I am fascinated by the way Chris Bernhardt uncovers quantum computing in his book. Explanation for Bell’s inequality is one of the gems in my humble opinion – just try to find a source that speaks human and not formulas. In this article I’d like to retell it so if you’re already skimming through the book – don’t bother reading further.
Introduction
Are you familiar with Einstein’s “He is not playing dice”? The whole idea of quantum physics is built upon probability: it is not possible to know the result of a measurement for sure – just the level of likeliness that certain variant of an event occurs. It does not relate to the macro-world though, such effects are observed only on the micro-level.
The classic physics, however, stands on the opposite grounds: if all parameters and natural laws are known, the outcome of objects’ interaction can be predicted with infinite precision. Probability is used only as a tool that is simple and precise enough, e.g., in thermodynamics.
As usual in scientific world, only an experiment can highlight which point of view better describes the reality. Note that science does not answer the “Why?” or “What?”: nobody can give a precise answer of what electron is or why it exists. The science describes the reality while also trying to predict the outcomes: although we don’t know what electron is, we can know for sure the way it interacts with electromagnetic field. So how do we stage such an experiment? Bell’s inequality gives an ingenious recipe for that.
Imagine that Alice and Bob have a bunch of entangled qubits:
Both Alice and Bob measure those qubits using 3 different orthogonal bases: A, B and C:
As you notice, B is derived from A by rotating it by 120° and C – by 240°. The first vector corresponds to binary 0, the second – to binary 1. Besides, Alice and Bob choose a basis independently from each other and in completely random way. What is the chance for Alice and Bob to yield the same result?
Quantum solution
As we know, if bases are the same, then Alice and Bob will get exactly the same result. The probability of such an event is 1/3. In order to understand what happens with different bases, we need to translate one basis to another. Here we invoke the power of trigonometry:
So if Alice selects A and Bob selects B, the probability for them to measure the same result 0 is 1/4; the same is valid for A and C, likewise for 1 as a measurement result.
Let’s count the total probability:
(1/3 × 1) + 2/3 × 1/4 = 1/3 + 1/6 = 1/2Quantum theory predicts that the probability for Alice and Bob to get the same result is equal to 1/2.
Classic solution
From the classic perspective, the system is predefined and no probability is present. Let’s use XYZ notation of qubit configuration:
- X – result of measurement in A;
- Y – result of measurement in B;
- Z – result of measurement in C.
Time to brute force all of the variants. ‘+’ means that measurement yields the same result, ‘-‘ – different result.
| A,A | A,B | A,C | B,A | B,B | B,C | C,A | C,B | C,C | |
| 000 | + | + | + | + | + | + | + | + | + |
| 001 | + | + | – | + | + | – | – | – | + |
| 010 | + | – | + | – | + | – | + | – | + |
| 011 | + | – | – | – | + | + | – | + | + |
| 100 | + | – | – | – | + | + | – | + | + |
| 101 | + | – | + | – | + | – | + | – | + |
| 110 | + | + | – | + | + | – | – | – | + |
| 111 | + | + | + | + | + | + | + | + | + |
Notice that in every row we have at least 5 ‘+’. It means that for any configuration of the qubit that chance to get the same measurement for Alice and Bob is at least 5/9. So the total probability must be larger than 5/9.
Inequality
The Bell’s inequality provides us with the way to determine which theory is correct to describe qubits – quantum or classic: if equality rate in the long run equals to 0.5 – quantum theory wins, if it is 5/9 – classic theory is correct. As you might imagine, creating entangled qubits and measuring them cannot be done in parents’ garage and it is an art of its own. However, several experiments so far show that the rate if 0.5 – quantum physics win (and I didn’t read the book for nothing)!
E91 protocol
Artur Ekert suggested a key distribution protocol that takes advantage of Bell’s inequality – E91 protocol. If you are familiar with BB84, you are going to notice a few similarities between the protocols.
Alice and Bob both receive a stream of entangled qubits and use the same set of bases (A, B and C) that we have discussed. If Alice and Bob choose the same basis to measure the pair of qubits, they will receive the same result. After a long series of measurement Alice and Bob should get an agreement on selected bases for 1/3 of total cases. They exchange the choices of basis over cleartext channel:
Alice:
B C B C A A B C B
A B A C A B B A C
Bob:In such a case Alice and Bob have a match at positions 4, 5, 7 so they could use corresponding bits to construct a shared key. However, Eve might be eavesdropping on the qubit stream so additional verification has to be put in place.
We know that, according to Bell’s inequality, the chance for Alice and Bob to get the same result with different bases equals to 1/4 if the qubits are entangled. However, what happens if Eve is listening? I would like to refer to the answer given by Frank Yellin:
We assume Eve, along with Alice and Bob, picks her basis at random. We are given that Alice and Bob chose different bases. So there is a 2/3 chance that Eve picks the same basis as either Alice or Bob, and a 1/3 chance that she picks the remaining basis.
If Eve picks the same one as either Alice or Bob, the odds remain at 1/4.
But what happens if she picks the remaining basis? She does her measurement, which disentangles the qubits, but gives them the same value. Alice now does her measurement. She has a 3/4 chance of getting a different answer from Eve and a 1/4 chance of getting the same answer as Eve. Likewise for Bob. So the probability that Alice and Bob get the same result is:
3/4 × 3/4 + 1/4 × 1/4 = 5/8
So all three have equal probability, the result is: 2/3 × 1/4 + 1/3 × 5/8 = 3/8
So we’ve got ourselves the way to verify if the Eve is listening or not:
- if the measurements with different bases match with probability of 0.500, then it’s safe to use the shared key;
- if the measurements match with probability of 0.375, then somebody is eavesdropping so the shared key cannot be used.
Kudos for review: Anastasiia Kuraleva
Networking and IT 