From 540d23bff03bedbc8f68287d71c8b5e7dc54b054 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Mon, 8 Mar 2021 15:04:06 +0100 Subject: Expand knowledge base --- .../know/concept/quantum-teleportation/index.pdc | 145 +++++++++++++++++++++ 1 file changed, 145 insertions(+) create mode 100644 content/know/concept/quantum-teleportation/index.pdc (limited to 'content/know/concept/quantum-teleportation/index.pdc') diff --git a/content/know/concept/quantum-teleportation/index.pdc b/content/know/concept/quantum-teleportation/index.pdc new file mode 100644 index 0000000..3287544 --- /dev/null +++ b/content/know/concept/quantum-teleportation/index.pdc @@ -0,0 +1,145 @@ +--- +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 $\ket{q}$. -- cgit v1.2.3