Oscillator Eigenfunctions and Hermite Polynomials

Derivation

Oscillator Eigenfunctions and Hermite Polynomials

D-146 Home PU-202 Threads energy · waves · matter Depends on The Harmonic Oscillator via Ladder Operators

Statement

For the one-dimensional quantum harmonic oscillator Ĥ = p̂²/2m + ½mω²x̂², the normalized position-space energy eigenfunctions are generated from the Gaussian ground state ψ₀ by repeated application of the raising operator, ψₙ = (â†)ⁿψ₀ / √(n!). Carrying this out in position space produces ψₙ(x) = Nₙ Hₙ(ξ) e⁻ξ²⁄² with ξ = √(mω/ℏ) x, where the polynomial prefactor Hₙ(ξ) is exactly the physicists' Hermite polynomial of degree n.

Why it matters

The oscillator is the single most reused exactly-solvable system in physics: every small oscillation, every normal mode, every free quantum field mode, and every coherent- or squeezed-state construction is built on these eigenfunctions. Knowing that the ladder algebra produces a definite, closed-form family of wavefunctions — and that this family is the classical Hermite polynomials times a Gaussian — is what lets the abstract operator solution be turned into probability densities, matrix elements, and expansion coefficients.

It also demonstrates a template that recurs across quantum mechanics: an algebraic ground-state condition (a first-order ODE) plus an algebraic raising map (a differential recursion) together deliver the entire spectrum's wavefunctions without ever solving the second-order Schrödinger ODE term-by-term. The Hermite polynomials appear not as an ansatz but as a forced consequence of the algebra.

Assumptions

The potential is exactly harmonic, V(x) = ½mω²x², over all of ℝ.If dropped, the raising operator no longer maps eigenstates to eigenstates with a uniform spacing and the closed Hermite form fails; anharmonic corrections mix the levels. The Hilbert space is L²(ℝ) with square-integrable, boundary-vanishing states.If dropped, the ground-state ODE admits the runaway solution e⁺ξ²⁄² and normalizability — the selection principle that fixes ψ₀ — is lost. The ladder commutator [â, â†] = 1 holds, i.e. x̂ and p̂ are canonically conjugate with [x̂, p̂] = iℏ.If dropped, the operator identity â† = (1/√2)(ξ − ∂_ξ) is not the true adjoint action and the generated functions are neither orthogonal nor normalized by √(n!).

Derivation

ξ ≡ √(mω/ℏ) x, â = (1/√2)(ξ + ∂_ξ), â† = (1/√2)(ξ − ∂_ξ)

Import the dimensionless position ξ and the position-space forms of the ladder operators from oscillator-ladder-operators; p̂ = −iℏ ∂_x becomes −i√(mωℏ) ∂_ξ. A

â ψ₀ = 0 ⇒ (ξ + ∂_ξ) ψ₀(ξ) = 0

The ground state is annihilated by lowering (no state lies below it); the 1/√2 divides out, leaving a first-order linear ODE. A

dψ₀/ψ₀ = −ξ dξ ⇒ ψ₀(ξ) = C e⁻ξ²⁄²

Separate variables and integrate; the second homogeneous solution e⁺ξ²⁄² is discarded as non-normalizable. B

1 = |C|²√(ℏ/mω) ∫ e⁻ξ² dξ = |C|²√(ℏ/mω)√π ⇒ C = (mω/πℏ)¹⁄⁴

Impose ∫|ψ₀|² dx = 1 with dx = √(ℏ/mω) dξ and the Gaussian integral ∫e⁻ξ²dξ = √π. B

ψₙ = (â†)ⁿ ψ₀ / √(n!)

Each raising step sends the normalized n-eigenstate to √(n+1) times the next; composing n times and dividing by √(n!) keeps ψₙ unit-normalized (from oscillator-ladder-operators). A

ψₙ(ξ) = (2ⁿ n!)⁻¹⁄² (ξ − ∂_ξ)ⁿ ψ₀(ξ)

Insert â† = (1/√2)(ξ − ∂_ξ); the n factors of 1/√2 combine with 1/√(n!) to give (2ⁿ n!)⁻¹⁄². B

(ξ − ∂_ξ) f = −eξ²⁄² ∂_ξ(e⁻ξ²⁄² f)

Verify by the product rule: −eξ²⁄²∂_ξ(e⁻ξ²⁄²f) = −eξ²⁄²(−ξ e⁻ξ²⁄²f + e⁻ξ²⁄²f′) = ξf − f′. The Gaussian conjugation turns multiplication-plus-derivative into a pure derivative. C

(ξ − ∂_ξ)ⁿ = (−1)ⁿ eξ²⁄² ∂_ξⁿ e⁻ξ²⁄²

Iterate step 7: adjacent factors e⁻ξ²⁄²·eξ²⁄² = 1 telescope, leaving one outer eξ²⁄², one inner e⁻ξ²⁄², and n bare derivatives. C

(ξ − ∂_ξ)ⁿ e⁻ξ²⁄² = e⁻ξ²⁄² ⋅ (−1)ⁿ eξ² ∂_ξⁿ e⁻ξ² ≡ e⁻ξ²⁄² Hₙ(ξ)

Set f = e⁻ξ²⁄² in step 8 so e⁻ξ²⁄²f = e⁻ξ²; the bracket is exactly the Rodrigues formula Hₙ(ξ) = (−1)ⁿ eξ² dⁿ/dξⁿ e⁻ξ², which is a degree-n polynomial. C

ψₙ(x) = (mω/πℏ)¹⁄⁴ (2ⁿ n!)⁻¹⁄² Hₙ(ξ) e⁻ξ²⁄², ξ = √(mω/ℏ) x

Substitute steps 4 and 9 into step 6. Orthonormality ∫HₙH_me⁻ξ²dξ = √π 2ⁿ n! δ_nm confirms the prefactor normalizes each state. B

Result

ψₙ(x) = (mω/πℏ)^1/4 · (2ⁿ n!)^−1/2 · Hₙ(ξ) · e^−ξ²/2, ξ = √(mω/ℏ) x, Eₙ = ℏω(n + ½)

Reading. Every oscillator eigenfunction is a fixed Gaussian envelope e⁻ξ²⁄² modulated by a Hermite polynomial Hₙ of degree n. The polynomial supplies exactly n real nodes (zeros), so ψₙ has n nodes and alternates parity: even in x for even n, odd for odd n. The Gaussian sets a universal width scale b = √(ℏ/mω) (the "oscillator length"), while raising n stretches probability outward toward the classical turning points.

Units check. In 1D a wavefunction carries dimension [length]⁻¹⁄² so that |ψ|² is a probability density. Here mω/ℏ has units kg·s⁻¹/(J·s) = m⁻², so (mω/πℏ)¹⁄⁴ ∼ m⁻¹⁄². The argument ξ = √(mω/ℏ) x is dimensionless, hence Hₙ(ξ) and e⁻ξ²⁄² are pure numbers, and ψₙ carries m⁻¹⁄² overall. Consistent.

Limiting cases

Ground state, n = 0: H₀ = 1, giving the minimum-uncertainty Gaussian ψ₀ = (mω/πℏ)¹⁄⁴ e⁻ξ²⁄² with Δx Δp = ℏ/2.
First excited, n = 1: H₁ = 2ξ, so ψ₁ ∝ ξ e⁻ξ²⁄² — one node at the origin, odd parity.
Large n (correspondence limit): the probability density |ψₙ|², smoothed over its rapid oscillations, approaches the classical distribution ∝ 1/√(A² − x²) that piles weight near the turning points ±A.
Free-particle limit, ω → 0: the oscillator length b = √(ℏ/mω) → ∞, the Gaussian flattens, and the discrete bound states spread into the continuum.
Classical limit, ℏ → 0 at fixed E: level spacing ℏω shrinks relative to E, node spacing collapses, and quantization becomes invisible.

Breaks when

Anharmonicity. Real potentials (e.g. Morse, or V with a quartic term) are only parabolic near the minimum. The ladder no longer closes, â† mixes would-be eigenstates, and the Hermite×Gaussian form is only the leading approximation for low-lying, small-amplitude states.
Relativistic regime. When ℏω becomes comparable to mc² the non-relativistic Ĥ is invalid; the Klein–Gordon/Dirac oscillator has a different spectrum and spinor structure, not these scalar Hermite functions.
Boundedness broken. On a half-line or with a wall/barrier the boundary condition changes; only the Hermite functions vanishing at the wall survive, so the naive full-line family is not the correct basis.
Time-dependent or driven ω(t). A time-varying frequency makes the instantaneous eigenfunctions non-stationary; the state becomes a squeezed superposition and no single Hₙ describes it.

Failure modes

Wrong Hermite convention. Using the probabilists' Heₙ(ξ) (weight e⁻ξ²⁄²) instead of the physicists' Hₙ(ξ) (weight e⁻ξ²); they differ by Hₙ(ξ) = 2^n/2Heₙ(ξ√2) and produce a wrong normalization and wrong node positions.
Forgetting to rescale x. Writing Hₙ(x) with dimensionful x rather than Hₙ(ξ) — the polynomial argument must be the dimensionless ξ = √(mω/ℏ) x.
Dropping the (2ⁿ n!)⁻¹⁄² factor. Applying â† without the compensating 1/√(n+1) at each step leaves states with norm √(n!), not 1.
Sign error in â†. Taking â† ∝ (ξ + ∂_ξ) instead of (ξ − ∂_ξ); that operator lowers and annihilates ψ₀, giving zero.
Keeping the growing solution. Retaining e⁺ξ²⁄² from the ground-state ODE; it satisfies the equation but is not normalizable and is unphysical.
Miscounting nodes. Claiming n+1 or n−1 nodes — a degree-n Hermite polynomial has exactly n real simple zeros, so ψₙ has exactly n interior nodes.

Discussion

The deep point is that solving a second-order ODE (the time-independent Schrödinger equation) has been replaced by two first-order operations: annihilate to pin the ground state, then raise. The Hermite polynomials are not guessed as a series ansatz; they are forced by the Gaussian conjugation identity in step 7, which converts the raising operator into a plain n-fold derivative and hands back the Rodrigues formula. This is why the same polynomials that Hermite studied for the heat kernel reappear in quantum mechanics — both are governed by the operator eξ²⁄²∂_ξe⁻ξ²⁄².

Physically, each Hermite function is a stationary standing wave in the parabolic well. The Gaussian is the ground-state "carrier" that guarantees decay into the classically forbidden region on both sides; the polynomial modulation supplies the oscillatory interior structure whose node count equals the excitation number, exactly mirroring the classical fact that higher energy means more wiggles per unit length. The turning-point pile-up at large n is the seed of the WKB and correspondence-principle bridge to classical motion.

These functions form a complete orthonormal basis of L²(ℝ), which is why they underlie so much machinery beyond the oscillator: they are the eigenfunctions of the Fourier transform (with eigenvalues (−i)ⁿ), the basis of Hermite–Gaussian optical modes, and the number-state basis in which coherent and squeezed states are expanded. The ladder structure directly gives the selection rule ∆n = ±1 for the dipole operator x̂ ∝ â + â†, the backbone of vibrational spectroscopy.

Two subtleties reward attention. First, the raising construction never invokes the differential equation for Hₙ or its three-term recurrence; those emerge as consequences — e.g. Hₙ‱ = 2n Hₙ₋₁ follows from âψₙ = √n ψₙ₋₁, and Hₙ₊₁ = 2ξHₙ − 2nHₙ₋₁ from â† plus â acting on ψₙ. Second, orthonormality ⟨ψ_m|ψₙ⟩ = δ_mn is guaranteed a priori because the ψₙ are non-degenerate eigenstates of the Hermitian Ĥ; the Hermite orthogonality integral is therefore not an independent assumption but a rediscovery of that spectral fact.

Common misconceptions. The Gaussian envelope is not unique to the ground state — it multiplies every ψₙ; what changes with n is only the polynomial, and the state's growing spatial extent comes from that polynomial reaching further out before the Gaussian cuts it off, not from a widening Gaussian. Also, "more energy" does not mean "narrower" — higher states spread out toward the turning points.

Worked examples

Build and normalize ψ₁ for an electron with ω = 1.00×10¹⁶ rad/s.

Apply â† once to ψ₀ and evaluate the oscillator length. B

ψ₁ = â†ψ₀ = (1/√2)(ξ − ∂_ξ)(mω/πℏ)¹⁄⁴e⁻ξ²⁄²

Use step 6 with n = 1; here 2ⁿn! = 2. A

(ξ − ∂_ξ)e⁻ξ²⁄² = ξe⁻ξ²⁄² − (−ξ)e⁻ξ²⁄² = 2ξ e⁻ξ²⁄²

Differentiate the Gaussian; this is H₁(ξ) = 2ξ as predicted. A

ψ₁(x) = (mω/πℏ)¹⁄⁴ (1/√2)(2ξ) e⁻ξ²⁄² = (mω/πℏ)¹⁄⁴ √2 ξ e⁻ξ²⁄²

The 1/√2 from â† and the 2 from H₁ leave the standard √2 prefactor. B

b = √(ℏ/mω) = √(1.055×10⁻³⁴ / (9.11×10⁻³¹ × 1.00×10¹⁶)) = √(1.158×10⁻²⁰) m

Insert ℏ = 1.055×10⁻³⁴ J·s, m_e = 9.11×10⁻³¹ kg. A

ψ₁(x) = √2 (mω/πℏ)^1/4(x/b) e^−x²/2b², b = 1.08×10⁻¹⁰ m ≈ 0.108 nm

Reading. The first excited electron state is an odd function with a single node at x = 0 and peaks near x = ±b. Its width scale of about one Ångström matches atomic dimensions, consistent with such a stiff (ℏω ≈ 6.6 eV) electronic oscillator.

Locate the nodes of ψ₃ for a proton with ω = 5.00×10¹⁴ rad/s.

Find the zeros of H₃(ξ) and convert to x. B

H₃(ξ) = 8ξ³ − 12ξ = 4ξ(2ξ² − 3)

The Rodrigues formula (step 9) with n = 3; factor to expose the roots. A

ξ = 0 or ξ = ±√(3/2) = ±1.2247

Three simple real zeros — exactly n = 3 nodes, confirming the node-count rule. A

b = √(ℏ/m_pω) = √(1.055×10⁻³⁴ / (1.673×10⁻²⁷ × 5.00×10¹⁴)) = √(1.261×10⁻²²) m

Insert m_p = 1.673×10⁻²⁷ kg. A

x = ξ b, b = 1.123×10⁻¹¹ m ⇒ x = 0, ±1.2247 × 1.123×10⁻¹¹ m

Undo the rescaling ξ = x/b. B

nodes at x = 0 and x = ±1.38×10⁻¹¹ m (≈ ±0.0138 nm)

Reading. The n = 3 proton state has odd parity with three nodes symmetric about the origin. The outer nodes sit at ±1.22 b, comfortably inside the classical turning points A = b√(2n+1) = b√7 ≈ 2.65 b, illustrating how the wavefunction oscillates between the turning points and decays beyond them.

Problems

Show explicitly that â†ψ₁ = √2 ψ₂ reproduces H₂(ξ) = 4ξ² − 2.
Solution
Start from ψ₁ ∝ H₁e⁻ξ²⁄² = 2ξ e⁻ξ²⁄². Then ψ₂ = (1/√2)â†ψ₁, and using the polynomial map, (ξ − ∂_ξ)[H₁ e⁻ξ²⁄²] = (ξH₁ − H₁′ + ξH₁)e⁻ξ²⁄² = (2ξH₁ − H₁′)e⁻ξ²⁄². With H₁ = 2ξ, H₁′ = 2: 2ξ(2ξ) − 2 = 4ξ² − 2 = H₂(ξ). The relation Hₙ₊₁ = 2ξHₙ − Hₙ′ is the raising recurrence; hence ψ₂ ∝ (4ξ²−2)e⁻ξ²⁄², an even state with two nodes at ξ = ±1/√2. ✓
Verify normalization of ψ₂ directly using ∫H₂²e⁻ξ²dξ = √π 2² 2! = 8√π.
Solution
∫|ψ₂|²dx = (mω/πℏ)¹⁄²(2²2!)⁻¹∫H₂²e⁻ξ² b dξ with b = √(ℏ/mω) and (mω/πℏ)¹⁄²b = √(mω/πℏ)√(ℏ/mω) = 1/√π. So the integral = (1/√π)(1/8)(8√π) = 1. ✓ The factor 2ⁿn! = 8 is precisely what the Hermite norm cancels.
An electron oscillator has ω = 2.00×10¹⁶ rad/s. Compute the oscillator length b and the classical amplitude A of the n = 4 state.
Solution
b = √(ℏ/mω) = √(1.055×10⁻³⁴/(9.11×10⁻³¹×2.00×10¹⁶)) = √(5.79×10⁻²¹) = 7.61×10⁻¹¹ m. Classical turning point from E₄ = ½mω²A² with E₄ = ℏω(4+½) = 4.5ℏω: A = b√(2n+1) = b√9 = 3b = 2.28×10⁻¹⁰ m (≈ 0.228 nm). ✓
Using x̂ = b(â + â†)/√2, evaluate the dipole matrix element ⟨ψ₃|x̂|ψ₂⟩ and state the selection rule it implies.
Solution
x̂|ψ₂⟩ = (b/√2)(â+â†)|ψ₂⟩ = (b/√2)(√2|ψ₁⟩ + √3|ψ₃⟩). Projecting onto ⟨ψ₃| keeps only the |ψ₃⟩ term: ⟨ψ₃|x̂|ψ₂⟩ = (b/√2)√3 = b√(3/2). Since x̂ only connects n to n±1, the selection rule is Δn = ±1. ✓
Show that ψₙ is an eigenfunction of the Fourier transform and find its eigenvalue for n = 2. (Take the transform in the dimensionless variable ξ.)
Solution
The Hermite functions φₙ(ξ) = Hₙ(ξ)e⁻ξ²⁄² satisfy F[φₙ] = (−i)ⁿφₙ because the Fourier transform commutes with Ĥ (it swaps ξ ↔ −i∂_ξ, under which â, â† pick up phases ∓i), so it must be diagonal on the non-degenerate eigenbasis; acting on â† shows each raising multiplies the eigenvalue by −i, giving (−i)ⁿ starting from +1 for φ₀ (a Gaussian is its own transform). For n = 2: (−i)² = −1. ✓ Thus ψ₂ transforms into itself up to the sign −1.