University Physics V · Quantum Uncertainty and Commutation Relations · 8.5
Minimum-Uncertainty States & Squeezing
Robertson's inequality says a product cannot fall below a floor. This lesson asks the harder question: which states actually sit on that floor, what you must give up to build one, and why a state that saturates the bound at noon has already left it by one o'clock.
Build the model
Connect the measurement to the mechanism.
Robertson's inequality is proved by throwing two things away, and this topic is the accounting for both. Cauchy–Schwarz keeps only the case of parallel vectors; splitting ⟨f|g⟩ into real and imaginary parts then discards the real half, the symmetrised covariance C = ½⟨Δx Δp + Δp Δx⟩. Put both back and equality becomes a condition on the state: (p − ⟨p⟩)|ψ⟩ = iλ(x − ⟨x⟩)|ψ⟩ with λ real and positive.
In the position basis that is a first-order ODE whose only normalisable solutions are Gaussians with a real quadratic exponent and a phase at most linear in x. So minimum uncertainty is not a vague ideal but a one-parameter family: choose σₓ and σₚ = ℏ/2σₓ follows, with λ = ℏ/2σₓ² naming which oscillator's ground state you have built. That free parameter is what squeezing exploits — S(r) rescales the two quadratures by exp(−r) and exp(+r), relocating noise at fixed product, which is how an interferometer buys phase resolution and pays in amplitude and in energy.
What no operation buys is a lower floor, because the theorem quantifies over every state. And saturation is fragile: let a free packet run and it acquires a phase quadratic in x, C stops vanishing, and the product climbs as √(1 + τ²) while σₚ never moves at all.
- Simple definition
- A minimum-uncertainty state is one for which Robertson's inequality is an equality, σA σB = ½|⟨[A, B]⟩|, which requires both that (B − ⟨B⟩)|ψ⟩ be parallel to (A − ⟨A⟩)|ψ⟩ and that the symmetrised covariance ½⟨ΔA ΔB + ΔB ΔA⟩ vanish.
- Example
- For x and p that picks out exactly the Gaussians: a packet with σₓ = 10 nm must carry σₚ = ℏ/2σₓ = 5.27 × 10⁻²⁷ kg m s⁻¹, and the product is 5.27 × 10⁻³⁵ J s, which is ℏ/2 to the digit.
The gap between the true product and Robertson's floor is exactly C², so any saturating state must have C = 0.
ΔA = A − ⟨A⟩. C is the symmetrised covariance, with units of A×B; Robertson keeps only the commutator term.
Converts 'which states saturate?' into a single eigenvalue equation, solvable in whichever basis is convenient.
Cauchy–Schwarz needs the two vectors parallel; C = 0 forces the constant purely imaginary. λ carries units [B]/[A].
The only normalisable solutions of the equality ODE, so the Gaussians are the complete saturating set for x and p.
σ = σₓ in metres and is free; then λ = ℏ/2σ² and σₚ = ℏ/2σ. Real quadratic exponent, phase linear in x.
The product stays at ℏ/2 for every r: squeezing relocates noise between conjugates, it never removes any.
r ≥ 0 and dimensionless. Noise reduction in decibels is 8.686 r, so 6.0 dB means r = 0.691.
The product grows as √(1 + τ²): saturation is an instant in a state's history, not a property it keeps.
σ₀ in m, m in kg, t in s. [p, H] = 0 for a free particle, so σₚ is frozen at its initial value.
The phase-space ellipse rotates at ω; Robertson is saturated only when its axes line up with x and p.
The product returns to ℏ/2 every quarter period and peaks at (ℏ/2) cosh 2r an eighth of a period later.
Robertson's proof throws two things away
Set |f⟩ = (A − ⟨A⟩)|ψ⟩ and |g⟩ = (B − ⟨B⟩)|ψ⟩, so σA² = ⟨f|f⟩ and σB² = ⟨g|g⟩. Cauchy–Schwarz gives ⟨f|f⟩⟨g|g⟩ ≥ |⟨f|g⟩|², and that is the first place slack enters: the inequality is tight only when the two vectors are parallel. Now expand the overlap as ⟨f|g⟩ = ½⟨ΔA ΔB + ΔB ΔA⟩ + ½⟨[A, B]⟩. For Hermitian A and B the anticommutator is Hermitian, so its expectation C is real; the commutator is anti-Hermitian, so its expectation is purely imaginary. Real and imaginary parts add in quadrature, giving |⟨f|g⟩|² = C² + ¼|⟨[A, B]⟩|². Keeping both terms is the Schrödinger relation. Robertson discards C², and that is the second place slack enters. Saturating Robertson therefore demands two separate things at once — parallel vectors and zero covariance — and most states fail the second for entirely mundane reasons.
Equality becomes an eigenvalue equation
Parallel vectors means |g⟩ = c|f⟩ for some complex c. Feed that back into the covariance: C = Re⟨f|g⟩ = Re(c)⋅⟨f|f⟩, so with |f⟩ non-zero the condition C = 0 forces Re c = 0, that is c = iλ with λ real. The whole content of 'minimum uncertainty' is then (B − ⟨B⟩)|ψ⟩ = iλ(A − ⟨A⟩)|ψ⟩. Rearranged, |ψ⟩ is an eigenvector of B − iλA, an operator that is deliberately not Hermitian — which is exactly why it may have a complex eigenvalue and why the equation has solutions at all. Taking norms of both sides gives σB = |λ| σA, so λ is not an extra unknown: it is the ratio of the two spreads, and the product is |λ|σA². For A = x and B = p the operator p − iλx is proportional to a lowering operator, so every minimum-uncertainty state of that pair is a coherent state of some oscillator — the one whose mω equals λ.
The ODE, and why only Gaussians survive
Project onto ⟨x| with p = −iℏ d/dx. The condition reads −iℏψ′ − ⟨p⟩ψ = iλ(x − ⟨x⟩)ψ, that is ψ′ = (i⟨p⟩/ℏ)ψ − (λ/ℏ)(x − ⟨x⟩)ψ. It is first order and separable, so the solution is unique up to normalisation: ψ(x) = N exp(i⟨p⟩x/ℏ) exp(−λ(x − ⟨x⟩)²/2ℏ). Normalisability is the only boundary condition available on the whole line, and it does real work — λ must be strictly positive, since λ ≤ 0 leaves a function that grows without bound or refuses to decay. Read off the moments: σₓ² = ℏ/2λ and σₚ = λσₓ = ℏ/2σₓ, so σₓσₚ = ℏ/2 for every λ. Two structural facts follow. The saturating set is one-parameter, not one state: pick any width and the conjugate width is fixed. And the phase matters as much as the modulus, since exp(i⟨p⟩x/ℏ) is a momentum boost and is allowed, while any term quadratic in x would make λ complex, break C = 0, and lift the product.
Squeezing changes λ, not the floor
Fix an oscillator of mass m and frequency ω. Its ground state is the minimum-uncertainty state with λ = mω, giving σₓ = √(ℏ/2mω). Choose λ = mω exp(2r) instead and you get σₓ = √(ℏ/2mω) exp(−r) with σₚ = √(mωℏ/2) exp(r): the same floor, a different share-out. The unitary that does this is S(r) = exp(r(a² − a†²)/2), under which a becomes a cosh r − a† sinh r; what it changes is the ratio of the two variances, never their product. Decibels count variance, so the position noise is down by 8.686 r dB — the 6 dB routinely injected into gravitational-wave interferometers is r = 0.691, a factor 0.501 on the amplitude. It costs energy: the squeezed vacuum carries n̄ = sinh²r = 0.558 quanta and ⟨H⟩ = (ℏω/2) cosh 2r = 1.06 ℏω, so it is emphatically not empty. Nothing here can push the product under ℏ/2, because the theorem quantifies over all states and a unitary maps states to states.
Saturation is an instant, not a property
Does a state stay on the floor once it is put there? For the oscillator with λ = mω it does: a coherent state's widths are constant, its phase-space contour is a circle, and rotating a circle changes nothing. Change λ and it does not. A squeezed state's ellipse rotates at ω, so σₓ²(t) = (ℏ/2mω)(exp(−2r)cos²ωt + exp(2r)sin²ωt) breathes at 2ω, the product returning to ℏ/2 only at ωt = 0, π/2, π and so on, and peaking at (ℏ/2) cosh 2r in between. The free particle is worse, because no restoring force brings the ellipse back: σₚ is constant since [p, H] = 0, while σₓ(t) = σ₀√(1 + τ²) with τ = ℏt/2mσ₀², so the product climbs monotonically as (ℏ/2)√(1 + τ²) and never returns. For an argon atom with σ₀ = 1.0 μm the width doubles in 2.2 ms. Throughout, the Schrödinger relation stays saturated, because a Gaussian under a quadratic Hamiltonian stays Gaussian; what is lost is Robertson's tightness, and what is gained is covariance.
Where nothing saturates the bound
The Gaussian answer belongs to x and p on the whole line, and three failures are worth carrying. Boundary conditions can rule it out: the infinite-well ground state must vanish at both walls, no Gaussian does, and the product comes out σₓσₚ = π√(1/12 − 1/2π²) ℏ = 0.568 ℏ — above the floor, and its own minimum rather than the theorem's. Excited states drift further off: oscillator state |n⟩ has σₓσₚ = (n + ½)ℏ, so only n = 0 sits on the bound and n = 1 is already three times it. And the bound itself can be empty: in the state |l, m = 0⟩ with l ≥ 1 the mean of Lz vanishes, so Robertson reads σLx σLy ≥ 0 while the actual product is ℏ²l(l+1)/2. A floor of zero forbids nothing. Before quoting any saturation claim, name the operator pair, the state, and the domain the operators act on.
Change one variable at a time
Make the relationship visible.
At r = 0, τ = 0 the shaded box sits exactly on the outline — the ground state is the floor. Sweep r with τ held at 0: the box changes shape, never area. Then nudge τ up, and the harder you squeezed x the faster the box widens, because the σₚ you paid for is what drives the spreading.
σₓ ÷ ZERO-POINT WIDTH0.55
σₚ ÷ ZERO-POINT WIDTH1.82
σₓ σₚ ÷ (ℏ/2)1.00
COVARIANCE C ÷ (ℏ/2)0.00
Live interpretationσₓ ÷ ZERO-POINT WIDTH: 0.55. σₚ ÷ ZERO-POINT WIDTH: 1.82. σₓ σₚ ÷ (ℏ/2): 1.00. COVARIANCE C ÷ (ℏ/2): 0.00
Catch the common trap
Explain before calculating.
A free particle is prepared at t = 0 in a Gaussian state with ⟨p⟩ = 0 and σₓ σₚ = ℏ/2. At a later time t the position density |ψ(x, t)|² is still exactly Gaussian, though wider. What is σₓ σₚ then?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is prepared in a minimum-uncertainty Gaussian with σₓ = 0.50 nm and ⟨p⟩ = 0. Find σₚ, check the product against the floor, and find the kinetic energy ⟨p²⟩/2m that the confinement forces on it. Take ℏ = 1.055 × 10⁻³⁴ J s and mₑ = 9.109 × 10⁻³¹ kg.
- Saturation fixes σₚ from σₓ alone, with no other input: σₚ = ℏ/2σₓ = 1.055 × 10⁻³⁴ ÷ (2 × 5.0 × 10⁻¹⁰ m) = 1.055 × 10⁻²⁵ kg m s⁻¹.
- Check the product: σₓ σₚ = 5.0 × 10⁻¹⁰ × 1.055 × 10⁻²⁵ = 5.28 × 10⁻³⁵ J s, and ℏ/2 = 5.28 × 10⁻³⁵ J s. The bound is met, not merely respected.
- With ⟨p⟩ = 0 the second moment is the variance: ⟨p²⟩ = σₚ² = 1.113 × 10⁻⁵⁰ kg² m² s⁻².
- So ⟨T⟩ = ⟨p²⟩/2m = 1.113 × 10⁻⁵⁰ ÷ (1.822 × 10⁻³⁰) = 6.11 × 10⁻²¹ J = 38 meV.
Answerσₚ = 1.06 × 10⁻²⁵ kg m s⁻¹, σₓ σₚ = ℏ/2 exactly, and ⟨T⟩ = 6.1 × 10⁻²¹ J = 38 meV — comparable to kBT at room temperature, 26 meV, which is why nanometre confinement is thermally visible.
MediumAn argon-40 atom (m = 6.63 × 10⁻²⁶ kg) leaves a source as a minimum-uncertainty Gaussian of width σ₀ = 1.0 μm with ⟨p⟩ = 0, then flies freely. How long until the packet is twice as wide, what is σₓ σₚ at that moment, and what is the covariance C = ½⟨Δx Δp + Δp Δx⟩ there?
- σₚ is set once and then frozen, since [p, H] = 0 for a free particle: σₚ = ℏ/2σ₀ = 1.055 × 10⁻³⁴ ÷ (2.0 × 10⁻⁶) = 5.28 × 10⁻²⁹ kg m s⁻¹.
- Free spreading gives σₓ(t) = σ₀√(1 + τ²) with τ = ℏt/2mσ₀². Doubling the width needs √(1 + τ²) = 2, so τ = √3 = 1.732.
- Invert for the time: t = τ · 2mσ₀²/ℏ = 1.732 × 2 × 6.63 × 10⁻²⁶ × 1.0 × 10⁻¹² ÷ (1.055 × 10⁻³⁴) = 2.18 × 10⁻³ s.
- The product tracks σₓ because σₚ cannot move: σₓ σₚ = (ℏ/2)√(1 + τ²) = 2 × (ℏ/2) = ℏ = 1.06 × 10⁻³⁴ J s, twice the floor.
- Covariance follows from x(t) = x + pt/m in the Heisenberg picture with C(0) = 0: C(t) = (t/m)σₚ² = (2.18 × 10⁻³ ÷ 6.63 × 10⁻²⁶) × (5.28 × 10⁻²⁹)² = 9.14 × 10⁻³⁵ J s.
- Check against Schrödinger: √(C² + ℏ²/4) = √(8.35 × 10⁻⁶⁹ + 2.78 × 10⁻⁶⁹) = 1.055 × 10⁻³⁴ J s, equal to σₓ σₚ. The state still saturates the sharper relation while missing Robertson's.
Answert = 2.2 ms, σₓ σₚ = ℏ = 1.06 × 10⁻³⁴ J s (twice the floor), and C = 9.1 × 10⁻³⁵ J s. The packet left the Robertson floor the moment it started spreading, and never leaves the Schrödinger one.
HardA mechanical mode of angular frequency ω is squeezed out of its ground state until the position quadrature shows 6.0 dB of noise reduction. Find r, the two quadrature widths in units of their zero-point values, the mean quantum number, the mean energy, and the largest value σₓ σₚ reaches as the state then evolves under the oscillator Hamiltonian.
- Decibels count variance, and the variance ratio is exp(−2r): 10 log₁₀(exp(−2r)) = −6.0 gives 2r = 0.60 ln 10 = 1.3816, so r = 0.691.
- Widths: σₓ = √(ℏ/2mω) exp(−r) = 0.501 × zero-point, σₚ = √(mωℏ/2) exp(r) = 1.995 × zero-point. Their product is 0.501 × 1.995 = 1.000, so the state sits exactly on ℏ/2.
- Occupation: n̄ = ⟨a†a⟩ = sinh²r, and sinh r = (1.9953 − 0.5012)/2 = 0.7470, so n̄ = 0.558 quanta. A squeezed vacuum is not empty.
- Energy: ⟨H⟩ = ℏω(n̄ + ½) = 1.058 ℏω, which is also (ℏω/2) cosh 2r with cosh 2r = (3.981 + 0.2512)/2 = 2.116. Squeezing is paid for.
- Evolution rotates the phase-space ellipse at ω: σₓ²(t) = (ℏ/2mω)(exp(−2r)cos²ωt + exp(2r)sin²ωt), and σₚ²(t) is the same with sine and cosine swapped.
- At ωt = π/4 both brackets equal cosh 2r, so σₓ σₚ = (ℏ/2) cosh 2r = 2.12 × ℏ/2, the maximum. It falls back to ℏ/2 at every quarter period.
Answerr = 0.691; σₓ = 0.501 and σₚ = 1.995 in zero-point units, product exactly ℏ/2; n̄ = 0.558 quanta; ⟨H⟩ = 1.06 ℏω; and σₓ σₚ peaks at (ℏ/2) cosh 2r = 1.06 ℏ an eighth of a period after each saturation.