# Free precession of rigid body from the quantum mechanical point of view

Suppose we have a triatomic molecole that has an inertia tensor of a symmetric top (two principal moments of inertia are equal, and the third is different). Therefore, if the triatomic molecole has no linear motion (only rotational motion), and there are no external forces/moments acting on it, than the hamiltonian operator in the Schrodinger equation can be written as: $$\hat H = \frac{{\hat L}_x^2+{\hat L}_y^2}{2I_1} + \frac{{\hat L}_z^2}{2I_3} = \frac{{\hat L}^2}{2I_1}+\left(\frac{1}{2I_3}-\frac{1}{2I_1}\right){\hat L}_z^2$$

Suppose we want to calculate, for example, the expectation value $$\langle\hat{L}_x\rangle$$ as a function of time. Since there are no moments acting on the molecule, we should expect this expectation value to be constant, both from the classical and quantum points of view. However, if we recall the Ehrenfest theorem, than one can say the time evolution of $${\hat L}_x$$ is controlled by the equation:

$$\frac{d}{dt}{\langle{\hat L}_x\rangle} = \frac{i}{\hbar}\langle[\hat H,\hat L_x]\rangle + \Bigl\langle\frac{d{\hat L}_x}{dt}\Bigr\rangle.$$

Therefore, let us calculate the commutator $$[\hat H,\hat L_x]$$:

$$[\hat{H},\hat{L_x}] = [\frac{\hat{L}^2}{2I_1}+(\frac{1}{2I_3}-\frac{1}{2I_1})\hat{L_z}^2,\hat{L_x}] = \frac{1}{2I_1}[\hat{L}^2,\hat{L_x}]+(\frac{1}{2I_3}-\frac{1}{2I_1})[\hat{L_z}^2,\hat{L_x}]$$

Now, let us use the canonical commutation relations: $$[\hat{L}^2,\hat{L_x}]=0$$ and $$[\hat{L_x},\hat{L_y}] = i\hbar \hat{L_z},[\hat{L_y},\hat{L_z}] = i\hbar \hat{L_x}, [\hat{L_z},\hat{L_x}] = i\hbar \hat{L_y}$$ to get:

$$[\hat{H},\hat{L_x}] = (\frac{1}{2I_3}-\frac{1}{2I_1})[\hat{L_z}^2,\hat{L_x}] = (\frac{1}{2I_3}-\frac{1}{2I_1})[\hat{L_z},[\hat{L_z},\hat{L_x}]] = (\frac{1}{2I_3}-\frac{1}{2I_1})i\hbar[\hat{L_z},\hat{L_y}] = (\frac{1}{2I_3}-\frac{1}{2I_1})(i\hbar)(-i\hbar)\hat{L_x} = (\frac{1}{2I_3}-\frac{1}{2I_1})\hbar^2\hat{L_x}$$

Let us put this result in the Ehrenfest equation to get:

$$\frac{d<\hat{L}_x>}{dt} = (\frac{1}{2I_3}-\frac{1}{2I_1})i\hbar\hat{L_x}+<\frac{d\hat{L_x}}{dt}>$$

and this is the point where I am stuck; classically, since the vector of angular momentum remains constant in the absence of external moments, the axis of rotation precesses in space (describe a cone) as a result of applying Euler's equations in rigid body theory to symmetric tops. The whole point of my attempt to solve this problem was to gain mathematical insight into the descriptions of rigid body dynamics in quantum physics, but I don't know how to evaluate the term $$<\frac{d\hat{L_x}}{dt}>$$ in the Ehrenfest equation in order to prove that the expectation value $$<\hat{L}_x>$$ is constant.

(I have a kind of intuition that there is an analogy between the term $$<\frac{d\hat{L_x}}{dt}>$$ which decribes the change in time of operator itself due to the rotation of of the principal axis cartesian system attached to the molecole (it is called "body system" in analytical mechanics), and one of the terms in Euler's equations. I would also like to know how to derive the precession of the molecole's axis of rotation in space in this quantum mechanical framework.)

Classically, the conserved angular momentum is the one measured in the inertial “laboratory” frame. However, in the frame of the rotating body, the angular momentum is not conserved.

It rather evolves according to Euler’s equation: $$\dot L +\Omega\times L=0$$ with $$\Omega$$ the angular velocity and $$L$$ the angular momentum which are all measured in the top’s (or in your case the molecule’s) frame. They are related by the inertia tensor $$I$$: $$L=I\Omega$$ Taking the principal directions as your basis $$I$$ will be time independent and diagonal. What is conserved is $$L^2$$ (angular momentum norm) and $$L\cdot \Omega$$ (kinetic energy up to a factor $$1/2$$).

In the quantum setting, you start from the Hamiltonian: $$H=\frac{L_1^2}{2I_1}+ \frac{L_2^2}{2I_2}+ \frac{L_3^2}{2I_3}$$ with the uncertainty relation: $$L\times L=i\hbar L$$ Heisenberg’s equations of motion give you Euler’s equations: \begin{align} i\hbar \dot L&=[L,H] \\ \dot L_1 &=(I_2^{-1}-I_3^{-1}) \frac{L_2L_3+L_3L_2}{2} \\ \dot L_2 &=(I_3^{-1}-I_1^{-1}) \frac{L_3L_1+L_1L_3}{2} \\ \dot L_3 &=(I_1^{-1}-I_2^{-1}) \frac{L_1L_2+L_2L_1}{2} \end{align} This is just Euler’s equation with a symmetrized product. Ehrenfest’s theorem is obtained by taking the expected value (btw in your formula consider using partial time derivatives for the second term for explicit time dependence): \begin{align} i\hbar \dot L&=[L,H] \\ \frac{d}{dt} \langle L_1 \rangle &=(I_2^{-1}-I_3^{-1}) \left\langle \frac{L_2L_3+L_3L_2}{2} \right\rangle \\ \frac{d}{dt} \langle L_2 \rangle &=(I_3^{-1}-I_1^{-1}) \left\langle \frac{L_3L_1+L_1L_3}{2} \right\rangle \\ \frac{d}{dt} \langle L_3 \rangle &=(I_1^{-1}-I_2^{-1}) \left\langle \frac{L_1L_2+L_2L_1}{2} \right\rangle \end{align}

Note that the expected values of the angular momentum does not quite follow Euler’s equation. Indeed, due to the uncertainty relation, you don’t have: $$\left\langle\frac{L_iL_j+L_jL_i}{2}\right\rangle =\langle L_i\rangle \langle L_j\rangle$$ However, in the limit of small $$\hbar$$ (or more realistically large angular momentum), they do match. Therefore, Ehrenfest’s theorem would give you exactly Euler’s equation.

If you want to take into account quantum fluctuations, then you’ll need to derive the EoM’ of the symmetrized second moemnts $$\left\langle\frac{L_iL_j+L_jL_i}{2}\right\rangle$$. Indeed, the previous set of equations is not closed. This in turn will require the knowledge of higher moments ad infinitum giving you a hierarchy similar to the BBGKY equations. The classical limit amounts to closing the equations using only first moments. A better approximation would be to close them at higher moments.

Hope this helps.

• Thanks! I had a mistake in my derivation since I assumed $[L_z^2,L_x] = [L_z,[L_z,L_x]]$, and that was not correct (I should have been more careful with the algebra). And thanks also for your detailed discussion of Euler's equations and their quantum analogy. Jul 10, 2023 at 6:13
• Another comment: in your statement that $\left\langle\frac{L_iL_j+L_jL_i}{2}\right\rangle \ne \langle L_i\rangle \langle L_j\rangle$, shouldn't it be $<L_i L_j>$ and not $<L_i><L_j>$ (in the right side of the equation)? Jul 10, 2023 at 6:37
• Glad it helped. No, you want to express the second moment in terms of the first moment to close the equations and get Euler’s equation. The difference between the term is given by the covariance which due to the uncertainty principle has a non zero lower bound.
– LPZ
Jul 10, 2023 at 8:57

To begin with, the final term in this expression should have a partial derivative with respect to time. This is how explicit time dependence is incorporated into the total derivative of your expectation value with respect to time.

$$\frac{d}{dt}{\langle{\hat L}_x\rangle} = \frac{i}{\hbar}\langle[\hat H,\hat L_x]\rangle + \Bigl\langle\frac{\partial{\hat L}_x}{\partial t}\Bigr\rangle.$$

For calculating the middle term, Let $$a = (\frac{1}{2I_3}-\frac{1}{2I_1})$$ for brevity.

$$[\hat{H},\hat{L_x}] = [\hat{L_z}^2,\hat{L_x}] = a(\hat{L_z}\hat{L_z}\hat{L_x}-\hat{L_x}\hat{L_z}\hat{L_z}) = a(\hat{L_z}\hat{L_z}\hat{L_x}-\hat{L_z}\hat{L_x}\hat{L_z}+\hat{L_z}\hat{L_x}\hat{L_z}-\hat{L_x}\hat{L_z}\hat{L_z}) = a(\hat{L_z},[\hat{L_z},\hat{L_x}] +[\hat{L_z},\hat{L_x}],\hat{L_z})= a i \hbar(\hat{L_z}\hat{L_y}+\hat{L_y}\hat{L_z})$$