Entanglement

We've studied how we can represent multiple qubits using the tensor product.

Sometimes, these multi-qubit systems are what we call "seperable". This means that the whole system can be represented as a tensor product of it's component qubits.

For example, the following state is seperable:

Because we can represent this 2-qubit state as the tensor product of two single qubits it is seperable.

A great question is: do non-separable separable states exist? If so, what do they look like?

Let's examine the following circuit:

We know how to work with these! We start in the state$\ket{00}$. We'll then apply the Hadamard gate to the first bit while leaving the second bit alone (i.e. applying the identity operation):

And then we apply the $\text{CNOT}$ gate. Remember that the $\text{CNOT}$ gate flips the second qubit only when the first qubit is in the state $\ket{1}$, so we have:

So we've ended in a 2-qubit state that is in the even superposition of $\ket{00}$ and $\ket{11}$. Is this seperable? Give it a go! See if you can find two single qubit states that can be tensored together to give this state.

Hopefully you don't find any, because this state is in fact non-separable! Because the two qubits have different values in both kets, we can't represent this state as a tensor product of two single qubit states.

Just take a moment to ponder what this means. In classical computation, we could have many bits, but we could always pick out a single bit and describe it's state independantly. This is not the case in quantum computation. The state of a qubit can be dependent on the state of another qubit. This is a very powerful concept that we will exploit in many quantum algorithms.

These 2-qubit entangled states are called EPR pairs, named after the three physicists who first described them in a famous paper in 1935: Einstein, Podolsky, and Rosen.

There are 4 of these pairs:

Hopefully you can see by what we've discussed before, that these states are all indeed, non-separable.

They have a slightly more succinct representation:

where $x$ and $y$ are either $0$ or $1$ and $\bar{y}$ is the "not" of $y$.

Also note the $(-1)^x$. This might seem a bit weird at first, but it's essentially saying: $1$ when $x$ is even (in our case just $0$) and $-1$ when $x$ is odd (in our case just $1$). I'm drawing special attention to this now, since things like this are quite common and will pop-up a lot with quantum algorithms. Take a second to familiarise yourself with it.

Taking a step back from the maths for a second, what could we achieve with this entanglement property? Although we draw it as a circuit, there is nothing that actually says the two qubits need to be nearby. This means that two entangled qubits could be many many light years apart.

Let's devise a hypothetical scenario where Alice and Bob get together to create an entangled pair $\ket{\beta_{00}}$, Alice takes the first qubit, Bob takes the other and then they travel far apart:

If Alice measures her qubit as $\ket{0}$ then the system of both their qubits collapses to $\ket{00}$ since the other possibility, $\ket{11}$, is not possible.

This means that Bob has a 100% chance to measure $\ket{0}$. But the outcome of Bob's measurement was decided by Alice's measurement, so did her measurement somehow influence Bob's? Did her measurement cause some information to be sent to Bob's qubit to tell it the measurement result? If Alice and Bob were light years apart, did this information travel faster than the speed of light?

The answer to all these questions is no. Them measuring the same value was a consequence of quantum correlation that isn't so intuitive in the normal classical world. We will see later on that this phenomenon doesn't let communication happen faster than the speed of light.

There are many people who have done a better job of explaining this phenomenon. It is a deep philosophical topic in the world of quantum mechanics and we won't get to side-tracked down that rabbit hole for now.

When dealing with unintuitive quantum effects like this, just remember one thing: "trust the math". As long as the math checks out, you can be confident that the quantum effects are real and can be exploited in quantum algorithms.

Exercises

Exercise 1

Exercise 2