# Spin state rotations and spinors rotations

I've tried to do the calculations to derive the SU(2) matrices that rotates spinors from the rotation of the spin eigenstates. The following is the procedure that I followed but at the end I didn't find the $$SU(2)$$ matrix that I expected. Anyway I don't understand why this idea should be wrong so I'd like if you could give me some insights about it.

The spin operator in the direction of the unitary vector $$\vec n$$ is $$\hat {\vec \sigma} \cdot \vec n=\hbar/2 \begin{bmatrix} n_z & n_x-in_y\\ n_x+in_y & -n_z \end{bmatrix}$$ Doing some calculation I found that the eigenstate with eigenvalue $$\hbar/2$$ of this operator is

$$e^{i\phi}\sqrt {\frac {1-n_z}2}\begin{bmatrix} \frac {-n_x+in_y}{n_z-1}\\ 1\end{bmatrix}$$ except for the case $$n_z=1$$ in that case it is $$e^{i\phi} \begin{bmatrix}1\\0\end{bmatrix}$$ where $$\phi$$ can be every real value

Now, if I rotate the unitary vector $$\vec n$$ of an angle $$\Delta\theta$$ around the z axis it will change in this way $$\begin{bmatrix}n'_x\\n'_y\\n'_z\end{bmatrix}=\begin {bmatrix} cos\Delta\theta & sen\Delta\theta & 0\\-sen\Delta\theta & cos\Delta\theta & 0\\0 &0&\ 1 \end {bmatrix} \begin{bmatrix}n_x\\n_y\\n_z\end{bmatrix}$$ thus the spin state will change in this way $$e^{i\phi}\sqrt {\frac {1-n'_z}2}\begin{bmatrix} \frac {-n'_x+in'_y}{n'_z-1}\\ 1\end{bmatrix}=\begin{bmatrix} cos\Delta\theta+isen\Delta\theta & 0 \\ 0&1\end{bmatrix}e^{i\phi}\sqrt {\frac {1-n_z}2}\begin{bmatrix} \frac {-n_x+in_y}{n_z-1}\\ 1\end{bmatrix}$$ so the matrix $$\begin{bmatrix} cos\Delta\theta+isen\Delta\theta & 0 \\ 0&1\end{bmatrix}$$ is the matrix that rotate the spin state when system is rotated. This matrix isn't the one that rotates spinors around the z axis and this confuses me, am I wrong with the calculations or is the idea wrong?

UPDATE

I've noticed that the matrix I found differs from the matrix that transform spinor just for a phase indeed $$\begin{bmatrix} cos\Delta\theta+isen\Delta\theta & 0 \\ 0&1\end{bmatrix}=e^{i\Delta\theta/2} \begin{bmatrix}e^{i\Delta\theta/2} & 0 \\ 0 & e^{-i\Delta\theta/2}\end{bmatrix}$$ Then, since it is possible to choose the form of the eigenstate up to a phase I can choose this: $$e^{i\phi}e^{i\theta/2}\sqrt {\frac {1-n_z}2}\begin{bmatrix} \frac {-n_x+in_y}{n_z-1}\\ 1\end{bmatrix}$$ where $$\theta=f(n_x,n_y)$$. In this case the matrix that transform the eigenstate is the one of $$SU(2)$$ $$\begin{bmatrix}e^{i\Delta\theta/2} & 0 \\ 0 & e^{-i\Delta\theta/2}\end{bmatrix}$$ But why should we use exact this phase choice? What is special in this choice?

I'll assume that it is about the case $$s = 1/2$$. (It is totally same for higher spin case.)

We can assume $$\hat{n} = \hat{z}$$ in general and rotate it to $$\hat{n}'= \sin \theta \cos \phi \hat{x} + \sin \theta \sin \phi \hat{y} + \cos \theta \hat{z}$$

Then, the inner product of spinor operator and $$\hat{n}'$$ will be $$\hat{\vec{\sigma}} \cdot \hat{n}' = \dfrac{\hbar}{2} \begin{pmatrix} \cos \theta & \sin\theta \exp[-i \phi]\\ \sin \theta \exp[i \phi] & -\cos \theta \end{pmatrix}$$

Then, the eigenspinor for $$\hbar/2$$ is given as $$\begin{pmatrix} \exp[-i\phi/2] \cos[\theta/2]\\ \exp[i\phi/2]\sin[\theta/2] \end{pmatrix}$$ and we can get a result for eigenspinor of $$-\hbar/2$$ easily as follow, $$\begin{pmatrix} -\exp[-i\phi/2]\sin[\theta/2]\\ \exp[i\phi/2]\cos[\theta/2] \end{pmatrix}$$

The transform matrix of eigenspniors is then, as you expected, an element of $$\mathrm{SU}(2)$$. Let us call this $$H \in \mathrm{SU}(2)$$.

$$H$$ will be given as $$H = \begin{pmatrix} \exp[-i\phi/2] \cos[\theta/2] & -\exp[-i\phi/2]\sin[\theta/2]\\ \exp[i\phi/2] \sin[\theta/2] & \exp[i \phi/2] \cos[\theta/2] \end{pmatrix}$$

Why can we get this? Because we always can find a $$\mathrm{SU}(2)$$-represetation of $$\hat{\vec{\sigma}}$$ and $$\mathrm{SO(3)}$$-transformation is properly interpreted as $$\mathrm{SU}(2)$$-action already in the procedure of $$\hat{\vec{\sigma}}\cdot \hat{n}$$ to $$\hat{\vec{\sigma}}\cdot{\hat{n}'}$$.

$$\hat{\vec{\sigma}}\cdot \hat{n} = H^{-1} \hat{\vec{\sigma}}\cdot{\hat{n}'} H$$

In abstract level, this is possible because $$\mathrm{SO}(3)$$ and $$\mathrm{SU}(2)$$ has same local structure and so they have a specific correspondence and share a generator but in their own language.

• I agree with yours calculations but it wasn't exactly what I meant, I update the post including my calculations. The result seems different but I don't understand why May 2, 2020 at 12:52
• @SimoBartz The transformation matrix of system should trasform all eigenspinor. Your result do not change another eigenspinor of $-\hbar/{2}$. May 2, 2020 at 14:14
• @SimoBartz and in addition, the correspondence between $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ is not really trivial in common sense! We can't interpret the element of $\mathrm{SU}(2)$ directly in our real-geometrical sense. I apologize if I don't understand your question well. May 2, 2020 at 14:20
• @SimoBartz Anyway, I found this article, arxiv.org/pdf/1312.3824.pdf . It explains why we choose these kinds of matrix well in the sense of Lie group. May 3, 2020 at 5:25
• @SimoBartz I think the choice of phase factor doesn't affect the system, and so I have a freedom of phase factor. Using this, I choose a phase factor to make such form of $\exp[-i \phi/2 \sigma_z]\exp[i \theta/2 \sigma_y]$. I think it is more natural way to think, because spinor $\sigma_x, \sigma_y, \sigma_z$ is an infinitesimal generator of $\mathrm{SU}(2)$! May 3, 2020 at 5:28