Real and Complex Orbitals

Introduction

Solving the Schrödinger equation for hydrogen gives angular functions that contain e^(imₗφ). For mₗ ≠ 0 these functions are complex , yet chemistry textbooks draw real pₓ, p y, d xy and similar orbitals with sharply defined lobes. Both sets are correct. They are simply different, equally valid choices of basis for the same degenerate energy level. This page shows how one set is converted into the other and what each choice is good for.

Core explanation

Where complex functions come from. The φ part of the hydrogen wavefunction is Φ(φ) = (1/√2π)e^(imₗφ), the particle-on-a-ring solution. It is an eigenfunction of L z = −iħ ∂/∂φ with eigenvalue mₗħ. Whenever mₗ ≠ 0, e^(imₗφ) = cos(mₗφ) + i sin(mₗφ) is complex.

For l = 1 the three angular functions, ignoring the normalisation constant (3/8π)^(1/2) and phase conventions, are

- mₗ = 0: Y ∝ √2 cos θ (real; this is already p z) - mₗ = +1: Y ∝ sin θ e^(+iφ) - mₗ = −1: Y ∝ sin θ e^(−iφ)

The probability density of a complex orbital. For mₗ = ±1, Y ² ∝ sin²θ, because e^(±iφ) ² = 1. The density does not depend on φ at all: it is a doughnut (torus) around the z-axis. Complex p orbitals do not point along x or y.

Forming real combinations. Because the mₗ = +1 and mₗ = −1 functions have the same energy, any linear combination of them is also an eigenfunction of the Hamiltonian with that energy. Using e^(±iφ) = cos φ ± i sin φ:

pₓ = (1/√2)[p₊₁ + p₋₁] ∝ sin θ cos φ = x/r

p y = (1/i√2)[p₊₁ − p₋₁] ∝ sin θ sin φ = y/r

The factors 1/√2 keep the new functions normalised, and the two are orthogonal to each other and to p z. (With the Condon–Shortley phase convention, p₊₁ carries an extra minus sign, so the combinations appear with different signs in some books; the resulting pₓ and p y are the same.)

What is kept and what is lost.

Property Complex p₊₁, p₋₁ Real pₓ, p y --- --- --- Energy (eigenvalue of H) same same L² = 2ħ² (l = 1) yes yes Definite L z yes, ±ħ no Real-valued no yes Shape of ψ ² torus about z dumbbell along x or y

A real orbital is an equal mixture of +ħ and −ħ states: measuring L z for an electron in pₓ gives +ħ or −ħ with 50% probability each, and the average is zero.

d orbitals. The same procedure pairs mₗ = ±1 into d xz and d yz, and mₗ = ±2 into d x²−y² and d xy. The mₗ = 0 function is d z².

Formulae

pₓ = (p₊₁ + p₋₁)/√2 and p y = (p₊₁ − p₋₁)/(i√2), with p z = p₀ (using the convention without the Condon–Shortley sign). Inverse: p±₁ = (pₓ ± i p y)/√2.

Step-by-step reasoning

To build a real orbital from complex ones:

1. Pick a pair with mₗ = +m and −m; they are degenerate. 2. Add them to make a function containing cos(mφ). 3. Subtract them and divide by i to make one containing sin(mφ). 4. Normalise with 1/√2 and translate cos φ, sin φ into x and y to obtain the Cartesian label.

Visual explanation

Picture p₊₁ and p₋₁ as two identical doughnuts lying in the xy-plane. Their phases wind around the ring in opposite directions. Adding them makes the phases reinforce along ±x and cancel along ±y, turning the doughnut into a dumbbell on the x-axis. Subtracting makes the dumbbell lie along y instead.

Real-world analogy

Two waves travelling in opposite directions around a circular track combine to make a standing wave with fixed crests and fixed quiet points. Complex orbitals are like the travelling waves, carrying circulation; real orbitals are like the standing waves, with fixed lobes and nodes.

Real-world example

In a strong magnetic field along z, the energy depends on mₗ (the Zeeman effect), so the complex orbitals are the natural states and spectral lines split into components. In a molecule or crystal, the surrounding atoms define x and y directions, and the real orbitals become the natural states. The ligand-field splitting of d orbitals in octahedral complexes is described using real d orbitals for this reason.

Why?

Why is it legitimate to mix states? The Schrödinger equation is linear. If Hψ₁ = Eψ₁ and Hψ₂ = Eψ₂ with the same E, then H(aψ₁ + bψ₂) = E(aψ₁ + bψ₂). Only degenerate states can be mixed in this way: a combination of 1s and 2p, which have different energies, would not be an eigenfunction of H at all.

Common misconception

"p₊₁ is pₓ and p₋₁ is p y." There is no one-to-one correspondence. Each real orbital is a 50:50 combination of both complex functions, and each complex function is a combination of pₓ and p y. Only p₀ equals a single real orbital, p z.

Worked example

Question: Show that pₓ as defined above is normalised, given that p₊₁ and p₋₁ are normalised and orthogonal.

Reasoning: ∫ pₓ ²dτ = ½∫(p₊₁ + p₋₁) (p₊₁ + p₋₁)dτ = ½[1 + 0 + 0 + 1] = 1, since the cross terms are overlap integrals of orthogonal functions.

Answer: The integral equals 1, so pₓ is normalised.

Quick check

1. What is the average value of L z for an electron in a real pₓ orbital, and why? Answer: Zero, because pₓ is an equal mixture of mₗ = +1 and mₗ = −1, which contribute +ħ and −ħ with equal probability.

Exam focus

Write the pₓ and p y combinations and explain that real orbitals are still eigenfunctions of H and L² but not of L z. Examiners also like the observation that ψ ² of a complex p orbital is a ring about the z-axis rather than a dumbbell.

Advanced insight

The choice between real and complex bases is a choice of which extra operator to diagonalise. Complex orbitals are eigenfunctions of L z and suit problems with cylindrical symmetry, such as linear molecules, where the π orbitals are naturally labelled by λ = ±1. Real orbitals transform simply under the reflections and rotations of molecular point groups, which is why they appear in character tables and in most quantum chemistry programs.

Summary

The mₗ ≠ 0 hydrogen orbitals are complex because of the factor e^(imₗφ). Since +m and −m states are degenerate, their normalised sum and difference give real orbitals such as pₓ ∝ sin θ cos φ and p y ∝ sin θ sin φ. The real orbitals keep the same energy and l but lose a definite L z; they are the natural choice in molecules, while complex orbitals suit magnetic fields and linear symmetry.

Practice questions

1. Why is p z already real while pₓ and p y must be constructed? Answer: p z has mₗ = 0, so its φ factor e^(0) = 1 is real; the mₗ = ±1 functions contain e^(±iφ) and are complex. 2. Express p₊₁ in terms of pₓ and p y. Answer: p₊₁ = (pₓ + i p y)/√2, using the convention without the Condon–Shortley sign. 3. Which pair of complex d functions combine to give d xy and d x²−y²? Answer: The mₗ = +2 and mₗ = −2 functions, which contain e^(±2iφ). 4. Could p₀ and p₊₁ be combined into a new orbital of the same energy in a free hydrogen atom? Explain. Answer: Yes in free hydrogen, because all n = 2 orbitals of a given l are degenerate, but the result would not have a definite L z and would simply be another orientation of a p orbital.