4.2.3 Quantum teleportation

While it is not possible to clone quantum bits, we can certainly move quantum bits from one place to another. Indeed, any qubit is stored inside some physical object or particle that carries the qubit. For example, if Alice stores her qubit as the polarization of a photon, she can simply send this photon to Bob. What is more surprising, however, is that one can move a quantum bit from one place to another by sending a finite number of classical bits (in fact, two bits suffice). The reason this is surprising is because a general qubit state |ψ⁢(θ)⟩\left|\psi(\theta)\right\rangle is specified by an arbitrary angle θ\theta, see Eq. 2.5, which generally cannot be encoded in a finite number of bits. Because of this surprising property this procedure is known as teleportation. We will see shortly that we need entanglement to make it work!

The starting point for teleportation is the following scenario. We imagine that Alice has two qubits and Bob has one qubit. Alice’s first qubit is the message qubit that she wants to send to Bob, and it starts out in some arbitrary state |ψ⟩\left|\psi\right\rangle – which may well be unknown to Alice herself! Her second qubit and Bob’s qubit are in a maximally entangled state |Φ+⟩\left|\Phi^{+}\right\rangle. Thus, the three qubits are in the following state:

|ψ⟩⊗|Φ+⟩,\displaystyle\left|\psi\right\rangle\otimes\left|\Phi^{+}\right\rangle,

where the first two qubits belong to Alice and the last qubit belongs to Bob (recall from Eq. 3.69 that |Φ+⟩\left|\Phi^{+}\right\rangle is a two-qubit state). The following Quirky circuit shows what this looks like in case that Alice wants to send a qubit in state |ψ⟩=|1⟩=NOT⁢|0⟩\left|\psi\right\rangle=\left|1\right\rangle=\mathrm{NOT}\left|0\right\rangle:

What should be the next step? Eventually, the goal is for Bob to obtain Alice’s qubit – since we cannot clone the qubit, this means that Alice has to perform some action that ‘destroys’ her qubit. A good way of going about this is to perform a measurement. But simply measuring Alice’s two qubits will do them no good, since we know that we cannot infer the state of |ψ⟩\left|\psi\right\rangle from a single measurement. This means that Alice should first apply some quantum operation on both of her qubits and then measure them. It turns out that the correct strategy for her is to perform the same operations that you used to discriminate the four Bell states in 3.13:

Note that in this example Alice’s four measurement outcomes occur with 25% each. Here is another example, corresponding to the situation where Alice tries to teleport the state |ψ⁢(0.42)⟩\left|\psi(0.42)\right\rangle instead:

It seems that, whatever the state of Alice’s message qubit, the four probabilities are always the same. This is already encouraging, since it means that the measurement gives Alice no information at all about her message qubit! Recall that we want her initial state |ψ⟩\left|\psi\right\rangle to end up completely with Bob, meaning that she should not extract or keep any information about this state around.

In general, the state right before Alice’s measurements looks as follows:

(H⊗I⊗I)⁢(CNOT1→2⊗I)⁢(|ψ⟩⊗|Φ+⟩)\displaystyle\left(H\otimes I\otimes I\right)\left(\mathrm{CNOT}_{1\to 2}% \otimes I\right)\left(\left|\psi\right\rangle\otimes\left|\Phi^{+}\right% \rangle\right)

Note that we have to be careful, since the last two tensor products are not aligned (cf. 4.3). Thus we compute:

(H⊗I⊗I)⁢(CNOT1→2⊗I)⁢(|ψ⟩⊗|Φ+⟩)\displaystyle\quad\left(H\otimes I\otimes I\right)\left(\mathrm{CNOT}_{1\to 2}% \otimes I\right)\left(\left|\psi\right\rangle\otimes\left|\Phi^{+}\right% \rangle\right)
=12⁢(H⊗I⊗I)⁢(CNOT1→2⊗I)⁢(ψ0⁢|000⟩+ψ0⁢|011⟩+ψ1⁢|100⟩+ψ1⁢|111⟩)\displaystyle=\frac{1}{\sqrt{2}}\left(H\otimes I\otimes I\right)\left(\mathrm{% CNOT}_{1\to 2}\otimes I\right)\left(\psi_{0}\left|000\right\rangle+\psi_{0}% \left|011\right\rangle+\psi_{1}\left|100\right\rangle+\psi_{1}\left|111\right% \rangle\right)
=12⁢(H⊗I⊗I)⁢(ψ0⁢|000⟩+ψ0⁢|011⟩+ψ1⁢|110⟩+ψ1⁢|101⟩)\displaystyle=\frac{1}{\sqrt{2}}\left(H\otimes I\otimes I\right)\left(\psi_{0}% \left|000\right\rangle+\psi_{0}\left|011\right\rangle+\psi_{1}\left|110\right% \rangle+\psi_{1}\left|101\right\rangle\right)
=12⁢(ψ0⁢|000⟩+ψ0⁢|100⟩+ψ0⁢|011⟩+ψ0⁢|111⟩+ψ1⁢|010⟩−ψ1⁢|110⟩+ψ1⁢|001⟩−ψ1⁢|101⟩)\displaystyle=\frac{1}{2}\left(\psi_{0}\left|000\right\rangle+\psi_{0}\left|10% 0\right\rangle+\psi_{0}\left|011\right\rangle+\psi_{0}\left|111\right\rangle+% \psi_{1}\left|010\right\rangle-\psi_{1}\left|110\right\rangle+\psi_{1}\left|00% 1\right\rangle-\psi_{1}\left|101\right\rangle\right)
=|00⟩⊗ψ0⁢|0⟩+ψ1⁢|1⟩2+|01⟩⊗ψ1⁢|0⟩+ψ0⁢|1⟩2\displaystyle=\left|00\right\rangle\otimes\frac{\psi_{0}\left|0\right\rangle+% \psi_{1}\left|1\right\rangle}{2}+\left|01\right\rangle\otimes\frac{\psi_{1}% \left|0\right\rangle+\psi_{0}\left|1\right\rangle}{2}
+|10⟩⊗ψ0⁢|0⟩−ψ1⁢|1⟩2+|11⟩⊗−ψ1⁢|0⟩+ψ0⁢|1⟩2.\displaystyle\,+\left|10\right\rangle\otimes\frac{\psi_{0}\left|0\right\rangle% -\psi_{1}\left|1\right\rangle}{2}+\left|11\right\rangle\otimes\frac{-\psi_{1}% \left|0\right\rangle+\psi_{0}\left|1\right\rangle}{2}.

This is the overall state right before Alice’s measurement. We can compute the probability of her measurement outcomes as discussed in Eq. 4.14. Namely, to compute the probability pa,bp_{a,b} of outcome [a⁢b][ab] we simply add the squared amplitudes of the relevant basis states |a⁢b⁢0⟩\left|ab0\right\rangle and |a⁢b⁢1⟩\left|ab1\right\rangle. In each case, one of the amplitudes is ψ0/2\psi_{0}/2 and the other is ±ψ1/2\pm\psi_{1}/2, so their squares always sum to the same result:

p00\displaystyle p_{00} =(ψ02)2+(ψ12)2=ψ02+ψ124=14,\displaystyle=\left(\frac{\psi_{0}}{2}\right)^{2}+\left(\frac{\psi_{1}}{2}% \right)^{2}=\frac{\psi_{0}^{2}+\psi_{1}^{2}}{4}=\frac{1}{4},
p01\displaystyle p_{01} =(ψ12)2+(ψ02)2=14,\displaystyle=\left(\frac{\psi_{1}}{2}\right)^{2}+\left(\frac{\psi_{0}}{2}% \right)^{2}=\frac{1}{4},
p10\displaystyle p_{10} =(ψ02)2+(−ψ12)2=14,\displaystyle=\left(\frac{\psi_{0}}{2}\right)^{2}+\left(\frac{-\psi_{1}}{2}% \right)^{2}=\frac{1}{4},
p11\displaystyle p_{11} =(−ψ12)2+(ψ02)2=14.\displaystyle=\left(\frac{-\psi_{1}}{2}\right)^{2}+\left(\frac{\psi_{0}}{2}% \right)^{2}=\frac{1}{4}.

Each outcome occurs with 25%. This confirms what we previously observed with Quirky.

After Alice’s measurement, the only remaining quantum bit is Bob’s qubit. Let us denote its state by |ψa,b′⟩\left|\smash[b]{\psi^{\prime}}_{a,b}\right\rangle, since it depends on Alice’s measurement outcome. We can determine it using Eq. 4.15 or, much easier, using the grouping and normalization method in Eq. 4.13. Either way, the result is that Bob’s qubit is in one of the following four states:

|ψ00′⟩\displaystyle\left|\psi^{\prime}_{00}\right\rangle =ψ0⁢|0⟩+ψ1⁢|1⟩,\displaystyle=\psi_{0}\left|0\right\rangle+\psi_{1}\left|1\right\rangle,
|ψ01′⟩\displaystyle\left|\psi^{\prime}_{01}\right\rangle =ψ1⁢|0⟩+ψ0⁢|1⟩,\displaystyle=\psi_{1}\left|0\right\rangle+\psi_{0}\left|1\right\rangle,
|ψ10′⟩\displaystyle\left|\psi^{\prime}_{10}\right\rangle =ψ0⁢|0⟩−ψ1⁢|1⟩,\displaystyle=\psi_{0}\left|0\right\rangle-\psi_{1}\left|1\right\rangle,
|ψ11′⟩\displaystyle\left|\psi^{\prime}_{11}\right\rangle =−ψ1⁢|0⟩+ψ0⁢|1⟩.\displaystyle=-\psi_{1}\left|0\right\rangle+\psi_{0}\left|1\right\rangle.

When [a⁢b]=[00][ab]=[00], Bob’s state coincides exactly with Alice’s original state |ψ⟩\left|\psi\right\rangle that she wanted to teleport:

|ψ00′⟩=|ψ⟩.\displaystyle\left|\psi^{\prime}_{00}\right\rangle=\left|\psi\right\rangle.

In the other three cases Bob’s quantum state is slightly garbled. However, if Alice sends her measurement outcomes (i.e., the two bits aa and bb) to Bob then he can apply an appropriate correction operation to ‘fix up’ his state:

NOT⁢|ψ01′⟩\displaystyle\mathrm{NOT}\left|\psi^{\prime}_{01}\right\rangle =|ψ⟩,\displaystyle=\left|\psi\right\rangle,
Z⁢|ψ10′⟩\displaystyle Z\left|\psi^{\prime}_{10}\right\rangle =|ψ⟩,\displaystyle=\left|\psi\right\rangle,
Z⁢NOT⁢|ψ11′⟩\displaystyle Z\,\mathrm{NOT}\left|\psi^{\prime}_{11}\right\rangle =|ψ⟩.\displaystyle=\left|\psi\right\rangle.

These four cases can be summarized in the following simple procedure for Bob:

  1. 1.
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    look at bit bb and if b=1b=1 then apply NOT\mathrm{NOT},

  2. 2.
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    look at bit aa and if a=1a=1 then apply ZZ.

We can implement this using a controlled-NOT and a controlled-ZZ operation, where the controls are classical bits. You can create the controlled-Z operation in the same way as the controlled NOT operation – we discussed this at the end of Section 3.2.4. Phew, this was quite a bit of work!

What does all this look like in Quirky? The end result is the following quantum circuit:

We added a gray box about the relevant part – the teleportation circuit – to separate it from the rest of the circuit that creates the input states. Here is a picture that shows only the teleportation circuit and the creation of the maximally entangled state, without specifying Alice’s first qubit:

We also cut off the wires for Alice’s two classical bits, since they are no longer of interest once Alice has sent the two measurement outcomes to Bob. We also removed the probability display, since it is not an actual quantum operation but just a way for us to inspect the circuit in Quirky. Thus, there is one input qubit for Alice’s message, two qubits for the maximally entangled state, and one output qubit on Bob’s side. The effect of the teleportation is simply to pass through an arbitrary state |ψ⟩\left|\psi\right\rangle from the input qubit on Alice’s side to the output qubit on Bob’s side. Crucially, inside the box, only classical bits are being sent from Alice to Bob!

Let’s make sure that we didn’t do a mistake. Since we expect that Bob’s qubit ends up in state |1⟩\left|1\right\rangle after the teleportation procedure, we can add a simple measurement to test if this is indeed what happened:

Indeed, we get outcome 1 with 100% probability, which confirms that we successfully teleported the |1⟩\left|1\right\rangle state! Now, |1⟩\left|1\right\rangle is not a particularly interesting state to teleport. How about the |+⟩\left|+\right\rangle state, which caused us trouble before when we discussed cloning? In the following homework you will test the teleportation circuit first for the |+⟩\left|+\right\rangle state and then for arbitrary single-qubit states.

Homework 4.2 (Testing teleportation).
  1. 1.
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    Why does the following circuit confirm that the |+⟩\left|+\right\rangle state was teleported correctly?

[Uncaptioned image]

  1. 2.
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    How you can you similarly test if a quantum state |ψ⁢(θ)⟩\left|\psi(\theta)\right\rangle was teleported correctly?

Hack.
  1. 1.
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    On the one hand, H⁢|0⟩=|+⟩H\left|0\right\rangle=\left|+\right\rangle, so Alice is indeed trying to teleport the right state. On the other hand, Bob’s probability display shows that the state right before the measurement is |0⟩\left|0\right\rangle. Since H⁢|+⟩=|0⟩H\left|+\right\rangle=\left|0\right\rangle, this means that the state before the Hadamard gate HH was |+⟩\left|+\right\rangle. This is what we wanted to check.

  2. 2.
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    We can use the same method as in the previous part of this problem, but this time Alice can apply U⁢(θ)U(\theta) to the state |0⟩\left|0\right\rangle to prepare the state |ψ⁢(θ)⟩\left|\psi(\theta)\right\rangle. If this state is correctly teleported, then Bob ends up with his qubit being in the state |ψ⁢(θ)⟩\left|\psi(\theta)\right\rangle, and hence if he applies U⁢(−θ)U(-\theta) to his qubit, he should end up with his qubit being in the state |0⟩\left|0\right\rangle. Using Quirky, we can confirm that Bob, after the application of U⁢(−θ)U(-\theta) indeed measures 0 with probaility 100%100\%, (or equivalently, that he measures 11 with 0%0\%). This confirms that Bob’s qubit just before the application of U⁢(−θ)U(-\theta) was in the state |ψ⁢(θ)⟩\left|\psi(\theta)\right\rangle, and hence proves that the state was correctly teleported.

    The diagram below shows the resulting circuit for θ=0.42\theta=0.42.

    [Uncaptioned image]

The teleportation circuit is quite remarkable. It allows to send a quantum bit from Alice to Bob by transmitting two classical bits only, provided that Alice and Bob share a maximally entangled state. However, it is not only useful for applications (we will discuss some of them below), but it also gives some interesting perspective on the difference between classical and quantum bits. Recall that at the end of last week we discussed superdense coding, which allowed us to send two bits by transmitting a single quantum bit, again using one maximally entangled state beween Alice and Bob. This shows that:

Given a free supply of quantum entanglement, sending two bits is

completely equivalent to sending one quantum bit!

Note that in neither teleportation nor superdense coding can we reuse the maximally entangled state once the procedure is finished – it is the ‘fuel’ that is consumed by either procedure.