University Physics IV · The Quantum Harmonic Oscillator · 8.7

Expectation Values & the Virial Split

Once x and p are written in ladder operators, every moment of an oscillator eigenstate is a counting exercise. Do it once and you get the spread in position, the spread in momentum, the even split of energy between kinetic and potential, and the exact uncertainty product — with no Hermite integral anywhere.

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Connect the measurement to the mechanism.

Every question about where an oscillator eigenstate sits and how fast it moves reduces to four numbers: ⟨x⟩, ⟨p⟩, ⟨x²⟩ and ⟨p²⟩. Computed as integrals over Hermite–Gaussian functions they are a chore; written in ladder operators they are bookkeeping. Put x̂ and p̂ in terms of â and â†, and the first moments vanish immediately, because those operators shift n by one and eigenstates of different n are orthogonal — so the variances are the second moments outright.

Keep only the diagonal pieces of x̂² and p̂², and both come out proportional to 2n + 1. Three consequences follow, and they are the content of the topic. The energy divides exactly in half, ⟨T⟩ = ⟨V⟩ = ½(n + ½)ħω, which the virial theorem both explains and limits: it holds because V is quadratic, and fails for the Coulomb potential and for the box.

The uncertainty product is exactly (n + ½)ħ, so the ground state alone sits on the Heisenberg floor and every excited state stands 2n + 1 times above it. And running the argument backwards — minimising ⟨p²⟩/2m + ½mω²⟨x²⟩ against that bound — returns ½ħω and the ground-state width without solving a differential equation at all. The cost is that all of it is exact only for a strictly quadratic potential, and only for stationary states.

Simple definition: In a harmonic-oscillator eigenstate the mean position and momentum are zero, the mean-square values grow as n + ½, the energy divides equally between kinetic and potential, and the uncertainty product is exactly (n + ½)ħ.
Example: For n = 2 with ħω = 0.20 eV: E₂ = 0.50 eV, so ⟨T⟩ = ⟨V⟩ = 0.25 eV, and Δx Δp = 2.5ħ = 2.6 × 10⁻³⁴ J s — five times the Heisenberg floor, which only the ground state reaches.

Position and momentum on the ladderx̂ = (b/√2)(â + â†), p̂ = i(ħ/b√2)(â† − â)

Every moment becomes a count of raising and lowering steps: no Hermite polynomial, no Gaussian integral.

b = √(ħ/mω) is the oscillator length in metres; â|n⟩ = √n |n−1⟩, â†|n⟩ = √(n+1) |n+1⟩

The first moments vanish⟨n|x̂|n⟩ = 0, ⟨n|p̂|n⟩ = 0 for every n

So Δx² = ⟨x²⟩ and Δp² = ⟨p²⟩ outright — the subtraction in the variance costs nothing in an eigenstate.

â and â† shift n by one, and eigenstates of different n are orthogonal.

Second moments⟨x²⟩ = (n + ½) ħ/mω, ⟨p²⟩ = (n + ½) mωħ

Linear in n, so the widths grow only as √(n + ½): the n = 8 state is just √17 ≈ 4.1 times as wide as the ground state.

m in kg, ω in rad s⁻¹; ⟨x²⟩ in m², ⟨p²⟩ in kg² m² s⁻². Both are diagonal and real.

The virial split2⟨T⟩ = ⟨x dV/dx⟩ ⟹ ⟨T⟩ = ⟨V⟩ = ½Eₙ = ½(n + ½)ħω

Find Eₙ and both averages follow with no integral; for V ∝ xᵏ the split is ⟨T⟩/E = k/(k+2), so only k = 2 halves it.

Stationary states only. V = ½mω²x² gives x dV/dx = 2V, which forces the equality.

Uncertainty product in the n-th stateΔx Δp = (n + ½) ħ ≥ ħ/2

Only n = 0 sits on the Heisenberg floor; the n-th state stands 2n + 1 times above it, and no state above it is Gaussian.

Δx = √((n+½)ħ/mω) in m, Δp = √((n+½)mωħ) in kg m s⁻¹.

Zero-point energy from the bound alone⟨H⟩(Δx) = ħ²/(8m Δx²) + ½mω²Δx² ≥ ½ħω

Gives ½ħω and the ground-state width before the Schrödinger equation is solved — confinement, not a leftover constant.

Uses Δp ≥ ħ/2Δx with ⟨x⟩ = ⟨p⟩ = 0; the minimum sits at Δx² = ħ/2mω.

Write x and p on the ladder, and the integrals disappear

Define â = (1/√2)(x̂/b + i b p̂/ħ) with b = √(ħ/mω), and invert it: x̂ = (b/√2)(â + â†) and p̂ = i(ħ/b√2)(â† − â). Now ⟨n|x̂|n⟩ is a sum of ⟨n|â|n⟩ and ⟨n|â†|n⟩, and â|n⟩ is proportional to |n−1⟩ while â†|n⟩ is proportional to |n+1⟩. Neither overlaps |n⟩, so both terms are zero: ⟨x̂⟩ = 0, and by the identical argument ⟨p̂⟩ = 0, for every n. Parity gets you the first of those — |ψₙ|² is even, so the odd integrand ∫x|ψₙ|²dx vanishes — but parity says nothing about ⟨p²⟩, and the ladder says everything. The payoff is immediate: since Δx² = ⟨x²⟩ − ⟨x⟩² and ⟨x⟩ = 0, the variance and the second moment are the same number, and the same holds for momentum.

Keep only the diagonal terms in x̂² and p̂²

Square the ladder forms. x̂² = (b²/2)(â² + â†² + ââ† + â†â). The first two terms connect |n⟩ to |n∓2⟩, so their diagonal elements vanish; the last two combine as ââ† + â†â = 2N̂ + 1, whose eigenvalue is 2n + 1. Hence ⟨x²⟩ = (b²/2)(2n + 1) = (n + ½)ħ/mω. The momentum square is p̂² = (ħ²/2b²)(2N̂ + 1 − â² − â†²), giving ⟨p²⟩ = (n + ½)mωħ — the same factor, with b² traded for ħ²/b². Put numbers on it. An electron (m = 9.11 × 10⁻³¹ kg) in a parabolic quantum dot with ħω = 10 meV has b = ħ/√(mħω) = 2.76 nm, so in the ground state Δx = b/√2 = 1.95 nm and Δp = (ħ/b)/√2 = 2.70 × 10⁻²⁶ kg m s⁻¹.

Half kinetic, half potential — and only here

For any stationary state the virial theorem reads 2⟨T⟩ = ⟨x dV/dx⟩. A quadratic potential has x dV/dx = mω²x² = 2V, so 2⟨T⟩ = 2⟨V⟩ and the energy divides exactly in half: ⟨T⟩ = ⟨V⟩ = ½(n + ½)ħω. Check it against the moments already in hand — ⟨p²⟩/2m = ½(n+½)ħω and ½mω²⟨x²⟩ = ½(n+½)ħω — and they agree. The same theorem tells you where the halving stops. For a homogeneous potential V ∝ xᵏ it gives 2⟨T⟩ = k⟨V⟩, so ⟨T⟩/E = k/(k+2), and only k = 2 splits evenly. The infinite well is the k → ∞ limit: V = 0 wherever the particle is, so ⟨V⟩ = 0 and ⟨T⟩ = E. Hydrogen is k = −1: in the ground state ⟨T⟩ = −E = 13.6 eV and ⟨V⟩ = 2E = −27.2 eV. The classical oscillator shares the 50:50 split, because cos²ωt and sin²ωt both time-average to ½.

The uncertainty product, and the one state that saturates it

With ⟨x⟩ = ⟨p⟩ = 0, Δx = √((n+½)ħ/mω) and Δp = √((n+½)mωħ), so Δx Δp = (n + ½)ħ exactly. Read that carefully. At n = 0 the product is ħ/2, sitting precisely on the Heisenberg bound; the ground state is a Gaussian, and a Gaussian is the only wavefunction that saturates Δx Δp ≥ ħ/2. At n = 1 the product is already 3ħ/2, three times the floor, and at n = 10 it is twenty-one times. So "the oscillator is a minimum-uncertainty system" is true of one state and false of all the rest, and what disqualifies the excited states is their nodes: a function with nodes cannot be Gaussian. Geometrically the product is an area in phase space. The classical orbit of energy Eₙ is an ellipse of semi-axes Aₙ and Pₙ enclosing πAₙPₙ = (n + ½)h — the Bohr–Sommerfeld condition, arriving here as a consequence rather than a postulate.

Zero-point energy is what the bound costs

Run the logic backwards and the ground state falls out without solving anything. In any state with ⟨x⟩ = ⟨p⟩ = 0 the mean energy is ⟨H⟩ = ⟨p²⟩/2m + ½mω²⟨x²⟩ = Δp²/2m + ½mω²Δx². The uncertainty principle forbids Δp from dropping below ħ/2Δx, so ⟨H⟩ ≥ ħ²/(8mΔx²) + ½mω²Δx². Differentiate with respect to Δx and set the derivative to zero: −ħ²/(4mΔx³) + mω²Δx = 0, giving Δx⁴ = ħ²/4m²ω², so Δx² = ħ/2mω — exactly the ground-state variance found above — and ⟨H⟩ = ħω/4 + ħω/4 = ħω/2. Squeezing the particle inward raises the kinetic term faster than it lowers the potential term, and letting it spread does the reverse. A ⁴⁰Ca⁺ ion trapped at ω/2π = 1.0 MHz therefore keeps 2.1 neV of motion, equal to k_B × 24 μK, which is the temperature the trap must beat.

How the spreads line up with the classical orbit

The classical turning point of the n-th state is Aₙ = √((2n+1)ħ/mω), and Δx = √((n+½)ħ/mω), so Δx = Aₙ/√2 at every n without exception. That is precisely the rms displacement of a classical oscillator of amplitude Aₙ, since the time average of cos²ωt is ½. The second moments therefore match the classical values even at n = 0, where no classical motion of that energy is ever observed. The shapes do not match. The ground state's probability density is a Gaussian peaked at the centre, while the classical density, proportional to 1/v(x), diverges at the turning points where the particle moves slowest — and about 16% of the ground-state probability lies beyond ±A₀, in the classically forbidden region. Correspondence in this topic is a claim about moments and about large-n averages, never about the shape of the distribution.

Change one variable at a time

Make the relationship visible.

Interactive model

Quantum number n2

Squeeze factor s (s = 1: eigenstate)1.0

Set s = 1: the box is the square inscribed in the classical circle, and equal sides mean ⟨T⟩ = ⟨V⟩. Slide s away and the product never moves, but ⟨H⟩ climbs above Eₙ — the eigenstate is the state of least energy at fixed uncertainty.

Δx · Δp2.50 ħ

⟨T⟩ kinetic1.25 ħω

⟨V⟩ potential1.25 ħω

⟨H⟩ total2.50 ħω

Live interpretationΔx · Δp: 2.50 ħ. ⟨T⟩ kinetic: 1.25 ħω. ⟨V⟩ potential: 1.25 ħω. ⟨H⟩ total: 2.50 ħω

Catch the common trap

Explain before calculating.

Misconception check

Students carry ⟨T⟩ = ⟨V⟩ = E/2 out of this topic and apply it to the next bound state they meet. It is a quadratic-potential result, not a general one. The virial theorem says 2⟨T⟩ = ⟨x dV/dx⟩, which for V ∝ xᵏ becomes 2⟨T⟩ = k⟨V⟩ — equal halves only at k = 2. In the infinite well, V = 0 wherever the particle is, so ⟨V⟩ = 0 and ⟨T⟩ = E. In hydrogen, k = −1, so in the ground state ⟨T⟩ = −E = +13.6 eV while ⟨V⟩ = 2E = −27.2 eV: the kinetic average is the whole binding energy, and the potential average is twice the total and negative. The theorem is also a statement about stationary states or time averages. In a superposition of |0⟩ and |2⟩ the split oscillates at 2ω and is equal only on average, never instant by instant.

A particle occupies the n = 2 eigenstate of a harmonic oscillator of angular frequency ω. What is the product Δx Δp for this state?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe CO stretching mode has ħω = 0.269 eV. A molecule sits in the n = 3 vibrational state. Find E₃, the average kinetic energy, the average potential energy, and the uncertainty product Δx Δp in units of ħ.

Eₙ = (n + ½)ħω, so E₃ = 3.5 × 0.269 eV = 0.9415 eV ≈ 0.942 eV.
The potential is quadratic, so x dV/dx = 2V and the virial theorem 2⟨T⟩ = ⟨x dV/dx⟩ becomes 2⟨T⟩ = 2⟨V⟩: the two averages are equal.
Since ⟨T⟩ + ⟨V⟩ = E₃, each is half the total: ⟨T⟩ = ⟨V⟩ = 0.9415/2 = 0.471 eV.
In any eigenstate ⟨x⟩ = ⟨p⟩ = 0, so Δx Δp = √(⟨x²⟩⟨p²⟩) = √[(n+½)ħ/mω × (n+½)mωħ] = (n + ½)ħ = 3.5ħ, which is seven times the ħ/2 floor.

AnswerE₃ = 0.942 eV; ⟨T⟩ = ⟨V⟩ = 0.471 eV; Δx Δp = 3.5ħ = 3.69 × 10⁻³⁴ J s.

MediumA ⁴⁰Ca⁺ ion of mass 6.64 × 10⁻²⁶ kg is held in a trap with ω/2π = 1.00 MHz and laser-cooled into the motional ground state. Find the oscillator length b = √(ħ/mω), then Δx and Δp, and verify the product. Finally give Δx if the cooling only reaches n = 10.

ω = 2π(1.00 × 10⁶ s⁻¹) = 6.283 × 10⁶ rad s⁻¹, so mω = 6.64 × 10⁻²⁶ × 6.283 × 10⁶ = 4.172 × 10⁻¹⁹ kg s⁻¹.
b = √(ħ/mω) = √(1.0546 × 10⁻³⁴ / 4.172 × 10⁻¹⁹) = √(2.528 × 10⁻¹⁶ m²) = 1.590 × 10⁻⁸ m = 15.9 nm.
Ground state, n = 0: Δx = b√(n + ½) = b/√2 = 11.2 nm, and Δp = (ħ/b)√(n + ½) = (1.0546 × 10⁻³⁴ / 1.590 × 10⁻⁸)/√2 = 4.69 × 10⁻²⁷ kg m s⁻¹.
Product: (1.124 × 10⁻⁸)(4.69 × 10⁻²⁷) = 5.27 × 10⁻³⁵ J s, which is ħ/2 to three figures — the ground state sits exactly on the bound.
At n = 10: Δx = b√10.5 = 15.90 × 3.240 = 51.5 nm, wider by √21 = 4.58, and Δx Δp = 10.5ħ, twenty-one times the floor.

Answerb = 15.9 nm; Δx = 11.2 nm, Δp = 4.69 × 10⁻²⁷ kg m s⁻¹, Δx Δp = ħ/2. At n = 10, Δx = 51.5 nm and Δx Δp = 10.5ħ.

HardAt t = 0 an oscillator is prepared in ψ = (|0⟩ + |2⟩)/√2. Using x̂ = (b/√2)(â + â†) and p̂ = i(ħ/b√2)(â† − â), find ⟨T⟩ and ⟨V⟩ at t = 0, show the equal split fails, and give their time averages.

x̂² = (b²/2)(â² + â†² + 2N̂ + 1). Diagonally, ⟨0|x̂²|0⟩ = b²/2 and ⟨2|x̂²|2⟩ = 5b²/2. The cross element is ⟨0|x̂²|2⟩ = (b²/2)⟨0|â²|2⟩ = (b²/2)√2, since â²|2⟩ = √2|0⟩.
The two states differ in energy by 2ħω, so the cross term beats at 2ω: ⟨x̂²⟩ = ½(b²/2 + 5b²/2) + (b²√2/2)cos 2ωt = (b²/2)(3 + √2 cos 2ωt).
p̂² = (ħ²/2b²)(2N̂ + 1 − â² − â†²) carries the same pieces with the cross term reversed in sign: ⟨p̂²⟩ = (ħ²/2b²)(3 − √2 cos 2ωt).
Using mω²b² = ħω and ħ²/mb² = ħω: ⟨V⟩ = ½mω²⟨x̂²⟩ = (ħω/4)(3 + √2 cos 2ωt) and ⟨T⟩ = ⟨p̂²⟩/2m = (ħω/4)(3 − √2 cos 2ωt).
At t = 0: ⟨V⟩ = (ħω/4)(4.414) = 1.104 ħω and ⟨T⟩ = (ħω/4)(1.586) = 0.396 ħω — a ratio of 2.78 : 1, not 1 : 1. Their sum is 1.5 ħω = ½(E₀ + E₂), as it must be.
cos 2ωt averages to zero over a period, so ⟨T⟩ = ⟨V⟩ = 0.75 ħω on average. The virial equality survives as a time average and fails instantaneously.

AnswerAt t = 0, ⟨V⟩ = 1.10 ħω and ⟨T⟩ = 0.40 ħω, a ratio of 2.78 : 1. Each oscillates at 2ω about the common time average 0.75 ħω = ½⟨H⟩.