University Physics IV · The Quantum Harmonic Oscillator · 8.5

Ladder Operators & the Algebraic Solution

The series route to the oscillator works, and it is a grind. This one factorises the Hamiltonian instead: two operators built out of x̂ and p̂, one commutator between them, and the whole spectrum, the ground state and every matrix element follow — with the algebra that later becomes the creation and annihilation of quanta.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A sum of two squares factorises: u² + v² = (u − iv)(u + iv). The oscillator Hamiltonian p̂²/2m + ½mω²x̂² is exactly that shape, with u = x̂ and v = p̂/mω, so try it. Multiply the factors back out and the cross terms refuse to cancel, because x̂ and p̂ do not commute: what is left over is i[x̂, p̂]/mω = −ħ/mω.

Absorb the scale √(mω/2ħ) into each bracket and that near-miss becomes two clean statements — [a, a†] = 1 and Ĥ = ħω(a†a + ½). The half-quantum is the failed cancellation written down; it is not a boundary condition and nobody chose it. Everything after that is bookkeeping on one commutator. [Ĥ, a†] = +ħω a† says a† carries an eigenstate one rung up, [Ĥ, a] = −ħω a says a carries it one rung down, and the fact that a norm can never be negative forces the descent to stop on a state with a|0⟩ = 0 — which is what makes n an integer and the spectrum Eₙ = (n + ½)ħω.

The cost is that all of it is welded to a strictly quadratic Ĥ. Add a cubic term and [Ĥ, a†] stops being proportional to a†: the rungs are no longer evenly spaced, and a and a† survive only as a convenient basis for perturbation theory.

Simple definition: The lowering operator a and the raising operator a† are the non-Hermitian combinations of position and momentum that obey [a, a†] = 1 and move an oscillator eigenstate one rung of ħω down or up the energy ladder.
Example: On the n = 2 state, a|2⟩ = √2 |1⟩ ≈ 1.414|1⟩ and a†|2⟩ = √3 |3⟩ ≈ 1.732|3⟩, so a†a|2⟩ = 2|2⟩ and E₂ = (2 + ½)ħω — which for ħω = 0.250 eV is 0.625 eV.

Lowering and raising operatorsa = √(mω/2ħ)(x̂ + ip̂/mω), a† = √(mω/2ħ)(x̂ − ip̂/mω)

Packs both observables into one object. Neither a nor a† is Hermitian, so neither is measurable — they move you between states, they are not quantities you read off a meter.

√(mω/2ħ) has unit m⁻¹ and p̂/mω has unit m, so a and a† are dimensionless; a† is the adjoint of a, not its inverse.

The commutator that runs the whole method[x̂, p̂] = iħ ⟹ [a, a†] = 1, [N̂, a†] = +a†, [N̂, a] = −a

The single physical input. Send ħ → 0 and every right-hand side collapses: no ladder, no zero-point term, just a classical sum of squares.

N̂ = a†a. All four brackets are dimensionless, the ħ having been scaled out by the prefactors.

Factorised Hamiltonian and the number operatorĤ = ħω(a†a + ½) = ħω(N̂ + ½), Eₙ = (n + ½)ħω

Turns an eigenvalue problem into counting. The ½ is the residue of the failed factorisation, not a constant anyone was free to choose.

N̂ = a†a is dimensionless with eigenvalues n = 0, 1, 2, …; ħω is the rung spacing in J, or in eV for molecules.

Normalised stepsa|n⟩ = √n |n−1⟩, a†|n⟩ = √(n+1) |n+1⟩, a|0⟩ = 0

Keeps every rung unit-normalised, and the √0 at n = 0 is what stops the ladder instead of letting the energy fall below ħω/2.

Dimensionless factors, fixed by ⟨n|a†a|n⟩ = n and ⟨n|aa†|n⟩ = n + 1 with the usual real-phase convention.

Ground state from a first-order equationa ψ₀ = 0 ⟹ ψ₀ = (mω/πħ)¹⁄⁴ exp(−x²/2x₀²), x₀ = √(ħ/mω)

The one differential equation the algebra still needs — first order, not second — after which every ψₙ comes from (a†)ⁿψ₀/√(n!).

x₀ is the oscillator length in m; for a proton at ω = 2.00 × 10¹⁴ rad s⁻¹ it is 17.8 pm.

Reading observables back off the ladderx̂ = √(ħ/2mω)(a + a†), p̂ = i√(mħω/2)(a† − a)

Turns every expectation integral into a step count: ⟨n|x̂²|n⟩ = (2n + 1)ħ/2mω drops out because only a†a and aa† survive on the diagonal.

√(ħ/2mω) carries unit m and √(mħω/2) carries kg m s⁻¹, so the operators recover their proper dimensions.

A sum of squares that will not quite factorise

Write the Hamiltonian as Ĥ = ½mω²(x̂² + p̂²/m²ω²) — a sum of two squares, with u = x̂ and v = p̂/mω. Over ordinary numbers that factorises as (u − iv)(u + iv). Try it on operators and the order starts to matter: (u − iv)(u + iv) = u² + v² + i(uv − vu) = u² + v² + i[x̂, p̂]/mω, and since [x̂, p̂] = iħ the leftover is i(iħ)/mω = −ħ/mω. The factorisation misses, and it misses by a constant rather than by an operator. Roll the scale factor √(mω/2ħ) into each bracket to get a and a†, and the miss becomes exactly one half: a†a = Ĥ/ħω − ½, or Ĥ = ħω(a†a + ½). Nothing has been assumed and nothing solved yet — this is a rearrangement of the same operator. But it has already produced the zero-point term and said where that term comes from: not from a wall, not from a boundary condition, but from two observables refusing to commute.

One commutator, and a ladder appears

Multiply in the other order: aa† = Ĥ/ħω + ½, so [a, a†] = aa† − a†a = 1. That one line is the entire engine. Define N̂ = a†a, so Ĥ = ħω(N̂ + ½), and then [N̂, a†] = a†[a, a†] = a† and [N̂, a] = −a, which in energies reads [Ĥ, a†] = ħω a†. Now let Ĥ|ψ⟩ = E|ψ⟩ and act on a†|ψ⟩: Ĥ(a†|ψ⟩) = (a†Ĥ + ħω a†)|ψ⟩ = (E + ħω)(a†|ψ⟩). So a†|ψ⟩ is again an eigenstate, with energy exactly one ħω higher, and the same computation with a lowers it by one ħω. No wavefunction has appeared, no differential equation has been solved, and no Hermite polynomial has been mentioned. The evenly spaced ladder is a consequence of an algebraic identity — which is why every strictly quadratic system, a lattice mode, a mode of the electromagnetic field, a trapped ion, inherits the same spectrum.

Why the ladder has a bottom rung and integer n

Raising can continue forever, since nothing bounds a bound state's energy above. Lowering cannot, and the reason is a norm. For a normalised eigenstate of energy E, ‖a|ψ⟩‖² = ⟨ψ|a†a|ψ⟩ = E/ħω − ½, and a squared norm is never negative, so E ≥ ħω/2 for every state whatsoever. Now suppose E = (ν + ½)ħω with ν not a whole number — say ν = 1.7. Then a|ψ⟩ is nonzero, since its squared norm is 1.7, and it sits at ν = 0.7; applying a again gives a state whose squared norm is 0.7, so it too is nonzero, and it sits at ν = −0.3, an energy of 0.2ħω — below the floor the norm has just established. Contradiction. The only escape is that the descent terminates, and it can terminate only on a state annihilated by a, so a|0⟩ = 0 and ν = 0 exactly. Every rung above is reached in whole steps, giving Eₙ = (n + ½)ħω with n = 0, 1, 2, …. Quantisation here comes from positivity of the inner product, not from fitting waves into a box.

The √n and √(n+1) that keep the rungs normalised

a|n⟩ is proportional to |n−1⟩, but with what number in front? Take the norm: ‖a|n⟩‖² = ⟨n|a†a|n⟩ = ⟨n|N̂|n⟩ = n, so a|n⟩ = √n |n−1⟩ once the free phase is fixed real. For the raise, use the commutator instead of a fresh integral: ‖a†|n⟩‖² = ⟨n|aa†|n⟩ = ⟨n|(a†a + 1)|n⟩ = n + 1, so a†|n⟩ = √(n+1)|n+1⟩. Two things follow at once. The lowering factor vanishes at n = 0, which is the bottom rung stated as an equation rather than argued for. And the raising factor beats the lowering factor by exactly one in the square, at every rung — that difference of one is [a, a†] itself. These roots are not decoration. The dipole matrix element ⟨n+1|x̂|n⟩ = √((n+1)ħ/2mω) is what makes vibrational transition strengths climb with n, so dropping the factor leaves the energies intact and the spectroscopy wrong.

The ground state costs one first-order equation

The algebra hands you energies but no wavefunctions, and it needs exactly one differential equation to get them. Put p̂ = −iħ d/dx into a|0⟩ = 0: √(mω/2ħ)(x + (ħ/mω) d/dx)ψ₀ = 0, so dψ₀/ψ₀ = −(mω/ħ)x dx and ψ₀ ∝ exp(−mωx²/2ħ). That is first order and separable — a line of work, against the second-order equation and two-step recursion the series method has to grind through. Normalising with the Gaussian integral gives ψ₀ = (mω/πħ)¹⁄⁴exp(−x²/2x₀²), where x₀ = √(ħ/mω) is the oscillator length. Every excited state then follows by differentiating and multiplying rather than by solving anything: ψₙ = (a†)ⁿψ₀/√(n!). One application of a† gives ψ₁ = √2 (x/x₀)ψ₀; two give ψ₂ ∝ (2x²/x₀² − 1)ψ₀, which is the Hermite polynomial H₂ = 4ξ² − 2 up to a factor of two, obtained without writing a recursion relation.

What the algebra buys, and where it stops

Expectation values become step counts. With x̂ = √(ħ/2mω)(a + a†), squaring gives terms in a², a†², a†a and aa†; the first two connect |n⟩ to |n±2⟩ and integrate to zero, so ⟨x̂²⟩ = (ħ/2mω)(n + (n+1)) = (2n+1)ħ/2mω with no integral performed. The same move gives ⟨p̂²⟩ = (2n+1)mħω/2, and with it the virial split and the uncertainty product. The cost is narrow but real. First, a and a† are not observables — they are not Hermitian, so 'measuring a' means nothing. Second, every step above used [Ĥ, a†] = ħω a†, and that identity holds only for a strictly quadratic Ĥ. Add a cubic term βx̂³ and the commutator picks up pieces in a², a†² and a†a, so a† no longer maps an eigenstate to an eigenstate: levels shift and the spacing closes up as n grows, which is the anharmonicity a real bond shows. What survives is the language — a† and a become the creation and annihilation operators for a quantum of a field, and the ladder index becomes particle number.

Change one variable at a time

Make the relationship visible.

Interactive model

Rung n2

Quantum ħω0.25 eV

Drag n to 0: the lowering arrow and its bar both vanish, which is a|0⟩ = 0 ending the ladder. Then check that the squares of the two bars differ by exactly one whatever rung you stand on — that difference is [a, a†] = 1.

ENERGY (n + ½)ħω0.625 eV

LOWERING √n1.414

RAISING √(n+1)1.732

aa† − a†a1

Live interpretationENERGY (n + ½)ħω: 0.625 eV. LOWERING √n: 1.414. RAISING √(n+1): 1.732. aa† − a†a: 1

Catch the common trap

Explain before calculating.

For the oscillator eigenstates, a|n⟩ = √n |n−1⟩ and a†|n⟩ = √(n+1)|n+1⟩. Acting on the n = 3 state with a† first and then with a, what is a a†|3⟩?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAn oscillator has ħω = 0.200 eV and sits in the state |5⟩. Write down a|5⟩ and a†|5⟩ with their numerical factors, confirm the eigenvalue of N̂ = a†a, and find E₅ in eV. Which part of this needed a wavefunction?

Lowering: a|5⟩ = √5 |4⟩ = 2.236|4⟩. The state drops one rung and the amplitude is √5, not 5.
Raising: a†|5⟩ = √(5+1)|6⟩ = √6 |6⟩ = 2.449|6⟩. The raising factor always beats the lowering factor by one in the square: 6 − 5 = 1 = [a, a†].
Number operator: a†a|5⟩ = a†(√5|4⟩) = √5 × √5 |5⟩ = 5|5⟩, so N̂ returns 5, as the label demands.
Energy: E₅ = (5 + ½)ħω = 5.5 × 0.200 eV = 1.10 eV.
None of it needed a wavefunction. Every line used only the ladder relations — no integral, no Hermite polynomial, and no ψ(x) written anywhere.

Answera|5⟩ = √5|4⟩ ≈ 2.236|4⟩, a†|5⟩ = √6|6⟩ ≈ 2.449|6⟩, N̂|5⟩ = 5|5⟩ and E₅ = 1.10 eV — all from the algebra alone.

MediumA proton (m = 1.67 × 10⁻²⁷ kg) is held in a harmonic trap of angular frequency ω = 2.00 × 10¹⁴ rad s⁻¹, in the n = 2 state. Use the ladder to find ⟨x̂²⟩ without evaluating an integral, then compare the root-mean-square displacement with the classical turning point at the same energy. Take ħ = 1.055 × 10⁻³⁴ J s.

Put position in ladder form: x̂ = √(ħ/2mω)(a + a†). Acting on |2⟩ gives x̂|2⟩ = √(ħ/2mω)(√2|1⟩ + √3|3⟩).
x̂ is Hermitian, so ⟨2|x̂²|2⟩ = ‖x̂|2⟩‖², and |1⟩ and |3⟩ are orthogonal, so the cross terms die: ⟨x̂²⟩ = (ħ/2mω)(2 + 3) = 5ħ/2mω. That is the general (2n+1)ħ/2mω at n = 2.
Numbers: ħ/2mω = 1.055 × 10⁻³⁴ ÷ (2 × 1.67 × 10⁻²⁷ × 2.00 × 10¹⁴) = 1.055 × 10⁻³⁴ ÷ 6.68 × 10⁻¹³ = 1.579 × 10⁻²² m². So ⟨x̂²⟩ = 7.90 × 10⁻²² m² and xᵣₘₛ = 2.81 × 10⁻¹¹ m = 28.1 pm.
Classical turning point at the same energy: ½mω²A² = (2 + ½)ħω gives A = √(5ħ/mω) = √(1.579 × 10⁻²¹ m²) = 3.97 × 10⁻¹¹ m = 39.7 pm.
Ratio: xᵣₘₛ/A = 28.1/39.7 = 0.707 = 1/√2 exactly, because ⟨x̂²⟩ = A²/2 at every n — the same mean square a classical sinusoid of amplitude A has.

Answer⟨x̂²⟩ = 5ħ/2mω = 7.90 × 10⁻²² m², so xᵣₘₛ = 28.1 pm against a classical turning point of A = 39.7 pm; the ratio is exactly 1/√2, since ⟨x̂²⟩ = A²/2 for every n.

HardBuild ψ₂ by raising the ground state twice. Work in ξ = x/x₀ with x₀ = √(ħ/mω), where a† = (ξ − d/dξ)/√2 and ψ₀ = N₀exp(−ξ²/2) with N₀ = (mω/πħ)¹⁄⁴. Show the result agrees with the Hermite form ψ₂ ∝ H₂(ξ)exp(−ξ²/2), H₂ = 4ξ² − 2, verify the normalisation, and locate the nodes.

First step: d/dξ exp(−ξ²/2) = −ξ exp(−ξ²/2), so a†ψ₀ = (1/√2)(ξ + ξ)N₀exp(−ξ²/2) = √2 ξ N₀ exp(−ξ²/2). Since a†|0⟩ = √1 |1⟩, this is ψ₁ itself.
Second step: a†ψ₁ = (1/√2)(ξ − d/dξ)(√2 ξ N₀exp(−ξ²/2)) = N₀(ξ² − 1 + ξ²)exp(−ξ²/2) = N₀(2ξ² − 1)exp(−ξ²/2).
Normalise by the ladder rather than by an integral: a†|1⟩ = √2|2⟩, so ψ₂ = (1/√2)N₀(2ξ² − 1)exp(−ξ²/2).
Compare with the series route: H₂(ξ) = 4ξ² − 2 = 2(2ξ² − 1), and the standard state is ψ₂ = (2²⋅2!)^(−1/2)N₀H₂(ξ)exp(−ξ²/2) = (1/(2√2))N₀ × 2(2ξ² − 1)exp(−ξ²/2) — the same function, reached without a recursion relation.
Check the norm with the Gaussian moments ∫exp(−ξ²)dξ = √π, ∫ξ²exp(−ξ²)dξ = √π/2 and ∫ξ⁴exp(−ξ²)dξ = 3√π/4: ∫(2ξ² − 1)²exp(−ξ²)dξ = 4(3√π/4) − 4(√π/2) + √π = 2√π. With N₀² = 1/(x₀√π) and dx = x₀dξ, the norm is (1/2)(1/√π)(2√π) = 1.
Nodes: 2ξ² − 1 = 0 at ξ = ±1/√2, that is x = ±x₀/√2 — two nodes, as n = 2 requires, and both well inside the classical turning points at ξ = ±√5.

Answerψ₂ = (1/√2)(mω/πħ)¹⁄⁴(2ξ² − 1)exp(−ξ²/2), identical to the Hermite result because H₂ = 2(2ξ² − 1); it is normalised, and its nodes lie at x = ±x₀/√2.