University Physics V · The Quantum Harmonic Oscillator · 7.3

The Number Operator & the Ladder's Floor

The oscillator's whole spectrum falls out of two facts about N = a†a: its expectation value is a squared length, and its commutator with a shifts eigenvalues by one. Here you learn to run that argument as a proof — and to see exactly which claim it never establishes.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

N = a†a is not a new postulate: it is what is left of H once ħω and the half-quantum are divided out, and the whole spectrum is already inside it. Two facts do all the work. First, ⟨ψ|N|ψ⟩ = ‖a|ψ⟩‖² ≥ 0 for every |ψ⟩ in the domain, so no eigenvalue of N is negative — a statement about the inner product, not about energy.

Second, [a, a†] = 1 forces [N, a] = −a and [N, a†] = +a†, so a and a† send an eigenvector of N to an eigenvector one rung lower or higher, or to the zero vector, and to nothing else. Together they close the argument: if ν were not a non-negative integer, the chain ‖aᵏ|ν⟩‖² = ν(ν−1)⋯(ν−k+1) would eventually go negative, which no Hilbert-space vector may do. The only escape is a product that lands exactly on zero, and that happens only for ν = 0, 1, 2, …, with a|0⟩ = 0 as a conclusion rather than an assumption.

Upward there is no escape at all, since ‖a†|n⟩‖² = n + 1 never vanishes, so the ladder is infinite and Fock space infinite-dimensional. What the argument does not buy is multiplicity: it fixes the spectrum as a set and is silent on how many independent vectors sit on each rung. Non-degeneracy is a separate input — irreducibility of the canonical commutation relation, or concretely that a|0⟩ = 0 is a first-order ODE with a one-dimensional normalisable solution space.

Simple definition: The number operator N = a†a is the Hermitian, positive operator whose eigenvalue n counts the quanta in a mode; H = ħω(N + ½) then makes energy a matter of counting rungs rather than solving a differential equation.
Example: For the CO stretch, ωₑ = 2170 cm⁻¹ gives ħω = 0.269 eV, so N|3⟩ = 3|3⟩ carries three quanta at E = 3.5ħω = 0.942 eV, while a|3⟩ = √3|2⟩ has squared norm exactly 3.

The number operatorN = a†a , N† = N , H = ħω(N + ½)

Every energy question about the oscillator becomes a question about the integer spectrum of one dimensionless operator.

N is dimensionless and its eigenvalue n counts quanta; ħω carries the joules.

Positivity, straight from the inner product⟨ψ|N|ψ⟩ = ⟨aψ|aψ⟩ = ‖a|ψ⟩‖² ≥ 0

The floor of the spectrum, fixed before any Hamiltonian is named — this is positive-definiteness, not energetics.

True for every |ψ⟩ in the domain of a, with equality only when a|ψ⟩ = 0.

The two ladder commutators[N, a] = −a , [N, a†] = +a†

N(a|ν⟩) = (ν − 1)(a|ν⟩): one rung down, or the zero vector, with no third possibility.

From [A, BC] = [A, B]C + B[A, C] with [a, a†] = 1; neither a nor a† is Hermitian.

Norms of the stepped vectors‖a|n⟩‖² = n , ‖a†|n⟩‖² = n + 1

The descent can stop, at n = 0 and nowhere else; the ascent never can, because n + 1 ≥ 1 always.

For ⟨n|n⟩ = 1. A phase convention then fixes a|n⟩ = √n|n−1⟩ and a†|n⟩ = √(n+1)|n+1⟩.

The descent chain that forces integers‖aᵏ|ν⟩‖² = ν(ν−1)(ν−2)⋯(ν−k+1)

A negative squared norm is impossible, so the product must land exactly on zero — which happens only for ν = 0, 1, 2, …

For non-integer ν > 0 the negative factor ν − ⌈ν⌉ enters at k = ⌈ν⌉ + 1: at ν = 2.5, k = 4 gives −0.9375.

Why no finite matrix holds the laddertr[a, a†] = 0 ≠ D = tr Itruncated: [a, a†] = I − D|D−1⟩⟨D−1|

The algebra has no finite-dimensional representation, so N is unbounded above — and every truncated basis lies in its top row.

D is the truncation dimension; the last diagonal entry of the commutator is 1 − D, so −39 at D = 40.

N is Hermitian, and its expectation value is a length

Build N = a†a and check the two properties that matter. It is symmetric, since N† = (a†a)† = a†(a†)† = a†a = N, so its eigenvalues are real. Then comes the move the whole topic rests on: ⟨ψ|N|ψ⟩ = ⟨ψ|a†a|ψ⟩ = ⟨aψ|aψ⟩ = ‖a|ψ⟩‖² ≥ 0, which uses only the defining property of the adjoint and the positive-definiteness of the inner product. For a normalised eigenvector, N|ν⟩ = ν|ν⟩ with ⟨ν|ν⟩ = 1 gives ν = ‖a|ν⟩‖². Read that as an identity, not an inequality: the eigenvalue is not merely non-negative, it is the squared length of the lowered vector, and that is what makes the rest of the argument quantitative rather than rhetorical. Equality ν = 0 holds if and only if a|ν⟩ = 0, so the vacuum is characterised rather than chosen. Nothing so far has mentioned m, ω or ħ. H = ħω(N + ½) enters only to convert a count into joules: for the CO stretch one rung is 0.269 eV, but the integer structure was settled before ω was named.

Two commutators, and an either/or most treatments skip

Expand with the product rule for commutators: [N, a] = [a†a, a] = a†[a, a] + [a†, a]a = 0 + (−1)a = −a, and identically [N, a†] = a†[a, a†] + [a†, a†]a = +a†. Now act on an eigenvector: N(a|ν⟩) = (aN + [N, a])|ν⟩ = (aN − a)|ν⟩ = (ν − 1)(a|ν⟩). The conclusion has two branches, and skipping the second is the standard sloppiness. Either a|ν⟩ is an eigenvector of N with eigenvalue ν − 1, or a|ν⟩ is the zero vector — which satisfies every eigenvalue equation and is an eigenvector of nothing. That either/or is the hinge of the topic, because it means the ladder can only ever stop by producing the zero vector; it cannot fade out, land on a non-eigenvector, or leave the space. The same statement runs upward for a†, giving eigenvalue ν + 1 or the zero vector. Notice what the commutators do not supply: nothing here makes ν real or positive, and nothing fixes the length of a|ν⟩. Those are separate questions, answered by Hermiticity and by the norm.

Chain the norms and the integers appear

With ⟨ν|ν⟩ = 1 the first link is ‖a|ν⟩‖² = ν, so the normalised lowered vector is a|ν⟩/√ν whenever ν > 0. Iterating, ‖a²|ν⟩‖² = ν(ν − 1), and in general ‖aᵏ|ν⟩‖² = ν(ν−1)(ν−2)⋯(ν−k+1). Every entry in that list is a squared norm and must therefore be non-negative. Take ν = 2.5 and read the chain: 2.5, then 2.5 × 1.5 = 3.75, then × 0.5 = 1.875, then × (−0.5) = −0.9375. The fourth entry is a negative squared norm — not a curiosity but a flat impossibility in an inner-product space — so no operator of the form a†a has 2.5 in its spectrum. The only way to dodge the negative factor is for the product to hit exactly zero first, and that requires some factor ν − j to vanish, which means ν = j, an integer. Zero is not a loophole either: ‖a^(j+1)|ν⟩‖² = 0 says a annihilates the last vector exactly, so the descent terminates at a|0⟩ = 0. The spectrum therefore lies inside (0, 1, 2, …), and the vacuum condition is a theorem rather than an axiom.

The ceiling that is not there

Run the same machinery upward and the symmetry breaks. ‖a†|n⟩‖² = ⟨n|aa†|n⟩ = ⟨n|(N + 1)|n⟩ = n + 1, and since n ≥ 0 this is at least 1 and never zero. So a†|n⟩ is never the zero vector, every rung has one above it, and the whole set (0, 1, 2, …) is realised: the spectrum is unbounded above and Fock space is infinite-dimensional. A second, entirely independent argument gives the same verdict in one line. Take the trace of [a, a†] = 1 in a space of finite dimension D: the left-hand side is tr(aa†) − tr(a†a) = 0 by cyclicity of the trace, while the right-hand side is D. No pair of D × D matrices satisfies the canonical relation, for any finite D. That is why a and a† must be unbounded operators with domains, and why the harmonic ladder cannot be squeezed into a qubit-sized space no matter how the matrices are chosen. The two-level systems that Pauli matrices describe obey a different algebra, not a truncated copy of this one.

What the algebra never proves: one vector per rung

The argument constrains the spectrum as a set. It says nothing about multiplicity, and that gap is real rather than pedantic. Let A = a ⊗ I act on F ⊗ ℂ², a single mode with a spin-½ spectator riding along untouched. Then [A, A†] = [a, a†] ⊗ I = I, so every line above survives verbatim and the spectrum of A†A is still (0, 1, 2, …) — yet ker A = (|0⟩ ⊗ |χ⟩) is two-dimensional, and (A†)ⁿ carries that whole plane upward, so every rung is doubly degenerate. The multiplicity of level n is simply dim ker a, and no commutator can report it. Restoring non-degeneracy takes an extra input. Abstractly it is irreducibility: by Stone–von Neumann, every irreducible representation of [x, p] = iħ by self-adjoint operators is unitarily equivalent to the Schrödinger one, whose vacuum is unique, and the spectator above makes the representation reducible. Concretely, project a|0⟩ = 0 onto ⟨x| and it becomes ψ₀′ + (mω/ħ)xψ₀ = 0, a first-order linear ODE whose solution space is one-dimensional. Uniqueness is an analysis fact about that equation, not an algebraic one.

Building the matrices without arguing in a circle

In NumPy, represent a by the D × D matrix with ⟨m|a|n⟩ = √n δ_(m, n−1) — a superdiagonal reading √1, √2, …, √(D−1). Then N = a†a comes out as diag(0, 1, …, D−1) and its eigenvalues are the integers you set out to derive, so printing them proves nothing; the construction assumed the answer. What the matrices are genuinely good for is checking the claims that were derived. Confirm column by column that ‖a|n⟩‖² = n and ‖a†|n⟩‖² = n + 1. Confirm that [a, a†] − I vanishes everywhere except the last diagonal slot, where it equals 1 − D: at D = 40 that entry is −39, not 1, because the truncation silently deleted the rung above. The practical rule follows. Any quantity computed in a truncated basis is trustworthy only while the state carries negligible weight on the top rungs, so convergence is established by raising D and watching a number stop moving — never by trusting one run. Add a quartic λx⁴ and this bites at once, since x⁴ couples |n⟩ to |n ± 4⟩ and so drags weight toward the truncation edge.

Change one variable at a time

Make the relationship visible.

Interactive model

Trial eigenvalue ν of N2.5

Lowering steps k applied4

Set ν to a whole number and every bar stands on or above the axis, the chain reaching exactly zero and stopping there. Now nudge ν to 2.5 and drag the step slider: the fourth bar drops below the axis to −0.9375, a squared norm no vector can have — which is the whole proof that ν is an integer.

TRIAL ν2.5

STEPS APPLIED k4

NEXT FACTOR ν − k-1.5

CHAIN DIES AT k =4

Live interpretationTRIAL ν: 2.5. STEPS APPLIED k: 4. NEXT FACTOR ν − k: −1.5. CHAIN DIES AT k =: 4

Catch the common trap

Explain before calculating.

Misconception check

The usual story is that the ladder stops because energy cannot be negative, or because you cannot have fewer than zero quanta, with a|0⟩ = 0 quietly assumed as the bottom rung. That is circular, and it is not what closes the argument. Nothing in [a, a†] = 1 forces integers by itself: on sequences indexed by all of ℤ, set c eₙ = eₙ₋₁ and b eₙ = (n + λ)eₙ₊₁, and [c, b] = 1 holds exactly while bc has spectrum λ − 1 + ℤ, unbounded below and generically non-integer. What that representation lacks is b = c†, an adjoint taken in a positive-definite inner product. Positivity is doing the work, and it does it quantitatively: ‖aᵏ|ν⟩‖² = ν(ν−1)⋯ may never go negative, so the product must land on zero, so ν is an integer, so a|0⟩ = 0. The vacuum condition is the conclusion, not the premise — and |0⟩ is a unit vector with ⟨0|H|0⟩ = ħω/2, not the zero vector.

A colleague reports a normalised single-mode state |ν⟩ obeying N|ν⟩ = 2.5|ν⟩, where N = a†a and [a, a†] = 1. Which statement correctly settles whether such a state can exist?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyFrom [a, a†] = 1 alone, derive [N, a†] = a† for N = a†a, show that a†|4⟩ is an eigenvector of N and name its eigenvalue, then compute ‖a†|4⟩‖² and ‖a|4⟩‖² and confirm their difference against the commutator.

Expand with [A, BC] = [A, B]C + B[A, C], applied as [a†a, a†] = a†[a, a†] + [a†, a†]a = a†(1) + 0 = a†. So [N, a†] = a†.
Act on the eigenvector: N(a†|4⟩) = (a†N + [N, a†])|4⟩ = (a†N + a†)|4⟩ = (4 + 1)(a†|4⟩). The eigenvalue is 5, provided a†|4⟩ is not the zero vector — which step 3 settles.
Norms, using ⟨4|4⟩ = 1 and nothing else: ‖a†|4⟩‖² = ⟨4|aa†|4⟩ = ⟨4|(N + 1)|4⟩ = 5, and ‖a|4⟩‖² = ⟨4|a†a|4⟩ = ⟨4|N|4⟩ = 4.
Difference: 5 − 4 = 1 = ⟨4|[a, a†]|4⟩, as it must be. Since 5 ≠ 0 the raised vector is genuine, and normalising gives a†|4⟩ = √5|5⟩ = 2.236|5⟩, while a|4⟩ = √4|3⟩ = 2|3⟩.

Answer[N, a†] = a†; a†|4⟩ has eigenvalue 5, with ‖a†|4⟩‖² = 5 and ‖a|4⟩‖² = 4 differing by exactly 1. Hence a†|4⟩ = √5|5⟩ = 2.236|5⟩ and a|4⟩ = 2|3⟩.

MediumA colleague claims a single-mode number operator with eigenvalue ν = 7/3. Compute the squared norms ‖aᵏ|ν⟩‖² for k = 1 to 4, identify the step at which the claim dies, and state in general the first k at which any non-integer ν > 0 is caught.

The chain is ‖aᵏ|ν⟩‖² = ν(ν−1)⋯(ν−k+1), and every entry is a squared norm, so every entry must be ≥ 0. Here ν = 7/3 = 2.3333.
k = 1: ν = 7/3 = 2.3333. k = 2: (7/3)(4/3) = 28/9 = 3.1111. Both positive, so the first two lowered vectors are legitimate normalisable states.
k = 3: multiply by ν − 2 = 1/3, giving 28/27 = 1.0370. Still positive, still no contradiction — the chain has not yet reached the dangerous factor.
k = 4: multiply by ν − 3 = −2/3, giving −56/81 = −0.6914. A negative squared norm, so ν = 7/3 cannot be an eigenvalue of any a†a.
In general the first negative factor is ν − ⌈ν⌉, which enters the product at k = ⌈ν⌉ + 1. Here ⌈7/3⌉ = 3, so k = 4, exactly as computed. For integer ν the same step k = ν + 1 delivers exactly zero instead, which is termination rather than contradiction.

AnswerThe chain reads 2.3333, 3.1111, 1.0370, −0.6914; the fourth entry is negative, so ν = 7/3 is excluded. Any non-integer ν > 0 is caught at k = ⌈ν⌉ + 1.

HardLet A = a ⊗ I act on F ⊗ ℂ², where a is a single-mode annihilation operator and ℂ² is a spin-½ spectator A never touches. Verify [A, A†] = 1, find the spectrum of N = A†A and the degeneracy of each level, identify which step of the ladder argument fails to deliver non-degeneracy, and name the extra input that restores it. Check the count by truncating at n ≤ 5.

[A, A†] = (a ⊗ I)(a† ⊗ I) − (a† ⊗ I)(a ⊗ I) = (aa† − a†a) ⊗ I = [a, a†] ⊗ I = I. Every commutator identity used in the ladder argument therefore holds verbatim.
So does positivity: ⟨Ψ|N|Ψ⟩ = ‖A|Ψ⟩‖² ≥ 0, and the chain ‖Aᵏ|Ψ⟩‖² = ν(ν−1)⋯(ν−k+1) still forces the spectrum into (0, 1, 2, …). Not one line of the derivation is weakened.
But ker A = (|0⟩ ⊗ |χ⟩ : |χ⟩ ∈ ℂ²) is two-dimensional. The map |χ⟩ ↦ (A†)ⁿ(|0⟩ ⊗ |χ⟩)/√(n!) = |n⟩ ⊗ |χ⟩ is linear and injective, so each eigenvalue n carries a two-dimensional eigenspace.
The failing step is one that was never taken. The ladder argument constrains the spectrum as a set; the multiplicity of level n equals dim ker a, and positivity, [N, a] = −a and the norm chain are all blind to that number.
The restoring input is irreducibility. Stone–von Neumann says any irreducible representation of [x, p] = iħ by self-adjoint operators is unitarily equivalent to the Schrödinger one; here spin operators commute with the whole algebra, so the representation is reducible. Concretely, ⟨x|a|0⟩ = 0 reads ψ₀′ + (mω/ħ)xψ₀ = 0, whose normalisable solutions form a one-dimensional space.
Count check: truncating at n ≤ 5 gives dimension 2 × 6 = 12 rather than 6, and tr N = 2(0 + 1 + 2 + 3 + 4 + 5) = 30 rather than 15. The spectrum is untouched while every multiplicity doubles — precisely the information the algebra never carried.

Answer[A, A†] = 1 and spec N = (0, 1, 2, …) exactly as before, but every level is doubly degenerate because dim ker A = 2. The ladder fixes the spectrum as a set only; non-degeneracy is the separate input dim ker a = 1.