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Measuring a qubit in a different basis

  1. Jan 30, 2013 #1
    Hi,
    I'm a student of computer science and am writing a simulator for Measurement-based quantum computing or one-way quantum computing (it's based on the paper "The Measurement Calculus" by Elham Kashefi et al.).

    Anyway, I'm still a bit confused when it comes to the calculations. If we write the state in let's say the z-basis that is in the Bloch sphere along the z-axis (up) is |0> and the opposite direction (down) is |1> a typical state might be [1/2, 1/2, 1/2, 1/2]T then if we measure the first qubit also in the z-basis, it has (1/2)2+(1/2)2 = 0.5 probability of collapsing into 0 (or 1). And now because one of the qubits has collapsed into a classical state we can remove it from the state and the new state will be [1/√2, 1/√2] (which includes only the second qubit). But what if we want to measure it in an arbitrary basis (as required for one-way quantum computers)? For example in the paper mentioned above, it says:
    Measurement Miα is defined by orthogonal projections on
    |+α> := 1/√2 (|0> + e |1>)
    |-α> := 1/√2 (|0> - e |1>)
    followed by a trace-out operator. The parameter α ∈ [0, 2π] is called the angle of the measurement. For α = 0, α = π , one obtains the X and Y Pauli measurements.

    My question is how do we do the calculations and what will the state after the measurement look like. I have found some materials on the Internet which basically says we should first change the basis, calculate the probability of the outcome of the measurement, do the measurement and remove the collapsed qubit from the state, and then change the basis back to whatever it was.

    But first, it seems like an awfully expensive operation (for a simulator) and if that's the only way, how can we change the basis of a system with more than 1 qubit to an arbitrary basis?

    Thanks for your help :)
     
  2. jcsd
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