University Physics V · The Quantum Harmonic Oscillator · 7.2

Annihilation & Creation Operators

Two Hermitian observables go in and one non-Hermitian object comes out, on purpose. This lesson builds a and a† from the quadratures and shows that everything the oscillator does afterwards is bookkeeping on the single number the factorisation misses by.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Over the complex numbers a sum of two squares always factorises, X² + P² = (X − iP)(X + iP), and the order of the two brackets never matters. Promote X and P to the dimensionless quadratures of a stable minimum, where [X, P] = i, and the identity breaks — not by a little, and not by an operator, but by exactly one. The cross term left behind is i(XP − PX) = i[X, P] = −1, so with a = (X + iP)/√2 and its adjoint a† = (X − iP)/√2 the two orderings read a†a = ½(X² + P²) − ½ and aa† = ½(X² + P²) + ½.

Subtract them and [a, a†] = 1; multiply by ℏω and Ĥ = ℏω(a†a + ½). The half-quantum is not a boundary condition, not a convention and not a small correction: it is half the commutator, written down. What the construction costs is that neither a nor a† is an observable.

Neither is Hermitian, so neither is measurable. a is not even normal, since [a, a†] is not zero, so the spectral theorem does not apply to it and it has no orthonormal eigenbasis. The pair is not symmetric either: the identity ‖a†ψ‖² = ‖aψ‖² + ‖ψ‖² makes a† injective, so it annihilates nothing, while a annihilates the state at the foot of the ladder. And both are unbounded, which is why tr[a, a†] = 0 rules out every matrix representation there is, and why a truncated Fock-space model of this algebra is broken at its top rung by construction.

Simple definition: The annihilation operator a and the creation operator a† are the non-Hermitian combinations a = (X + iP)/√2 and a† = (X − iP)/√2 of the dimensionless quadratures, chosen so that [a, a†] = 1 and Ĥ = ℏω(a†a + ½).
Example: For ¹H³⁵Cl, ℏω = 0.358 eV, so the ordering residue ℏω(aa† − a†a)/2 = ℏω/2 = 0.179 eV. It is the same 0.179 eV in every state of the molecule, because [a, a†] = 1 carries no state label at all.

The pair, and its inversea = (X + iP)/√2a† = (X − iP)/√2X = (a + a†)/√2P = i(a† − a)/√2

Packs two Hermitian observables into one non-Hermitian object and back again, so any x̂ or p̂ moment becomes a count of ladder steps.

X = √(mω/ℏ) x̂ and P = p̂/√(mℏω) are dimensionless, so a and a† are too. The dagger is an adjoint, never an inverse.

The one commutator, rescaled[a, a†] = −i[X, P] = 1[a, a] = [a†, a†] = 0

The single physical input. Send it to zero and a, a† commute, Ĥ collapses to ℏω a†a, and the classical factorisation is restored.

Dimensionless: ℏ was left behind with the quadratures. The √2 in each definition is what turns a 2 into a 1.

The two orderings, and their gapa†a = ½(X² + P²) − ½aa† = ½(X² + P²) + ½aa† − a†a = 1

Names the half-quantum as an ordering residue rather than an energy: half a commutator, the same in every state.

Both a†a and aa† are Hermitian and positive; the gap ⟨aa†⟩ − ⟨a†a⟩ is 1 in every normalised state.

The factorised HamiltonianĤ = ½ℏω(X² + P²) = ℏω(a†a + ½) = ℏω(aa† − ½) = ½ℏω(a†a + aa†)

Four spellings of one operator. Only the symmetric last one is manifestly kinetic plus potential; normal ordering drops a real term.

ℏω is the quantum, in J or eV; for ¹H³⁵Cl it is 0.358 eV, so the ordering residue is 0.179 eV.

The norm identity behind the asymmetry‖a†ψ‖² = ‖aψ‖² + ‖ψ‖²

One line, three results: a† annihilates nothing, a can, and ‖a†|n⟩‖ = √(n+1) grows without bound, so neither operator is bounded.

Straight from ⟨ψ|aa†|ψ⟩ = ⟨ψ|(a†a + 1)|ψ⟩. Every term is a squared Hilbert-space norm, so all three are real and non-negative.

Why no finite matrix will dotr[a, a†] = 0, but tr 1_d = d[a_d, a_d†] = 1_d − d |d−1⟩⟨d−1|

Says exactly where a Fock-space NumPy model lies to you, and that no smarter truncation moves the lie somewhere safer.

a_d is the d × d truncation carrying √(n+1) on the first superdiagonal; the defect is rank one and sits on the top rung.

The factorisation that numbers allow and operators do not

Over the complex numbers, u² + v² = (u − iv)(u + iv), and the two brackets may be written in either order because numbers commute. The dimensionless oscillator is exactly that shape: Ĥ = ½ℏω(X² + P²), with X = √(mω/ℏ) x̂ and P = p̂/√(mℏω) both Hermitian and obeying [X, P] = i. Try the factorisation anyway. Expanding (X − iP)(X + iP) gives X² + P² + i(XP − PX) = X² + P² + i[X, P], and i[X, P] = i⋅i = −1. The identity misses — but it misses by a number, not by an operator, which is the whole reason the method survives. Reversing the order flips the sign of that residue, so the two products differ by 2 before any scaling. Absorb a √2 into each bracket and the residue becomes ½ in each direction: a†a = ½(X² + P²) − ½ and aa† = ½(X² + P²) + ½. Nothing has been solved. One operator has been rewritten in two ways, and the difference between them is a number.

Why the definition needs dimensionless quadratures

You cannot add x̂ to ip̂: one is a length, the other a momentum. The quadratures fix that by dividing out the two scales the Hamiltonian itself supplies, a length √(ℏ/mω) and a momentum √(mℏω), so X and P are pure numbers and X + iP is legal. In laboratory clothing the same operator reads a = √(mω/2ℏ)(x̂ + ip̂/mω). For ¹H³⁵Cl, with reduced mass 1.614 × 10⁻²⁷ kg and ω = 5.436 × 10¹⁴ rad s⁻¹, that length is 10.96 pm and √(ℏ/2mω) is 7.75 pm, set against a bond of 127 pm. The √2 is not decoration either. Define b = X + iP without it and you get [b, b†] = 2 and Ĥ = ½ℏω(b†b + 1): the same physics with every later factor doubled. Demanding [a, a†] = 1 is what fixes the normalisation once and for all.

Neither operator is an observable, and they fail differently

Hermiticity is the licence to measure, and a fails it: a† is not a, so the value of a is not something an apparatus returns. What is Hermitian is built from the pair — a + a† = √2 X, i(a† − a) = √2 P, and both a†a and aa†, since (a†a)† = a†(a†)† = a†a. Worse than non-Hermitian, a is not normal: normality means [a, a†] = 0, and here that commutator is 1, the largest failure on offer. So the spectral theorem does not apply and a has no orthonormal eigenbasis; its eigenvectors, met later as coherent states, are non-orthogonal and over-complete. The pair is lopsided too. Take norms in aa† = a†a + 1 to get ‖a†ψ‖² = ‖aψ‖² + ‖ψ‖², so a† is bounded below and can never send a state to zero, while a can, and does, at the foot of the ladder. That single asymmetry is why the spectrum has a bottom and no top.

Unbounded, so never a matrix and never bounded

Suppose a and a† were d × d matrices. The trace is cyclic, so tr(aa†) = tr(a†a) and tr[a, a†] = 0, while tr 1_d = d, never zero. No finite representation of [a, a†] = 1 exists, in any dimension and any basis. Boundedness fails as well: induction on the commutator gives [a, (a†)ⁿ] = n(a†)ⁿ⁻¹, no power of a† can vanish, and taking operator norms leaves n ≤ 2‖a‖‖a†‖ for every integer n, which is the Wintner–Wielandt theorem. So a and a† are unbounded, defined not on the whole Hilbert space but on the dense domain of states of finite mean energy. In practice that rule governs every Fock-space computation: the d × d truncation obeys [a_d, a_d†] = 1_d − d |d−1⟩⟨d−1|, exact on every rung but the last and wrong by d on that one, so convergence is tested by raising d, never by trusting the top of the basis.

Reading the observables back off the pair

Inverting the definitions costs one line: X = (a + a†)/√2 and P = i(a† − a)/√2, or in laboratory units x̂ = √(ℏ/2mω)(a + a†) and p̂ = i√(mℏω/2)(a† − a). The payoff is structural. Because a and a† each shift the rung index by one, x̂ and p̂ are tridiagonal in the energy basis while x̂² and p̂² are pentadiagonal, so a moment that would have been a Gaussian integral becomes a count of which terms survive orthogonality. The same inversion is what makes the pair worth defining far from any spring: in a field theory the mode amplitude is the observable, and a†, a are the operators that add and remove one quantum of it. The algebra travels; the picture of a mass on a spring does not have to.

The ordering you may choose, and its price

Because ½(X² + P²) = a†a + ½ = aa† − ½, you may write the Hamiltonian in whichever ordering suits you, and normal ordering — pushing every a† to the left, then discarding the leftover number — is standard in field theory, where summing ½ℏω over infinitely many modes otherwise diverges. What you may not do is forget that a choice was made. Ĥ and ℏω a†a are different operators, differing by ℏω/2 times the identity, and they stop differing by a constant the moment ω changes. There is a numerical version of the same trap. In a d = 4 truncation, a†a + ½ returns diag(0.5, 1.5, 2.5, 3.5) exactly, while ½(X² + P²) built from the truncated X and P returns diag(0.5, 1.5, 2.5, 1.5). Identical operators, different matrices: the normal-ordered form only ever reaches down the ladder, so it never asks for the rung you deleted.

Change one variable at a time

Make the relationship visible.

Interactive model

Rung n on the ladder3

Rungs kept above n2

Hold n and drag k down to 0. The middle bar collapses, because the rung above n has been deleted and a† has nowhere to send the state, so the difference bar reads −n instead of +1 — the one place a finite matrix is forced to break [a, a†] = 1.

‖a|n⟩‖² = ⟨a†a⟩3

‖a†|n⟩‖² = ⟨aa†⟩4

⟨[a, a†]⟩ at this rung1

⟨Ĥ⟩ / ℏω3.5

Live interpretation‖a|n⟩‖² = ⟨a†a⟩: 3. ‖a†|n⟩‖² = ⟨aa†⟩: 4. ⟨[a, a†]⟩ at this rung: 1. ⟨Ĥ⟩ / ℏω: 3.5

Catch the common trap

Explain before calculating.

Misconception check

The half is not a convention you can normal-order away. Because the classical factorisation is blind to order, students write Ĥ = ℏω a†a, call the ½ unmeasurable, and quietly treat the two as the same operator. They are not: they differ by ℏω/2 times the identity, and that difference is [a, a†], the single physical input in the whole construction. Redefining a Hamiltonian by a c-number is allowed, and field theory does it every day, but you must then also accept that a†a is no longer ½(X² + P²), so the operator you diagonalised is not kinetic plus potential energy, and the moment ω is free to vary the discarded term stops being a constant. The slip runs the other way too, when 'a is not Hermitian' is stretched into 'so a†a cannot be an observable'. (a†a)† = a†a. The number operator is as Hermitian as they come; it is a and a† alone that are not.

Working in dimensionless quadratures, where X and P are Hermitian and [X, P] = i, a student factorises ½(X² + P²) as a†a, concludes that the lowest energy is zero, and blames the observed ℏω/2 on the wavefunction having to vanish far from the minimum. Where exactly does that argument fail?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyWith X and P Hermitian and [X, P] = i, set a = (X + iP)/√2 and a† = (X − iP)/√2. (a) Expand a†a and aa† in terms of X² + P². (b) Read off [a, a†]. (c) Say which of a, a†, a + a†, i(a† − a) and a†a are Hermitian.

a†a = ½(X − iP)(X + iP) = ½(X² + iXP − iPX + P²) = ½(X² + P²) + ½i[X, P]. Substituting [X, P] = i gives a†a = ½(X² + P²) − ½.
aa† = ½(X + iP)(X − iP) = ½(X² + P²) − ½i[X, P] = ½(X² + P²) + ½. Only the sign of the cross term moved, because only the order did.
Subtract: [a, a†] = aa† − a†a = ½ − (−½) = 1. The quadrature part cancels completely, so what is left is a number rather than an operator, and no state was ever mentioned.
Adjoints: (a)† = a† is not a, and (a†)† = a is not a†, so neither is Hermitian. (a + a†)† = a† + a, and a + a† = √2 X. (i(a† − a))† = −i(a − a†) = i(a† − a), and it equals √2 P. (a†a)† = a†(a†)† = a†a.

Answera†a = ½(X² + P²) − ½, aa† = ½(X² + P²) + ½, and [a, a†] = 1. Hermitian: a + a†, i(a† − a) and a†a. Not Hermitian: a and a† themselves.

MediumBuild the d = 4 truncation of a as the 4 × 4 matrix carrying √1, √2 and √3 on its first superdiagonal and zeros elsewhere. Report the diagonals of aa† − a†a, of a†a + ½, and of ½(X² + P²) with X = (a + a†)/√2 and P = i(a† − a)/√2, then say which of the last two you should trust.

The superdiagonal entries are √1, √2, √3 = 1, 1.41421, 1.73205, which is one call to numpy.diag with offset 1. Then a†a is diagonal with entries 0, 1, 2, 3, and aa† is diagonal with entries 1, 2, 3, 0 — the last is zero only because the rung above n = 3 was deleted.
Subtracting, diag(aa† − a†a) = (1, 1, 1, −3). The relation [a, a†] = 1 holds on the three lower rungs and misses by 4 on the top one. The trace is 3 − 3 = 0, as cyclicity forces for any finite matrices.
a†a + ½ returns diag(0.5, 1.5, 2.5, 3.5), which is n + ½ exactly on every rung kept.
In ½(X² + P²) the a² and a†² pieces cancel between the two squares, leaving a diagonal. Each square picks up n/2 from the rung below and (n+1)/2 from the rung above, giving n + ½ on an interior rung. At n = 3 the rung above is gone, so each square gives 1.5 and the half-sum is 1.5, not 3.5: diag(0.5, 1.5, 2.5, 1.5).
Two spellings of one operator are not one matrix once truncated. Trust a†a + ½: normal ordering only ever reaches down the ladder, so it never asks for the rung you deleted, and its error on the kept rungs is exactly zero.

Answerdiag(aa† − a†a) = (1, 1, 1, −3), trace 0. a†a + ½ = diag(0.5, 1.5, 2.5, 3.5) is exact, while ½(X² + P²) = diag(0.5, 1.5, 2.5, 1.5) is wrong by 2 on the top rung.

HardShow that no pair of d × d matrices satisfies [A, A†] = 1, and that no bounded operators do either. Then state what the d-dimensional truncation of a actually satisfies, and evaluate the operator norm of a_d for d = 6 and d = 100.

Finite matrices: the trace is cyclic, so tr(AA†) = tr(A†A) and tr[A, A†] = 0 for any A whatsoever, while tr 1_d = d is never zero. The relation therefore has no representation in any finite dimension and any basis. This is a fact about the algebra, not about a poor choice of matrices.
Bounded operators: induction on [a, a†] = 1 gives [a, (a†)ⁿ] = n(a†)ⁿ⁻¹. No power of a† can vanish, since if the smallest vanishing power were the m-th, that identity would force the one below it to vanish too. Taking operator norms, n‖(a†)ⁿ⁻¹‖ ≤ 2‖a‖‖a†‖‖(a†)ⁿ⁻¹‖, so n ≤ 2‖a‖‖a†‖ for every integer n, which no finite pair of norms survives. Both operators are unbounded: the Wintner–Wielandt theorem.
What the truncation does obey: with √(n+1) on the first superdiagonal, aa† has diagonal entry n + 1 for n ≤ d − 2 and 0 at n = d − 1, while a†a has diagonal entry n throughout. Hence [a_d, a_d†] = 1_d − d |d−1⟩⟨d−1|, a rank-one defect of size d sitting on the top rung, of trace d − d = 0 exactly as step 1 demands.
Norm: a_d†a_d = diag(0, 1, ..., d − 1), so the operator norm of a_d is √(d − 1). At d = 6 that is √5 = 2.236, and at d = 100 it is √99 = 9.950. It diverges with d, which is what unboundedness looks like on a computer.
Practical rule: results are trustworthy only on rungs well below d − 1. Converge a Fock-space calculation by raising d until the quantity of interest stops moving, and never quote a number that leans on the top of the basis.

AnswerNo finite or bounded representation exists: tr[A, A†] = 0 while tr 1_d = d, and boundedness would force n ≤ 2‖a‖‖a†‖ for all n. The truncation obeys [a_d, a_d†] = 1_d − d |d−1⟩⟨d−1|, and the norm of a_d is √(d−1): 2.236 at d = 6, 9.950 at d = 100.