Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rotation operators on Bloch sphere

  1. Mar 31, 2015 #1
    Can anyone explain to me why the following operators are rotation operators:
    \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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook