Quantum teleportation transmits the exact qubit even when the state $\alpha\vert 1\rangle+\beta\vert 0\rangle$ has arbitrary $\alpha,\beta$?

Question

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.

Answer 1

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.