Rotations in Bloch Sphere about an arbitrary axis

[polyChron]

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.

 

[jfizzix]

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

$\sigma_{\hat{n}} = \sigma_{x}n_{x} + \sigma_{y} n_{y} + \sigma_{z} n_{z}$.

If you know what $R_{x}(\theta)$ is in terms of $\sigma_{x}$, then I think you can work out what $R_{\hat{n}}(\theta)$ is in terms of $\sigma_{\hat{n}}$.

 

[polyChron]

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

$\sigma_{\hat{n}} = \sigma_{x}n_{x} + \sigma_{y} n_{y} + \sigma_{z} n_{z}$.

If you know what $R_{x}(\theta)$ is in terms of $\sigma_{x}$, then I think you can work out what $R_{\hat{n}}(\theta)$ is in terms of $\sigma_{\hat{n}}$.

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 $R_x(\theta) = e^{-i\sigma_x /2}$ and the rotation matrices for $\sigma_y$ and $\sigma_z$ follows in the same manner. I could also prove that $R_{\hat{n}}(\theta) = cos(\frac{\theta}{2})I - i sin(\frac{\theta}{2})(n_x\sigma_x+n_y\sigma_y+n_z\sigma_z)$ using the Taylor expansion. But the difficulties for me start from here. How do I show that $R_{\hat{n}}(\theta)$ is infact a rotation about $\hat{n}$ axis by $\theta$. How can I construct a concrete proof?

 

[jfizzix]

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 $R_x(\theta) = e^{-i\sigma_x /2}$ and the rotation matrices for $\sigma_y$ and $\sigma_z$ follows in the same manner. I could also prove that $R_{\hat{n}}(\theta) = cos(\frac{\theta}{2})I - i sin(\frac{\theta}{2})(n_x\sigma_x+n_y\sigma_y+n_z\sigma_z)$ using the Taylor expansion. But the difficulties for me start from here. How do I show that $R_{\hat{n}}(\theta)$ is infact a rotation about $\hat{n}$ axis by $\theta$. How can I construct a concrete proof?

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

To find the angle, you need to project $\hat{u}$ and $\hat{u}'$ onto the plane perpendicular to $\hat{n}$. 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 $\theta$.