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 ∣q⟩A′\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⟩=α∣0⟩A′+β∣1⟩A′\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 (for which she would need 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 A′A' being Alice’ qubit, AA her side of the Bell state, and BB Bob’s side:

∣q⟩A′⊗∣Φ+⟩AB=12(α∣0⟩+β∣1⟩)A′⊗(∣00⟩+∣11⟩)AB=12(α∣000⟩+β∣100⟩+α∣011⟩+β∣111⟩)A′AB\begin{aligned} \Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB} &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \otimes \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+⟩+∣Φ−⟩2∣11⟩=∣Φ+⟩−∣Φ−⟩2∣01⟩=∣Ψ+⟩+∣Ψ−⟩2∣10⟩=∣Ψ+⟩−∣Ψ−⟩2\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:

∣q⟩A′∣Φ+⟩AB=α2(∣Φ+⟩+∣Φ−⟩)A′A∣0⟩B+β2(∣Ψ+⟩−∣Ψ−⟩)A′A∣0⟩B+α2(∣Ψ+⟩+∣Ψ−⟩)A′A∣1⟩B+β2(∣Φ+⟩−∣Φ−⟩)A′A∣1⟩B\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:

∣q⟩A′∣Φ+⟩AB=12(∣Φ+⟩A′A(α∣0⟩+β∣1⟩)B+∣Φ−⟩A′A(α∣0⟩−β∣1⟩)B+∣Ψ+⟩A′A(α∣1⟩+β∣0⟩)B+∣Ψ−⟩A′A(α∣1⟩−β∣0⟩)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}

Therefore, thanks to entanglement, Bob’s qubit BB is in a superposition of the following states, where σ^x\hat{\sigma}_x and σ^z\hat{\sigma}_z are Pauli matrices (see quantum gate):

∣q⟩=α∣0⟩+β∣1⟩σ^z∣q⟩=α∣0⟩−β∣1⟩σ^x∣q⟩=α∣1⟩+β∣0⟩σ^xσ^z∣q⟩=α∣1⟩−β∣0⟩\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 seeing different sides of this entangled three-qubit state:

∣q⟩A′∣Φ+⟩AB=12(∣Φ+⟩A′A(∣q⟩)B +∣Φ−⟩A′A(σ^z∣q⟩)B+∣Ψ+⟩A′A(σ^x∣q⟩)B+∣Ψ−⟩A′A(σ^xσ^z∣q⟩)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 ∣q⟩A′\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 A′AA'A in the Bell basis. Consequently, A′AA'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^z∣q⟩\hat{\sigma}_z \Ket{q}, σ^x∣q⟩\hat{\sigma}_x \Ket{q}, or σ^xσ^z∣q⟩\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 A′A' 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^z∣q⟩\hat{\sigma}_z \Ket{q}), applies σ^x\hat{\sigma}_x (for σ^x∣q⟩\hat{\sigma}_x \Ket{q}), or applies σ^zσ^x\hat{\sigma}_z \hat{\sigma}_x (for σ^xσ^z∣q⟩\hat{\sigma}_x \hat{\sigma}_z \Ket{q}). Then, thanks 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 A′A' 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}.

References

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 J.B. Brask, Quantum information: lecture notes, 2021, unpublished.