Teleportation

We saw at the end of the previous article how the idea of entanglement can be used to coordinate two qubits even if they are very far apart. However we also mentioned that this does not allow us to send information faster than the speed of light.

In this article, we'll delve into why that is by constructing communication channels using our entangled qubits and analysing their limitations.

Let's start by setting up a hypothetical scenario where Alice and Bob get together to create an entangled pair: $\ket{\beta_{00}} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$. Alice will keep one qubit and Bob will take the other. Alice will also have some arbitrary third qubit $\ket{\psi} = a\ket{0} + b\ket{1}$ that she wishes to send to Bob.

Her goal is to find some way to transfer the state of her qubit to Bob somehow, using their entangled pair as a sort of communication channel.

So we have a 3 qubit system. Let's express it mathematically. We'll put Alice's 2 qubits on the left and Bob's qubit on the right:

We'll apply a few operations on this system, so to keep track of each step, we'll label the whole state after $i$ steps as $\ket{\phi_i}$ beginning with our starting state:

In our starting configuration, the qubit alice wants to send is completely separate to the entangled pair. So any operations we perform to it won't have any effect on Bob's qubit. We might therefore guess that the first thing Alice needs to do is interact $\ket{\psi}$ with her half of the entangled pair in some way.

The simplest way we already know to do this is using a $\text{CNOT}$ gate. So let's see what happens when we apply a $\text{CNOT}$ to the first 2 qubits:

In order to collapse Bob's qubit into some definite state, Alice will need to perform some measurement, just like we've seen before. So, what would happen if Alice measured her qubits now?

Well, we know she would either get $\ket{00}$, $\ket{01}$, $\ket{10}$, or $\ket{11}$. But in either of these scenarios, we can see that Bob's qubit would collapse into a superposition of just a single ket. e.g. if Alice measured $\ket{00}$ Bob's qubit would simply become $\ket{0}$.

Our state is currently a superposition of 4 kets, so when alice measures her bits, there is only 1 ket left over to determine Bob's qubit. The original qubit we want to send is a superposition of 2 kets, so a good guess here would be to apply some operation that would increase the number of kets in our superposition to 8.

We know one such gate that can achieve this. It's the Hadamard gate. It splits a single ket into a superposition of 2 kets. So let's apply the Hadamard gate to Alice's qubit:

Now let's re-arrange the brackets:

This form is very helpful, because we can quite clearly see the outcome for Bob's qubit based on what Alice measures.

Before moving on, let's just take a look at what we've done so far and how it looks in a circuit diagram:

Those 2 symbols on the right represent measurement of the qubit.

Ok, so now going back to our state of $\ket{\phi_2}$, we can see that if Alice measures $\ket{00}$, Bob's qubit will collapse into $a\ket{0} + b\ket{1}$ which is exactly $\ket{\psi}$, the state we originally tried to send, so we are done!

Unfortunately, we can also see that Alice has a $(1/2)^2 = 1/4$ to measure each of the 4 possible kets. So what do we do in the scenario where Alice doesn't measure $\ket{00}$?

Well, we can see that the resulting state of Bob's qubit will be almost $\ket{\psi}$, but not quite. Perhaps Bob can apply some operation on his end to transform his qubit into the correct form?

If Alice measures $\ket{01}$, then Bob has: $a\ket{1} + b\ket{0}$. So the coefficients are the wrong way round. We know we can fix this by simply applying an $X$ gate. So if Bob applies an $X$ gate to his qubit, he will end up with $a\ket{0} + b\ket{1}$, which is the state we wanted to send.

If Alice measures $\ket{10}$, then Bob has $a\ket{0} - b\ket{1}$. So the coefficients are the right way round, but the wrong sign. We can fix this by applying a $Z$ gate. So if Bob applies a $Z$ gate to his qubit, he will end up with $a\ket{0} + b\ket{1}$.

If Alice measures $\ket{11}$, then Bob has $a\ket{1} - b\ket{0}$. So the coefficients are the wrong way round and the wrong sign. We can fix this by applying an $X$ gate and a $Z$ gate. So if Bob applies $ZX$his qubit, he will end up with $a\ket{0} + b\ket{1}$.

Observing this pattern, we see that the first bit of Alice's measurement tells Bob whether or not he must apply a $Z$ gate and the second bit tells Bob whether or not to apply an $X$. (Note that the $X$ gate must always be applied first and the $Z$ gate second)

So after measurement, Alice must send her 2 measured bits to Bob, in order for him to decode his entangled bit. We can represent all this in our circuit diagram:

So we can see how the outcomes of Alice's measurement can be applied to Bob's qubit as a $\text{CNOT}$ and $\text{cZ}$ (controlled-Z) gate.

And here we see the exact reason why this does NOT allow faster than light communication. Alice must communicate her 2 measured bits to Bob in order for Bob to know what operations to apply to his qubit. Either that, or Alice's qubits must be brought physically close to Bob's qubit in order to apply the two controlled gates.

Both of these tasks (communicating 2 bits of information or bringing physical qubits together) we know cannot be done faster than the speed of light. And therefore, it is impossible for Bob to know how to recover $\ket{\psi}$ quicker than the time it takes light to travel between him and Alice.