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I Rotations in Bloch Sphere about an arbitrary axis

  1. Mar 21, 2016 #1
    (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.

  2. jcsd
  3. Mar 21, 2016 #2


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    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


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    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].
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