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To measure only one qubit of two

  1. Mar 11, 2015 #1


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    If you have a two-qubit system |x>[(|0>-|1>)/√2], whereby x∈{0,1}, and you want to just know x, do you
    (a) assume that this means that |x> belongs to one particle, (|0>-|1>)/√2 to another particle, so you just measure the first particle? or
    (b) if you can't tell the particles apart, pass this through a gate (Identity ⊗ Hadamard) to get either 2 of the same (in which case x=1) or 2 different (in which case x =0),
    (c) other?
  2. jcsd
  3. Mar 11, 2015 #2


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    You just measure the first qubit.

    If you can't tell them apart, you wouldn't be able to apply an operation like I x H, because that applies a different gate to each of them (and thus requires telling them apart). If you can apply different gates to them, but can't measure them individually in-place for some reason, just do a controlled-not off of the first one onto a third qubit then measure the third qubit.

    You may find playing with a quantum circuit simulator like this one helpful. Drag gates onto the wires and see how it affects the probabilities (shown at the right of the wires), the final quantum state before measuring (shown bottom right), and the intermediate quantum states (shown next to bottom left when you hover over a column).
  4. Mar 11, 2015 #3


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    Thanks very much, Strilanc. The explanation makes sense, and the simulator link will be very helpful.
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