Categories: Quantum information.

Quantum teleportation

Quantum teleportation is a method to transfer quantum information between systems without the use of a quantum channel. It is based on quantum entanglement.

Suppose that Alice has a qubit qA\Ket{q}_{A'} that she wants to send to Bob. Since she has not measured it yet, she does not know α\alpha or β\beta; she just wants Bob to get the same qubit:

q=α0A+β1A\begin{aligned} \Ket{q} = \alpha \Ket{0}_{A'} + \beta \Ket{1}_{A'} \end{aligned}

She can only directly communicate with Bob over a classical channel. This is not enough: even if Alice did know α\alpha and β\beta exactly (which would need her having infinitely many copies to measure), sending an arbitrary real number requires an infinite amount of classical data.

However, between them, she and Bob also have an entangled Bell state, e.g. Φ+AB\ket{\Phi^+}_{AB} (it does not matter which Bell state it is) The state of the composite system is then as follows, with AA' being Alice’ qubit, AA her side of the Bell state, and BB Bob’s side:

qAΦ+AB=12(α0+β1)A(00+11)AB=12(α000+β100+α011+β111)AAB\begin{aligned} \Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB} &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \Big( \Ket{00} + \Ket{11} \Big)_{AB} \\ &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{000} + \beta \Ket{100} + \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB} \end{aligned}

Now, observe that we can write any combination of 0\Ket{0} and 1\Ket{1} in the Bell basis like so:

00=Φ++Φ211=Φ+Φ201=Ψ++Ψ210=Ψ+Ψ2\begin{aligned} \Ket{00} &= \frac{\Ket{\Phi^{+}} + \Ket{\Phi^{-}}}{\sqrt{2}} \qquad \quad \Ket{11} = \frac{\Ket{\Phi^{+}} - \Ket{\Phi^{-}}}{\sqrt{2}} \\ \Ket{01} &= \frac{\Ket{\Psi^{+}} + \Ket{\Psi^{-}}}{\sqrt{2}} \qquad \quad \Ket{10} = \frac{\Ket{\Psi^{+}} - \Ket{\Psi^{-}}}{\sqrt{2}} \end{aligned}

Using this, we can rewrite our previous result in terms of the Bell states as follows:

qAΦ+AB=α2(Φ++Φ)AA0B+β2(Ψ+Ψ)AA0B+α2(Ψ++Ψ)AA1B+β2(Φ+Φ)AA1B\begin{aligned} \Ket{q}_{A'} \ket{\Phi^+}_{AB} &= \frac{\alpha}{2} \Big( \ket{\Phi^{+}} + \ket{\Phi^{-}} \Big)_{A'A} \Ket{0}_B + \frac{\beta}{2} \Big( \ket{\Psi^{+}} - \ket{\Psi^{-}} \Big)_{A'A} \Ket{0}_B \\ &+ \frac{\alpha}{2} \Big( \ket{\Psi^{+}} + \ket{\Psi^{-}} \Big)_{A'A} \Ket{1}_B + \frac{\beta}{2} \Big( \ket{\Phi^{+}} - \ket{\Phi^{-}} \Big)_{A'A} \Ket{1}_B \end{aligned}

If we group all terms according to the Bell states, we end up with an interesting expression:

qAΦ+AB=12(Φ+AA(α0+β1)B+ΦAA(α0β1)B+Ψ+AA(α1+β0)B+ΨAA(α1β0)B)\begin{aligned} \Ket{q}_{A'} \ket{\Phi^+}_{AB} = \frac{1}{2} \bigg( &\ket{\Phi^{+}}_{A'A} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{B} + \ket{\Phi^{-}}_{A'A} \Big( \alpha \Ket{0} - \beta \Ket{1} \Big)_{B} \\ + &\ket{\Psi^{+}}_{A'A} \Big( \alpha \Ket{1} + \beta \Ket{0} \Big)_{B} + \ket{\Psi^{-}}_{A'A} \Big( \alpha \Ket{1} - \beta \Ket{0} \Big)_{B} \bigg) \end{aligned}

Thus, purely due to entanglement, Bob’s qubit BB is in a superposition of the following states:

q=α0+β1σ^zq=α0β1σ^xq=α1+β0σ^xσ^zq=α1β0\begin{aligned} \Ket{q} &= \alpha \Ket{0} + \beta \Ket{1} \qquad \quad \quad \hat{\sigma}_z \Ket{q} = \alpha \Ket{0} - \beta \Ket{1} \\ \hat{\sigma}_x \Ket{q} &= \alpha \Ket{1} + \beta \Ket{0} \qquad \quad \hat{\sigma}_x \hat{\sigma}_z \Ket{q} = \alpha \Ket{1} - \beta \Ket{0} \end{aligned}

Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of) the following entangled three-qubit state:

qAΦ+AB=12(Φ+AA(q)B+ΦAA(σ^zq)B+Ψ+AA(σ^xq)B+ΨAA(σ^xσ^zq)B)\begin{aligned} \Ket{q}_{A'} \ket{\Phi^+}_{AB} = \frac{1}{2} \bigg( &\ket{\Phi^{+}}_{A'A} \Big( \Ket{q} \Big)_B \quad\, + \ket{\Phi^{-}}_{A'A} \Big( \hat{\sigma}_z \Ket{q} \Big)_B \\ + &\ket{\Psi^{+}}_{A'A} \Big( \hat{\sigma}_x \Ket{q} \Big)_B + \ket{\Psi^{-}}_{A'A} \Big( \hat{\sigma}_x \hat{\sigma}_z \Ket{q} \Big)_B \bigg) \end{aligned}

The point is that, thanks to the initial entanglement between Alice and Bob, adding qA\Ket{q}_{A'} into the mix somehow “teleports” that information to Bob, although it is not in a usable form yet.

To finish the process, Alice measures her side AAA'A in the Bell basis. Consequently, AAA'A collapses into one of Φ+\ket{\Phi^{+}}, Φ\ket{\Phi^{-}}, Ψ+\ket{\Psi^{+}}, Ψ\ket{\Psi^{-}} with equal probability, and she knows which. This collapse leaves Bob’s side BB in q\Ket{q}, σ^zq\hat{\sigma}_z \Ket{q}, σ^xq\hat{\sigma}_x \Ket{q}, or σ^xσ^zq\hat{\sigma}_x \hat{\sigma}_z \Ket{q}, respectively. The entanglement between AA and BB is thus broken, and instead Alice has local entanglement between AA' and AA.

She then uses the classical channel to tell Bob her result, who then either does nothing (for q\Ket{q}), applies σ^z\hat{\sigma}_z (for σ^zq\hat{\sigma}_z \Ket{q}), applies σ^x\hat{\sigma}_x (for σ^xq\hat{\sigma}_x \Ket{q}), or applies σ^zσ^x\hat{\sigma}_z \hat{\sigma}_x (for σ^xσ^zq\hat{\sigma}_x \hat{\sigma}_z \Ket{q}). Then, due to the fact that σ^x2=σ^z2=I^\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}, he recovers q\Ket{q} in his local qubit BB.

This is not violating the no-cloning theorem because Alice does not require any knowledge of q\Ket{q}, and after the measurement, her qubit AA' will no longer be in that state. In other words, quantum teleportation moves states, rather than copying them.

Nor does this conflict with Einstein’s relativity, since the information travels no faster than light: the entangled Φ+AB\ket{\Phi^{+}}_{AB} state must be distributed in advance, and Alice’ declaration of her result is sent classically. Before receiving that, Bob only sees his side of the maximally entangled Bell state Φ+AB\ket{\Phi^{+}}_{AB}, which contains nothing of q\Ket{q}.


  1. J.B. Brask, Quantum information: lecture notes, 2021, unpublished.