I'm trying to understand quantum teleportation and I was wondering if anyone could provide an intuition about it. I have seen the derivation but it still bugs me.

You start with an entangled pair of qbits in a Bell state: $$\frac{|00\rangle+|11\rangle}{\sqrt{2}}$$

You want to 'teleport' the qbit $\psi$ to the second entangled qbit. $\psi$ is in an arbitrary quantum state.

In the end, the quantum teleportation protocol requires you to perform only one of four deterministic operations on the resulting qbit and it will become $\psi$.

Here is the problem I'm having: This whole procedure puts the resulting qbit in one of 4 predetermined states even though there is a continuum of possibilities for the original state $\psi$.

Any guesses to what I'm missing?

  • 1
    $\begingroup$ A diagram would help clarify what you are asking. In your last but one paragraph the 4 predetermined states do depend on $\psi$, so it doesn't really matter that you have a continuum of possibilities for $\psi$. $\endgroup$ Commented Mar 4, 2018 at 19:52
  • $\begingroup$ The procedure puts the first entangled qubit and the qubit in the state $\psi$ in one of four states. It doesn't put the second entangled qubit in one of the four states. $\endgroup$
    – gautampk
    Commented Mar 4, 2018 at 21:56

2 Answers 2


What is missing in the derivation you looked at, and in most discussions of entanglement, is a discussion of how information is transferred from one member of the entangled pair of qubits to the other.

The information is transferred between the entangled qubits by being carried as locally inaccessible quantum information in decoherent systems. Locally inaccessible information is information that is present in a system but cant be accessed by any measurement on that system alone. Entanglement makes some quantum information inaccessible unless you use both members of the entangled pair to read it out. When you measure one half of the entangled pair the information in that qubit is contained in the measuring instrument and in any other system to which it transfers information, such as you, or a device designed to apply different operations depending on the measurement outcome. The device that applies those operations transfers the quantum information from the first measured qubit to the other qubit.

For a detailed account of how this happens, see:



  • $\begingroup$ I was under the impression that entanglement is just strong correlation and that it cannot be used to affect one member of a pair using the other. However, this is exactly what seems to happen in quantum teleportation. The details on the information transfer seem to fill the gaps. $\endgroup$
    – bvkaradz
    Commented Mar 4, 2018 at 20:27
  • $\begingroup$ Measuring the first qubit doesn't affect the second qubit at the time of the measurement. The information is transferred from one qubit to the other by being local means - it is transferred in decoherent systems. $\endgroup$
    – alanf
    Commented Mar 4, 2018 at 20:53

I do not know if this can be regarded as an answer. Anyway this is my picture to understand it.

If we check the computation procedure of the teleportation. Starting from ABC are initially in a product state:

(1) AB are entangled to $|00+11\rangle$ by a gate $U_{entangle}$.

(2) C is transformed to an unknown state $|a0+b1\rangle$.

(3) AC carries a scrambling operation $U_{scramble}$.

(4) AC are measured to get 4 possible results.

(5) B is transformed to recover C.

From the recent GM=GR idea, these operation changes the state of the 3 qubit and therefore constructs new geometry. After the measurement of AC, the geometry of the state of ABC is:

AC is cut off from the geometry due to the measurement. B is connected with the original state of C $|a0+b1\rangle$ by two 'gates', one is the gate to entangle AB, $U_{entangle}$, the other is the gate of $U_{scramble}$. $U_{entangle}$ generate a wormhole and its length is 0 (AB are completely entangled), $U_{scramble}$ is a wormhole with a length 1 (Without the information of AC, $U_{scramble}$ can not be inversed). With the measurement result, the 2 classical bits, this wormhole can be shrinked (by the post-operation on B, which is essentially to inverse $U_{scramble}$ using the two classical bits) to a 0 length and therefore the information $|a0+b1\rangle$ appears from the other end of the wormhole at the position of B.

So the information of C is there, teleportation just digs it out.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.