Δx Δp ≥ ħ/2 and the price of confinement
A floor under every state, and a price for confinement.
This is the uncertainty relation with no escape clause: its right side is the same number for every state. Use it to say what a position run and a momentum run on one preparation are allowed to show, find the only states that meet it exactly, and turn it into the floor under the zero-point energy that every trap charges.
Build the model
One constant commutator, one floor under every preparation, and a bill for holding still.
1The big idea
The general relation of UQMI-04.07, σA σB ≥ ½|⟨[Â, B̂]⟩|, usually has a right side that depends on the state and vanishes in some states. Position and momentum are the exception that made the relation famous: [x̂, p̂] = iħ (UQMI-04.05) is a number, so its mean is iħ∫|ψ|² dx = iħ in every normalised state, and σₓσₚ ≥ ħ/2 holds with no exceptions. Split the copies of any preparation between a position run and a momentum run and the two histograms can never both be narrow; neither can be a spike, which is why no normalisable state is an eigenfunction of x̂ or of p̂. Equality needs both inequalities in the proof to be tight, and that condition is a first-order differential equation whose only normalisable solutions are Gaussians carrying at most a linear phase. Squared and divided by 2m, the bound becomes a price list: a particle held to a spread σₓ carries ⟨T⟩ ≥ ħ²/(8mσₓ²) even when it is at rest on average, the zero-point energy of every well and oscillator ahead. It is a floor, not a value: the infinite well's ground energy sits π² times above the floor ħ²/(2mL²) that its width L sets, while the oscillator, whose ground state is the saturating Gaussian, lands exactly on its floor. The relation needs a Cartesian coordinate running over the whole line, the energy price needs p²/2m to be valid, and neither says anything about how accurately one detector reads one particle.
2In one sentence
For every normalisable state, the standard deviations of the position outcomes and of the momentum outcomes, each measured on separate copies of that state, obey σₓσₚ ≥ ħ/2, so narrowing one histogram forces the other to widen.
Example. Electrons prepared with σₓ = 0.50 nm must show σₚ ≥ 1.05 × 10⁻²⁵ kg m s⁻¹ in a momentum run: a velocity spread of at least 1.16 × 10⁵ m s⁻¹ and a mean kinetic energy of at least 38.1 meV.
3A constant commutator gives one floor for every state
Two results are quoted here and proved elsewhere: the general relation σA σB ≥ ½|⟨[Â, B̂]⟩| (UQMI-04.07) and the canonical commutator [x̂, p̂] = iħ (UQMI-04.05). Because iħ is a number, ⟨[x̂, p̂]⟩ = iħ∫|ψ|² dx = iħ, and normalisation is the only property of ψ the step uses. So σₓσₚ ≥ ħ/2 = 5.27 × 10⁻³⁵ J s for every normalisable state; the units check, since m × kg m s⁻¹ = J s. Read it as two runs on copies of one preparation. An electron source whose position histogram has σₓ = 0.10 nm, about an atom's size, must give a momentum histogram with σₚ ≥ 5.27 × 10⁻²⁵ kg m s⁻¹, a velocity spread of at least 5.8 × 10⁵ m s⁻¹. A 1.0 μg dust grain held to σₓ = 1.0 μm needs only σᵥ ≥ 5.3 × 10⁻²⁰ m s⁻¹, far below anything a laboratory resolves. The velocity floor ħ/(2mσₓ) is that small for every everyday mass and size, which is why classical mechanics could treat position and velocity as sharp together.
4Neither histogram can be a spike
In UQMI-04.07 a bound went empty wherever ⟨[Â, B̂]⟩ = 0, so an energy eigenstate could have σE = 0 beside a finite σₓ. Here the right side never drops below ħ/2, so σₓ = 0 would need σₚ infinite and σₚ = 0 would need σₓ infinite. That is why x̂ and p̂ have no normalisable eigenfunctions: e(ip₀x/ħ) has ∫|ψ|² dx infinite, and a spike at x₀ is not square-integrable, so neither is a state anyone can prepare (UQMI-05.02 returns to this). What an experiment can do is select. A filter that passes only momentum outcomes inside a window of width δp leaves a state whose momentum outcomes all lie in that window; a spread confined to a width w can be at most w/2, so σₚ ≤ δp/2 and σₓ ≥ ħ/δp. Passing electrons within δp = 1.0 × 10⁻²⁷ kg m s⁻¹, a velocity window of 1.1 km s⁻¹, leaves packets with σₓ of at least 105 nm. Only the product is constrained: either histogram alone can be made as narrow as you like, provided the other widens to pay.
5Equality is a first-order differential equation
The proof in UQMI-04.07 used two inequalities, and equality needs both tight. The Schwarz step is tight only when one deviation is a constant multiple of the other, (p̂ − ⟨p⟩)ψ = λ(x − ⟨x⟩)ψ. Discarding the real part of the overlap costs nothing only when that real part is zero, which forces λ = ia with a real. Write p̂ = −iħ d/dx and the condition is a first-order ODE, dψ/dx = [i⟨p⟩ − a(x − ⟨x⟩)]ψ/ħ, which integrates at once: ψ = A exp[−a(x − ⟨x⟩)²/(2ħ)] e(i⟨p⟩x/ħ). For negative a it grows without limit and for a = 0 it is a plane wave, so only positive a gives a state, and matching |ψ|² to a Gaussian of standard deviation σₓ fixes a = ħ/(2σₓ²), in kg s⁻¹ as λ must be. The states that reach ħ/2 are exactly the Gaussians, with any centre, width and mean momentum, and each has σₚ = ħ/(2σₓ). With a = mω the same equation returns in the harmonic-oscillator unit as the condition on its ground state.
6Shape and phase decide how far above the floor a state sits
Every other state lands strictly above ħ/2, and by how much is a fact about ψ. The sine that fills an infinite well gives σₓσₚ = (ħ/2)√(π²/3 − 2) = 0.568ħ. The cusped e(−|x|/b) gives ħ/√2 = 0.707ħ, its sharp point costing momentum spread. A state cut off abruptly, uniform across a slit, has ⟨p²⟩ = ħ²∫|dψ/dx|² dx infinite, because each abrupt edge makes dψ/dx a δ function: the bound holds with infinite room to spare. Phase matters as much as shape. Give a Gaussian a curved phase, ψ ∝ exp[−(1 − ic)x²/(4σₓ²)], and |ψ|², hence the whole position histogram, is unchanged. But now p̂ψ = λxψ with λ = ħ(c + i)/(2σₓ²): the Schwarz step is still tight, the real part is not zero, and σₓσₚ = (ħ/2)√(1 + c²), which is 0.901ħ at c = 1.5. The momentum run sees what the position run cannot, so a position histogram alone never tells you how close a preparation sits to the floor.
7Confinement is paid for in kinetic energy
The mean kinetic energy is the average of p²/2m over the momentum run, and ⟨p²⟩ = σₚ² + ⟨p⟩². Insert the floor: ⟨T⟩ ≥ ħ²/(8mσₓ²) + ⟨p⟩²/2m. The second term is the cost of moving as a whole. The first is charged even when ⟨p⟩ = 0, as in every bound stationary state, where ⟨x⟩ is constant and Ehrenfest's ⟨p⟩ = m d⟨x⟩/dt gives zero. A trap or wall that makes ψ vanish outside a length L leaves σₓ ≤ L/2, so ⟨T⟩ ≥ ħ²/(2mL²). For an electron that is 38.1 meV at L = 1.0 nm and 0.42 eV at 0.30 nm; halve L and the price quadruples. A neutron is 1839 times heavier, and held within 10 fm it pays 0.21 MeV along each axis, about 5.4 million times the electron's price at 1.0 nm, which is why nuclear energies come in MeV rather than eV. Check every use against the assumption inside p²/2m, that σₚc ≪ mc². An electron squeezed to a few femtometres fails it, and the estimate fails with it.
8A floor, never a value
The bound forecasts how zero-point energies scale, as 1/(mL²), not what they are. The infinite well's ground energy E₁ = π²ħ²/(2mL²) (UQMI-06.03) sits π² = 9.87 times above the floor ħ²/(2mL²), and the gap splits exactly into the two places the estimate gave ground. The sine's σₓ is 0.1808L rather than L/2, a factor (0.5/0.1808)² = 7.65, and its shape sits at 0.568ħ rather than 0.5ħ, a factor (0.568/0.5)² = 1.29; together 9.87. In the oscillator the floor is the value. There ⟨V⟩ = ½mω²⟨x²⟩ ≥ ½mω²σₓ², so every state obeys ⟨H⟩ ≥ ħ²/(8mσₓ²) + ½mω²σₓ², whose smallest value over σₓ is ħω/2, reached at σₓ² = ħ/(2mω). That is the exact ground energy, because the oscillator's ground state is the Gaussian that makes both steps tight. Elsewhere, trust the 1/(mL²) scaling and treat the number as a floor only: the true energy lies above it, by a factor of π² in the well, and inserting the full width L in place of σₓ falls 4π² ≈ 39 times short.
Key equations
What each relation says, and the units it is written in.
Position–momentum uncertainty relation
σₓ σₚ ≥ ħ/2
⟨[x̂, p̂]⟩ = iħ∫|ψ|² dx = iħ in every state, so, unlike the bound for most pairs, this one never goes empty and neither spread can reach zero.
σₓ in m, σₚ in kg m s⁻¹, ħ = 1.055 × 10⁻³⁴ J s; any normalisable ψ, each spread from its own run on copies of ψ
Condition for equality
(p̂ − ⟨p⟩)ψ = ia(x − ⟨x⟩)ψ, with a real and positive
With p̂ = −iħ d/dx this is a first-order ODE for ψ. Only positive a gives a normalisable solution, and then a = ħ/(2σₓ²).
a in kg s⁻¹. Schwarz is tight only for a constant multiple λ; dropping the real part of the overlap is free only if λ = ia is imaginary
The minimum-uncertainty packet
ψ(x) = (2πσₓ²)⁻¹⁄⁴ exp[−(x − x₀)²/(4σₓ²)] e(ip₀x/ħ)
Any centre, any mean momentum, any width, and nothing else: a different shape, or a phase that curves across the packet, lands above ħ/2.
x₀ = ⟨x⟩ in m, p₀ = ⟨p⟩ in kg m s⁻¹, ψ in m⁻¹⁄²; its momentum spread is exactly σₚ = ħ/(2σₓ)
A curved phase lifts the product
σₓ σₚ = (ħ/2)√(1 + c²) for ψ ∝ exp[−(1 − ic)x²/(4σₓ²)]
A position run cannot see c; only a momentum run can. The Schwarz step stays tight, but λ = ħ(c + i)/(2σₓ²) now has a real part.
c dimensionless; |ψ|² and σₓ do not depend on c. Free flight for a time t from c = 0 gives c = ħt/(2mσ₀²) (UQMI-05.09)
Kinetic-energy floor
⟨T⟩ ≥ ħ²/(8mσₓ²) + ⟨p⟩²/(2m)
The first term is the price of confinement. It is charged even when ⟨p⟩ = 0, as in every bound stationary state, where nothing moves on average.
m in kg, ⟨T⟩ in J; from ⟨p²⟩ = σₚ² + ⟨p⟩². Nonrelativistic only: needs σₚc ≪ mc²
Confined to a length L
⟨T⟩ ≥ ħ²/(2mL²)
Forecasts the 1/(mL²) scaling of every zero-point energy ahead. The infinite well's E₁ = π²ħ²/(2mL²) is 9.87 times this floor.
L in m; ψ zero outside an interval of length L forces σₓ ≤ L/2. An electron with L = 1.0 nm: 38.1 meV
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. At c = 0 the box sits on the dashed floor lines and σₓσₚ = 0.500 ħ. Narrow σₓ: the ellipse stands up and ⟨T⟩ climbs. Curve the phase: 2σₓ and the area stay put while the ellipse tilts and σₚ rises off the floor. Shift ⟨p⟩: the ellipse slides unchanged and ⟨T⟩ gains ⟨p⟩²/2m.
Live results
- MOMENTUM SPREAD σₚ
- 1.05 10⁻²⁵ kg m s⁻¹
- PRODUCT σₓσₚ
- 0.500 ħ
- ELECTRON ⟨T⟩
- 38.1 meV
- FLOOR FROM σₓ AND ⟨p⟩
- 38.1 meV
Catch the common trap
The mistake this lesson is built to prevent.
The commonest slip reads the inequality as an equation and σₚ as a momentum: “the particle is confined to L, so its momentum is ħ/(2L) and its energy is ħ²/(8mL²).” Three things are wrong. The relation bounds a product of spreads from below, and only a Gaussian reaches it; the infinite well's ground state is a sine and sits above it. σₓ is not L: for that sine it is 0.181L. And σₚ is the width of a momentum histogram centred on zero, not a momentum every particle carries; outcomes near zero are perfectly allowed, and what the bound limits, from below, is ⟨p²⟩, the mean of the squared outcomes. Put together, the recipe returns ħ²/(8mL²), which is 1/(4π²) ≈ 0.025 of the true E₁ = π²ħ²/(2mL²). Use the bound to say what a state cannot do, and compute what it does do from ψ.
Test it: what changes?
- σₚ doubles, staying at ħ/(2σₓ)
- σₓσₚ stays at 0.500ħ
- ⟨T⟩ quadruples and stays on its floor
- Ellipse stands up; area stays πħ/2
- Position histogram and 2σₓ unchanged
- σₓσₚ rises from 0.500ħ to 0.901ħ
- ⟨T⟩ lifts above its unchanged floor
- Ellipse tilts; area stays πħ/2
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
Scattering data on a medium-mass nucleus put the position spread of a bound neutron at σ = 2.0 fm along each Cartesian axis. Find the smallest momentum spread per axis that a knockout experiment could reveal, and the smallest mean kinetic energy the neutron can have. Take mₙ = 1.675 × 10⁻²⁷ kg.
View solution →
- Each Cartesian coordinate and its own momentum component obey [x̂, p̂ₓ] = iħ, so the bound holds axis by axis: σₚ ≥ ħ/(2σ) = 1.055 × 10⁻³⁴ J s ÷ (2 × 2.0 × 10⁻¹⁵ m) = 2.636 × 10⁻²⁰ kg m s⁻¹.
- Per axis, ⟨pₓ²⟩/2mₙ ≥ σₚ²/2mₙ = (2.636 × 10⁻²⁰ kg m s⁻¹)² ÷ (2 × 1.675 × 10⁻²⁷ kg) = 2.07 × 10⁻¹³ J = 1.30 MeV.
- The kinetic energy sums the three axes: ⟨T⟩ ≥ 3 × 1.30 MeV = 3.9 MeV.
- Checks: (kg m s⁻¹)²/kg = J. Nonrelativistic: σₚc = 49 MeV against mₙc² = 940 MeV, about 5%, so p²/2m holds. Classical limit: the floor falls as 1/m, and for a macroscopic mass held to the same size it is negligible.
Answer. σₚ ≥ 2.6 × 10⁻²⁰ kg m s⁻¹ along each axis, and ⟨T⟩ ≥ 3.9 MeV: a floor the true kinetic energy will exceed, not an estimate of it.
MediumExample 2
An electron is released from a trap in a minimum-uncertainty Gaussian state with σ₀ = 1.0 nm and ⟨p⟩ = 0, and flies freely. Since p̂ commutes with Ĥ = p̂²/2m, its momentum histogram never changes; its position histogram stays Gaussian with σₓ(t) = σ₀√(1 + τ²), where τ = ħt/(2mσ₀²). At t = 20 fs find σₓ, σₚ and σₓσₚ, compare ⟨T⟩ with the floor set by the new σₓ, and say what a position run alone would conclude.
View solution →
- The momentum spread is the one prepared at release: σₚ = ħ/(2σ₀) = 1.055 × 10⁻³⁴ J s ÷ (2.0 × 10⁻⁹ m) = 5.27 × 10⁻²⁶ kg m s⁻¹, at every later time.
- τ = ħt/(2mσ₀²) = (1.055 × 10⁻³⁴ J s × 2.0 × 10⁻¹⁴ s) ÷ (2 × 9.109 × 10⁻³¹ kg × 1.0 × 10⁻¹⁸ m²) = 1.158, so 1 + τ² = 2.340 and σₓ = 1.0 nm × 1.530 = 1.53 nm.
- σₓσₚ = (ħ/2)√(1 + τ²) = 1.53 × ħ/2 = 0.765ħ. The histogram is still Gaussian, yet the state has left the floor: free flight has written a curved phase with c = τ into ψ, faster components running ahead.
- ⟨T⟩ = σₚ²/2m = ħ²/(8mσ₀²) = 9.52 meV and does not change, while the floor from the present width is ħ²/(8mσₓ²) = 9.52 meV ÷ 2.340 = 4.07 meV. A position run alone, read as a minimum-uncertainty packet of width 1.53 nm, would predict σₚ = 3.45 × 10⁻²⁶ kg m s⁻¹ and ⟨T⟩ = 4.07 meV, both too low.
- Checks: at t = 0, τ = 0 and the product returns to ħ/2. At large t, σₓ → ħt/(2mσ₀) = (σₚ/m)t, the width of a classical cloud whose velocities are spread by σₚ/m = 5.8 × 10⁴ m s⁻¹.
Answer. σₓ = 1.53 nm, σₚ = 5.27 × 10⁻²⁶ kg m s⁻¹ unchanged, σₓσₚ = 0.765ħ; ⟨T⟩ = 9.52 meV against a floor of 4.07 meV. A position run alone would underestimate σₚ by a factor 1.53.
HardExample 3
An electron sits in a trap tuned so that its potential is purely quartic, V = βx⁴ with β = 2.0 eV nm⁻⁴; take ħ²/m = 0.0762 eV nm². Using only σₓσₚ ≥ ħ/2 and the fact that no variance is negative, find a lower bound on the ground-state energy and the σₓ at which it is reached. A numerical solution gives E₀ = 0.151 eV, σₓ = 0.311 nm and ⟨T⟩ = 0.101 eV. Which step of your bound gave the most away?
View solution →
- Kinetic floor: ⟨T⟩ = ⟨p²⟩/2m ≥ σₚ²/2m ≥ ħ²/(8mσₓ²), in any state.
- Potential floor: ⟨x⁴⟩ − ⟨x²⟩² is the variance of x², so ⟨x⁴⟩ ≥ ⟨x²⟩², and ⟨x²⟩ = σₓ² + ⟨x⟩² ≥ σₓ². Hence ⟨V⟩ ≥ βσₓ⁴, and every state obeys ⟨H⟩ ≥ f(σₓ) = ħ²/(8mσₓ²) + βσₓ⁴.
- The ground state has some σₓ, so E₀ is at least the smallest value of f. Setting df/dσₓ = −ħ²/(4mσₓ³) + 4βσₓ³ = 0 gives σₓ⁶ = ħ²/(16mβ) = 0.0762 eV nm² ÷ 32 eV nm⁻⁴ = 2.38 × 10⁻³ nm⁶, so σₓ = 0.365 nm.
- There βσₓ⁴ = ħ²/(16mσₓ²), half the kinetic term, so fₘᵢₙ = 3ħ²/(16mσₓ²) = 3 × 0.0762 eV nm² ÷ (16 × 0.1335 nm²) = 0.107 eV. In general form, fₘᵢₙ = 0.472(ħ⁴β/m²)¹⁄³.
- Compare: 0.151 eV ÷ 0.107 eV = 1.41. The kinetic step is nearly tight: ħ²/(8mσₓ²) at σₓ = 0.311 nm is 0.0985 eV against ⟨T⟩ = 0.101 eV, so σₓσₚ = 0.506ħ. The potential step is not: ⟨V⟩ = 0.151 − 0.101 = 0.050 eV, while βσₓ⁴ = 0.0187 eV, a factor of about 2.7, because x² itself has a large spread in the ground state.
- Checks: (ħ⁴β/m²)¹⁄³ is an energy, and fₘᵢₙ scales as β¹⁄³ m⁻²⁄³. As β → 0 the bound falls to zero, as it must with no trap; as ħ → 0 it vanishes, the classical minimum at rest at x = 0.
Answer. E₀ ≥ 0.107 eV, with the bound's optimum at σₓ = 0.365 nm. The true 0.151 eV is 1.41 times higher, and most of the gap lies in ⟨x⁴⟩ ≥ ⟨x²⟩², very little in the uncertainty relation.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
σₓσₚ ≥ ħ/2 in every state; only Gaussians reach it.
Use σₚ ≥ ħ/(2σₓ) for spreads from separate runs on one preparation, and ⟨T⟩ ≥ ħ²/(8mσₓ²) + ⟨p⟩²/(2m) as a floor, never a value. Only Gaussians with at most a linear phase reach equality. The energy price needs p²/2m, so check σₚc ≪ mc².