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Molecular vibrational states associated with an adiabatic electronic state within the Born-Oppenheimer Approximation are typically defined by doing the harmonic approximation for the potential at an equilibrium geometry. This allows us to define normal coordinates in which the harmonically approximated potential is diagonal. The vibrational Hamiltonian in these coordinates is then a sum of uncoupled 1D harmonic oscillators. This allows us to obtain vibrational states of a molecule.

Is there a similar procedure to obtain molecular rotational states for polyatomic molecules? How are molecular rotational states constructed/modeled?

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  • $\begingroup$ I'm migrating this question to Matter Modeling at your request. $\endgroup$
    – ACuriousMind
    May 1 at 9:31
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    $\begingroup$ Is it true that you understand how it's done for diatomics and are only looking for an answer for polyatomics? Also, could you please show us what you've looked at so far? I don't want to type up things that you might have already seen, only to find out that you already read those and wanted something different. $\endgroup$ May 1 at 19:32
  • $\begingroup$ @NikeDattani I am familiar with the rigid rotor model for a diatomic. But i don't know how that transfers exactly to nonlinear molecules. Is the same quantization procedure done simply in the same way for the three principle axes of the inertia tensor ? Do i have 3 independent rotational quantum numbers in these cases ? These are points that i liked clarified. $\endgroup$
    – Hans Wurst
    May 1 at 21:23
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The Hamiltonian

Just as for vibrations we have the harmonic oscillator approximation, for rotations we often use the rigid rotor approximation, where bond lengths are fixed. Recall the rigid-rotor Hamiltonian (in this case the kinetic energy operator) for a diatomic, which is often written as follows:

$$\tag{1} \hat{H} = \hat{T} = \frac{J_x^2}{2I_x} + \frac{J_y^2}{2I_y} + \frac{J_z^2}{2I_z}, $$

where $\mathbf{\hat{J}} = \mathbf{\hat{r}}\times \mathbf{\hat{p}}$ is the angular momentum operator, and $(I_x,I_y,I_z)$ are the diagonal elements of the moment of inertia tensor when its written in the $(\mathbf{x},\mathbf{y},\mathbf{z})$ coordinate system..

The Hamiltonian for polyatomics can be written in the same way, but instead of using the lab-fixed frame which has the fixed axes $(\mathbf{x},\mathbf{y},\mathbf{z})$, it's easier to use a body-fixed or molecule-fixed frame with axes that rotate along with the molecule: these axes (known as the "principal axes") will be the eigenvectors $(\mathbf{a},\mathbf{b},\mathbf{c})$ of the moment of inertia tensor, and the corresopnding eigenvalues are denoted by $(I_a,I_b,I_c)$. The Hamiltonian (kinetic energy operator) is now:

$$\tag{2} \hat{H} = \hat{T} = \frac{J_a^2}{2I_a} +\frac{J_b^2}{2I_b}+\frac{J_c^2}{2I_c}. $$

The solution

As you noted correctly in your comment we have three quantum numbers, let's call them $(j,k,m)$.

To solve the Schroedinger equation for this Hamiltonian, we can model the wavefunction in terms of the three Euler angles:

$$\tag{3}\psi_{jkm}(\theta,\chi,\varphi) \equiv |jkm\rangle = P_{jkm}(\theta) e^{\textrm{i}k\chi}e^{\textrm{i}m\varphi}, $$

which has the following eigenvalues for various operators:

\begin{align} \hat{J}^2|jkm\rangle &= j(j+1)|jkm\rangle, ~~~~~\left(\hat{J}^2 \equiv \hat{J}_x^2 + \hat{J}_y^2 + \hat{J}_z^2 = \hat{J}_a^2 + \hat{J}_b^2 + \hat{J}_c^2 \right), \tag{4}\\ \hat{J}_z|jkm\rangle &= m|jkm\rangle,\tag{5}\\ \hat{J}_c|jkm\rangle &=k|jkm\rangle,\tag{6}\\ \end{align}

and the following results when raising your positions on the $m$ and $k$ ladders respectively:

\begin{align} \hat{J}_{m\pm}|jkm\rangle &= \sqrt{(j \mp m)(j \pm m +1)}|j,k,m\pm1\rangle, ~~~~ \hat{J}_{m\pm} \equiv \hat{J}_x \pm \textrm{i}\hat{J}_y,\tag{7}\\ \hat{J}_{k\pm}|jkm\rangle &=\sqrt{(j \mp k)(j \pm k +1)}|j,k\pm1,m\rangle, ~~~~~~~ \hat{J}_{k\pm} \equiv \hat{J}_a \mp \textrm{i}\hat{J}_b.\tag{8}\\ \end{align}

Now with the following definitions:

\begin{align} \alpha &\equiv \frac{1}{4}\left( \frac{1}{I_a} + \frac{1}{I_b}\right), \tag{9}\\ \gamma &\equiv \frac{1}{8}\left( \frac{1}{I_a} - \frac{1}{I_b}\right), \tag{10}\\ \beta &\equiv \frac{1}{2}\left( \frac{1}{I_c} - \frac{1}{2}\left(\frac{1}{I_a} + \frac{1}{I_b}\right)\right), \tag{11} \end{align}

we can write Eq (2). in the following algebraicly equivalent form for which our wavefunction ansatz $|jkm\rangle$ is already an eigenvector for many cases:

$$ \hat{H} = \alpha \hat{J}^2 + \beta \hat{J}_c^2 + \gamma ( {\hat{J}^+}^2 + {\hat{J}^-}^2 ).\tag{12} $$

Now we're ready to get the solutions for various cases based on the symmetry present in the molecule:

Spherical top molecules (e.g. $\ce{CH4}$):

All eigenvalues of the moment of inertia tensor will be equal to each other: $I_a = I_b = I_c \equiv I$, so we have $\beta = \gamma = 0$ and $\alpha=\frac{1}{2I}$:

\begin{align} H &= \frac{ J^2}{2I} \tag{13}\\ H|jkm\rangle &= \frac{j(j+1)}{2I}|jkm\rangle \tag{14}\\\\ E&= \frac{ j(j+1)}{2I} \tag{15}\\ |\psi\rangle&=|jkm\rangle.\tag{16}\\ \end{align}


Symmetric top molecules (e.g. $\ce{NH3}$):

Two eigenvalues of the moment of inertia tensor will be equal to each other, so if we pick them to be the ones with eigenvectors $\mathbf{a}$ and $\mathbf{b}$ we will get $\gamma=0$:

\begin{align} H &= \alpha J^2 + \beta J^2_c \tag{17}\\ H|jkm\rangle &= \alpha j(j+1)|jkm\rangle + \beta k^2|jkm\rangle \tag{18}\\ &= \left( \alpha j(j+1) + \beta k^2 \right) |jkm\rangle \tag{19}\\\\ E&= \alpha j(j+1) + \beta k^2 \tag{20}\\ |\psi\rangle&=|jkm\rangle.\tag{21}\\ \end{align}

Linear molecules (e.g. $\ce{CO2}$):

The situation is the same as for spherical tops.

Asymmetric top molecules (e.g. $\ce{H2O}$):

Diagonalizing the matrix is not as simple, but it's not too bad if you use $J_\pm$ to build the Hamiltonian for each $(j,m)$ pair and for $k$ from $-j$ to $j$.

General non-symmetric molecules (e.g. $\ce{CHBrClF}$):

The eigenvalues and wavefunctions would typically be obtained by numerically diagonalizing submatrices of size $(2j+1)$ × $(2j+1)$ .


Some helpful resources:

Much of what I wrote here can be found here, and I also very much liked this and for the relation to the diatomic case: this and this.

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  • $\begingroup$ That is the kind of answer is was looking for, thank you. I have one question regarding the ansatz for the wavefunction, equation (3). How is $P_{jkm}(\theta)$ defined. Is it a simple coefficient or does it stand for a more complicated function like a Legendre polynomial or spherical harmonics ? $\endgroup$
    – Hans Wurst
    May 2 at 8:17

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