# Quantum teleportation

Quantum states are in many ways different from information stored in classical systems – quantum states cannot be cloned and quantum information cannot be erased. However, it turns out that quantum information can be transmitted and replicated by combining a quantum channel and a classical channel – a process known as quantum teleportation.

## Bell states

Before we explain the teleportation algorithm, let us quickly recall some definitions. Suppose that we are given a system with two qubits, say A and B. Consider the unitary operator $U = C_{NOT} (H \otimes I)$

i.e. the operator that applies a Hadamard operator to qubit A and then a CNOT gate with control qubit A and target qubit B. Let us calculate the action of this operator on the basis  state $|0 \rangle |0 \rangle$. $U |0 \rangle |0 \rangle = \frac{1}{\sqrt{2}} C_{NOT} (|0 \rangle |0 \rangle + |1 \rangle |0 \rangle) = \frac{1}{\sqrt{2}} (|00 \rangle + |11 \rangle)$

It is not difficult to show that this state cannot be written as a product – it is an entangled state. Traditionally, this state is called a Bell state.

Now the operator U is unitary, and it therefore maps a Hilbert space basis to a Hilbert space basis. We can therefore obtain a basis of our two-qubit Hilbert space that consists of entangled states by applying the transformation U to the elements of the computational basis. The resulting states are easily calculated and are as follows (we use the notation from ).

Computational basis vector Bell basis vector $|00 \rangle$ $|\beta_{00} \rangle = \frac{1}{\sqrt{2}} (|00 \rangle + |11 \rangle)$ $|01 \rangle$ $|\beta_{01} \rangle = \frac{1}{\sqrt{2}} (|01 \rangle + |10 \rangle)$ $|10 \rangle$ $|\beta_{10} \rangle = \frac{1}{\sqrt{2}} (|00 \rangle - |11 \rangle)$ $|11 \rangle$ $|\beta_{11} \rangle = \frac{1}{\sqrt{2}} (|01 \rangle - |10 \rangle)$

Of course we can also reverse this process and express the elements of the computational basis in terms of the Bell basis. For later reference, we note that we can write the computational basis as $|00\rangle = \frac{1}{\sqrt{2}} (\beta_{00} + \beta_{10})$ $|01\rangle = \frac{1}{\sqrt{2}} (\beta_{01} + \beta_{11})$ $|10\rangle = \frac{1}{\sqrt{2}} (\beta_{01} - \beta_{11})$ $|11\rangle = \frac{1}{\sqrt{2}} (\beta_{00} - \beta_{10})$

## Quantum teleportation

Let us now turn to the real topic of this post – quantum teleportation. Suppose that Alice is in possession of a qubit that captures some quantum state $|\psi \rangle = a |0 \rangle + b|1 \rangle$, stored in some sort of quantum device, which could be a superconducting qubit, a trapped ion, a polarized photon or any other physical implementation of a qubit. Bob is operating a similar device. The task is to create a quantum state in Bob’s device which is identical to the state stored in Alice device.

To be able to do this, Alice and Bob will need some communication channel. Let us suppose that there is a classical channel that Alice can use to transmit a finite number of bits to Bob. Is that sufficient to transmit the state of the qubit?

Obviously, the answer is no. Alice will not be able to measure both coefficients a and b, and even if she would find a way to do this, she would be faced with the challenge of transmitting two complex numbers with arbitrary precision using a finite string of bits.

At the other extreme, if Alice and Bob were able to fully entangle their quantum devices, they would probably be able to transmit the state. They could for instance implement a swap gate to move the state from Alice device onto Bob’s device.

So if Alice and Bob are able to perform arbitrary entangled quantum operations on the combined system consisting of Alice qubit and Bob’s qubit, a transmission is possible. But what happens if the devices are separated? It turns out that a transmission is still possible based on the classical channel provided that Alice and Bob had a chance to prepare a common state before Alice prepares $|\psi \rangle$. In fact, this common state does not depend at all on $|\psi \rangle$, and at no point during the process, any knowledge of the state $|\psi \rangle$ is needed.

The mechanism first described in  works as follows. We need three qubits that we denote by A, B and C. The qubit C contains the unknown state $|\psi \rangle_C$ (we use the lower index to denote the qubit in which the state lives). We also assume that Alice is able to perform arbitrary measurements and operations on A and C, and that B similarly controls qubit B.

The first step in the algorithm is to prepare an entangled state $|\beta_{00}^{AB} \rangle = \frac{1}{\sqrt{2}} \big[ |0\rangle _A |0 \rangle _B+ |1\rangle_A |1 \rangle_B) \big]$

Here the upper index at $\beta$ indicates the two qubits in which the Bell state vector lives. This is the common state mentioned above, and it can obviously be prepared upfront, without having the state $|\psi \rangle_C$. Bob and Alice could even prepare an entire repository of such states, ready for being used when the actual transmission should take place.

Now let us look at the state of the combined system consisting of all three qubits. $|\beta_{00}^{AB} \rangle |\psi\rangle_C = \frac{1}{\sqrt{2}} \big[ a |000 \rangle + a |110 \rangle + b |001\rangle + b |111 \rangle \big]$

For a reason that will become clear in a second, we will now adapt the tensor product order and choose the new order A-C-B. Then our state is (simply swap the last two qubits) $\frac{1}{\sqrt{2}} \big[ a |000 \rangle + a |101 \rangle + b |010\rangle + b |111 \rangle \big]$

Now instead of using the computational basis for our three-qubit system, we could as well use the basis that is given by the Bell basis of the qubits A and C and the computational basis of B, i.e. $|\beta^{AC}_{ij} \rangle |k\rangle$. Using the table above, we can express our state in this basis. This will give us four terms that contain a and four terms that contain b. The terms that contain a are $\frac{a}{2} \big[ |\beta^{AC}_{00}\rangle |0 \rangle + |\beta^{AC}_{10}\rangle |0 \rangle + |\beta^{AC}_{01}\rangle |1 \rangle - |\beta^{AC}_{11} \rangle |1 \rangle \big]$

and the terms that contain b are $\frac{b}{2} \big[ |\beta^{AC}_{01}\rangle |0 \rangle + |\beta^{AC}_{11}\rangle |0 \rangle + |\beta^{AC}_{00}\rangle |1 \rangle - |\beta^{AC}_{10} \rangle |1 \rangle \big]$

So far we have not done anything to our state, we have simply re-written the state as an expansion into a different basis. Now Alice conducts a measurement – but she measures A and C in the Bell basis. This will project the state onto one of the basis vectors $|\beta_{ij}^{AC}\rangle$. Thus there are four different possible outcomes of this measurement, which we can read off from the expansion in terms of the Bell basis above.

Measurement State of qubit B $\beta_{00}$ $a|0\rangle + b|1\rangle = |\psi \rangle$ $\beta_{01}$ $a|1\rangle + b|0\rangle = X|\psi \rangle$ $\beta_{10}$ $a|0\rangle - b|1\rangle = Z|\psi \rangle$ $\beta_{11}$ $b|0\rangle - a|1\rangle = -XZ|\psi \rangle$

Now Alice sends the outcome of the measurement to Bob using the classical channel. The table above shows that the state in Bob’s qubit B is now the result of a unitary transformation which depends only on the measurement outcome. Thus if Bob receives the measurement outcomes (two bits), he can do a lookup on the table above and apply the inverse transformation on his qubit. If, for instance, he receives the measurement outcome 01, he can apply a Pauli X gate to his qubit and then knows that the state in this qubit will be identical to the original state $|\psi \rangle$. The teleportation process is complete.

## Implementing teleportation as a circuit

Let us now build a circuit that models the teleportation procedure. The first part that we need is a circuit that creates a Bell state (or, more generally, transforms the elements of the computational basis into the Bell basis). This is achieved by the circuit below (which again uses the Qiskit convention that the most significant qubit q1 is the leftmost one in the tensor product). It is obvious how this circuit can be reversed – as both individual gates used are there own inverse, we simply need to reverse their order to obtain a circuit that maps the Bell basis to the computational basis.

Let us now see how the overall teleportation circuit looks like. Here is the circuit (drawn with Qiskit). Let us see how this relates to the above description. First, the role of the qubits is, from the top to the bottom

C = q
A = q
B = q

The first two gates (the Hadamard and the CNOT gate) act on qubits A and B and create the Bell basis vector $\beta_{00}$ from the initial state. This state is the shared entangled state that Alice and Bob prepare upfront.

The next section of the circuit operates on the qubits A and C and realizes the measurement in the Bell basis. We first use a combination of another CNOT gate and another Hadamard gate to map the Bell basis to the computational basis and then perform a measurement – these steps are equivalent to doing a measurement in the Bell basis directly.

Finally, we have a controlled X gate and a controlled Z gate. As described above, these gates apply the conditional corrections to Bob’s state in qubit B that depend on the outcome of the measurement. As a result, the state of qubit B will be the same as the initial state of qubit C.

How can we test this circuit? One approach could be to add a pre-processing part and a post-processing part to our teleportation circuit. The pre-processing gate acts with one or more gates on qubit C to put this qubit into a defined state. We then apply the teleportation circuit. Finally, we apply a post-processing circuit, namely the inverse sequence of gates to the “target” of the teleportation, i.e. to qubit B. If the teleportation works, the pre- and post-steps act on the same logical state and therefore cancel each other. Thus the final result in qubit B will be the initial state of qubit C, i.e. $|0 \rangle$. Therefore, a measurement of this qubit will always yield zero. Thus our final test circuit looks as follows. When we run this on a simulator, the result will be as expected – the value in the classical register c2 after the measurement will always be zero. I have not been able to test this on real hardware as the IBM device does not (yet?) support classical conditional gates, but the result is predictable – we would probably get the same output with some noise. If you want to play with the circuits yourself, you can, as always, find the code in my GitHub repository.

So this is quantum teleportation – not quite as mysterious as the name suggests. In particular, we do not magically teleport particles and actual matter, but simply quantum information, based on a clever split of the information contained in an unknown state into a classical part, transmitted in two classical bit, and a quantum part that we can prepare upfront. We also do not violate special relativity – to complete the process, we depend on the measurement outcomes that are transmitted via a classical channel and therefore not faster than the speed of light. It is also important to note that quantum teleportation does also not violate the “no-cloning”-principle – the measurements let the original state collapse and therefore we do not end up with two copies of the same state.

Quantum teleportation has already been demonstrated in reality several times. The first experimental verification was published in  in 1997. Since then, the distance between the involved parties “Alice” and “Bob” has been gradually increased. In 2017, a chinese team reported a teleportation of a single qubit between a ground observatory to a low-Earth-orbit satellite (see ).

The term quantum teleportation is also used in the context of quantum error correction for a circuit that uses a measurement on an entangled state to bring one of the involved qubits into a specific state. When using a surface code, for instance, this technique – also called gate teleportation – is used to implement the T gate on the level of logical qubits. We refer to  for a more detailed discussion of this pattern.

## References

1. M. Nielsen, I. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, Cambridge 2010
2. C.H. Bennett, G. Brassard, C. Crepeau, R. Jozsa, A Peres, W.K. Wootter, Teleporting an Unknown Quantum State via Dual Classical and EPR Channels, Phys. Rev. Lett. 70, March 1993
3. D. Bouwmeester, Jian-Wei Pan, K. Mattle, M. Eibl, H. Weinfurter, A. Zeilinger, Experimental quantum teleportation, Nature Vol. 390, Dec. 1997
4. Ji-Gang Ren et. al., Ground-to-satellite quantum teleportation, Nature Vol. 549, September 2017
5. D. Gottesman, I. L. Chuang, Quantum Teleportation is a Universal Computational Primitive, arXiv:quant-ph/9908010