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Quantum computing - bloch sphere coordinates

  1. Feb 23, 2007 #1
    I have one critical problem with quantum computing.
    When I have one quantum state on bloch sphere, I do some transformation on that state for example PauliX transformation.

    As far as I know, we can present quantum state as
    |y> = cos (theta/2) |0> + e^(i*phi) sin (theta/2) |0>

    So if we have theta=PI/4 and phi=3/2PI
    cos PI/8 = 0.924; sin PI/8=0.383 and e^(i*(3/2PI))= -i
    When I want to visualize it on Bloch sphere, the coordinates I get from:
    x = sin(theta)*cos(phi)
    y = sin(theta)*sin(phi)
    z = cos(theta)

    For these parameters quantum state looks like that: |y> = 0.924 |0> +
    (-0.383i) |1>
    I can write it as a matrix
    [ 0.924 ]
    [ -0.383i ]

    But problem for me appears when I do an PauliX transformation
    [ 0 1 ]
    [ 1 0 ]
    and I want to get phi and theta angles from new state.
    After PauliX new state is:
    [ -0.383i ]
    [ 0.924 ]

    I thought that when matrix is again:
    [ cos(theta/2) ]
    [ e^(i*phi)*sin(theta/2) ]

    then I can take theta from arccos of upper -0.383i (which is 0.75PI)
    so theta is calculated for me (0.75PI)
    Then I thought I can divide lower part of matrix by sin(theta/2) to get
    only e^(i*phi),
    but when I divide 0.924 by sin(0.75/2 * PI)=0.924 I get "1".
    As we know
    but arccos(1) = 0
    Phi cannot be 0, because when I enter these data (theta=45 and phi=270)
    in some applet on
    result after PauliX = theta'=135 (which is in fact
    0.75PI) and phi'=90 (PI/2).

    I really don't know how to calculate phi angle. I'm confused. No one can
    help me. I know that it can be really simple solution to my problem.

    Could You please help me and answer me how can I get these angles?
  2. jcsd
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