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+---
+title: "Quantum teleportation"
+firstLetter: "Q"
+publishDate: 2021-03-07
+categories:
+- Quantum information
+
+date: 2021-03-07T20:30:30+01:00
+draft: false
+markup: pandoc
+---
+
+# 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](/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,
+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:
+
+$$\begin{aligned}
+    \ket{q}
+    &= \alpha \ket{0} + \beta \ket{1}
+    \qquad \quad
+    \quad \hat{\sigma}_z \ket{q}
+    = \alpha \ket{0} - \beta \ket{1}
+    \\
+    \hat{\sigma}_x \ket{q}
+    &= \alpha \ket{1} + \beta \ket{0}
+    \qquad \quad
+    \hat{\sigma}_x \hat{\sigma}_z \ket{q}
+    = \alpha \ket{0} - \beta \ket{1}
+\end{aligned}$$
+
+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 $\ket{q}$.