Acceptable Wavefunctions

Introduction

The Schrödinger equation is a differential equation, and like most differential equations it has infinitely many mathematical solutions for any chosen energy. Most of them are physically meaningless. The Born interpretation places strict demands on the wavefunction, and only functions that meet them can describe a real particle. Remarkably, these apparently modest demands are exactly what produces quantised energy levels — the defining feature of quantum chemistry.

Core explanation

An acceptable wavefunction must satisfy four conditions.

1. Single-valued. At each point in space, ψ must have one value only. If it had two, the probability density ψ ² at that point would be ambiguous. This matters when a coordinate returns to its starting point, as for a particle moving on a ring: the angles φ and φ + 2π describe the same position, so ψ(φ + 2π) must equal ψ(φ). For ψ = e^(imφ) this demands that m be an integer — a quantisation condition that later gives the quantisation of angular momentum.

2. Continuous. ψ must not jump abruptly from one value to another. A discontinuity would make the probability density ambiguous at the jump and would make the second derivative — which appears in the kinetic-energy term of the Schrödinger equation — infinite. For a particle in a box with impenetrable walls, ψ is zero outside the box, so continuity forces ψ = 0 at the walls. This boundary condition selects only those sine waves that fit a whole number of half-wavelengths into the box.

3. Continuous first derivative. The slope dψ/dx must also be continuous wherever the potential energy is finite. A sudden kink in ψ would imply an infinite second derivative and hence infinite kinetic energy at that point. Kinks are permitted only where the potential itself is infinite: at the walls of an infinitely deep box, or at a nucleus where the Coulomb potential −e²/4πε₀r diverges. The hydrogen 1s orbital e^(−r/a₀) indeed has a cusp at r = 0.

4. Finite and square-integrable. The integral ∫ ψ ² dτ over all space must be finite, so that ψ can be normalised. A function that grows without bound, such as e^(+r/a₀), is rejected. A function may become infinite at an isolated point provided the integral stays finite, but for most chemical problems ψ is finite everywhere.

How the conditions produce quantisation. For an electron in an atom, the Schrödinger equation has a solution for every value of E. For almost all energies, however, the solution grows exponentially at large r and violates condition 4. Only at special energies does the solution die away properly. These special values are the allowed energy levels. Quantisation is therefore not an extra assumption, as it was in the Bohr model; it emerges from requiring an acceptable wavefunction. The same logic applies to the particle in a box (conditions 2 and 4), the particle on a ring (condition 1) and the harmonic oscillator (condition 4).

Additional symmetry requirement. For identical particles such as electrons, the total wavefunction must also change sign when the coordinates of any two electrons are exchanged. This antisymmetry requirement leads to the Pauli exclusion principle and is essential for many-electron atoms.

Step-by-step reasoning

To test whether a function is acceptable over a given region:

1. Check that it has one value at every point, including where coordinates repeat. 2. Look for jumps: is ψ continuous everywhere? 3. Look for kinks: is dψ/dx continuous where the potential is finite? 4. Examine behaviour at the boundaries and at infinity: does ∫ ψ ² dτ converge? 5. Accept the function only if every test is passed.

Visual explanation

Sketch five curves across an x-axis: a smooth hump that falls to zero at both ends (acceptable); a curve that rises exponentially to the right (not square-integrable); a step with a vertical jump (discontinuous); a V-shape with a sharp point in a region of finite potential (discontinuous slope); and a tangent curve with vertical asymptotes (infinite). Only the first passes.

Real-world analogy

A guitar string is clamped at both ends. Its vibration must be zero at the clamps, must not tear (continuous) and must not bend at a sharp angle (smooth). These physical constraints allow only certain vibration patterns — the fundamental and its harmonics — just as the conditions on ψ allow only certain energies.

Real-world example

Computational chemists choose basis functions that automatically obey the acceptability conditions. Gaussian functions such as e^(−αr²) are smooth, finite and square-integrable, which is why they form the backbone of widely used quantum chemistry programs, even though they lack the correct cusp at the nucleus.

Why?

Why must the slope of ψ be continuous where the potential is finite? The kinetic energy depends on d²ψ/dx². A sudden change of slope would make this second derivative infinite, and a finite total energy cannot contain an infinite kinetic-energy contribution.

Common misconception

"Quantised energies are assumed at the start of quantum mechanics." The Schrödinger equation contains no quantisation rule; quantisation appears because only certain energies give wavefunctions that satisfy the acceptability conditions.

Worked example

Question: Which of these is acceptable over the stated range? (a) e^(−x²) for all x; (b) e^(−x) for all x; (c) e^(iφ/2) for 0 ≤ φ ≤ 2π on a ring; (d) sin(x) for all x.

Reasoning: (a) Smooth, single-valued and ∫e^(−2x²)dx = (π/2)^½ is finite: acceptable. (b) Grows without limit as x → −∞: not square-integrable. (c) At φ = 2π it equals e^(iπ) = −1 but at φ = 0 it equals 1, yet these are the same point: not single-valued. (d) Oscillates forever, so ∫sin²x dx is infinite: not normalisable.

Answer: Only (a) is acceptable.

Quick check

1. Which acceptability condition forces a particle-in-a-box wavefunction to be zero at the walls? Answer: Continuity, because ψ is zero outside the box and must not jump in value at the walls.

Exam focus

Learn the four conditions as a list and be ready to apply each to a given function, stating the reason for rejection. Link the conditions explicitly to quantisation: examiners reward the statement that boundary conditions select the allowed energies.

Advanced insight

The electron–nucleus cusp is quantified by Kato's cusp condition: at a nucleus of charge Z, the spherical average of ψ satisfies (dψ/dr)/ψ = −Z/a₀ at r = 0. The hydrogen 1s function obeys this exactly. Gaussian basis functions have zero slope at the nucleus, so many must be combined to imitate the cusp.

Summary

An acceptable wavefunction is single-valued, continuous, has a continuous first derivative wherever the potential is finite, and is square-integrable. For electrons it must also be antisymmetric under exchange. These conditions reject most mathematical solutions of the Schrödinger equation, and the energies that survive are exactly the quantised energy levels observed in spectra.

Practice questions

1. Is ψ = 1/x acceptable on the region 1 ≤ x < ∞? Explain. Answer: Yes; it is single-valued, continuous and smooth there, and ∫₁^∞ x⁻² dx = 1 is finite. 2. Why is ψ = e^(imφ) acceptable on a ring only when m is an integer? Answer: Single-valuedness requires e^(im(φ+2π)) = e^(imφ), so e^(2πim) = 1, which holds only for integer m. 3. Explain why the hydrogen 1s orbital may have a cusp at the nucleus but a box wavefunction may not have a kink in the middle of the box. Answer: Kinks are allowed only where the potential is infinite; the Coulomb potential diverges at the nucleus, but inside the box the potential is finite (zero). 4. State why a solution of the hydrogen-atom Schrödinger equation with an arbitrary energy is usually rejected. Answer: For most energies the solution grows exponentially at large r, so it is not square-integrable and cannot be normalised.