I am interested in how to do a rotation about the $x$-axis in QM for spin $s = 1$ system. In an answer to the post we have that for a general rotation in QM where spin $s = 1$ we have the equation: \begin{equation} \begin{aligned} \exp(i\alpha \mathbf{J}\cdot\hat{\mathbf{n}}) & = 1 + i\hat{\mathbf{n}}\cdot\mathbf{J}\sin\alpha + (\hat{\mathbf{n}}\cdot\mathbf{J})^2(\cos\alpha-1) \\ & = 1 + \left[2i\hat{\mathbf{n}}\cdot\mathbf{J}\sin(\alpha/2)\right]\cos(\alpha/2) + \frac{1}{2}\left[2i\hat{\mathbf{n}}\cdot\mathbf{J}\sin(\alpha/2)\right]^2, \end{aligned} \end{equation} Questions: Should the LHS not be $\exp(i\alpha \mathbf{J}\cdot\hat{\mathbf{n}}/2)$ as in the $s = 1/2$ case, where we have $$\exp(-i\frac{\alpha}{2}\vec{\sigma}\cdot\textbf{n}) = \cos\biggl(\frac{\alpha}{2}\biggr)-i\vec{\sigma}\cdot\textbf{n}\sin\biggl(\frac{\alpha}{2}\biggr)?$$ Also, would the idea be to then be to write $J_x = J_{+}+J_{-}$ where we have the raising and lowering operators, and then to express this as a matrix in the basis of $J_z$ eigenstates?


No. When writing $\exp(i\alpha \hat n\cdot \vec J)$ one must use matrices $\hat J_x,\hat J_y,\hat J_z$ with the standard commutation relations: $$ [\hat J_x,\hat J_y]=i\hbar \hat J_z\, , \hbox{etc} $$ For $s=1/2$, the matrices that satisfy the commutation relations are $\{\textstyle\frac{1}{2}\sigma_x,\textstyle\frac{1}{2}\sigma_y,\textstyle\frac{1}{2}\sigma_z\}$ rather than $\{\sigma_x,\sigma_y,\sigma_z\}$, hence the need for the $\textstyle\frac{1}{2}$ factor.

Yes in general one would obtain the matrices for $\hat J_x$ and $\hat J_y$, lump them with $\hat J_z$ to construct $\exp(i\alpha \hat n\cdot \vec J)$ and exponentiate. The result does not depend on the basis but the basis of eigenstates of $\hat J_z$ is convenient since the $\hat J_\pm$ in this basis are well known and easy to compute.

Edit: in answer to a comment, the rotation matrices are usually of the form $$ R_z(\alpha)R_y(\beta)R_z(\gamma)=e^{-i \alpha L_z}e^{-i\beta Ly} e^{-i\gamma L_z} $$ To get $R_x$ one should choose $\alpha=-\pi/2$ and $\gamma=\pi/2$.

In a basis of eigenstates of $\hat J_z$, the rotation $R_x(\beta)=e^{i \pi L_z/2} R_y(\beta) e^{-i\pi L_z/2}$ for $s=1/2$ is given by $$ R_x(\beta)=\left( \begin{array}{cc} \cos \left(\frac{\beta }{2}\right) & -i \sin \left(\frac{\beta }{2}\right) \\ -i \sin \left(\frac{\beta }{2}\right) & \cos \left(\frac{\beta }{2}\right) \\ \end{array} \right)=e^{-i\beta \sigma_x/2}\, , $$ with states ordered as $\vert 1/2,1/2\rangle,\vert 1/2,-1/2\rangle$.

For $\ell=1$ the corresponding result is $$ R_x(\beta)=\left( \begin{array}{ccc} \cos ^2\left(\frac{\beta }{2}\right) & -\frac{i \sin (\beta )}{\sqrt{2}} & -\sin ^2\left(\frac{\beta }{2}\right) \\ -\frac{i \sin (\beta )}{\sqrt{2}} & \cos (\beta ) & -\frac{i \sin (\beta )}{\sqrt{2}} \\ -\sin ^2\left(\frac{\beta }{2}\right) & -\frac{i \sin (\beta )}{\sqrt{2}} & \cos ^2\left(\frac{\beta }{2}\right) \\ \end{array} \right) $$ for the ordering $\vert 1,1\rangle, \vert 1,0\rangle, \vert 1,-1\rangle$

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  • $\begingroup$ Are the signs for the above terms correct? It seems that he made a mistake that the second term is '+' for the $s = \frac{1}{2}$ case where it should be '-'. $\endgroup$ – user110903 Jun 23 '17 at 16:34
  • $\begingroup$ Can I ask your advice about something. I have a python code which I wrote using that formula to calculate the rotation of a state $|s = 1; J_z \rangle$ about the $x$ axis. I have another code which does the same thing but gives a different answer, but the difference is that the signs and the order of components are different (the numbers are the same). So it's like one of them is a permutation of the other with a sign difference at times for the components. I am using $J_x = \frac{1}{2}J_{+} + \frac{1}{2}J_{-}$. $\endgroup$ – user110903 Jun 23 '17 at 17:11
  • $\begingroup$ Thanks that is quite useful, been doing that by hand...for a while... For the last matrix, are you sure you don't get $i\frac{sin(\beta)}{\sqrt{2}}$ (without the negative sign that you have)? $\endgroup$ – user110903 Jun 23 '17 at 17:53
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    $\begingroup$ @JohnJack this looks ok. $\endgroup$ – ZeroTheHero Jun 24 '17 at 13:23
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    $\begingroup$ @JohnJack The scale on axes is different. If you use $0\le\theta\le \pi$ and $0\le\phi\le 2\Pi$ this should help with the symmetry. $\endgroup$ – ZeroTheHero Jun 24 '17 at 13:35

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