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Proof: 3 Reversible evolutions -- Hermitian Conjugate

  1. Nov 4, 2015 #1

    RJLiberator

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    Gold Member

    1. The problem statement, all variables and given/known data
    Consider a qubit whihc undergoes a sequence of three reversible evolutions of 3 unitary matrices A, B, and C (in that order). Suppose that no matter what the initial state |v> of the qubit is before the three evolutions, it always comes back to the sam state |v> after the three evolutions.
    Show that we must have C=(BA)†

    2. Relevant equations
    † = hermitian conjugate

    3. The attempt at a solution

    The diagram of the reversible evolution allows us to see that the process |v> --> A --> B --> C = |v> results in the equation:

    C(BA)|v> = |v>
    From here we see that: C(BA) = I (where I is the identity matrix)
    We multiple both sides by (BA)†
    C(BA)(BA)†=I(BA)†
    By definition of unitary we see
    C=(BA)†


    This was quite easy, we see it only took 3-4 steps.

    Have I successfully completed this proof? Recall, I needed to show that this works for all possible |v>.
    |v> seems irrelevant, however, since it could be 'cancelled' out in the second step.

    Success? Or failure?
     
  2. jcsd
  3. Nov 5, 2015 #2

    DrClaude

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    Staff: Mentor

    I would give more details here, as you can only take that A is unitary and B is unitary, not suppose it for BA.


    Again, to give more details: ##C(BA)|v\rangle = |v\rangle \, \forall \, |v\rangle \Rightarrow C(BA) = I##. You can only go from the first to the second line if you say that this has to hold for all states.
     
  4. Nov 5, 2015 #3

    RJLiberator

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    Gold Member

    Absolutely excellent point. I glossed over that.

    Indeed, what you suggest seems to bring more clarity.
     
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