Reduced mass[edit]
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[edit]
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[edit]
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,l1,...,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[edit]
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}. Failed to parse (syntax error): {\displaystyle \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