Second-Order Differential Equations — Maths for Quantum Chemistry

Oscillations and the birth of quantization

Where quantum chemistry begins

A second-order ODE involves the second derivative $d^{2} y / d x^{2}$ . The most important example in chemistry sets the second derivative proportional to the negative of the function:

\frac{d ^{2} ψ}{d x ^{2}} = - k^{2} ψ \Rightarrow ψ = A sin k x + B cos k x

Its solutions are sine and cosine — they oscillate. When the wave is confined and must vanish at the boundaries (the box walls), only special wavelengths survive: the boundary conditions $ψ (0) = 0$ forces $B = 0$ , and $ψ (L) = 0$ requires $k L = nπ$ for integer $n$ . Only these standing waves fit — and energy becomes quantized.

The same equation, with the physical constants re-inserted, is the time-independent Schrödinger equation for a free particle in a box:

- \frac{ℏ ^{2}}{2 m} \frac{d ^{2} ψ}{d x ^{2}} = E ψ \Rightarrow \frac{d ^{2} ψ}{d x ^{2}} = - \frac{2 m E}{ℏ ^{2}} ψ

Comparing with $d^{2} ψ / d x^{2} = - k^{2} ψ$ shows that $k^{2} = 2 m E / ℏ^{2}$ , and the boundary conditions give $k = nπ / L$ , so the allowed energies are $E_{n} = n^{2} h^{2} / (8 m L^{2})$ .

For a diatomic vibrating bond, the restoring force is $- κ x$ (Hooke's law), giving the simple harmonic oscillator ODE $d^{2} x / d t^{2} = - ω^{2} x$ with $ω = κ / μ$ . The IR vibrational frequency is $ν = (1/2 π) κ / μ$ , so measuring the frequency tells us the bond force constant.

Key second-order ODE results

General solution of y″ = −k²y

y = A sin k x + B cos k x

Particle-in-a-box energies

E_{n} = \frac{n ^{2} h ^{2}}{8 m L ^{2}}, n = 1, 2, 3, \dots

SHO frequency (IR)

ν = \frac{1}{2 π} \frac{κ}{μ}

Boundary conditions

A 2nd-order ODE needs 2 conditions to fix

A

and

B

— e.g.

ψ (0) = 0

and

ψ (L) = 0

for the box.

Quantum + vibrations

The time-independent Schrödinger equation is a second-order ODE; so is the equation for a vibrating bond (IR spectroscopy). Boundary conditions are what turn continuous physics into discrete energy levels. Open the quantization lab — drag the quantum number away from an integer and watch the wave fail to vanish at the wall.

Quantised harmonic oscillator energy. The quantum version of the SHO (solved by ladder operators, not the classical separation) gives equally spaced levels: $E_{v} = (v + \frac{1}{2}) h ν$ where $v = 0, 1, 2, \dots$ and $ν = (1/2 π) κ / μ$ . The $\frac{1}{2} h ν$ zero-point energy is a purely quantum effect: even at 0 K the oscillator retains this residual energy, confirmed experimentally by the fact that bonds do not break at absolute zero.

Damped oscillations arise when a frictional force proportional to velocity is added: $d^{2} x / d t^{2} + 2 γ d x / d t + ω_{0}^{2} x = 0$ , giving solutions of the form $x = e^{- γ t} (A cos ω_{d} t + B sin ω_{d} t)$ with damped frequency $ω_{d} = ω_{0}^{2} - γ^{2}$ . NMR relaxation and overdamped chemical oscillators (like the Belousov–Zhabotinsky reaction near a bifurcation) are described by this family of solutions.

Quantised SHO levels

E_{v} = (v + \frac{1}{2}) h ν, v = 0, 1, 2, \dots

Damped oscillator

x = e^{- γ t} (A cos ω_{d} t + B sin ω_{d} t)

ω_{d} = ω_{0}^{2} - γ^{2}

Frequency vs wavenumber

ν (Hz), ω = 2 π ν (rad s^{- 1}), \tilde{ν} = ν / c (c m^{- 1})

Box model for conjugated π systems

Fill electrons in pairs; the HOMO–LUMO gap

Δ E = E_{N /2 + 1} - E_{N /2}

gives the absorption wavelength.

Common pitfalls

Dropping one independent solution. The general solution of a second-order ODE requires two linearly independent terms ( $A sin k x + B cos k x$ ). Taking only one term gives an incomplete family that cannot satisfy arbitrary boundary conditions.
Misapplying boundary conditions. $ψ (0) = 0$ forces $B = 0$ (it eliminates the cosine term, not the sine). Always substitute the boundary value into the full general solution before concluding which constants vanish.
Confusing ω, ν and $\tilde{ν}$ . Angular frequency $ω = 2 π ν$ (rad s⁻¹), ordinary frequency $ν = ω /2 π$ (Hz), wavenumber $\tilde{ν} = ν / c$ (cm⁻¹). Using ω in place of ν in $E = h ν$ gives an answer wrong by a factor of $2 π$ .
Forgetting the zero-point energy. $v = 0$ does not give $E_{0} = 0$ ; it gives $E_{0} = h ν /2$ . Adding the v = 0 energy to vibration problems requires including this term.

Quantization lab

Drag the quantum number — whole integers give standing waves that vanish at both walls; non-integers fail the boundary condition and are forbidden.

Snap to a level:

ψ_{n} = sin (\frac{nπ x}{L}), E_{n} = \frac{n ^{2} h ^{2}}{8 m L ^{2}}

quantum number n (try non-integers)2.00

ψ at the wall (x = L)-0.000

allowed?yes — fits!

energy ∝ n²4

The wave must be zero at both walls. Only whole-number n manage it (green dot) — drag to a non-integer and the wave misses the wall (red). That single requirement is the origin of quantized energy levels.

Worked example 1Verifying the sine wavefunction satisfies the ODE

Show that $ψ_{n} (x) = A sin (nπ x / L)$ satisfies $d^{2} ψ / d x^{2} = - k^{2} ψ$ with $k = nπ / L$ , and identify the resulting energy $E_{n}$ .

Worked example 2Energy level spacing for π-electrons in butadiene

Model the four π-electrons in 1,3-butadiene as a particle in a box of length $L = 5.78 \overset{˚}{A} = 5.78 \times 1 0^{- 10} m$ . Calculate the energy gap $Δ E = E_{3} - E_{2}$ and convert to a wavelength (in nm) using $λ = h c /Δ E$ . ( $m_{e} = 9.109 \times 1 0^{- 31} kg$ , $h = 6.626 \times 1 0^{- 34} J s$ )

Worked examples

From SHO vibrational frequencies to Schrödinger boundary conditions — try each before revealing the solution.

Worked example 3IR vibrational frequency of H–Cl

The force constant of the H–Cl bond is $κ = 516 N m^{- 1}$ . The reduced mass of $^{1} H^{35} Cl$ is $μ = 1.626 \times 1 0^{- 27} kg$ . Calculate the fundamental IR frequency $ν$ and wavenumber $\tilde{ν}$ .

Worked example 4Simple harmonic oscillator — verifying the solution

Show that $x (t) = A cos (ω t + ϕ)$ satisfies $d^{2} x / d t^{2} = - ω^{2} x$ and state the physical meanings of $A$ and $ϕ$ .

Worked example 5Zero-point energy of H–Cl from the quantised SHO

Using the result from the worked example above, calculate the zero-point energy of the H–Cl oscillator (the $v = 0$ level of the quantised SHO, energy $E_{0} = \frac{1}{2} h ν$ ).

Worked example 6Applying boundary conditions to fix A and B — particle in a box

The general solution of the Schrödinger equation inside a box of length $L$ (where the potential is zero) is $ψ (x) = A sin k x + B cos k x$ . Apply the two boundary conditions $ψ (0) = 0$ and $ψ (L) = 0$ to determine $A$ and $B$ (up to normalisation) and hence the allowed values of $k$ .

Worked example 7v = 1 → v = 0 vibrational energy and the C=O stretching frequency

The C=O bond in acetone has a force constant $κ \approx 1560 N m^{- 1}$ . The reduced mass (C and O atoms) is $μ = \frac{12 \times 16}{12 + 16} \times 1.661 \times 1 0^{- 27} = 1.139 \times 1 0^{- 26} kg$ . Calculate (a) the fundamental frequency $ν$ , (b) the wavenumber $\tilde{ν}$ , and (c) the $v = 1 \to v = 0$ transition energy in kJ mol⁻¹.

ChallengeChallenge — butadiene π→π* transition wavelength from the particle-in-a-box

1,3-Butadiene has 4 π-electrons. Model the π system as a particle in a box of length $L = 5.78 \overset{˚}{A}$ . Place the 4 electrons in the two lowest levels (2 per level). Calculate the wavelength of the $n = 2 \to n = 3$ (HOMO→LUMO) transition and compare with the experimental value of 217 nm. ( $m_{e} = 9.109 \times 1 0^{- 31} kg$ , $h = 6.626 \times 1 0^{- 34} J s$ , $c = 2.998 \times 1 0^{8} m s^{- 1}$ )

Check yourself

Four questions from ODE structure to quantized energy levels.

Question 1 of 4 · Score 0

A second-order differential equation contains a:

Choose an answer.