# Quantum mechanics/Rigid rotor
## Reduced mass

It is a well known result of classical mechanics that a system with only two particles (with coordinates x$_1$, y$_1$, z$_1$ and x$_2$, y$_2$, z$_2$ and masses m$_1$ and m$_2$) can be described in terms of the relative positions between the two particles, i.e. using the coordinates x = x$_1$ - x$_2$, y = y$_1$ - y$_2$, z = z$_1$ - z$_2$. To do so you have to used the reduced mass of the system μ = m$_1$m$_2$ ÷ (m$_1$ + m$_2$). Intuitively it is like sitting on one of the particles (e.g. the nucleus of a hydrogen atom) and seeing the motion of the other from that point of view.

## Potential only dependent on the distance

Consider two particles whose interaction depends only on the distance r between them. The Hamiltonian for such a system is
${\hat {H}}=\left\{-{\frac {\hbar ^{2}}{2\mu }}\left\{{\frac {\partial }{\partial x^{2}}}+{\frac {\partial }{\partial y^{2}}}+{\frac {\partial }{\partial z^{2}}}\right\}+V(r)\right\}$ (Eq. 1)
It makes sense to use polar coordinates in this case. The partial derivatives of the kinetic energy operator take the following form when expressed in polar coordinates:
$\left\{{\frac {\partial }{\partial x^{2}}}+{\frac {\partial }{\partial y^{2}}}+{\frac {\partial }{\partial z^{2}}}\right\}=\left\{{\frac {\partial }{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}+{\frac {1}{r^{2}}}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial }{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}}{\partial \phi ^{2}}}\right]\right\}$ (Eq. 2)
It is customary to give a special symbol ${\hat {\Lambda }}(\theta ,\phi )$ to the operator in the square brackets and involving all the angular variables:
$-{\frac {\hbar ^{2}}{2\mu }}\left\{{\frac {\partial }{\partial x^{2}}}+{\frac {\partial }{\partial y^{2}}}+{\frac {\partial }{\partial z^{2}}}\right\}=-{\frac {\hbar ^{2}}{2\mu }}\left\{{\frac {\partial }{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}+{\frac {1}{r^{2}}}{\hat {\Lambda }}(\theta ,\phi )\right\}$ (Eq. 3)
The Schrödinger equation for two particles whose potential energy only depends on the distance between them is therefore:
$\left\{-{\frac {\hbar ^{2}}{2\mu }}\left({\frac {\partial }{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial }{\partial r}}+{\frac {1}{r^{2}}}{\hat {\Lambda }}(\theta ,\phi )\right)+V(r)\right\}\psi (r,\theta ,\phi )=E\psi (r,\theta ,\phi )$ (Eq. 4)
We will consider two important cases: the rigid rotor, and the hydrogen atom.

## Rigid rotor

Consider two particles with reduced mass μ whose distance is fixed to the value R. This is named a rigid rotor and is a model for the rotation in space of a biatomic molecule (if we neglect its vibration). Equation 4 becomes:
$\left\{-{\frac {\hbar ^{2}}{2\mu R^{2}}}{\hat {\Lambda }}(\theta ,\phi )\right\}\psi (\theta ,\phi )=E\psi (\theta ,\phi )$ (Eq. 5)
which is obtained by removing all derivatives with respect to r and ignoring V(r). Equation 5 also represents the motion of a particle with mass μ constrained on a sphere of radius R.
The eigenvalues and eigenfunctions of equation 5 will be presented without proof as their calculation is too complicated. They are called spherical harmonics, and are indicated as Y$_{lm}$(θ, Φ)
$\left\{-{\frac {\hbar ^{2}}{2\mu R^{2}}}{\hat {\Lambda }}(\theta ,\phi )\right\}Y_{lm}(\theta ,\phi )=E_{l}Y_{lm}(\theta ,\phi )$ (Eq. 5')
The spherical harmonics depend on two quantum numbers:
* l - the orbital angular momentum quantum number
* m - the magnetic quantum number
l can take the values 0,1,2,... . m can take values that depend on l: m=l,l-1,...,-l
The eigenvalues depend only on the quantum number l.
$E_{lm}=E_{l}={\frac {\hbar ^{2}}{2I}}l(l+1)$ (Eq. 6)
I = μR$^2$ is the moment of inertia. For each quantum number l, there are 2l+1 different wavefunctions (one for each value of m) with the same energy, i.e. that are degenerate. The level with quantum number l has degeneracy 2l+1. The eigenfunctions have the following form:
$Y_{lm}(\theta ,\phi )=N_{lm}\Theta _{lm}(\theta )\Phi _{m}(\phi )$ (Eq. 7)
N$_{lm}$ is a normalization constant. Θ$_{lm}$(θ) is a polynomial of cos(θ) and sin(θ) which depends on both quantum numbers l and m. Φ$_m$ = e$^{imφ}$ is identical to the wavefunction of the particle in a ring.

## Classical and quantum angular momentum

The total energy of a classical rigid rotor is $E={\frac {L^{2}}{2I}}$ where L$^2$ is the square of the total angular momentum. By comparison with Equation 6 we can argue that the quantum angular momentum is quantized as L$^2$ ↔ ħ$^2$l(l+1). We have already seen that the projection of the angular momentum on the z axis is related to the quantum number m as L$^z$ ↔ ħm.
Exercise
Calculate the first 3 energy levels of the hydrogen molecule (H$_2$) as a rigid rotor (the required data has to be searched yourself!) and give their degeneracy.
Repeat the calculation with the deuterium molecule (D$_2$).
Different spectroscopic measurements use different units. Find the way to convert J, eV, cm$^{-1}$, GHz and identify good units to be used for the rotational energy levels of a small molecule.
Show that if ψ$_1$ and ψ$_2$ are different eigenfunctions of Ĥ corresponding the same eigenvalue E (i.e. they are degenerate eigenfunctions), any linear combination of them ψ = aψ$_1$ + bψ$_1$ is still an eigenfunction of Ĥ corresponding to the eigenvalue of E.
Since Y$_{l=1,m=-1}$ and Y$_{l=1,m=1}$ are degenerate, we can consider the linear combinations Y$_{px}$ = (Y$_{l=1,m=-1}$ + Y$_{l=1,m=1}$) and Y$_{py}$ = i(Y$_{l=1,m=-1}$ - Y$_{l=1,m=-1}$) as alternative eigenfunctions of the rigid rotor. Show that Y$_{px}$ and Y$_{py}$ are real (i.e. not complex) and therefore easier to plot (they will be the angular part of the p$_x$ and p$_y$ orbitals).
Solutions
m$_1$ = m$_2$ = 1.67x10$^{-27}$ Kg; μ = 0.835x10$^{-27}$ Kg; R = 0.74 Å. ${\frac {\hbar ^{2}}{2I}}={\frac {(1.05\times 10^{-34})^{2}}{2\times 0.835\times 10^{-27}\times (0.74\times 10^{-10})^{2}}}=1.21\times 10^{-21}{\mbox{J}}=0.0075{\mbox{eV}}=60.8{\mbox{cm}}^{-1}$. E$_{l=0,m=0}$ = 0 (degeneracy 1); E$_{l=1,m=-1}$ = E$_{l=1,m=0}$ = E$_{l=1,m=2}$ = 121.6 cm$^{-1}$ (degeneracy 3); E$_{l=2,m=-2}$ = E$_{l=2,m=-1}$ = E$_{l=2,m=0}$ = E$_{l=2,m=2}$ = E$_{l=2,m=2}$ = 364.8 cm$^{-1}$ (degeneracy 5).
The reduced mass of D$_2$ is twice larger than that of H$_2$.
3
Ĥψ$_1$ = Eψ$_1$ and Ĥψ$_2$ = Eψ$_2$. ${\hat {H}}(a\psi _{1}+b\psi _{2})=E(a\psi _{1}+b\psi _{2})\to {\hat {H}}(a\psi _{1}+b\psi _{2})=^{\#}{\hat {H}}a\psi _{1}+{\hat {H}}b\psi _{2}=^{\#}a{\hat {H}}\psi _{1}+b{\hat {H}}\psi _{2}=^{*}aE\psi _{1}+bE\psi _{2}=E(a\psi _{1}+b\psi _{2})$ # using the fact the Ĥ is a linear operator. * using the fact that ψ$_1$ and ψ$_2$ are eigenfunctions of Ĥ.
Y$_{px}$ = (Y$_{l=1,m=-1}$ + Y$_{l=1,m=1}$) = Nsinθcosφ and Y$_{py}$ = i(Y$_{l=1,m=-1}$ - Y$_{l=1,m=1}$) = Nsinθsinφ :
Next: Lesson 7 - The Hydrogen Atom