University Physics V · The Quantum Harmonic Oscillator · 7.7
Expectation Values, the Virial Split & Uncertainty
The ladder does more than fix the spectrum. Rewrite x and p through a and a†, and every expectation value in a Fock state collapses to a count of rungs — the vanishing first moments, the equal kinetic and potential shares, and the uncertainty product (n + ½)ħ that only the ground state drives to the floor.
Build the model
Connect the measurement to the mechanism.
In a Fock state |n⟩ no integral is needed to find a moment. Invert the ladder definitions to get x = √(ħ/2mω)(a + a†) and p = i√(mħω/2)(a† − a); every power of x or p is then a polynomial in a and a†, and ⟨n|⋅|n⟩ keeps only the terms that raise exactly as often as they lower. Odd powers keep nothing, so ⟨x⟩ = ⟨p⟩ = 0 on every rung.
The squares keep the number-operator term, giving ⟨x²⟩ = (ħ/mω)(n + ½) and ⟨p²⟩ = mħω(n + ½), so both spreads grow as √(n + ½) and their product is exactly (n + ½)ħ — the Robertson bound at n = 0 and a multiple of it above, which is to say an energy eigenstate is not automatically a minimum-uncertainty state. The same two numbers split the energy: ⟨T⟩ = ⟨V⟩ = Eₙ/2, which is the quantum virial theorem 2⟨T⟩ = ⟨x dV/dx⟩ evaluated on a quadratic V. The cost is that both results belong to the quadratic model alone.
Change the exponent and the split changes with it — hydrogen gives 2⟨T⟩ = −⟨V⟩ — and add a quartic term to the oscillator and the exact statement becomes ⟨T⟩ − ⟨V⟩ = ⟨λx⁴⟩, which is not zero.
- Simple definition
- An expectation value in a Fock state is the diagonal matrix element ⟨n|Â|n⟩; writing  through a and a† reduces it to whichever term raises and lowers equally often, so the calculation becomes counting rather than integration.
- Example
- For x² = (ħ/2mω)(a² + a†² + 2N + 1), only 2N + 1 is diagonal, so ⟨3|x²|3⟩ = (ħ/mω)(3.5). With CO's ħω = 269 meV and μ = 1.139 × 10⁻²⁶ kg that is 7.93 × 10⁻²³ m², an rms stretch of 8.90 pm on a 113 pm bond.
Turns every moment into an orthonormality count — no wavefunction, no integral.
m in kg, ω in rad s⁻¹; a and a† are dimensionless, and neither is Hermitian
The same lines give the electric-dipole selection rule Δn = ±1 on a harmonic ladder.
a|n⟩ ∝ |n−1⟩ and a†|n⟩ ∝ |n+1⟩, while ⟨n|n±1⟩ = 0
Because ⟨x⟩ = ⟨p⟩ = 0, these square-root straight into Δx and Δp.
Units m² and kg² m² s⁻²; the (n + ½) is the same one that sits in Eₙ
An eigenstate of H need not be a minimum-uncertainty state — only |0⟩ is.
ħ = 1.0546 × 10⁻³⁴ J s; m and ω cancel out entirely, equality only at n = 0
Halves the energy without an integral, and names the one exponent where halving works.
For V ∝ xᵏ this reads 2⟨T⟩ = k⟨V⟩; k = 2 here, k = −1 for Coulomb
Gives the first-order shift of a λx⁴ term and the size of the virial imbalance it opens.
Sum of |⟨m|x²|n⟩|² over m = n−2, n, n+2; units m⁴
Invert the ladder before you reach for an integral
The definitions a = (X + iP)/√2 and a† = (X − iP)/√2 invert to X = (a + a†)/√2 and P = i(a† − a)/√2, and restoring the scales x₀ = √(ħ/mω) and p₀ = √(mħω) gives x = √(ħ/2mω)(a + a†) and p = i√(mħω/2)(a† − a). Every power of x or p is now a polynomial in a and a†, and the diagonal element ⟨n|⋅|n⟩ survives only on the terms that raise exactly as often as they lower. An odd power has no such term, so ⟨n|x|n⟩ = ⟨n|p|n⟩ = 0 on every rung. Parity would give the same two zeros, but the algebra gives strictly more: it also hands you the off-diagonal elements, ⟨m|x|n⟩ = √(ħ/2mω)(√n δ_(m, n−1) + √(n+1) δ_{m, n+1}), which is exactly the Δn = ±1 selection rule for an electric-dipole transition on a harmonic ladder.
Square it, and keep the one balanced term
x² = (ħ/2mω)(a + a†)² = (ħ/2mω)(a² + a†² + a a† + a† a). The first two terms move the ket two rungs and drop out of the diagonal element; the last two combine through a a† = a† a + 1 into 2N + 1. So ⟨n|x²|n⟩ = (ħ/2mω)(2n + 1) = (ħ/mω)(n + ½). The same move on p² = −(mħω/2)(a† − a)² leaves ⟨n|p²|n⟩ = mħω(n + ½), the sign flip supplied by the i². Two checks are worth doing. Recombining them gives ⟨p²⟩/2m + ½mω²⟨x²⟩ = (n + ½)ħω = Eₙ, as it must. And at n = 0 the position result reads ⟨x²⟩ = ħ/2mω, precisely the variance of the Gaussian |ψ₀|² ∝ exp(−mωx²/ħ) — the ladder has reproduced an integral you never performed.
The virial theorem is a statement about d⟨xp⟩/dt
Nothing in a stationary state changes, so d⟨xp⟩/dt = 0. Ehrenfest turns that into a commutator: d⟨xp⟩/dt = (i/ħ)⟨[H, xp]⟩. With [p², x] = −2iħp and [V(x), p] = iħ dV/dx, the commutator is [H, xp] = −iħ(p²/m − x dV/dx), so the condition reads 2⟨T⟩ = ⟨x dV/dx⟩. That is the quantum virial theorem, and note what it needs: a stationary state, not a time average. For a homogeneous potential V ∝ xᵏ it collapses to 2⟨T⟩ = k⟨V⟩. The oscillator has k = 2, hence ⟨T⟩ = ⟨V⟩ = Eₙ/2 = (n + ½)ħω/2 — which reproduces the two second moments of the previous section with no extra work. Hydrogen has k = −1 and therefore 2⟨T⟩ = −⟨V⟩: in the ground state ⟨T⟩ = +13.6 eV while ⟨V⟩ = −27.2 eV. Equal halves are a property of k = 2, not of bound states.
Why only the ground state sits on the uncertainty floor
Because ⟨x⟩ = ⟨p⟩ = 0, the spreads are just the square roots of the second moments, so Δx Δp = √[(ħ/mω)(n + ½)] · √[mħω(n + ½)] = (n + ½)ħ. Mass and frequency cancel completely; only the rung number survives. The Robertson floor is ħ/2 for every state, and the Robertson–Schrödinger refinement adds a covariance term ½⟨{x, p}⟩ − ⟨x⟩⟨p⟩ that vanishes here: xp = (iħ/2)(1 − a² + a†²) gives ⟨n|xp|n⟩ = iħ/2, so ⟨{x, p}⟩ = 0 on every rung. The bound therefore stays at ħ/2 for all n, and only n = 0 attains it. The reason is the saturation condition itself — equality in Cauchy–Schwarz demands (p − ⟨p⟩)|ψ⟩ = iλ(x − ⟨x⟩)|ψ⟩ with real λ, a first-order equation whose normalisable solutions are Gaussians. Of the Fock states only |0⟩ is Gaussian; |1⟩ already carries a node.
Put numbers on a real bond
For CO, ħω = 269 meV and the reduced mass is μ = 1.139 × 10⁻²⁶ kg, so ω = 4.086 × 10¹⁴ rad s⁻¹ and the length scale is x₀ = √(ħ/μω) = 4.76 pm. In the ground state Δx = x₀/√2 = 3.37 pm and Δp = 1.57 × 10⁻²³ kg m s⁻¹, whose product is 5.27 × 10⁻³⁵ J s = ħ/2 to the digit. The energy split is ⟨T⟩ = ⟨V⟩ = 67.3 meV, half of the 134.5 meV zero-point energy. Against a 113 pm bond a 3.37 pm spread is 3.0%, small enough that a quadratic fit to the true internuclear potential is honest. Climb to n = 10 and Δx = x₀√10.5 = 15.4 pm, or 14% of the bond, where the cubic term dropped from the Taylor expansion is no longer negligible. One coincidence deserves naming: the classical turning point is Aₙ = √2 Δx, and a classical oscillator's time-averaged x² is also A²/2, so the quantum and classical second moments agree at every n even though the distributions do not.
Add a quartic term and the split opens
Completeness gives the next moment in one line: ⟨n|x⁴|n⟩ = Σₘ |⟨m|x²|n⟩|², where only m = n − 2, n, n + 2 contribute, so with s = ħ/2mω the result is s²[(2n + 1)² + n(n − 1) + (n + 1)(n + 2)] = 3(2n² + 2n + 1)s². Perturb the Hamiltonian by λx⁴ and the first-order shift is 3(2n² + 2n + 1)λs², which makes the spacing E_{n+1} − Eₙ = ħω + 12λs²(n + 1) — the uniform ladder is gone, and with it the flat spectrum that the commutator alone produced. The virial statement degrades in step: 2⟨T⟩ = ⟨mω²x² + 4λx⁴⟩ = 2⟨V₂⟩ + 4⟨V₄⟩, so subtracting ⟨V⟩ = ⟨V₂⟩ + ⟨V₄⟩ leaves the exact identity ⟨T⟩ − ⟨V⟩ = ⟨λx⁴⟩. Hellmann–Feynman fixes its size, since ∂E/∂λ = ⟨x⁴⟩, so at leading order the ground-state imbalance is 3λs². Equal shares were never a quantum principle; they were the k = 2 special case.
Change one variable at a time
Make the relationship visible.
Hold n at 0 and sweep ħω: the box turns tall and narrow, but its area never changes — that fixed area is Δx Δp = ħ/2. Then raise n and watch the area climb as 1, 3, 5, ... times ħ/2, so no rung above the ground state saturates the bound.
SPREAD Δx9.06 pm
SPREAD Δp4.07 10⁻²³ kg m s⁻¹
PRODUCT Δx Δp3.5 ħ
KINETIC = POTENTIAL455 meV
Live interpretationSPREAD Δx: 9.06 pm. SPREAD Δp: 4.07 10⁻²³ kg m s⁻¹. PRODUCT Δx Δp: 3.5 ħ. KINETIC = POTENTIAL: 455 meV
Catch the common trap
Explain before calculating.
An oscillator is prepared in the Fock state |4⟩. Which statement about ⟨x⟩ and the product Δx Δp is correct?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA CO molecule vibrates with ħω = 269 meV and reduced mass μ = 1.139 × 10⁻²⁶ kg. For the ground state n = 0, find Δx and Δp from the ladder algebra and check the uncertainty product.
- x is linear in a and a†, so ⟨0|x|0⟩ = 0, and likewise ⟨0|p|0⟩ = 0. The spreads are therefore just Δx = √⟨x²⟩ and Δp = √⟨p²⟩.
- ω = ħω/ħ = (269 × 10⁻³ × 1.602 × 10⁻¹⁹ J)/(1.0546 × 10⁻³⁴ J s) = 4.086 × 10¹⁴ rad s⁻¹.
- ⟨0|x²|0⟩ = (ħ/μω)(½) = (1.0546 × 10⁻³⁴)/(1.139 × 10⁻²⁶ × 4.086 × 10¹⁴) × 0.5 = 1.133 × 10⁻²³ m², so Δx = 3.37 × 10⁻¹² m.
- ⟨0|p²|0⟩ = μħω(½) = 1.139 × 10⁻²⁶ × 4.309 × 10⁻²⁰ × 0.5 = 2.454 × 10⁻⁴⁶ kg² m² s⁻², so Δp = 1.567 × 10⁻²³ kg m s⁻¹.
- Δx Δp = 3.366 × 10⁻¹² × 1.567 × 10⁻²³ = 5.27 × 10⁻³⁵ J s, which is ħ/2 exactly — the value (n + ½)ħ demands at n = 0, and the only rung that reaches the floor.
AnswerΔx = 3.37 pm, Δp = 1.57 × 10⁻²³ kg m s⁻¹, and Δx Δp = 5.27 × 10⁻³⁵ J s = ħ/2.
MediumFor the same CO oscillator (ħω = 269 meV, μ = 1.139 × 10⁻²⁶ kg, equilibrium bond length 113 pm), take the n = 3 state. Find ⟨T⟩ and ⟨V⟩, the rms stretch √⟨x²⟩ and the classical turning point, then say what their ratio to the bond length implies.
- E₃ = (3 + ½)ħω = 3.5 × 269 meV = 941.5 meV. The virial theorem with V ∝ x² gives 2⟨T⟩ = 2⟨V⟩, so ⟨T⟩ = ⟨V⟩ = E₃/2 = 470.8 meV.
- ⟨x²⟩ = (ħ/μω)(n + ½) = 2.266 × 10⁻²³ × 3.5 = 7.93 × 10⁻²³ m², so √⟨x²⟩ = 8.90 × 10⁻¹² m = 8.90 pm.
- Check the split the long way: ⟨V⟩ = ½μω²⟨x²⟩ = 0.5 × 1.139 × 10⁻²⁶ × (4.086 × 10¹⁴)² × 7.93 × 10⁻²³ = 7.54 × 10⁻²⁰ J = 470.8 meV. It agrees, as the virial theorem said it must.
- The classical turning point at the same energy satisfies ½μω²A² = E₃, so A = √[(2n + 1)ħ/μω] = √(7 × 2.266 × 10⁻²³) = 1.259 × 10⁻¹¹ m = 12.59 pm — exactly √2 times the rms stretch, the same relation a classical oscillator obeys.
- 8.90 pm on a 113 pm bond is 7.9%, and the turning points reach 11.1%. That is far enough out that the cubic term dropped from the Taylor expansion is already measurable as anharmonicity, so the harmonic numbers are a first estimate rather than a prediction.
Answer⟨T⟩ = ⟨V⟩ = 470.8 meV; √⟨x²⟩ = 8.90 pm; A = 12.59 pm = √2 √⟨x²⟩ — that is 7.9% and 11.1% of the 113 pm bond.
HardAdd a stiffening quartic term λx⁴ to the same CO oscillator (ħω = 269 meV), with strength set so that λs² = 1.0 meV, where s = ħ/2μω. Using ⟨n|x⁴|n⟩ = 3(2n² + 2n + 1)s², find the first-order shifts of n = 0 and n = 1, the new 0→1 spacing, and the virial imbalance ⟨T⟩ − ⟨V⟩ in the ground state.
- First-order shift: Eₙ⁽¹⁾ = λ⟨n|x⁴|n⟩ = 3(2n² + 2n + 1)λs². For n = 0 the bracket is 3(1) = 3, giving +3.0 meV; for n = 1 it is 3(2 + 2 + 1) = 15, giving +15.0 meV.
- So E₀ = 0.5 × 269 + 3.0 = 137.5 meV and E₁ = 1.5 × 269 + 15.0 = 418.5 meV. The 0→1 spacing is 281.0 meV against the harmonic 269.0 meV — a rise of 12.0 meV, which is 12λs².
- In general E_{n+1} − Eₙ = ħω + 12λs²(n + 1), so the rungs are no longer equally spaced. The commutator argument that produced a uniform spectrum has lost its footing the moment the potential stops being quadratic.
- For the virial, 2⟨T⟩ = ⟨x dV/dx⟩ = ⟨μω²x² + 4λx⁴⟩ = 2⟨V₂⟩ + 4⟨V₄⟩, where V₂ = ½μω²x² and V₄ = λx⁴. Subtracting ⟨V⟩ = ⟨V₂⟩ + ⟨V₄⟩ leaves the exact result ⟨T⟩ − ⟨V⟩ = ⟨V₄⟩ = ⟨λx⁴⟩.
- At leading order ⟨V₄⟩ = 3λs² = 3.0 meV, and ⟨T⟩ + ⟨V⟩ = E₀ = 137.5 meV, so ⟨T⟩ = 70.3 meV and ⟨V⟩ = 67.3 meV. Compare the harmonic case, where both shares were 67.25 meV: at this order the surplus is carried entirely by the kinetic share.
- The 3.0 meV imbalance is 2.2% of E₀ — small, but non-zero, and it is precisely what kills the equal split.
AnswerE₀ shifts by +3.0 meV to 137.5 meV and E₁ by +15.0 meV to 418.5 meV; the 0→1 spacing rises from 269.0 to 281.0 meV; and ⟨T⟩ − ⟨V⟩ = ⟨λx⁴⟩ = 3.0 meV, with ⟨T⟩ = 70.3 meV against ⟨V⟩ = 67.3 meV.