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.
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:
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:
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:
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)
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:
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.
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).
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.