7. Angular momentum and rotational states
Position in the course: Lesson 7 of 10. Complete the preceding derivation and use the explained exercises to check understanding.
Prerequisites: Commutators, eigenvectors and spherical coordinates
1. Rotations have generators
Orbital angular momentum is \(L=r\times p\). Applying the canonical commutator gives \([L_x,L_y]=i\hbar L_z\) and cyclic permutations. Components cannot generally be sharp together, but \(L^2=L_x^2+L_y^2+L_z^2\) commutes with each component. Choose simultaneous eigenstates of \(L^2\) and \(L_z\). This choice exploits symmetry; it does not make the other components zero.
The length associated with the squared angular momentum is \(\hbar\sqrt{l(l+1)}\), whereas the chosen projection is \(m\hbar\). The classical picture of a rigid vector with all three definite coordinates cannot reproduce these operator relations.
2. Derive the ladder and its endpoints
Define \(L_\pm=L_x\pm iL_y\). Commutators imply \([L_z,L_\pm]=\pm\hbar L_\pm\), so application changes \(m\) by one while leaving \(l\) fixed. The useful product identities are
Norms are nonnegative. The ladder must stop above and below, giving \(m=-l,-l+1,\ldots,l\). Normalization supplies
General angular momentum permits integer or half-integer \(l\). Orbital scalar wavefunctions on ordinary three-dimensional space require integer \(l\) through their single-valued angular dependence. Spin will provide half-integer representations in the next lesson.
3. Spherical harmonics and parity
The coordinate functions \(Y_{lm}(\theta,\phi)\) are angular momentum eigenfunctions normalized over solid angle. Their azimuthal dependence is \(e^{im\phi}\). The polar equation is an associated Legendre equation. Under spatial inversion the parity is \((-1)^l\). Labels such as \(s,p,d\) correspond to \(l=0,1,2\), not to electron paths. Real chemical orbitals are linear combinations of the complex \(m\) eigenfunctions; their familiar lobes display amplitude or density in a selected representation.
4. Worked rigid rotor
A linear molecule with fixed bond length has moment of inertia \(I=\mu R^2\). Its rotational Hamiltonian is \(L^2/(2I)\), so
For \(J=0,1,2\) the levels are \(0,2B,6B\), with degeneracies \(1,3,5\) before external fields. Doubling the bond length at fixed reduced mass quadruples \(I\) and quarters every rotational spacing. The rigid rotor neglects vibration, centrifugal distortion and electronic spin couplings. Observed transition intensities also require matrix elements and populations; energy differences alone do not determine a spectrum.
5. Exercises and solutions
Find \(L_+|1,0\rangle\). Substitution gives \(\hbar\sqrt2|1,1\rangle\). Applying once more gives zero because \(l(l+1)-m(m+1)=2-2=0\) at \(m=1\).
For \(|1,0\rangle\), find \(\langle L_x^2\rangle\) and \(\langle L_y^2\rangle\). Axial symmetry makes them equal; their sum is \(\langle L^2-L_z^2\rangle=2\hbar^2\), so each is \(\hbar^2\). A zero \(z\) projection therefore does not mean no angular momentum.
6. Connect symmetry to computation
Angular basis functions organize atomic orbitals and selection rules. Truncating at a maximum angular momentum is a representation approximation and must be converged. Rotational symmetry can be broken by crystal fields or a molecular environment, mixing different atomic labels. Keep the exact symmetry of the actual Hamiltonian distinct from labels inherited from an isolated atom.
Further conceptual check
Parity also constrains dipole transitions. Since position is odd under inversion, a matrix element between two states of equal definite parity vanishes. This is a symmetry statement before any radial integral is evaluated. An environment without inversion symmetry can mix the parity labels and remove that prohibition.
7. References and study connections
The derivations and toy arithmetic are original teaching constructions. No molecular simulation is reported here.
8. Related theory and practice
Molecular methods · Quantum Monte Carlo · Gaussian · VASP