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?


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?


1 Answer 1


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.

  • $\begingroup$ 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 $\endgroup$
    – SimoBartz
    May 2, 2020 at 12:52
  • $\begingroup$ @SimoBartz The transformation matrix of system should trasform all eigenspinor. Your result do not change another eigenspinor of $-\hbar/{2}$. $\endgroup$
    – ChoMedit
    May 2, 2020 at 14:14
  • $\begingroup$ @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. $\endgroup$
    – ChoMedit
    May 2, 2020 at 14:20
  • 1
    $\begingroup$ @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. $\endgroup$
    – ChoMedit
    May 3, 2020 at 5:25
  • 1
    $\begingroup$ @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 $\mathrm{SU}(2)$! $\endgroup$
    – ChoMedit
    May 3, 2020 at 5:28

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