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+---
+title: "Quantum teleportation"
+date: 2021-03-07
+categories:
+- Quantum information
+layout: "concept"
+---
+
+**Quantum teleportation** is a method to transfer quantum information
+between systems without the use of a quantum channel.
+It is based on [quantum entanglement](/know/concept/quantum-entanglement/).
+
+Suppose that Alice has a qubit $\Ket{q}_{A'}$ that she wants to send to Bob.
+Since she has not measured it yet, she does not know $\alpha$ or $\beta$;
+she just wants Bob to get the same qubit:
+
+$$\begin{aligned}
+    \Ket{q}
+    = \alpha \Ket{0}_{A'} + \beta \Ket{1}_{A'}
+\end{aligned}$$
+
+She can only directly communicate with Bob over a classical channel.
+This is not enough: even if Alice did know $\alpha$ and $\beta$ 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](/know/concept/bell-state/),
+e.g. $\ket{\Phi^+}_{AB}$ (it does not matter which Bell state it is)
+The state of the composite system is then as follows,
+with $A'$ being Alice' qubit, $A$ her side of the Bell state, and $B$ Bob's side:
+
+$$\begin{aligned}
+    \Ket{q}_{A'} \otimes \ket{\Phi^+}_{AB}
+    &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{A'} \Big( \Ket{00} + \Ket{11} \Big)_{AB}
+    \\
+    &= \frac{1}{\sqrt{2}} \Big( \alpha \Ket{000} + \beta \Ket{100}
+    + \alpha \Ket{011} + \beta \Ket{111} \Big)_{A'AB}
+\end{aligned}$$
+
+Now, observe that we can write any combination of $\Ket{0}$ and $\Ket{1}$
+in the Bell basis like so:
+
+$$\begin{aligned}
+    \Ket{00}
+    &= \frac{\Ket{\Phi^{+}} + \Ket{\Phi^{-}}}{\sqrt{2}}
+    \qquad \quad
+    \Ket{11}
+    = \frac{\Ket{\Phi^{+}} - \Ket{\Phi^{-}}}{\sqrt{2}}
+    \\
+    \Ket{01}
+    &= \frac{\Ket{\Psi^{+}} + \Ket{\Psi^{-}}}{\sqrt{2}}
+    \qquad \quad
+    \Ket{10}
+    = \frac{\Ket{\Psi^{+}} - \Ket{\Psi^{-}}}{\sqrt{2}}
+\end{aligned}$$
+
+Using this, we can rewrite our previous result in terms of the Bell states as follows:
+
+$$\begin{aligned}
+    \Ket{q}_{A'} \ket{\Phi^+}_{AB}
+    &= \frac{\alpha}{2} \Big( \ket{\Phi^{+}} + \ket{\Phi^{-}} \Big)_{A'A} \Ket{0}_B
+    + \frac{\beta}{2} \Big( \ket{\Psi^{+}} - \ket{\Psi^{-}} \Big)_{A'A} \Ket{0}_B
+    \\
+    &+ \frac{\alpha}{2} \Big( \ket{\Psi^{+}} + \ket{\Psi^{-}} \Big)_{A'A} \Ket{1}_B
+    + \frac{\beta}{2} \Big( \ket{\Phi^{+}} - \ket{\Phi^{-}} \Big)_{A'A} \Ket{1}_B
+\end{aligned}$$
+
+If we group all terms according to the Bell states,
+we end up with an interesting expression:
+
+$$\begin{aligned}
+    \Ket{q}_{A'} \ket{\Phi^+}_{AB}
+    = \frac{1}{2} \bigg( &\ket{\Phi^{+}}_{A'A} \Big( \alpha \Ket{0} + \beta \Ket{1} \Big)_{B}
+    + \ket{\Phi^{-}}_{A'A} \Big( \alpha \Ket{0} - \beta \Ket{1} \Big)_{B}
+    \\
+    + &\ket{\Psi^{+}}_{A'A} \Big( \alpha \Ket{1} + \beta \Ket{0} \Big)_{B}
+    + \ket{\Psi^{-}}_{A'A} \Big( \alpha \Ket{1} - \beta \Ket{0} \Big)_{B} \bigg)
+\end{aligned}$$
+
+Thus, purely due to entanglement,
+Bob's qubit $B$ is in a superposition of the following states:
+
+$$\begin{aligned}
+    \Ket{q}
+    &= \alpha \Ket{0} + \beta \Ket{1}
+    \qquad \quad
+    \quad \hat{\sigma}_z \Ket{q}
+    = \alpha \Ket{0} - \beta \Ket{1}
+    \\
+    \hat{\sigma}_x \Ket{q}
+    &= \alpha \Ket{1} + \beta \Ket{0}
+    \qquad \quad
+    \hat{\sigma}_x \hat{\sigma}_z \Ket{q}
+    = \alpha \Ket{1} - \beta \Ket{0}
+\end{aligned}$$
+
+Consequently, Alice and Bob are sharing (or, to be precise, seeing different sides of)
+the following entangled three-qubit state:
+
+$$\begin{aligned}
+    \Ket{q}_{A'} \ket{\Phi^+}_{AB}
+    = \frac{1}{2} \bigg( &\ket{\Phi^{+}}_{A'A} \Big( \Ket{q} \Big)_B \quad\, + \ket{\Phi^{-}}_{A'A} \Big( \hat{\sigma}_z \Ket{q} \Big)_B
+    \\
+    + &\ket{\Psi^{+}}_{A'A} \Big( \hat{\sigma}_x \Ket{q} \Big)_B + \ket{\Psi^{-}}_{A'A} \Big( \hat{\sigma}_x \hat{\sigma}_z \Ket{q} \Big)_B \bigg)
+\end{aligned}$$
+
+The point is that, thanks to the initial entanglement between Alice and Bob,
+adding $\Ket{q}_{A'}$ 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 $A'A$ in the Bell basis.
+Consequently, $A'A$ collapses into one of
+$\ket{\Phi^{+}}$, $\ket{\Phi^{-}}$, $\ket{\Psi^{+}}$, $\ket{\Psi^{-}}$
+with equal probability, and she knows which.
+This collapse leaves Bob's side $B$ in $\Ket{q}$, $\hat{\sigma}_z \Ket{q}$,
+$\hat{\sigma}_x \Ket{q}$, or $\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$, respectively.
+The entanglement between $A$ and $B$ is thus broken,
+and instead Alice has local entanglement between $A'$ and $A$.
+
+She then uses the classical channel to tell Bob her result,
+who then either does nothing (for $\Ket{q}$),
+applies $\hat{\sigma}_z$ (for $\hat{\sigma}_z \Ket{q}$),
+applies $\hat{\sigma}_x$ (for $\hat{\sigma}_x \Ket{q}$),
+or applies $\hat{\sigma}_z \hat{\sigma}_x$ (for $\hat{\sigma}_x \hat{\sigma}_z \Ket{q}$).
+Then, due to the fact that $\hat{\sigma}_x^2 = \hat{\sigma}_z^2 = \hat{I}$,
+he recovers $\Ket{q}$ in his local qubit $B$.
+
+This is not violating the [no-cloning theorem](/know/concept/no-cloning-theorem)
+because Alice does not require any knowledge of $\Ket{q}$,
+and after the measurement, her qubit $A'$ 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 $\ket{\Phi^{+}}_{AB}$ 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 $\ket{\Phi^{+}}_{AB}$, which contains nothing of $\Ket{q}$.
+
+
+## References
+1.  J.B. Brask,
+    *Quantum information: lecture notes*,
+    2021, unpublished.