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Quantum teleportation

  1. May 5, 2014 #1
    I have some confusion on the Bell-state measurement step of quantum teleportation, it's following:

    Let particle B and C be entangled, created by the EPR source, my understanding is, they are created with certain known superposition state, like this |ψBC>=√1/2 (|0A>|1B>+|1A>|0B>) which in this case, A can be 0 or 1 while B is 1 or 0, can be other possibilities, such as A=1 B=1 or A=0 B=0, but this state is known as the entangle pair is created.

    And particle A is another particle. As Alice holds A and B, she will perform a Bell state measurement on A and B to project them into certain Bell state (similar to the one B and C has, four possibilities of the outcome), assume this measured Bell state is orthogonal on A and B, like B and C, then we know A must have the same state as C, since B and C are orthogonal, which the receiver of particle C will gain the state of A.

    Please correct me if there is any mistake in my understanding.

    Here is the part I failed to understand
    1:The entangled known state between B and C will remain as the BSM performed on A and B, is this right?
    2: It says that in the entire process, Alice will not know the state of A, is it assumed that Alice doesn't know the entangled state between B and C? If she knows, after the BSM, Alice should be able to know the state of A, is this right?
    And I assume this is the reason that extra classical information about the measurement is required from Alice to Bob?

    Thanks a lot
  2. jcsd
  3. May 5, 2014 #2


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    Maybe tracking the state of the system throughout will help.

    It starts off with Alice having a qubit she wants to send, and with both Alice and Bob having part of a bell pair. (Warning: I'm ignoring factors of [itex]\sqrt{2}[/itex] wherever possible.) The unknown qubit is [itex]\alpha \left| 0 \right> + \beta \left| 1 \right>[/itex]. The bell pair is [itex]\left| 00 \right> + \left| 11 \right>[/itex]. So our system starts in this state:

    [itex](\alpha \left| 0 \right> + \beta \left| 1 \right>) ( \left| 00 \right> + \left| 11 \right> )[/itex]

    The combined state is the tensor product of those two states, which is:

    [itex]\alpha \left| 000 \right> + \alpha \left| 011 \right> + \beta \left| 100 \right> + \beta \left| 111 \right>[/itex]

    Now Alice applies a conditional NOT, inverting her half of the bell pair in the parts of the superposition where the qubit to be transmitted is true. Our states are of the form [itex|\left| TAB \right> [/itex] where T is the qubit to teleport, A is Alice's half of the bell pair, and B is Bob's half of the bell pair. So the operation is: flip A when T. Applying it gives this result:

    [itex]\alpha \left| 000 \right> + \alpha \left| 011 \right> + \beta \left| 110 \right> + \beta \left| 101 \right>[/itex]

    Then Alice applies a Hadamard to the bit to be teleported. When it's 0 it becomes |0>+|1>, and when it's 1 it becomes |0>-|1>. So we get:

    [itex]\alpha \left| 000 \right> + \alpha \left| 100 \right> + \alpha \left| 011 \right> + \alpha \left| 111 \right> + \beta \left| 010 \right> - \beta \left| 110 \right> + \beta \left| 001 \right> + \beta \left| 101 \right>[/itex]

    Normally the explanation has Alice measure here, but I'm going to defer doing that. It won't change the outcome, because the measurements are used for nothing except determining if other operations are applied or not.

    Now Bob applies an X rotation to his half of the bell pair, which is a NOT operation, but conditioned on Alice's part of the bell pair. Flip B when A:

    [itex]\alpha \left| 000 \right> + \alpha \left| 100 \right> + \alpha \left| 010 \right> + \alpha \left| 110 \right> + \beta \left| 011 \right> - \beta \left| 111 \right> + \beta \left| 001 \right> + \beta \left| 101 \right>[/itex]

    Then Bob applies another conditioned operation, but this time it's a Z rotation conditioned on the qubit to be teleported. Z operations negate the phase when the bit is true. So when T and B are true, we negate:

    [itex]\alpha \left| 000 \right> + \alpha \left| 100 \right> + \alpha \left| 010 \right> + \alpha \left| 110 \right> + \beta \left| 011 \right> + \beta \left| 111 \right> + \beta \left| 001 \right> + \beta \left| 101 \right>[/itex]

    And suddenly we can factor out Bob's qubit:

    [itex]\left( \left| 00 \right> + \left| 10 \right> + \left| 01 \right> + \left| 11 \right> \right) \left( \alpha \left| 0 \right> + \beta \left| 1 \right> \right)[/itex]

    Hey, that alpha + beta thing is the state T started off with! It was swapped to Bob's qubit. And because later operations don't use the other two qubits, except for conditioning on them, they can be measured half-way through! Neat.

    In answer to your questions:

    1. No. There is still entanglement, but it no longer factors out as a bell state pair.

    2. Not sure what you mean by knowing. If Alice measured her bell pair before operating on it, the teleportation wouldn't work. Before she measures she knows everything about the superposition except alpha, beta, and the measurement outcomes. After she measures she knows everything about the superposition except the values of alpha and beta.
  4. May 12, 2014 #3


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    skip to p89 in exploring the quantum (2.86)
    You see that when Alice apply the BSM on u-a Bob receive the correct transported qbit copy of u only when the output of the BSM is the state of the previous a-b. If it is not the same Alice has to send him the result (4 possibilities) by a classical channel.
    After the BSM photons u and a are destroyed
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