Bell’s Experiment

I am studying the online class on quantum computing (coursera). I did a quantum mechanic course more than a decade ago but I didn’t go into the detail of the Bell’s experiment then. The explanation from Prof. Vazirani is great and I feel that I finally understand the setup after all these years! Here, I just want to jot down some notes that I can review the material quickly.

As many of you knew already, Einstein didn’t believe in quantum mechanics in his last years. He believed that quantum mechanics is incomplete and there were some hidden variables that accounted for the spooky non-local behavior from quantum mechanics. After a decade of Einstein’s death, John Bell devised an experiment that can verify whether quantum mechanics or hidden variable theory is correct.

The simplified experiment described by Prof. Vazirani contains two boxes. Each box receives one bit input and returns one bit output. Let the inputs of the two boxes be X and Y, and the outputs of the two boxes be A and B. The boxes are assumed to be very far apart so that communication between them is not possible.  Let us use the natural shorthand notation 00/01/10/11 to represent the inputs and outputs. For example, an input 10 means that X=1 and Y=0 and an output 01 means that A=0 and B=1. The goal is to design a device that will generate different outputs (i.e., 01 or 10) when the input is 11 and the same outputs (i.e., 00 or 11) for all other inputs (00,10,01). Assume that the outcomes of inputs (00,01,10,11) occur with equal probability. Interestingly, the probability of success varies significantly for the hidden variable theory and quantum mechanics. And this allows one to verify which one is actually correct. More precisely, the Bell’s Theorem stated that if local hidden variable theory is correct, it is impossible to construct a device with success rate more than 0.75. On the other hand, if quantum mechanics is correct, it is possible to construct such a device with success rate more than 0.85.

Let’s first see why the probability of success cannot be larger than 3/4 for the hidden variable theory. For the probability of success to be larger than 3/4, since the hidden variable theory is deterministic, we need to have the outputs to be the desired ones for all inputs. Therefore, for input 00, the output can only be 00 or 11. Without loss of generality, let’s assume that the output is 00. Now, for input 01, or when X=0 and Y=1, since the box is local and deterministic, A has to still be 0 because X has not been changed. And for the outcome to be the desired one, i.e., A=B, B has to be 0 as well. Therefore, the output has to be 00. By the same token, the output for input 10 has to be 00 as well. Now, for input 11, i.e., X=1 and Y=1, from the fact that A=0 for the case when X=1 and Y=0, A still has to be 0 since A only depends on X and X hasn’t changed. Similarly, we have B=0 for the same reason. Therefore, we have the output for input 11 be 00 and is not a desired one. Thus, it is impossible to construct a local deterministic scheme (that’s an assumption of hidden variable theory) that results in the desired outputs for all inputs and hence the probability of success cannot be larger than 3/4.

To see why quantum mechanics will allows us to achieve a success probability larger than 3/4, we need to first introduce the Bell state, which is given by $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$.

One very nice property of the Bell state is that it has the same form for any orthogonal basis. Actually, it is very easy to verify that for any real numbers $a$ and $b$ and $a^2+b^2=1$, we have $\frac{1}{\sqrt{2}}(|u u\rangle+|u^\bot u^\bot\rangle)$

being also the Bell state, where $u=a|0\rangle+b|1\rangle$ and $u^\bot=b|0\rangle-a|1\rangle$. (Note that if we want to relax $a$ and $b$ to be complex numbers, we would like to modify the Bell state to $\frac{1}{\sqrt{2}}(|01\rangle -|10\rangle)$ instead. But ignoring this does not affect our discussion here.)

Now, let say we have another basis $|v\rangle$ and $|v^\bot\rangle$, assume that the angle between $|u\rangle$ and $|v\rangle$ is $\theta$.  If after measuring the first qubit with the basis $|u\rangle$ and $|u^\bot\rangle$ and obtaining the result $|u\rangle$, what will be the probability of getting $|v\rangle$ when measuring the second qubit with the basis $|v\rangle$ and $|v^\bot\rangle$? This question becomes very simple after knowing the aforementioned fact that Bell state has an invariant form regarding the basis. Thus, since the Bell state can be written as $\frac{1}{\sqrt{2}}(|u u\rangle+|u^\bot u^\bot\rangle)$, after measuring the first qubit as $|u\rangle$, the quantum state will collapse and the new state will be $|uu\rangle$. Therefore, the probability of getting $|v\rangle$ when measuring the second qubit will simply be $cos^2(\theta)$ as the angle between $|u\rangle$ and $|v\rangle$ is $\theta$.

The key trick of Bell’s proof is in carefully constructing the measuring basis as shown below (note that the blue $u$ is 45 degrees from the x-axis and the red and blue $v$ are + and – 22.5 degrees from the x-axis). When the input of box 1 is 0 (i.e., X=0), the red basis will be used. Otherwise, the blue basis is used. The same rule applies to box 2 also. With this setup, one can readily see that the angles between $u$ and $v$ are always 22.5 degrees except when the input is 00,01, and 10. For these cases, the probability of success, i.e., A=B, thus will be $cos^2(22.5)=0.85$.  And when input is 11, the angle between $u$ and $v$ is 67.5 degree. Note that in this case, we want A to be different from B. Thus, the probability of not success, i.e., A=B, will be $cos^2(67.5)=sin^2(22.5)$.  And so the probability of success is again $1-sin^2(22.5)=cos^2(22.5)=0.85$. Therefore, regardless the input, the success rate is always 0.85 and this concludes the second part of the Bell’s Theorem.