Schrödinger Wave Equation and Wave Functions

The Schrödinger equation, Ĥψ = Eψ, is the fundamental equation of quantum mechanics: its solutions are wave functions ψ, whose squares ψ2 give the probability density of finding a particle, and energies E that are quantized by the boundary conditions. In brief: for an electron in a one-dimensional box of length 1.0 nm the allowed energies are En = n2h2/(8mL2) = 0.38 n2 eV, and for hydrogen the same approach gives the energy levels −13.6/n2 eV.

Chemistry describes electrons in atoms and molecules with wave functions. The orbitals, the shapes of molecules and the energies of bonds are all consequences of solutions of the Schrödinger equation. This article explains what the equation says, what a wave function is and how to interpret it, solves the simplest case, the particle in a box, with worked examples and a table, and shows how the quantum numbers of the hydrogen atom arise.

What is the Schrödinger equation?

The time-independent Schrödinger equation is:

Ĥψ = Eψ

Here Ĥ is the Hamiltonian operator, which stands for the total energy of the system (kinetic plus potential energy), ψ is the wave function and E is the energy. For a single particle moving in one dimension in a potential V(x):

−(ħ2/2m) d2ψ/dx2 + V(x)ψ = Eψ

It is an eigenvalue equation: only for certain values of E (the eigenvalues) are there acceptable solutions ψ (the eigenfunctions). Acceptable solutions must be continuous, finite, single-valued and fulfill the boundary conditions of the problem. The restriction to acceptable solutions is the source of quantization: energy levels are not assumed, but come out of the mathematics.

What is a wave function?

The wave function ψ by itself has no direct physical meaning; it can be positive, negative or complex. Max Born proposed in 1926 that the square of its magnitude, |ψ|2, is the probability density: the probability of finding the particle in a small volume dV is |ψ|2dV. Because the particle must be found somewhere, the total probability is 1, which gives the normalization condition ∫|ψ|2dV = 1. The places where ψ passes through zero are nodes, where the probability of finding the particle is zero. An electron in an atom is thus described by a “probability cloud”, and an orbital is a one-electron wave function.

How does the particle in a box work?

Consider a particle of mass m confined to a one-dimensional box of length L, with V = 0 inside and infinitely high walls. The wave function must be zero at the walls (x = 0 and x = L), since the particle cannot be outside. The solutions are standing waves, like those of a string fixed at both ends:

ψn(x) = √(2/L) sin(nπx/L), En = n2h2 / (8mL2), n = 1, 2, 3, …

Important features follow:

n = 1 E1 n = 2 E2 = 4E1 n = 3 E3 = 9E1 0 L

Energies and nodes in a box

Electron in a box of length 1.0 nm (E1 = 6.02 × 10−20 J = 0.376 eV)
n Energy En in units of E1 Energy (eV) Nodes between the walls
110.380
241.501
393.382
4166.023
5259.404

Worked examples

Example 1: ground-state energy

For an electron in a box with L = 1.0 nm: E1 = h2 / (8mL2) = (6.626 × 10−34)2 / (8 × 9.109 × 10−31 × (1.0 × 10−9)2) = 4.390 × 10−67 / 7.287 × 10−48 = 6.02 × 10−20 J = 0.376 eV.

Example 2: a transition

The electron makes a jump from n = 2 to n = 1. ΔE = E2 − E1 = (4 − 1) × 6.02 × 10−20 = 1.81 × 10−19 J. The photon emitted has λ = hc/ΔE = 1.988 × 10−25 / 1.81 × 10−19 = 1.10 × 10−6 m = 1100 nm, in the infrared.

Example 3: probability in part of the box

What is the probability of finding the particle in the left third of the box (0 ≤ x ≤ L/3) in the ground state? P = ∫0L/3 (2/L) sin2(πx/L) dx = [x/L − sin(2πx/L) / (2π)] evaluated from 0 to L/3 = 1/3 − sin(2π/3)/(2π) = 0.3333 − 0.8660/6.2832 = 0.3333 − 0.1378 = 0.196. The probability is 19.6 %, which is less than 33 %, because the particle is most likely to be found in the middle of the box.

0 L/3 L 0.196 |ψ|2

Example 4: the conjugated polyene model

The π electrons of a conjugated molecule can be approximated as particles in a box. For a molecule with 4 π electrons (such as butadiene) and a box length of L = 0.58 nm (the length of the carbon chain plus half a bond on each side), the electrons fill the levels n = 1 and n = 2 with two electrons each. The lowest excitation is from n = 2 to n = 3: ΔE = (9 − 4) × h2/(8mL2), where h2/(8mL2) = 4.390 × 10−67 / (8 × 9.109 × 10−31 × (5.8 × 10−10)2) = 1.79 × 10−19 J. ΔE = 5 × 1.79 × 10−19 = 8.95 × 10−19 J, giving λ = 222 nm in the ultraviolet. The measured absorption of butadiene is at about 217 nm, and the model correctly predicts that longer conjugated chains absorb at longer wavelengths, and so dyes with long chains of conjugated double bonds are colored.

E n = 1 n = 2 n = 3 n = 4 ΔE = (9 − 4)E1 = 5E1 E1 = 1.79 × 10−19 J ΔE = 8.95 × 10−19 J λ = 222 nm

What does the equation give for hydrogen?

For the electron in a hydrogen atom, the potential is the Coulomb attraction V = −e2/(4πε0r). In three dimensions, with spherical coordinates, the solutions can be separated into a radial part R(r) and an angular part Y(θ, φ). The requirement that the wave function is acceptable gives three quantum numbers:

The ground state wave function is ψ1s = (1/√(πa03)) e−r/a0, with a0 = 52.9 pm. It is largest at the nucleus and decays exponentially. The fourth quantum number, the spin ms, does not come from this equation, but from the relativistic Dirac equation.

Why does this matter for chemistry?

Who developed the equation?

Erwin Schrödinger, an Austrian physicist, published his equation in 1926 in a series of papers, motivated by de Broglie’s idea of matter waves. At about the same time Werner Heisenberg had formulated quantum mechanics in terms of matrices, and Schrödinger soon proved that the two approaches are equivalent. Max Born gave the probability interpretation in 1926. Schrödinger and Paul Dirac shared the Nobel Prize in Physics in 1933.

Which mistakes are common?