Computing rotation matrices - non-commuting operators

Question

A rotation matrix parametrized by Euler ZYZ angles, $\alpha, \beta, \gamma$ can be written as: $$ \hat{R}(\alpha, \beta, \gamma) = \exp{\left( -i\alpha\hat{J}_{z} \right)} \cdot \exp{\left( -i\beta\hat{J}_{y} \right)} \cdot \exp{\left( -i\gamma\hat{J}_{z} \right)}. $$

Computationally speaking, the matrix exponential $\exp(\hat{A})$ can be carried out relatively straightforwardly in a basis where the operator $\hat{A}$ is diagonal. However, since the angular momentum operators that define the rotation operator do not commute and hence there is no basis which diagonalizes all operators simultaneously, how would we go about calculating $\hat{R}(\alpha, \beta, \gamma)$? Would it be necessary to calculate $$\exp(\hat{A})=\sum_{k=0}^{\infty}\hat{A}^{k}/k!$$ via other means?

Answer 1

Yes, you will have to calculate the exponential factors explicitly using the infinite series. Fortunately, there is an easier way to do this. There are standard form of rotation matrices for rotations around $x,y$ and $z$ axis. Rotation around the $z,y$ axis by an angle $\alpha,\beta$ respectively is implemented by $$ \hat{R}_z(\alpha)=\begin{pmatrix} \cos\alpha &-\sin\alpha & 0\\\sin\alpha &\cos\alpha &0\\0&0&1\end{pmatrix},\quad \hat{R}_y(\beta)=\begin{pmatrix} \cos\beta &0 &\sin\beta\\0&1&0\\-\sin\beta&0 &\cos\beta \end{pmatrix}. $$ You can now multiply the matrices and get the full rotation: $$ \hat{R}(\alpha, \beta, \gamma) = \hat{R}_z(\alpha)\hat{R}_y(\beta)\hat{R}_z(\gamma)=$$

In terms of the generators of angular momentum, if you use the explicit matrix representation of the generators $$ \hat{J}_z=\begin{pmatrix} 0&0&0\\0&0&-1\\0&1&0\end{pmatrix},\quad \hat{J}_y=\begin{pmatrix} 0&0&1\\0&0&0\\-1&0&0\end{pmatrix},\quad \hat{J}_z=\begin{pmatrix} 0&-1&0\\1&0&0\\0&0&0\end{pmatrix} $$ then the exponential series simplifies and you get $\exp\left(-i\alpha \hat{J}_z\right)=\hat{R}_z(\alpha), \exp\left(-i\alpha \hat{J}_y\right)=\hat{R}_y(\alpha)$.

Answer 2

Since $L_z$ is diagonal in the $\vert jm\rangle$ basis, the only nontrivial rotation is about the $y$-axis. Denote $$ d^j_{m’m}(\beta):= \langle jm’\vert R_y(\beta)\vert jm\rangle $$

For this yes you can compute the exponential although there are tricks.

1. First these functions have a closed form expression and some of them are tabulated on the same wiki page.
2. Since the eigenvalues of $L_y$ must be those of $L_z$, one can explicitly and fairly easily construct the similarity transformation $T$ such that $T L_y T^{-1}$ is diagonal, exponentiate the diagonal form and sandwich between $T^{-1}$ and $T$. This is usually simpler than exponentiating the non-diagonal $L_y$.
3. Mathematica has the $D$-functions built-in although the definition is not the standard physics one,
4. Use a recursion relation as described in

Wolters GF, Simple method for the explicit calculation of d-functions. Nuclear Physics B. 1970 May 1;18(2):625-53.