox). It takes into account two particles A and B that have some "same" characteristics (states) and by messuring some of this characteristics of particle A implies that the same characteristic particle B has. Quantum physics says that there exist some complementary characteristics (as e.g. position and velocity) - let's call them C1 and C2 - that cannot be messured together with arbitrary precision (so called uncertainity principle). (With this come also another question - what does mean "to be measured together"?) EPR paradox operates with measuring characteristics only on one particle (say A) which I don't quite understand - when measuring C1 on A than I in fact can not meassure C2 on A with arbitrary precision (or is it false?). But I think that it would make sense to measure C1 for A and C2 for B - both with arbitrary precision. And because A and B have C1 and C2 same, that can break uncertainity principle - or not?

Thank you for answering my questions and also for explaining (or sending link) the thought of EPR paradox.

Honzik

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EPR paradox operates with measuring characteristics only on one particle (say A) which I don't quite understand - when measuring C1 on A than I in fact can not meassure C2 on A with arbitrary precision (or is it false?). But I think that it would make sense to measure C1 for A and C2 for B - both with arbitrary precision. And because A and B have C1 and C2 same, that can break uncertainity principle - or not?
You can measure C1 for A and C2 for B, but you can't actually assume that A and B have the "same" value for all their unmeasured properties (i.e. that A had the same value as B did for C2 even though you didn't measure C2 for A), and that this is the explanation for why they always have the same value when you do measure the same properties for both (if you measure C2 for both A and B, you are guaranteed to get the same measured value). This idea would be a "local hidden variables" explanation for why they always give the same results when you measure the same properties, but it implies certain conclusions about the expected statistics when you measure different properties--conclusions that can be written as a "Bell inequality" of some type--and it turns out that these conclusions are falsified by the actual statistics predicted (and experimentally verified) in QM. Here's an analogy I came up with a while ago that illustrates this:
Suppose we have a machine that generates pairs of scratch lotto cards, each of which has three boxes that, when scratched, can reveal either a cherry or a lemon. We give one card to Alice and one to Bob, and each scratches only one of the three boxes. When we repeat this many times, we find that whenever they both pick the same box to scratch, they always get the same result--if Bob scratches box A and finds a cherry, and Alice scratches box A on her card, she's guaranteed to find a cherry too.

Classically, we might explain this by supposing that there is definitely either a cherry or a lemon in each box, even though we don't reveal it until we scratch it, and that the machine prints pairs of cards in such a way that the "hidden" fruit in a given box of one card always matches the hidden fruit in the same box of the other card. If we represent cherries as + and lemons as -, so that a B+ card would represent one where box B's hidden fruit is a cherry, then the classical assumption is that each card's +'s and -'s are the same as the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must also have been created with the hidden fruits A+,B+,C-.

The problem is that if this were true, it would force you to the conclusion that on those trials where Alice and Bob picked different boxes to scratch, they should find the same fruit on at least 1/3 of the trials. For example, if we imagine Bob and Alice's cards each have the hidden fruits A+,B-,C+, then we can look at each possible way that Alice and Bob can randomly choose different boxes to scratch, and what the results would be:

Bob picks A, Alice picks B: opposite results (Bob gets a cherry, Alice gets a lemon)

Bob picks A, Alice picks C: same results (Bob gets a cherry, Alice gets a cherry)

Bob picks B, Alice picks A: opposite results (Bob gets a lemon, Alice gets a cherry)

Bob picks B, Alice picks C: opposite results (Bob gets a lemon, Alice gets a cherry)

Bob picks C, Alice picks A: same results (Bob gets a cherry, Alice gets a cherry)

Bob picks C, Alice picks picks B: opposite results (Bob gets a cherry, Alice gets a lemon)

In this case, you can see that in 1/3 of trials where they pick different boxes, they should get the same results. You'd get the same answer if you assumed any other preexisting state where there are two fruits of one type and one of the other, like A+,B+,C- or A+,B-,C-. On the other hand, if you assume a state where each card has the same fruit behind all three boxes, so either they're both getting A+,B+,C+ or they're both getting A-,B-,C-, then of course even if Alice and Bob pick different boxes to scratch they're guaranteed to get the same fruits with probability 1. So if you imagine that when multiple pairs of cards are generated by the machine, some fraction of pairs are created in inhomogoneous preexisting states like A+,B-,C- while other pairs are created in homogoneous preexisting states like A+,B+,C+, then the probability of getting the same fruits when you scratch different boxes should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if 100% of all the pairs were created in inhomogoneous preexisting states, it wouldn't make sense for you to get the same answers in less than 1/3 of trials where you scratch different boxes, provided you assume that each card has such a preexisting state with "hidden fruits" in each box.

But now suppose Alice and Bob look at all the trials where they picked different boxes, and found that they only got the same fruits 1/4 of the time! That would be the violation of Bell's inequality, and something equivalent actually can happen when you measure the spin of entangled photons along one of three different possible axes. So in this example, it seems we can't resolve the mystery by just assuming the machine creates two cards with definite "hidden fruits" behind each box, such that the two cards always have the same fruits in a given box.
And you can modify this example to show some different Bell inequalities, see post #8 of this thread if you're interested.

But I think that it would make sense to measure C1 for A and C2 for B - both with arbitrary precision. And because A and B have C1 and C2 same, that can break uncertainity principle - or not?

Thank you for answering my questions and also for explaining (or sending link) the thought of EPR paradox.

Honzik

Welcome to PhysicsForums, Honzik!

The answer is YES - you can measure to arbitrary precision Alice for C1 and Bob for C2; and NO - this does not violate the HUP. The explanation is: a measurement of C2 on Bob after a measurement of C1 on Alice doesn't tell you anything about C1 on Bob or C2 on Alice. This is in violation of common sense, and is also in violation of what EPR assumed. But nonetheless, this has been experimentally confirmed many times.

It will help you to read about Bell's Theorem along with EPR as they go together and are almost never considered separately anymore.

Welcome to PhysicsForums, Honzik!

The explanation is: a measurement of C2 on Bob after a measurement of C1 on Alice doesn't tell you anything about C1 on Bob or C2 on Alice. This is in violation of common sense, ...

Thanks for answering - Bell's Theorem is really very fascinating result.

I have some more questions about measurement and characteristics discussed above. Please tell me whether my questions (answers on them) are right or wrong:

1) first measurement - i.e. C1 on Alice - tells me that the same value of C1 has also Bob (because Alice and Bob are entangled)

2) important question: are Alice and Bob still entangled after arbitrary number (especially the first) of measurements on them or will they at some point in time loose this property of "being entangled"?

3) after second measurement (suppose it was done just after the first measurement in ponit 1)) - i.e. C2 on Bob - has still Alice "retained" its measured value of C1 (from point 1) or will this measure of C2 on Bob somehow destroy the value of C1 on Alice? And another question connected to the second measurement (in case Alice and Bob are still entangled): will Alice have the same value of C2 as Bob after this second measurement (of C2 on Bob)?

4) One common question: It is commonly known that measurement causes collapse of wave function, but: what in fact is the "measurement" from the physic's point of view? What kind of physic's interaction is it? And after we answer this question can we admit that the same physic's interaction spontaneously taking place in the nature can also cause collapse of the wave function? (and are we able to detect it?...)

Thank you in answering this questions, they will help me to understand many nuances of quantum physics.

Honzik

Thanks for answering - Bell's Theorem is really very fascinating result.

I have some more questions about measurement and characteristics discussed above. Please tell me whether my questions (answers on them) are right or wrong:

1) first measurement - i.e. C1 on Alice - tells me that the same value of C1 has also Bob (because Alice and Bob are entangled)

2) important question: are Alice and Bob still entangled after arbitrary number (especially the first) of measurements on them or will they at some point in time loose this property of "being entangled"?

3) after second measurement (suppose it was done just after the first measurement in ponit 1)) - i.e. C2 on Bob - has still Alice "retained" its measured value of C1 (from point 1) or will this measure of C2 on Bob somehow destroy the value of C1 on Alice? And another question connected to the second measurement (in case Alice and Bob are still entangled): will Alice have the same value of C2 as Bob after this second measurement (of C2 on Bob)?

4) One common question: It is commonly known that measurement causes collapse of wave function, but: what in fact is the "measurement" from the physic's point of view? What kind of physic's interaction is it? And after we answer this question can we admit that the same physic's interaction spontaneously taking place in the nature can also cause collapse of the wave function? (and are we able to detect it?...)

Thank you in answering this questions, they will help me to understand many nuances of quantum physics.

Honzik

1. Yes, that is correct.

2. No, they are no longer entangled after the first measurement. Note that the order of the measurements of Alice and Bob does not affect the observed outcome.

3. Yes, repeated measurement of C1 on Alice will yield the same results. But no, Alice will not have the same C2 as Bob (except by coincidence) as a result of his measurement.

4. We don't really know what a measurement is, or what physically happens with collapse. This leads to what is called the "measurement problem", which some feel is a deficiency in the standard interpretation of quantum mechanics.