Density matrix, change of basis, I don't understand the basics

[Matterwave]

$\alpha$ and $\beta$ are not completely arbitrary if the matrix is to be unitary. For the matrix to be unitary at all, you have the condition on their square modulus, which reduces the degrees of freedom to 3. The point being that the 4-degrees-of-freedom matrix is not unitary without this condition. However, if the matrix is unitary, then it is also special unitary and you need to add the overall phase to get a general unitary matrix.

I see what you mean. So to clarify for my own sake, the matrix I wrote down can ONLY be used to specify a SU(2) matrix, and not a U(2) one because it is either SU(2) (given $|\alpha|^2+|\beta|^2=1$) or not unitary at all (if the previous condition is violated), am I right? A general U(2) matrix has the form of the matrix I wrote down, along with the requirement $|\alpha|^2+|\beta|^2=1$, AND multiplied by an overall phase $e^{i\varphi}$ correct?

 

[Matterwave]

He asked if what I had said meant that simply writing down the general form of a unitary matrix solves the problem. I'm assuming that this was meant to be an argument against my approach, since it clearly doesn't solve the problem. So I explained how and why my approach solves the problem.

I was not really arguing "against" your approach, so to speak. In fact, your approach was the way I was trying to lead fluid in the first place (and also why I wrote in my post #33 such a matrix in the first place, and remarked that the first column was basically $\left|\chi\right>$ and that it was orthogonal to the second column). I was hoping that fluid would realize that such a matrix solves his problem (I could not figure out a way to guide him to that matrix...).

But what I did mean was that such an approach might not be enough for what the question was asking for, since it asked for a "rotation" (otherwise, I would just be giving fluid the answer to his homework problem).

Of course, a "rotation" on 2-component spinor space IS just some matrix in SU(2). But, I thought the question also wanted you to correspond such a rotation with a rotation in real space. I.e. I assumed the question also wanted you to make a correspondence with a matrix (find a matrix) in SO(3) as well.

So I would do everything with your approach, and then construct also the SU(2) matrix $U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}$ and make a correspondence between the two matrices.

 

[Fredrik]

Of course, a "rotation" on 2-component spinor space IS just some matrix in SU(2).

Right, so maybe the word "rotation" in the problem statement is just telling us to choose that $r$ in post #56 so that the matrix gets the determinant 1.

But, I thought the question also wanted you to correspond such a rotation with a rotation in real space. I.e. I assumed the question also wanted you to make a correspondence with a matrix (find a matrix) in SO(3) as well.

So I would do everything with your approach, and then construct also the SU(2) matrix $U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}$ and make a correspondence between the two matrices.

Have you found a way to do that (other than the complicated method that I described but didn't carry out)? I haven't, but I haven't really tried yet.

 

[Matterwave]

Right, so maybe the word "rotation" in the problem statement is just telling us to choose that $r$ in post #56 so that the matrix gets the determinant 1.Have you found a way to do that (other than the complicated method that I described but didn't carry out)? I haven't, but I haven't really tried yet.

I mean at the end of that calculation you have a matrix exactly of the form in post #33, so you can make an easy identification with $\alpha,\beta$. Frist we want to construct $U=\cos(\eta/2)\hat{I}-i\sin(\eta/2)\hat{u}\cdot\vec{\sigma}$. We can use spherical coordinates for $\hat{u}$ so that it has components $\hat{u}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ where $\theta$ is the azimuthal angle, and $\phi$ is the polar angle.

Thus, assuming I did all the calculations right, the final matrix you get at the end is:

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

Therefore, we simply make the identifications:

$$\alpha=\cos(\eta/2)-i\cos(\theta)sin(\eta/2),~~\beta= -i\sin(\eta/2)\sin(\theta)e^{-i\phi}$$

So that now we know how a real space rotation given by a rotation of angle $\eta$ around an axis of rotation given by $\theta,\phi$ corresponds to a rotation in spin space from an eigenstate $\left|z_+\right>$ to $\left|\chi\right>$, some arbitrary spin state.