1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Rotations in Bloch Sphere about an arbitrary axis

  1. Mar 21, 2016 #1
    Hey,
    (I have already asked the question at http://physics.stackexchange.com/questions/244586/bloch-sphere-interpretation-of-rotations, I am not sure this forum's etiquette allows that!)

    I am trying to understand the following statement. "Suppose a single qubit has a state represented by the Bloch vector ##\vec{\lambda}##. Then the effect of the rotation ##R_{\hat{n}}(\theta)## on the state is to rotate it by an angle $\theta$ about the ##\hat{n}## axis of the Bloch sphere. This fact explains the rather mysterious looking factor of two in the definition of the rotation matrices."
    I could work out that the rotation operators ##R_x(\theta)##, ##R_y(\theta)## and ##R_z(\theta)## are infact rotations about the ##X,Y## and ##Z## axis. But how do I extend this for ##R_{\hat{n}}(\theta)## and prove the above statement. Please point me in the right direction.

    Thanks.
     
  2. jcsd
  3. Mar 21, 2016 #2

    jfizzix

    User Avatar
    Science Advisor
    Gold Member

    I think what you want is what the pauli spin observable is along the "[itex]\hat{n}[/itex]" axis.
    Where [itex]\hat{n}[/itex] is a unit vector with components [itex](n_{x},n_{y},n_{z})[/itex],

    [itex]\sigma_{\hat{n}} = \sigma_{x}n_{x} + \sigma_{y} n_{y} + \sigma_{z} n_{z}[/itex].

    If you know what [itex]R_{x}(\theta)[/itex] is in terms of [itex]\sigma_{x}[/itex], then I think you can work out what [itex]R_{\hat{n}}(\theta)[/itex] is in terms of [itex]\sigma_{\hat{n}}[/itex].
     
  4. Mar 21, 2016 #3
    Thanks for your reply.
    Yes I have figured that out as well. Let me explain.
    I know the rotation matrices in terms of the Pauli matrices, i.e [itex]R_x(\theta) = e^{-i\sigma_x /2}[/itex] and the rotation matrices for [itex]\sigma_y[/itex] and [itex]\sigma_z[/itex] follows in the same manner. I could also prove that [itex]R_{\hat{n}}(\theta) = cos(\frac{\theta}{2})I - i sin(\frac{\theta}{2})(n_x\sigma_x+n_y\sigma_y+n_z\sigma_z)[/itex] using the Taylor expansion. But the difficulties for me start from here. How do I show that [itex]R_{\hat{n}}(\theta)[/itex] is infact a rotation about [itex]\hat{n}[/itex] axis by [itex]\theta[/itex]. How can I construct a concrete proof?
     
  5. Mar 22, 2016 #4

    jfizzix

    User Avatar
    Science Advisor
    Gold Member

    I think what you should do is a proof by demonstration. Compare a bloch vector before [itex]\hat{u}[/itex] and after [itex]\hat{u}'[/itex] a rotation about the n-axis.
    Both [itex]\hat{u}[/itex] and [itex]\hat{u}'[/itex] dotted with [itex]\hat{n}[/itex] should give the same value, and the vectors themselves should have the same magnitude. This proves that [itex]R_{n}(\theta)[/itex] is at least some sort of rotation about the [itex]\hat{n}[/itex]-axis.

    To find the angle, you need to project [itex]\hat{u}[/itex] and [itex]\hat{u}'[/itex] onto the plane perpendicular to [itex]\hat{n}[/itex]. The dot product of these projected vectors will be the magnitude of each times the cosine of the angle between them, and hopefully that angle will be none other than [itex]\theta[/itex].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Rotations in Bloch Sphere about an arbitrary axis
Loading...