Let me pick a particularly simple version of the EPR paradox. We have a source of anti-correlated electron-positron pairs. Alice has a device that measures spins along one of three possible axes:
- a: Along the y-axis.
- b: Along the line that makes a 120 degree angle with the y-axis in the x-y plane.
- c: Along the line that makes a 240 degree angle with the y-axis in the x-y plane.
We'll map "spin-up" to the result +1 and "spin-down" to the result -1.
Bob has the same three choices.
Alice's and Bob's results are anti-correlated, meaning if they both choose to measure the spin along the same axis, they always get opposite results.
So here is a model for what is happening:
- Associated with each particle pair, there are three numbers, A,B,CA,B,CA, B, C. Each is either +1 or -1.
- If Alice measures her particle along axis a, she will get result A. If she measures her particle along axis b, she will get B. If she measures along axis c, she will get C.
- Bob always gets the opposite: If he measures along axis a, he will get result -A, etc.
The awkward situation is that even though this model has 3 numbers associated with each pair, A,B,CA,B,CA, B, C, Alice can only measure one of them, and Bob can only measure one. So at best, they can only measure 2 out of the 3 numbers. But the model assumes that there is a result for all three directions, even if you can only measure two of them.
Let's assume that Alice and Bob perform lots and lots of measurements on twin pairs. Let's define some statistical quantities:
- E(a,b)=1N∑nAnBnE(a,b)=1N∑nAnBnE(a, b) = \frac{1}{N} \sum_n A_n B_n where AnAnA_n is the value of AAA for pair number nnn, and NNN is the number of pairs produced.
- Similarly, E(a,b)E(a,b)E(a, b) and E(b,c)E(b,c)E(b, c)
Here's where we use some pure mathematics to get some inequalities on these quantities.
1. E(a,b)+E(a,c)=1N∑n(AnBn+AnCn)E(a,b)+E(a,c)=1N∑n(AnBn+AnCn)E(a,b) + E(a, c) = \frac{1}{N} \sum_n (A_n B_n + A_n C_n)
2. Since Bn=±1Bn=±1B_n = \pm 1, BnBn=1BnBn=1B_n B_n = 1. So we can rewrite the right-hand side of equation 1 as
1N∑n(AnBn+AnBnBnCn)1N∑n(AnBn+AnBnBnCn)\frac{1}{N} \sum_n (A_n B_n + A_n B_n B_n C_n)
=1N∑n(AnBn(1+BnCn))=1N∑n(AnBn(1+BnCn))= \frac{1}{N} \sum_n (A_n B_n (1 + B_n C_n))
3. Taking absolute values, we get:
1N|∑n(AnBn(1+BnCn))|≤1N∑n|AnBn||1+BnCn|1N|∑n(AnBn(1+BnCn))|≤1N∑n|AnBn||1+BnCn|\frac{1}{N} |\sum_n (A_n B_n (1 + B_n C_n))| \leq \frac{1}{N} \sum_n |A_n B_n| |1+B_n C_n|
4. Since |AnBn|=1|AnBn|=1|A_n B_n| = 1, we get:
1N∑n|AnBn||1+BnCn|1N∑n|AnBn||1+BnCn|\frac{1}{N} \sum_n |A_n B_n| |1+B_n C_n|
=1N∑n|1+BnCn|=1N∑n|1+BnCn|= \frac{1}{N} \sum_n |1+B_n C_n|
=1N∑n(1+BnCn)=1N∑n(1+BnCn)= \frac{1}{N} \sum_n (1+B_n C_n)
=1+1N∑nBnCn=1+1N∑nBnCn = 1 + \frac{1}{N} \sum_n B_n C_n
=1+E(b,c)=1+E(b,c)= 1 + E(b, c)
5. So we conclude that:
|E(a,b)+E(a,c)|≤1+E(b,c)|E(a,b)+E(a,c)|≤1+E(b,c)|E(a,b) + E(a, c) | \leq 1 + E(b,c)
But experimentally, we find that:
- E(a,b)=E(a,c)=−1/2E(a,b)=E(a,c)=−1/2E(a,b) = E(a, c) = -1/2
Technically, we can prove that E(a,b)=cos(120)=−1/2E(a,b)=cos(120)=−1/2E(a,b) = cos(120) = -1/2
That violates the inequality:
|E(a,b)+E(a,c)|=|−1/2+−1/2|=1|E(a,b)+E(a,c)|=|−1/2+−1/2|=1|E(a,b) + E(a,c)| = |-1/2 + -1/2| = 1
1+E(b,c)=1+−1/2=1/21+E(b,c)=1+−1/2=1/21 + E(b,c) = 1 + -1/2 = 1/2
There is one technical assumption that may or may not be worrisome. Alice and Bob can't actually measure E(a,b)E(a,b)E(a,b) for the entire run, because some of the runs, they will measure the spins along axes aaa and ccc. For that run, they have no idea what the value of BBB is. For other runs, they will have no idea what the value of AAA or CCC is. What's assumed (and I'm not sure if there is a name for this assumption) is that the correlation E(a,b)E(a,b)E(a,b) computed using only those runs where AAA and BBB are measured gives the same result as if we had computed E(a,b)E(a,b)E(a,b) using all the runs. That is, we're assuming that the statistics for unmeasured quantities is the same as for the quantities that were actually measured.