Rotating a wavefunction to get an exponential of $L^2$ and $L_z$

Question

With the canonical choices of directions etc, assuming the quantum mechanical wavefunction $\Psi$ is an eigenfunction of the angular momentum operator $L_z$, we can express a rotation of angle $\alpha$ around the $z$-axis in terms of an exponential of $L_z$:

$R_z(\alpha)\Psi=\exp(-i\alpha L_z)\Psi$.

Further assuming that $\Psi$ is an eigenfunction of $L^2$, is there a rotation (potentially including all three directions) that lead to an exponential of $L^2$ (and possibly $L_z$)?

I assumed there would be a way to express a general rotation in this way, i.e. something like

$R_x(\alpha)R_y(\beta)R_z(\gamma)\Psi=\exp\big(-if(\alpha,\beta,\gamma,L^2,L_z)\big)\Psi$,

where $f$ is linear in $L_z$ and $L^2$, but I haven't been able to find it.

Note that the question is not about finding an expression for the general rotation but about finding any rotation that can be expressed as an exponential that includes $L^2$ and possibly $L_z$ but no other operators or powers of these (and nothing like $0\cdot L^2$).

Also, if it exists, getting the rotation in terms of spherical coordinates $\theta$ and $\phi$ is great if that is easier.

Answer 1

You can't do what quite what you are trying. Rotations are generated by the Lie algebra $\mathfrak{so}(3)$. That is, any rotation matrix can be expressed as $R(\theta_x,\theta_y,\theta_z) = \exp(\theta_x L_x + \theta_y L_y +\theta_z L_z)$ where the $L_i$ are elements of the algebra. This means that for any $X,Y$ in the Lie algebra, $e^X e^Y = e^Z$ where $Z$ is also an element of the algebra. This fact is known as the Baker-Campbell-Hausdorff lemma.

$L^2 = L_x^2 +L_y^2 +L_z^2$ is not an element of the Lie algebra. It is a hermitian operator, but its exponential doesn't have any generally accepted meaning (though you might see it in the time-evolution operator if it appears in the Hamiltonian).

You might be interested in the Wigner D Functions. If $\Psi$ is a vector or spinor with $L^2 = j(j+1)$, then it has components corresponding to different $L_z$ eigenvalues. If $\Psi$ has components $\Psi_s$ then the components after the rotation $R_z(\alpha)R_y(\beta)R_z(\gamma)$ are $\Psi'_{s} = \sum_{t} D^{(j)}_{s,t}(\alpha,\beta,\gamma) \Psi_t$. (Note, $\alpha,\beta,\gamma$ are Euler angles, slightly different from the angles you define)