Quantum mechanics/Rigid rotor

Subject classification: this is a chemistry resource.

Type classification: this is a lesson resource.

Reduced mass

It is a well known result of classical mechanics that a system with only two particles (with coordinates x₁, y₁, z₁ and x₂, y₂, z₂ and masses m₁ and m₂) can be described in terms of the relative positions between the two particles, i.e. using the coordinates x = x₁ - x₂, y = y₁ - y₂, z = z₁ - z₂. To do so you have to used the reduced mass of the system μ = m₁m₂ ÷ (m₁ + m₂). 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 = μR² 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 L² is the square of the total angular momentum. By comparison with Equation 6 we can argue that the quantum angular momentum is quantized as L² ↔ ħ²l(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 (H2) as a rigid rotor (the required data has to be searched yourself!) and give their degeneracy. Repeat the calculation with the deuterium molecule (D2). 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 ψ1 and ψ2 are different eigenfunctions of Ĥ corresponding the same eigenvalue E (i.e. they are degenerate eigenfunctions), any linear combination of them ψ = aψ1 + bψ1 is still an eigenfunction of Ĥ corresponding to the eigenvalue of E. Since Yl=1,m=-1 and Yl=1,m=1 are degenerate, we can consider the linear combinations Ypx = (Yl=1,m=-1 + Yl=1,m=1) and Ypy = i(Yl=1,m=-1 - Yl=1,m=-1) as alternative eigenfunctions of the rigid rotor. Show that Ypx and Ypy are real (i.e. not complex) and therefore easier to plot (they will be the angular part of the px and py orbitals). Solutions m1 = m2 = 1.67x10-27 Kg; μ = 0.835x10-27 Kg; R = 0.74 Å. ${\frac {\hbar ^{2}}{2I}}={\frac {(1.05\times 10^{-34})^{2}}{2\times 0.835\times 10^{-27}\times (0.74\times 10^{-10})^{2}}}=1.21\times 10^{-21}{\mbox{J}}=0.0075{\mbox{eV}}=60.8{\mbox{cm}}^{-1}$. El=0,m=0 = 0 (degeneracy 1); El=1,m=-1 = El=1,m=0 = El=1,m=2 = 121.6 cm-1 (degeneracy 3); El=2,m=-2 = El=2,m=-1 = El=2,m=0 = El=2,m=2 = El=2,m=2 = 364.8 cm-1 (degeneracy 5). The reduced mass of D2 is twice larger than that of H2. 3 Ĥψ1 = Eψ1 and Ĥψ2 = Eψ2. ${\hat {H}}(a\psi _{1}+b\psi _{2})=E(a\psi _{1}+b\psi _{2})\to {\hat {H}}(a\psi _{1}+b\psi _{2})=^{\#}{\hat {H}}a\psi _{1}+{\hat {H}}b\psi _{2}=^{\#}a{\hat {H}}\psi _{1}+b{\hat {H}}\psi _{2}=^{*}aE\psi _{1}+bE\psi _{2}=E(a\psi _{1}+b\psi _{2})$ # using the fact the Ĥ is a linear operator. * using the fact that ψ1 and ψ2 are eigenfunctions of Ĥ. Ypx = (Yl=1,m=-1 + Yl=1,m=1) = Nsinθcosφ and Ypy = i(Yl=1,m=-1 - Yl=1,m=1) = Nsinθsinφ

Next: Lesson 7 - The Hydrogen Atom