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Complementing a Qubit in the Bloch-Sphere

  1. Nov 6, 2015 #1
    Hi :-)

    for my master thesis I'm working with qubits in the Bloch-sphere representation ##|q\rangle = cos(\frac{\theta}{2})|0\rangle + e^{i\phi}sin(\frac{\theta}{2})|1\rangle##.
    Side question: why is only the second amplitude complex?

    But lets move to my main question. I need to know how the complement qubit is written in this representation; so the qubit wich is placed at the opposite side of the bloch sphere. There is a paper about this problem, that states that it is impossible to flip an unknown qubit ideally but that's not what I want. There they say, that the function complementing a qubit would be
    ##NOT(\alpha |0\rangle + \beta |1\rangle)= \beta^*|0\rangle - \alpha^*|1\rangle##

    That would lead to:
    ##|q^\perp\rangle = e^{-i\phi}sin(\frac{\theta}{2})|0\rangle - cos(\frac{\theta}{2})|1\rangle##

    Now, how do I get back to the representation where the first factor is a Cosinus function an the second factor is the only complex factor?
    Is there a representation like ##|q^\perp\rangle = cos(\frac{\theta'}{2})|0\rangle + e^{i\phi'}sin(\frac{\theta'}{2})|1\rangle## with a mapping ##(\theta,\phi) \rightarrow (\theta',\phi')## ?

    Looking forward to your answers!

    Edit: I found the word for what I mean: Antipodes xD For a given point on the Blochsphere I want the representation of the antipode of this point.
     
    Last edited: Nov 6, 2015
  2. jcsd
  3. Nov 6, 2015 #2

    Strilanc

    User Avatar
    Science Advisor

    Only one of the terms needs a phase factor because only relative phase factors matter. Keeping track of both can be mathematically convenient, and does matter when entangled with other qubits (because your global phase factors become relative as more cases are added), but is not necessary in this simple case.

    Because global phase factors don't matter, you can always multiply a quantum state by ##e^{ix}## for any ##x## you want. So ##|q^\perp\rangle = e^{-i\phi}sin(\frac{\theta}{2})|0\rangle - cos(\frac{\theta}{2})|1\rangle## is congruent to ##|q^\perp\rangle = sin(\frac{\theta}{2})|0\rangle - e^{i\phi}cos(\frac{\theta}{2})|1\rangle##.

    There is no operation that takes the complement of a state. Quantum operations are always unitary, but you can't represent the complement operation as a unitary matrix.

    I'm pretty sure that assuming you can perform the complement gives you super powers, like erasing qubits and performing FTL communication, but I can't seem to recall how to make that happen.
     
    Last edited: Nov 6, 2015
  4. Nov 9, 2015 #3
    The fact that I can always multiply a quantum state by ##e^{ix}## is very interesting and useful, thanks!

    I came to another solution, by taking the 3D-Koordinates of a point on the Bloch sphere ##
    \left(\begin{array}{c}
    sin(\theta)cos(\phi)\\
    sin(\theta)sin(\phi)\\
    cos(\theta)\\\end{array}\right)
    ##

    negating them to ##
    \left(\begin{array}{c}
    -sin(\theta)cos(\phi)\\
    -sin(\theta)sin(\phi)\\
    -cos(\theta)\\\end{array}\right) =
    \left(\begin{array}{c}
    sin(\theta+\pi)cos(\phi)\\
    sin(\theta+\pi)sin(\phi)\\
    cos(\theta+\pi)\\\end{array}\right)
    ##
    wich lead me to ##|q^\perp\rangle=cos\left(\frac{\theta}{2}+\frac{\pi}{2}\right)|0\rangle+e^{i\phi}sin\left(\frac{\theta}{2}+\frac{\pi}{2}\right)|1\rangle##
    wich is the same solution if you switch ##sin## and ##cos## functions and add a phase factor of ##e^{i\pi}=-1##

    But I'm not yet sure if the global phase is really irrelevant for me. On the one hand I'm working with two qubits and apply an operation on both of them, on the other hand, in the beginning they are totally unentangled. Can I take these two pure states and ignore their initial global phase because they are pure at that moment, or is the global phase really just ignorable, if I work with one qubit all the time?

    Maybe you know some literature about global and relative phase? I don't think I really have a mental concept of what a phase really is yet.
     
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