0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't understand the question. A rotation of angle $\alpha$ around the $i$-th axis is always given by the exponential $\exp(\mathrm{i}\alpha L_i)$, no matter what state it acts on. $\endgroup$ – ACuriousMind Aug 11 '16 at 21:05
  • $\begingroup$ True, but I'm looking for a rotation that will lead to an exponential that looks something like $\exp(i(\alpha L^2+\beta L_z))$ where $\alpha$ is nonzero. $\endgroup$ – jorgen Aug 11 '16 at 21:06
  • 3
    $\begingroup$ You cannot do this. The Baker-Campbell-Hausdorff lemma guarantees that for $X,Y$ elements of a Lie algebra, $exp(X)exp(Y) = exp(Z)$ where $Z$ is also an element of the Lie algebra. $L^2$ is not an element of the Lie algebra $SO(3) = span(L_x,L_y,L_z)$. $\endgroup$ – Luke Pritchett Aug 11 '16 at 21:16
3
$\begingroup$

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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.