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Orthogonality on Inner Product (Quantum Mechanics also)

  1. Nov 6, 2015 #1


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    Gold Member

    1. The problem statement, all variables and given/known data
    Consider a qubit in the state |v> ∈ ℂ^2. Suppose that a measurement of δn is made on the qubit. Show that the probability of obtaining the result "+1" in the measurement is equal to 0 if and only if |v> and |n,+> are orthogonal.

    2. Relevant equations
    Inner product axioms
    |v>|w> are orthogonal if |v>|w> = 0

    3. The attempt at a solution
    First things first, that this is a if and only if proof.
    a. Prob(+1) = 0 implies that |v> and |n,+> are orthogonal
    b. |v> and |n,+> are orthogonal implies Prob(+1) = 0

    Proof of b: Prob(+1) = | |n,+> |v> |^2 , but since |n,+> and |v> are orthogonal as assumed, by definition of orthogonality, this is equal to 0. And so Prob(+1) = 0.

    Proof of a: Similarly, Prob(+1) = | |n,+> |v> |^2 and since this equals 0, we see that |n,+>|v> = 0. But this is precisely the definition of orthogonality.

    My concern: My concern is mainly with the proof of a, it seems very weak. Would you consider this a proper proof?
  2. jcsd
  3. Nov 8, 2015 #2


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    One little bump before the week starts. :)
  4. Nov 9, 2015 #3


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    Staff: Mentor

    To make the proof less "handwavy", I would start from the fact that ##\left|\pm n \right\rangle## form a complete basis, i.e., one can always write
    \left|\psi \right\rangle = a \left| +n \right\rangle + b \left|-n \right\rangle
    and deduce things about the values of ##a## and ##b##.

    Also, please note that the inner product is noted <v|w>, not |v>|w>. |v> is a column vector, and <v| a row vector.
  5. Nov 9, 2015 #4


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    Gold Member

    Thanks for the help here. That's a good idea and may make it a stronger proof.
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