Eigenvalue Equations and Quantization

Eigenvalue equations are the computational backbone of essentially all of modern quantum chemistry. Electronic structure programs (Gaussian, ORCA, Psi4, VASP) solve the Hartree–Fock or Kohn–Sham equations, which are themselves large matrix eigenvalue problems — diagonalizing a Fock or Kohn–Sham matrix yields the discrete molecular orbital energies and coefficients that predict UV-Vis spectra, redox potentials, and reaction barriers. Infrared and Raman spectroscopy, used routinely to fingerprint organic molecules and monitor industrial reactions, directly measure the evenly spaced vibrational eigenvalue ladder derived above; anharmonic corrections to that ladder are now standard in predicting overtone bands in atmospheric and combustion chemistry. Nuclear magnetic resonance spectroscopy depends on the eigenvalues of the nuclear spin Hamiltonian in a magnetic field, and MRI machines exploit the same discrete spin eigenvalue structure to generate medical images. In materials science, the discreteness (or, in metals, near-continuity) of electronic eigenvalues explains the band gaps of semiconductors used in solar cells and LEDs — a band gap is literally a forbidden zone between two clusters of allowed eigenvalues. Quantum dots, nanocrystals a few nanometers across, are essentially three-dimensional particles-in-a-box whose tunable, size-dependent eigenvalue spacing is engineered directly into commercial QLED televisions and biological fluorescent tags.

Ask the tutor: How does imposing boundary conditions on the hydrogen atom's radial equation produce the $E_n = -13.6\,\text{eV}/n^2$ eigenvalue ladder?

Every measurable quantity in quantum mechanics — energy, momentum, angular momentum, position — is represented by a linear operator. An eigenvalue equation asks a very specific question: are there special wavefunctions that, when acted on by this operator, come back completely unchanged in shape, only rescaled by a constant? Formally,

If such a $\psi_n$ exists, it is an eigenfunction of $\hat{A}$, and $a_n$ is the corresponding eigenvalue — the number you would obtain if you measured the observable $A$ while the system is in state $\psi_n$. The subscript $n$ anticipates that there may be many such special functions, each with its own eigenvalue, forming a spectrum. The crucial chemistry-relevant twist is that $\psi_n$ cannot be just any solution of the differential equation $\hat{A}\psi = a\psi$ — it must also obey the boundary conditions appropriate to the physical situation. For a particle trapped in a box, $\psi$ must vanish at the impenetrable walls. For an electron bound to a nucleus, $\psi$ must decay to zero as $r \to \infty$ so that the wavefunction is normalizable. For an electron circling a ring or nucleus, the wavefunction must be single-valued as the angle advances by $2\pi$. Each of these conditions eliminates almost all mathematically possible solutions, leaving behind only a discrete, countable family — and that family is indexed by what chemists call quantum numbers. Contrast this with an operator applied without such confinement — a free particle roaming all of space. There, no boundary condition is imposed (the particle never has to "come back" or "vanish"), and the eigenvalue equation is solved by plane waves $\psi_k(x) = e^{ikx}$ for any real $k$. The momentum eigenvalue $\hbar k$ and energy eigenvalue $\hbar^2k^2/2m$ then vary continuously. Quantization is therefore not a universal law but a specific consequence of confinement.

The same logic explains orbital shapes. The angular part of the hydrogen-like wavefunction must be single-valued on the sphere — going once around a latitude line must bring you back to the same value. This periodicity condition alone forces the projection of orbital angular momentum onto an axis, $L_z$, to take only integer multiples of $\hbar$:

and demanding that the full angular wavefunction be smooth and finite everywhere on the sphere (not just periodic in $\phi$, but well-behaved at the poles too) forces the total angular momentum eigenvalue to be

These integers $\ell$ and $m_\ell$ are exactly the orbital angular momentum quantum number and magnetic quantum number that label $s$, $p$, $d$, $f$ orbitals — a direct, unavoidable consequence of an eigenvalue equation combined with single-valuedness and smoothness on a sphere.

The general eigenvalue equation for an observable $A$ represented by the linear operator $\hat{A}$ is

where:

• $\hat{A}$ — a linear operator built from position, momentum, or their combinations, representing the observable of interest,
• $\psi_n$ — the $n$-th eigenfunction of $\hat{A}$, a specific, normalizable wavefunction,
• $a_n$ — the $n$-th eigenvalue, the single real number that measurement of $A$ would yield if the system is in state $\psi_n$.

Reading: the operator does not scramble $\psi_n$ into something qualitatively different — it returns the exact same function, just rescaled. This is a very restrictive requirement, and only special functions satisfy it for a given operator. The energy eigenvalue equation — the time-independent Schrödinger equation — is the central case for chemistry:

where:

• $\hat{H}$ — the Hamiltonian operator, kinetic plus potential energy,
• $\psi_n(\mathbf{r})$ — the $n$-th stationary-state wavefunction,
• $E_n$ — the $n$-th allowed total energy, discrete when $V(\mathbf r)$ confines the particle.

Reading: solving this differential equation subject to the boundary conditions dictated by $V(\mathbf{r})$ is precisely what produces the ladder of atomic and molecular energy levels. For a particle confined to a one-dimensional box of length $L$ with infinite walls, the boundary conditions $\psi(0) = \psi(L) = 0$ turn the general solution $\psi(x) = A\sin(kx) + B\cos(kx)$ into a discrete family:

where:

• $n$ — a positive integer quantum number, forced to be an integer by the requirement that $\sin(kL)=0$,
• $h$ — Planck's constant, $m$ — the particle mass, $L$ — the box length.

Reading: only wavelengths that "fit" an integer number of half-wavelengths between the walls survive; every other trial solution fails the boundary condition and is discarded, so energy jumps in discrete steps as $n^2$. For the quantum harmonic oscillator with force constant $k_f$ and potential $V(x) = \tfrac12 k_f x^2$, requiring $\psi(x) \to 0$ as $x \to \pm\infty$ (rather than a hard wall) forces the same kind of discreteness, but through the mathematics of Hermite polynomials:

where:

• $v$ — the vibrational quantum number,
• $\mu$ — the reduced mass of the vibrating pair of atoms,
• $\omega$ — the classical vibrational angular frequency.

Reading: even without hard walls, the requirement that the wavefunction decay rather than blow up at large $|x|$ is itself a boundary condition, and it is exactly as restrictive as a box wall — it too permits only a discrete ladder of eigenvalues, evenly spaced by $\hbar\omega$.

Example 1 — Foundational: energy levels of an electron in a 1-D box (verifying the eigenvalue ladder). Consider an electron ($m_e = 9.109\times10^{-31}\,\text{kg}$) confined to a box of length $L = 0.500\,\text{nm}$ (a rough model for a conjugated $\pi$-system). Find $E_1$, $E_2$, $E_3$ and confirm the ratio predicted by the eigenvalue equation. Using $E_n = n^2h^2/(8mL^2)$ with $h = 6.626\times10^{-34}\,\text{J s}$:

Compute the numerator: $(6.626\times10^{-34})^2 = 4.390\times10^{-67}\,\text{J}^2\text{s}^2$. Compute the denominator: $8(9.109\times10^{-31})(0.500\times10^{-9})^2 = 8(9.109\times10^{-31})(2.500\times10^{-19}) = 1.822\times10^{-48}\,\text{kg m}^2$.

So $E_1 = 1\times(2.410\times10^{-19}\,\text{J}) = 2.41\times10^{-19}\,\text{J} = 1.505\,\text{eV}$.

The eigenvalue equation predicts the exact ratio $E_1 : E_2 : E_3 = 1 : 4 : 9$, and indeed $9.64/2.41 = 4.00$ and $21.69/2.41 = 9.00$ — confirming the $n^2$ dependence that fell directly out of the boundary condition $\sin(kL)=0$. The $1\to2$ transition energy, $\Delta E = 6.02 - 1.505 = 4.52\,\text{eV}$, corresponds to $\lambda = hc/\Delta E \approx 274\,\text{nm}$, in the UV — a sensible order of magnitude for a small conjugated chromophore. Limiting case check: as $L \to \infty$ (the box becomes unconfined), $E_n = n^2h^2/8mL^2 \to 0$ for every fixed $n$, and the level spacing $E_{n+1}-E_n \to 0$ as well. The discrete ladder collapses smoothly into the continuous free-particle spectrum, exactly as expected when the boundary condition responsible for quantization is removed.

Example 2 — Applied: vibrational quantum number and the IR spectrum of HCl (asymmetric masses). The H–Cl bond has force constant $k_f = 516\,\text{N/m}$. Using the actual isotopic masses $m_H = 1.008\,\text{u}$ and $m_{Cl} = 34.97\,\text{u}$ (not the symmetric equal-mass case), find the reduced mass, the vibrational eigenvalue spacing $\hbar\omega$, and the wavenumber of the fundamental absorption ($v=0\to1$). The reduced mass:

Convert to kg using $1\,\text{u} = 1.6605\times10^{-27}\,\text{kg}$:

The angular frequency from the eigenvalue formula's building block $\omega=\sqrt{k_f/\mu}$:

The eigenvalue spacing between adjacent vibrational levels is $\Delta E = \hbar\omega$ (constant, per the derivation above):

Converting to a spectroscopic wavenumber via $\tilde\nu = \Delta E/(hc)$:

This matches the experimentally observed HCl fundamental stretch near $2886\,\text{cm}^{-1}$ (the small discrepancy is anharmonicity — real bonds are not perfectly parabolic, so the true eigenvalue ladder is not perfectly evenly spaced, but the harmonic eigenvalue equation captures the dominant physics). Limiting case check: if $m_{Cl}\to\infty$ (an infinitely heavy "wall" atom), $\mu \to m_H = 1.008\,\text{u}$ exactly — the reduced-mass formula correctly reduces to the light-particle limit, and the vibrational frequency becomes that of hydrogen oscillating against an immovable anchor, as physically expected.

Example 3 — Exam-level: angular momentum eigenvalues and the number of degenerate states for $\ell=3$. An electron in an $f$ orbital has orbital angular momentum quantum number $\ell = 3$. (a) Find the eigenvalue of $\hat L^2$ in SI units. (b) List the allowed $m_\ell$ eigenvalues of $\hat L_z$. (c) Compute the angle between $\mathbf L$ and the $z$-axis for the state with $m_\ell = 2$ (an off-special, non-maximal value, to test the general vector-model relation, not the trivial $m_\ell=\ell$ case). (a) By the eigenvalue equation $\hat L^2 Y_\ell^{m_\ell} = \ell(\ell+1)\hbar^2 Y_\ell^{m_\ell}$ with $\ell=3$:

Numerically, with $\hbar = 1.0546\times10^{-34}\,\text{J s}$:

so $|\mathbf L| = \sqrt{12}\,\hbar = 3.4641\,\hbar = 3.654\times10^{-34}\,\text{J s}$. (b) The eigenvalue equation $\hat L_z Y_\ell^{m_\ell} = m_\ell\hbar\, Y_\ell^{m_\ell}$ permits $m_\ell = -\ell,\ldots,+\ell$ in integer steps, so for $\ell=3$:

(c) For $m_\ell=2$: $L_z = 2\hbar$. The vector-model angle between $\mathbf L$ and $z$ satisfies

Note this is not $0^\circ$ even for the largest available $m_\ell=3$ — check: $\cos\theta_{\max} = 3/\sqrt{12} = 0.8660$, giving $\theta_{\min}=30.0^\circ$, never zero. This is the eigenvalue-based origin of the fact that orbital angular momentum can never point exactly along the $z$-axis, a genuinely quantum result invisible in the classical vector picture. Limiting case check: as $\ell \to \infty$ (correspondence-principle / classical limit), $\theta_{\min} = \cos^{-1}\!\big(\ell/\sqrt{\ell(\ell+1)}\big) \to \cos^{-1}(1) = 0^\circ$, recovering the classical expectation that angular momentum can align arbitrarily closely with an axis for a macroscopic rotor — the strictly nonzero minimum angle is a small-quantum-number effect.