Rotation operators on Bloch sphere

[jimmycricket]

Can anyone explain to me why the following operators are rotation operators:
<br /> \begin{align*}R_x(\theta) &amp;= e^{-i\theta X/2}=\cos(\frac{\theta}{2})I-i\sin(\frac{\theta}{2})X=<br /> \left(\!\begin{array}{cc}\cos(\frac{\theta}{2}) &amp; -i\sin(\frac{\theta}{2}) \\ -i\sin(\frac{\theta}{2})&amp; \cos(\frac{\theta}{2}) \end{array}\!\right)\\<br /> R_y(\theta) &amp;= e^{-i\theta Y/2}=\cos(\frac{\theta}{2})I-i\sin(\frac{\theta}{2})Y=\left(\!\begin{array}{cc}\cos(\frac{\theta}{2}) &amp; -\sin(\frac{\theta}{2}) \\ \sin(\frac{\theta}{2})&amp; \:\:\cos(\frac{\theta}{2}) \end{array}\!\right)\\<br /> R_z(\theta) &amp;= e^{-i\theta Z/2}=\cos(\frac{\theta}{2})I-i\sin(\frac{\theta}{2})Z=\left(\!\begin{array}{cc}e^{-i\theta/2} &amp; 0\\ 0 &amp; e^{-i\theta/2} \end{array}\!\right)\end{align*}.

I understand that when considering the 2-d case, any complex number z can be rotated anti-clockwise by an angle \theta with the transformation z\mapsto ze^{i\theta}. This has no factor of 1/2 so where does it come from in the rotation operators?

 

[JorisL]

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)

 

[stevendaryl]

You have the wrong R_z. It should be:

R_z(\epsilon) = \left( \begin{array}\\ e^{-i \frac{\epsilon}{2}} &amp; 0 \\ 0 &amp; e^{+i \frac{\epsilon}{2}} \end{array} \right)

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

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

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

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

Then the significance of those rotation matrices is this:

If \left( \begin{array}\\ \alpha \\ \beta \end{array} \right) is the spinor representation of the vector (x,y,z), then
R_x(\epsilon) \left( \begin{array}\\ \alpha \\ \beta \end{array} \right) is the spinor representation of the vector (x&#039;,y&#039;,z&#039;) resulting from rotating (x,y,z) by an angle \epsilon about the x-axis. (and similarly for R_y and R_z).

 

[Strilanc]

This has no factor of 1/2 so where does it come from in the rotation operators?

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.