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Delayed choice as a quantum logic circuit

  1. Aug 20, 2015 #1

    Strilanc

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    The recent thread about the delayed choice experiment made me want to understand the experiment as a quantum circuit. I made this:

    XOKyenD.png

    (contrast with the optical setup diagram from a relevant paper)

    The left hand side is the qubit-holding wires and gates to apply, with Alice and Bob each owning one of the top wires and Eve owning the bottom two wires.

    The right hand side shows a representation of the final state, with each cell corresponding to an amplitude in the state space. The size and orientation of the circles shows the value of the amplitude for the corresponding basis state (the amplitudes are all +1/sqrt(8) or -1/sqrt(8) in this case).

    The "interesting" thing is that:

    a) If you trace over Eve's bits in order to measure the entanglement between A and B, you will get a result of "no entanglement".
    b) But each individual row, each possible value of Eve's bits, does have entanglement between A and B (but you have to ignore the other rows).

    So the whole is unentangled, but it is made up of parts that are entangled. By measuring or conditioning or post-selecting on Eve's bits, you can force yourself into one of the rows and thus, suddenly, de-facto entanglement between A and B!

    The main caveat is that A and B need to know which type of entanglement exists between them (same values vs opposite values, same phase vs opposite phase) before they can actually take advantage of it. E.g. that's why quantum teleportation requires sending some classical bits to tell the receiver which case they're in. If they didn't need to be told which type of entanglement they had, you could do FTL signalling and other too-powerful-to-be-allowed things.
     

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    Last edited: Aug 20, 2015
  2. jcsd
  3. Aug 25, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Aug 27, 2015 #3
    Is it possible to give more explanation of what this all means, or is that too much like asking for a QM textbook in the form of a forum post?

    I'm not sure how a photon polarization and a qbit state relate... you can only measure the qbit in one way while you can measure the polarization in any angle. So what do the input/output qbits represent in practice?

    I assume H and X are logic gates of some sort? But maybe something more related to 0 and 45 deg measurements and not a Hadamard and whatever gate?
    And how do you apply a logic gate before the copying/entangling of a qbit? I thought they are created entangled and can not be copied afterwards...

    And I completely do not understand the right image/table/thing...

    Apologies for being such a layman... Maybe I should start a separate topic for these questions instead of hijacking this one, but it seems a bit slow here anyway.
     
  5. Aug 27, 2015 #4

    Strilanc

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    You can play around with a quantum circuit simulator here.

    As far as I know, a photon's polarization is a qubit. Any two-level quantum system is a qubit.

    The Pauli X Gate is basically just a NOT gate. It maps ##\left| 0 \right\rangle## to ##\left| 1 \right\rangle## and ##\left| 1 \right\rangle## to ##\left| 0 \right\rangle##. It's a 180 degree turn around the X axis of the bloch sphere.

    The dot connected to the X gates indicates that they are controlled operations; they only apply in the parts of the superposition where the dotted wire's qubit is ##\left| 1 \right\rangle##.

    The Hadamard Gate is basically a special kind of beam splitter; it maps ##\left| 0 \right\rangle## to ##\frac{1}{\sqrt{2}} \left| 0 \right\rangle + \frac{1}{\sqrt{2}} \left| 1 \right\rangle## and ##\left| 1 \right\rangle## to ##\frac{1}{\sqrt{2}} \left| 0 \right\rangle - \frac{1}{\sqrt{2}} \left| 1 \right\rangle##. It is its own inverse. It's a 180 degree turn around the diagonal X+Z axis of the bloch sphere, so it basically swaps the X and Z axies.

    Quantum computers are probably a bit more flexible than you're used to. They can freely setup and breakdown bell pairs (as is done in the circuit I showed, via the Hadamard and controlled-not operations).

    It's a visual representation of the final superposition, what comes out of the circuit (assuming the input was all-qubits-off). Each cell in the table represents the amplitude of one of the classical basis states (e.g. A=True,B=False,1=False,2=True is under the "A_" column and in the "_2" row). The orientation of the circle in a cell indicates the phase, the radius of the circle indicates the magnitude, and the filled in area at the bottom has height proportional to the associated probability (squared magnitude). For the purposes of this post, all that really matters is has-circle vs not-has-circle.
     
  6. Aug 27, 2015 #5
    Just when I relaxed that the dot connections are not literally making a copy of a qbit and I don't need to call the no-copy police, you started me thinking how the controlled-not output seems to be a copy of the control when the input is 0...
    I'll need more processing time for the full picture, but you definitely got me on the right track. Thank you for the explanation.
     
  7. Aug 27, 2015 #6

    Strilanc

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    The author of the de-facto standard quantum computing textbook has an accessible video series that you can watch: Quantum Computing for the Determined.
     
  8. Aug 27, 2015 #7

    Strilanc

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    Correct. However, it's not an independent copy (a "clone") which is what the no-cloning theorem forbids. It's an entangled copy, where if you measure one value you know the other value.
     
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