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Rotation operators on Bloch sphere

  1. Mar 31, 2015 #1
    Can anyone explain to me why the following operators are rotation operators:
    [tex]
    \begin{align*}R_x(\theta) &= e^{-i\theta X/2}=\cos(\frac{\theta}{2})I-i\sin(\frac{\theta}{2})X=
    \left(\!\begin{array}{cc}\cos(\frac{\theta}{2}) & -i\sin(\frac{\theta}{2}) \\ -i\sin(\frac{\theta}{2})& \cos(\frac{\theta}{2}) \end{array}\!\right)\\
    R_y(\theta) &= e^{-i\theta Y/2}=\cos(\frac{\theta}{2})I-i\sin(\frac{\theta}{2})Y=\left(\!\begin{array}{cc}\cos(\frac{\theta}{2}) & -\sin(\frac{\theta}{2}) \\ \sin(\frac{\theta}{2})& \:\:\cos(\frac{\theta}{2}) \end{array}\!\right)\\
    R_z(\theta) &= e^{-i\theta Z/2}=\cos(\frac{\theta}{2})I-i\sin(\frac{\theta}{2})Z=\left(\!\begin{array}{cc}e^{-i\theta/2} & 0\\ 0 & e^{-i\theta/2} \end{array}\!\right)\end{align*}.[/tex]

    I understand that when considering the 2-d case, any complex number z can be rotated anti-clockwise by an angle [itex]\theta[/itex] with the transformation [itex]z\mapsto ze^{i\theta}[/itex]. This has no factor of 1/2 so where does it come from in the rotation operators?
     
  2. jcsd
  3. Mar 31, 2015 #2
    With the current amount of information the best I can do is giving you a push in the right direction, as I perceive it.

    First of, your result is potentially confusing since those matrices don't allow for 3 axes as I see it. Also how did you define the matrices? (forgive me my laziness)
    Second, you want to get a little bit of feeling for generating groups from an algebra.
    Third, as you are talking about the Bloch Sphere which is S^2 embedded in ##\mathbb{R}^3## you can see what happens.

    I would really like to give some context as to how you defined stuff (and what you can conclude from those)

    Do it like WannabeNewton, I always enjoy reading his/her threads because they have a great opening post. (First one I found, might not be very suitable for you but important is the amount of text used to describe the problem encountered)
     
  4. Mar 31, 2015 #3

    stevendaryl

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    You have the wrong [itex]R_z[/itex]. It should be:

    [itex]R_z(\epsilon) = \left( \begin{array}\\ e^{-i \frac{\epsilon}{2}} & 0 \\ 0 & e^{+i \frac{\epsilon}{2}} \end{array} \right)[/itex]

    Okay, well, the connection between these matrices and rotations is via spinors.

    The 3-D vector [itex](x,y,z)[/itex] can be written in spherical coordinates as:
    [itex]x = r sin(\theta) cos(\phi)[/itex]
    [itex]y = r sin(\theta) sin(\phi)[/itex]
    [itex]z = r cos(\theta)[/itex]

    We can combine these three real numbers into a 2-component complex spinor by letting:

    [itex]\alpha = \sqrt{2r} cos(\frac{\theta}{2}) e^{-i \frac{\phi}{2}}[/itex]
    [itex]\beta= \sqrt{2r} sin(\frac{\theta}{2}) e^{+i \frac{\phi}{2}}[/itex]

    Then the significance of those rotation matrices is this:

    If [itex]\left( \begin{array}\\ \alpha \\ \beta \end{array} \right)[/itex] is the spinor representation of the vector [itex](x,y,z)[/itex], then
    [itex]R_x(\epsilon) \left( \begin{array}\\ \alpha \\ \beta \end{array} \right)[/itex] is the spinor representation of the vector [itex](x',y',z')[/itex] resulting from rotating [itex](x,y,z)[/itex] by an angle [itex]\epsilon[/itex] about the x-axis. (and similarly for [itex]R_y[/itex] and [itex]R_z[/itex]).
     
  5. Apr 2, 2015 #4

    Strilanc

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    The division by 2 has to do with how axis-angle rotations get converted into quaternion rotations. Quaternions have the property that ##q## and ##-q## represent the same rotation, so to avoid ending up back where you started half-way through you need to cut the speed in half. 2x2 unitary matrices are isomorphic to unit quaternions (i.e. the ones used for rotation), plus a phase factor, and thus have the same quirk.

    When working with unitary matrices you can get rid of the factor of 2, though. You just apply a phase correction. You end up with something like ##U(\hat{v}, \theta)=\frac{1}{2} I (1+e^{is \theta})−\frac{1}{2} s\hat{v}σ(1−e^{is \theta})##. The main issue is that you're forced to choose ##s = \pm 1##, and no matter how you pick it you're forced to introduce a discontinuity in the phase as the rotation axis is changed. Also it's not longer the case that you're just exponentiating ##e^{i \sigma \hat{v} \theta}##. On the other hand, you actually go from ##I## to ##X## and back, instead of ##iX##... so it's a mixed bag.
     
    Last edited: Apr 2, 2015
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