SO(3) rotation of eigenvectors

[spaghetti3451]

Consider the eigenvectors $(0, 1)$ and $(1, 0)$ for the quantum system described the magnetic field $\vec{B} = (0,0,B)$.

Say I now rotate the magnetic field as $\vec{B} = (B\sin\theta\cos\phi,B\sin\theta\sin\phi,B\cos\theta)$.

Then the eigenvectors are supposed to change as $(\cos(\theta/2),e^{i\phi}\sin(\theta/2))$ and $(-\sin(\theta/2),e^{i\phi}\cos(\theta/2))$.

How can I prove that under SO(3) rotation, this is how the eigenvectors change?

P.S. : I've written the eigenvectors as row vectors, but they should be column vectors.

 

[hilbert2]

Here's how to calculate the matrix representation of spin operator for arbitrary axis: http://quantummechanics.ucsd.edu/ph130a/130_notes/node269.html . That should help you solve the problem.

 

[spaghetti3451]

I was wondering if you could explain why a general rotation matrix is given by:

$\begin{pmatrix} -\sin(\theta/2) & \cos(\theta/2) \\ e^{i\phi}\cos(\theta/2) & e^{i\phi}\sin(\theta/2)\\ \end{pmatrix}$?

 

[hilbert2]

Use the angles $\theta$ and $\phi$ to parametrize the 2x2 spin matrix $S_u$ (along axis u) and solve its eigenvectors. Both the 3D spatial rotation and the 2D state space rotation have two parameters in them.

 

[spaghetti3451]

Some steps would be helpful.

 

[vanhees71]

Note that the full rotation group is a three-parameter group. Most common are two ways to parametrize it. One is to use a vector $\vec{\phi}$ in a sphere of radius $\pi$ (with diametral points on the boundary identified). Then the spin-1/2 representation is given by
$$\hat{D}(\vec{\phi})=\exp(-\mathrm{i} \vec{\phi} \cdot \hat{\vec{\sigma}}/2),$$
where $\hat{\vec{\sigma}}$ are the Pauli matrices.

The other way are Euler angles. In quantum theory one usually uses the variant, where you rotate around the 3- and the 2-axis:
$$\hat{R}(\alpha,\beta,\gamma)=\hat{D}_3(\alpha) \hat{D}_2(\beta) \hat{D}_3(\gamma), \quad \alpha,\gamma \in [0,2 \pi[, \quad \beta \in [0,\pi[.$$
The matrices $\hat{D}_j$ ($j \in \{1,2,3 \}$) are given by
$$\hat{D}_j(\phi)=\exp(-\mathrm{i} \sigma_j \phi/2).$$

 

[hilbert2]

Some steps would be helpful.

You will find the matrices representing $S_{x}, S_{y}$ and $S_z$ from the site I linked to. Now create the spin operator $S_u = \sin\theta\cos\phi S_x + \sin\theta\sin\phi S_y + \cos\theta S_z$ and calculate its eigenvalues and eigenvectors with the usual method of characteristic polynomial, linear pair of equations...

 

[spaghetti3451]

Note that the full rotation group is a three-parameter group. Most common are two ways to parametrize it. One is to use a vector $\vec{\phi}$ in a sphere of radius $\pi$ (with diametral points on the boundary identified). Then the spin-1/2 representation is given by
$$\hat{D}(\vec{\phi})=\exp(-\mathrm{i} \vec{\phi} \cdot \hat{\vec{\sigma}}/2),$$
where $\hat{\vec{\sigma}}$ are the Pauli matrices.

In my case, I have a two-sphere. So, I was wondering which Pauli matrices correspond to $\phi$ and $\theta$.

Also, what does it mean for the diametral points on the boundary to be identified?

 

[vanhees71]

I don't know, what you want to achieve with your rotation. If you want a point on the two-sphere the meaning of the usual angles (polar and azimuthal) is to rotate the point on the northpole first by $\theta$ ($\theta \in (0,\pi)$) around the $x$ axis and then the result by $\phi$ ($\phi \in [0,2 \pi)$) around the (new) $z$ axis, i.e.,
$$\hat{D}(\theta,\phi)=\hat{D}_3(\phi) \hat{D}_1(\theta).$$