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I saw the concept of quantum teleportation, and they say you have a pre-shared state between Alice and Bob, such as $\alpha\vert 1\rangle+\beta\vert 0\rangle$, with $\alpha=\beta=\frac{1}{\sqrt{2}}$. Then you apply a little circuit with a couple of CNOTs and Hs, then do a measurement which gives you two classical bits as a result. And then you send the two classical bits, and then bob recovers the state according to those two classical bits and the pre-shared state. However, I don't see how this would work when the state to transmit has, for example, $\alpha=\sqrt{0.1345}$ and $\beta=\sqrt{0.8655}$, because, you can actually get different measurements repeating the same experiment, and I don't see how you would get the example numbers from a maximally entangled state and two classical bits, seems impossible.

Probably I am misunderstanding the purpose of quantum teleportation or missing something big. Thanks for any help.

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  • $\begingroup$ Short answer: yes, quantum teleportation teleports arbitrary state. $\endgroup$
    – kludg
    Aug 2, 2022 at 17:20

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Here's Quirk's example circuit for quantum teleportation. It has some helpful visuals to show how the state from the top qubit gets perfectly transferred to the bottom qubit regardless of the input state. You can play around with e.g. removing the controlled X gate at the end to see how different parts of the circuit make different things happen.

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  • $\begingroup$ that is a very nice demo. when the transmission is done, the measurement is only done once? or as usual in quantum you measure several times to get the result? (maybe this was what i was missing) $\endgroup$
    – Sfp
    Aug 2, 2022 at 17:46
  • $\begingroup$ The teleportation circuit just sends a quantum state to another place; what you do with the state afterwards is up to you. The measurements in this circuit are part of the transmission, and afterwards the complete state of the particle has been sent to the bottom bit of the circuit. Afterwards you could use it in a different circuit and eventually do some measurements, or whatever else you want. $\endgroup$
    – jchadwick
    Aug 2, 2022 at 18:40
  • $\begingroup$ I mean the two classical bits that are sent, not the whole result of the thing transfered $\endgroup$
    – Sfp
    Aug 2, 2022 at 19:30
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    $\begingroup$ @Samuel In order to teleport one qubit, you only need to make $2$ one-qubit measurements on the sender side (or $1$ two-qubit measurement) and send the two-bit outcome over a classical channel to the receiver. Once the receiver applies the necessary gates on their qubit depending upon the $2$ classical bits received over the classical channel, the precise quantum state that the sender sent is recovered on the receiver side. You don't need to perform multiple measurements to somehow statistically recover the $\alpha,\beta$ from the distributions of several measurement outcomes. $\endgroup$ Aug 2, 2022 at 20:13
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    $\begingroup$ Just to be clear, even if the receiver recovers the precise quantum state, the state is still unknown to them. They can't recover $\alpha,\beta$ from this state. This is the Holevo's bound which says that a single qubit can be used to recover only $1$ bit of classical information. Along the lines the you are thinking, if the sender indeed wants to communicate the $\alpha,\beta$ then they'd have to teleport multiple qubits prepared in the same state by repeating the teleportation procedure multiple times (using a fresh EPR pair every time). $\endgroup$ Aug 2, 2022 at 20:15

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