CSIR-NET Quantum Chemistry — Operators and Eigenvalues
Quantum chemistry questions in CSIR-NET Chemical Sciences almost never ask you to solve a differential equation from scratch. They ask something much narrower: is this function an eigenfunction of that operator, do these two operators commute, what is the expectation value of this observable. All three are mechanical once the definitions are clean. This article fixes the definitions and then does the algebra in full.
What an operator actually is
An operator is an instruction. Â acting on a function gives another function. The special case that matters physically is when the output is the same function multiplied by a number:
ψ is an eigenfunction of Â, and the number a is the corresponding eigenvalue — the value the observable takes with certainty in that state.
Two conditions turn this piece of mathematics into physics. First, Â must be linear: Â(c₁ψ₁ + c₂ψ₂) = c₁Âψ₁ + c₂Âψ₂. Second, Â must be Hermitian, ∫ψ*Âφ dτ = ∫φ(Âψ)* dτ, which guarantees that every eigenvalue is real and that eigenfunctions belonging to different eigenvalues are orthogonal. Physical observables are always represented by Hermitian operators — that is not a convention, it is the reason measured energies are real numbers.
| Observable | Operator (one dimension unless noted) |
|---|---|
| Position x | x̂ = x × (multiply) |
| Linear momentum px | p̂x = −iħ ∂/∂x |
| Kinetic energy T | T̂ = −(ħ²/2m) ∂²/∂x² |
| Potential energy V | V̂ = V(x) × (multiply) |
| Total energy | Ĥ = T̂ + V̂ |
| z-component of angular momentum | L̂z = −iħ ∂/∂φ (spherical polars) |
When the observable does not have a definite value
If ψ is not an eigenfunction of Â, a single measurement still returns one of Â's eigenvalues — but which one is not predictable. What is predictable is the average over many measurements:
(the denominator is 1 for a normalised ψ)
Commutators — the algebra behind uncertainty
[x̂, p̂x] = iħ · [L̂x, L̂y] = iħL̂z · [L̂², L̂z] = 0
Useful identity: [Â, B̂Ĉ] = B̂[Â, Ĉ] + [Â, B̂]Ĉ
If two operators commute they share a complete set of eigenfunctions, so both observables can be known simultaneously and exactly. That is why L̂² and L̂z both have good quantum numbers — l and ml — while L̂x and L̂y do not, and why the vector model draws angular momentum as a cone rather than an arrow.
Worked example 1 — is a Gaussian an eigenfunction?
Q. For what value of a is ψ = e−ax² an eigenfunction of the harmonic-oscillator Hamiltonian Ĥ = −(ħ²/2m) d²/dx² + ½kx², and what is the eigenvalue?
Step 1 — differentiate twice.
dψ/dx = −2ax e−ax²
d²ψ/dx² = (−2a + 4a²x²) e−ax² = (4a²x² − 2a) ψ
Step 2 — apply Ĥ.
Ĥψ = −(ħ²/2m)(4a²x² − 2a)ψ + ½kx²ψ
= (ħ²a/m)ψ + [½k − 2ħ²a²/m] x²ψ
Step 3 — kill the x² term. For ψ to be an eigenfunction, the bracket must
vanish, because an eigenvalue cannot depend on x:
½k = 2ħ²a²/m → a² = mk/(4ħ²) →
a = √(mk)/(2ħ) = mω/(2ħ), using ω = √(k/m).
Step 4 — read off the eigenvalue.
E = ħ²a/m = (ħ²/m) × mω/(2ħ) = ½ħω
That is the zero-point energy of the harmonic oscillator, derived without ever solving the Schrödinger equation. Note the structure of the argument: an eigenvalue must be a constant, so any surviving x-dependence is the condition you solve.
Worked example 2 — expectation values in a box
Q. For a particle in a one-dimensional box of length L in the state n = 1, with ψ₁ = √(2/L) sin(πx/L), verify the uncertainty principle.
Position. By symmetry about the box centre, ⟨x⟩ = L/2. The standard integral gives
⟨x²⟩ = L²[1/3 − 1/(2n²π²)] = L²[0.33333 − 1/(2 × 9.8696)] = L²[0.33333 − 0.05066] = 0.28267 L²
σx = √(⟨x²⟩ − ⟨x⟩²) = √(0.28267 − 0.25000) L = √0.03267 L = 0.1808 L
Momentum. ⟨p⟩ = 0 (the particle is equally likely to be moving either way). Since V = 0 inside the box, ⟨p²⟩ = 2mE₁, and E₁ = h²/(8mL²), so
⟨p²⟩ = 2m × h²/(8mL²) = h²/(4L²) = (2πħ)²/(4L²) = π²ħ²/L²
σp = πħ/L
Product. σxσp = 0.1808 L × πħ/L = 0.568 ħ.
Since 0.568 ħ ≥ ħ/2 = 0.500 ħ, the Heisenberg relation holds — and the ground state of the box comes close to, but does not reach, the minimum-uncertainty limit. Only a Gaussian wavefunction (the harmonic-oscillator ground state) achieves σxσp = ħ/2 exactly.
Worked example 3 — a commutator by identity
Q. Evaluate [x̂, p̂x²].
Working. Use [Â, B̂Ĉ] = B̂[Â, Ĉ] + [Â, B̂]Ĉ with B̂ = Ĉ = p̂x:
[x̂, p̂x²] = p̂x[x̂, p̂x] + [x̂, p̂x]p̂x = p̂x(iħ) + (iħ)p̂x = 2iħ p̂x
An immediate corollary: [x̂, T̂] = [x̂, p̂x²/2m] = iħp̂x/m, so position and kinetic energy are not simultaneously measurable — which is another way of saying that a particle with a sharply defined energy is delocalised.
The three model systems worth memorising cold
| System | Energy levels | Spacing behaviour | Degeneracy |
|---|---|---|---|
| Particle in a 1-D box | En = n²h²/(8mL²), n = 1, 2, 3… | Widens as n increases | None |
| Harmonic oscillator | Ev = (v + ½)ħω, v = 0, 1, 2… | Uniform, ħω | None (1-D) |
| Rigid rotor | EJ = J(J + 1)ħ²/2I | Widens as 2B(J+1) | 2J + 1 |
For the rotor, L̂²Ylm = l(l+1)ħ²Ylm and L̂zYlm = mlħYlm. The magnitude of angular momentum is therefore √(l(l+1))ħ, never simply lħ — a distinction examiners test almost every year.
Common mistakes that cost marks
- Calling a function an eigenfunction because the operator "works" on it. Every operator acts on every function. Eigenfunction means the output is a constant multiple of the input, with no leftover x-dependence.
- Treating operators as if they multiply. ÂB̂ψ means apply B̂ first, then Â. Order matters whenever the commutator is non-zero.
- Writing |L| = lħ. The magnitude is √(l(l+1))ħ; only the z-projection is an integer multiple of ħ.
- Forgetting the volume element. In spherical polars dτ = r² sin θ dr dθ dφ; dropping r² sin θ makes normalisation and every expectation value wrong.
- Assuming ⟨A²⟩ = ⟨A⟩². They are equal only when ψ is an eigenfunction of  — otherwise the difference is exactly the variance you are asked to find.
Check the numbers, not just the algebra. Quantum questions end in arithmetic — energy gaps, de Broglie wavelengths, photon energies. The ABC Chemistry Calculator Suite has the derivative, integral and physical-constants tools you need beside your problem set.
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