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Which operator is suitable to define a qubit?

  1. Nov 3, 2008 #1
    What requirements must an observable A meet to be usefull as a qubit for
    quantum computers?

    I think that the wave function must have time independent coefficients in the base
    of A (when we are not applying quantum gates). This means that expected value <A> must be constant, so the necessary condition is [A,H]=0.

    But is this enough? What if the eigenstates of A have different energies (eigenvalues of H)? Then the phase of the two states is changing with different frequencies. Is quantum computing possible anyway or must both eigenstates have the same energy?
     
  2. jcsd
  3. Nov 3, 2008 #2
    I suppose you mean: for what A are two eigenstates of A useful as basis states for a qubit?

    So yes, preferably you have [H,A]=0.

    And indeed, that's not quite enough, as you observe, but almost: as long as you know the energy difference precisely enough, then it's okay. In practice, the energy difference (divided by \hbar) is often matched to the difference between two laser frequencies which are stabilized.
     
  4. Nov 5, 2008 #3
    Thanks for the reply. I suppose a precisely known energy difference would not be a big problem, since we could correct the phase with phase shifter gates. Or maybe we could even
    use controled energy difference to create phase shifter gates?
     
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