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If K=|a><b| where |a> and |b>

  1. Oct 1, 2005 #1
    if K=|a><b| where |a> and |b> are two vectors of the state space.
    I'm trying to show that K can always be written in the form K=gPQ where g is a constant and P and Q are projectors.
    this is what I get:
    K=|a><b|
    <a|b>K=<a|b>|a><b| multiplying by the number <a|b>
    <a|b>K=|a><a|b>|<b| inserting that number in between
    P=|a><a|
    Q=|b><b|
    g=<a|b>^-1
    so I proved if the vectors are not orthogonal, how to prove it when they are???? :confused:
     
  2. jcsd
  3. Oct 1, 2005 #2

    vanesch

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    Intuitively I'd have a hard time imagining how you should be able to do this!
    After all, what the original operator does, is to take a random vector, find its b component, and then reproduce an a component with the same length.
    What you did in the non-orthogonal case was in fact to use Thales' theorem: you found the b component, which has (because of the non-orthogonality) itself of course still a small a component ; so you re-projected on a, and adjusted for the loss (the angle between a and b).

    With 3 operators, I can see a solution: you project first on b, then on something like (a+b) and then finally on b, adjusting for the losses. But I have a hard time imagining, with 2 projectors, how to do so. I'm not saying it is impossible, just that I don't intuitively see how to do so.
    You could also simply try to write down an equation, having each of the projectors to have as general a form as possible, like
    P = ( u|a) + v|b) ) (u*(a| + v*(b| ) etc... and see if you can find a condition that works out.

    cheers,
    Patrick.
     
  4. Oct 2, 2005 #3

    CarlB

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    Dear cire,

    Your work is correct, and your comment that there is a problem when the vectors is orthogonal is also correct. If <a| and <b| are orthogonal, there is no possible way to solve the problem as stated.

    Carl
     
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