University Physics V · Quantum Uncertainty and Commutation Relations · 8.4
Δx Δp ≥ ħ/2 and the Cost of Confinement
Robertson's inequality turns into a number the moment you substitute [x̂, p̂] = iħ. This lesson spends that number: it prices confinement in electronvolts, forces the oscillator's ground state up to ħω/2, and keeps electrons out of the nucleus — while refusing the ΔE Δt usually smuggled in beside it.
Build the model
Connect the measurement to the mechanism.
Every other Robertson bound you will meet depends on the state; this one does not. Because [x̂, p̂] = iħ is a multiple of the identity rather than an operator, ⟨[x̂, p̂]⟩ = iħ in every normalised state, and the general bound σA σB ≥ ½|⟨[Â, B̂]⟩| collapses to a constant floor, σₓ σₚ ≥ ħ/2. That universality is what makes it usable as physics rather than as a slogan, and the use is always the same three-step move: a length you can name bounds σₓ from above, the relation converts that into a lower bound on σₚ, and the variance identity ⟨p̂²⟩ = ⟨p̂⟩² + σₚ² ≥ σₚ² converts that into a lower bound on kinetic energy.
A geometric fact has become an energy in joules. Confine an electron to a box of width L and it costs at least ħ²/2mL²; put it in a harmonic well and minimising ⟨Ĥ⟩ against the bound returns ħω/2 exactly, the ground state, with no differential equation solved. What it costs in return is precision.
The bound is a floor, and floors are usually not touched: the infinite well's own ground state sits at 0.568ħ, above ħ/2, because saturation demands a Gaussian and no Gaussian vanishes at two walls. And it prices nothing at all about time, because t̂ is not an operator you can put in the commutator.
- Simple definition
- The position–momentum uncertainty relation says that for one prepared state, the standard deviation of x̂ and the standard deviation of p̂ — each read from its own ensemble of identical copies — satisfy σₓ σₚ ≥ ħ/2, whatever the state and whatever the apparatus.
- Example
- Localise an electron to σₓ = 0.10 nm and σₚ is at least ħ/2σₓ = 5.3 × 10⁻²⁵ kg m s⁻¹, so ⟨p̂²⟩/2m ≥ 1.5 × 10⁻¹⁹ J = 0.95 eV of kinetic energy that no cooling can remove.
The commutator is a c-number, so the right-hand side is the same in every state — unlike σLx σLy, which this same bound lets vanish in |l, 0⟩.
σₓ in m, σₚ in kg m s⁻¹, ħ = 1.0546 × 10⁻³⁴ J s, so the floor is 5.27 × 10⁻³⁵ J s
The working form of the whole topic: name a length the particle is held inside, and read off an energy nothing can go below.
⟨T̂⟩ in J, σₓ in m, m in kg; uses ⟨p̂²⟩ = ⟨p̂⟩² + σₚ² and ⟨p̂⟩² ≥ 0
A rigorous floor sitting π² below the exact ground state π²ħ²/2mL² — right scaling, honest slack, no wavefunction used.
Popoviciu: any density supported on an interval of length L has variance at most L²/4
Returns the exact oscillator ground state from an inequality, because that state really is a Gaussian and saturates it.
ω in rad s⁻¹; at the minimum the two terms are equal at ħω/4 each — the virial split of a quadratic well
Separates the theorem's floor from a state's actual product — quoting ħ/2 as a value is wrong for every state but a Gaussian.
n = 1 gives 0.568ħ and n = 2 gives 1.670ħ; the product grows without limit as n rises
Mandelstam–Tamm bounds how fast ⟨Â⟩ can move. There is no time operator, so it is proved separately, not read off Robertson.
Γ is the FWHM linewidth; with ħ = 6.582 × 10⁻¹⁶ eV s, τ = 1.6 ns gives Γ = 4.1 × 10⁻⁷ eV
Substitute the commutator and the state drops out
Robertson's theorem gives σA σB ≥ ½|⟨[Â, B̂]⟩|, and in general the right-hand side is an expectation value that moves as the state moves. The position–momentum case is the exception. [x̂, p̂] = iħ⋅1̂ is a c-number times the identity, so ⟨[x̂, p̂]⟩ = iħ⟨ψ|ψ⟩ = iħ for every normalised ψ, and the bound reads σₓ σₚ ≥ ħ/2 with no ψ left in it. Compare angular momentum, where [L̂ₓ, L̂y] = iħL̂z: in the state |l, m = 0⟩ the bound reads σLx σLy ≥ 0 and says nothing at all, even though the actual product there is l(l+1)ħ²/2, which is ħ² for l = 1. Notice also what the two σ's are. Both are computed from one state vector — ⟨ψ|x̂²|ψ⟩ − ⟨ψ|x̂|ψ⟩², and likewise for p̂. Testing the inequality means preparing that state twice over and measuring x̂ on one ensemble, p̂ on the other. Nothing in the derivation ever measures both on the same copy.
Turning a spread into an energy
The bound as written constrains statistics, not energies; one identity converts it. Write ⟨p̂²⟩ = ⟨p̂⟩² + σₚ², drop the first term because it cannot be negative, and you have ⟨T̂⟩ = ⟨p̂²⟩/2m ≥ σₚ²/2m ≥ ħ²/8mσₓ². Every step is an inequality pointing the same way, so the result is a genuine floor and never an estimate that might come out too high. What the chain needs from you is an upper bound on σₓ, and that is geometry rather than quantum mechanics: a density living inside an interval of length L has variance at most L²/4, so σₓ ≤ L/2 and ⟨T̂⟩ ≥ ħ²/2mL². For an electron in L = 0.20 nm that is (1.0546 × 10⁻³⁴)² ÷ (2 × 9.109 × 10⁻³¹ × 4.0 × 10⁻²⁰) = 1.53 × 10⁻¹⁹ J = 0.95 eV. Read the direction carefully: this floor applies to every state in the box, not only the ground state, and it is why holding an electron at atomic size costs of order an electronvolt — the scale of chemistry.
A floor is not a value: the well never touches it
Compute the well's actual product. Its ground state ψ₁ = √(2/L) sin(πx/L) has ⟨x̂⟩ = L/2 and ⟨x̂²⟩ = L²(1/3 − 1/2π²), so σₓ = L√(1/12 − 1/2π²) = 0.1808 L; and with ⟨p̂⟩ = 0 and ⟨p̂²⟩ = 2mE₁ = π²ħ²/L², σₚ = πħ/L. The product is 0.1808 × π ħ = 0.568ħ, fourteen per cent above the floor, and level n gives ħ√(n²π²/12 − ½), which grows without limit. The reason is a boundary condition. Saturating Robertson requires (p̂ − ⟨p̂⟩)|ψ⟩ = iλ(x̂ − ⟨x̂⟩)|ψ⟩ for real λ, a first-order equation whose only normalisable solutions are Gaussians — and no Gaussian vanishes at x = 0 and x = L. The walls forbid saturation before any integral is done. Nothing about being the ground state makes a state minimum-uncertainty; that is a separate property, taken up in the next topic.
The oscillator: minimising against the bound is exact
Take Ĥ = p̂²/2m + ½mω²x̂². The ground state of a symmetric potential has definite parity, so ⟨x̂⟩ = ⟨p̂⟩ = 0 and the expectation is built from variances alone: ⟨Ĥ⟩ = σₚ²/2m + ½mω²σₓ². The bound says σₚ cannot fall below ħ/2σₓ, so the best any state of width σₓ can do is E(σₓ) = ħ²/8mσₓ² + ½mω²σₓ². Differentiate: −ħ²/4mσₓ³ + mω²σₓ = 0 gives σₓ = √(ħ/2mω), and substituting back returns ħω/4 from each term, so Eₘᵢₙ = ħω/2. That is not an estimate but the exact ground-state energy, and the reason is the previous section run backwards: the oscillator's ground state is a Gaussian, so it saturates the bound the calculation assumed, while the box's is a sine and leaves a factor π² on the table. For carbon monoxide, ν̃ = 2143 cm⁻¹ gives ω = 4.04 × 10¹⁴ rad s⁻¹ and a zero-point energy of 0.133 eV that survives to absolute zero.
Confinement as a scale test: who fits inside a nucleus
Because the floor scales as ħ²/mL², it sorts physics by mass and size faster than any model does. Work in ħc = 197.3 MeV fm. Confine anything to a nucleus, σₓ ≤ 5.0 fm, and its momentum spread obeys σₚ c ≥ ħc/(10 fm) = 19.7 MeV. For an electron that dwarfs mₑ c² = 0.511 MeV, so it is ultra-relativistic, E ≈ pc, and its energy scale is about 20 MeV — against β⁻ electrons observed at roughly 1 MeV and a nuclear well only some 8 MeV per nucleon deep. Nothing binds a 20 MeV particle in an 8 MeV well, which is why nuclei contain no electrons: the β electron is created at the decay, n → p + e⁻ + ν̄ₑ. Put a proton in the identical box and the identical 19.7 MeV of momentum spread costs only (19.7)²/(2 × 938) = 0.21 MeV, comfortably bound. Same box, same ħ; the mass in the denominator decides.
Why ΔE Δt is not an instance of this theorem
Robertson takes two operators on one Hilbert space. Time is not one of them: in non-relativistic quantum mechanics t is a parameter labelling which state vector you hold, and Pauli's argument shows it cannot be promoted — a self-adjoint T̂ with [T̂, Ĥ] = iħ would make exp(iεT̂/ħ) shift every energy eigenvalue by ε, forcing the spectrum of Ĥ to fill the whole real line and destroying any ground state. There is simply nothing to substitute. Two true statements wear the name instead. Mandelstam–Tamm: for any observable Â, σH · σA/|d⟨Â⟩/dt| ≥ ħ/2, a bound on how long an expectation value takes to move by one of its own standard deviations. And the spectral one: an amplitude decaying as exp(−t/2τ) has a Lorentzian transform of width Γ = ħ/τ. Hydrogen's 2p level lives 1.6 ns, so Γ = 4.1 × 10⁻⁷ eV, about 100 MHz on a 2.47 × 10¹⁵ Hz line. Both are proved on their own terms; neither is σₓ σₚ ≥ ħ/2 in disguise.
Change one variable at a time
Make the relationship visible.
Leave the product at its floor 0.50 ħ and drag σx to 0.70, the step nearest √q: the solid curve bottoms out at 0.500 ħω, the true ground state. Now raise the product to 1.00 ħ and no width recovers it — the lowest ⟨H⟩ any state can reach is just σx σp times ω.
MOMENTUM SPREAD σp0.42 √(mħω)
KINETIC ⟨T⟩0.087 ħω
TOTAL ⟨H⟩0.807 ħω
FLOOR ON ⟨H⟩ = σxσp ω0.50 ħω
Live interpretationMOMENTUM SPREAD σp: 0.42 √(mħω). KINETIC ⟨T⟩: 0.087 ħω. TOTAL ⟨H⟩: 0.807 ħω. FLOOR ON ⟨H⟩ = σxσp ω: 0.50 ħω
Catch the common trap
Explain before calculating.
An electron is prepared many times in the ground state of an infinite square well of width L. Position is measured on one batch of identical copies and momentum on another. What is σₓ σₚ?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron is confined to a one-dimensional box of width L = 0.20 nm. Use σₓ σₚ ≥ ħ/2 to find the smallest mean kinetic energy any state in the box can have, then compare it with the exact ground-state energy E₁ = h²/8mL² = 9.40 eV.
- Bound σₓ first, from geometry rather than from any wavefunction: the density lives inside an interval of length L, and a distribution on an interval of length L has variance at most L²/4. So σₓ ≤ L/2 = 1.0 × 10⁻¹⁰ m.
- Feed that into the relation: σₚ ≥ ħ/2σₓ ≥ ħ/L = 1.0546 × 10⁻³⁴ ÷ 2.0 × 10⁻¹⁰ = 5.27 × 10⁻²⁵ kg m s⁻¹.
- Convert the spread to an energy with ⟨p̂²⟩ = ⟨p̂⟩² + σₚ² ≥ σₚ²: ⟨T̂⟩ ≥ σₚ²/2m = (5.27 × 10⁻²⁵)² ÷ (2 × 9.109 × 10⁻³¹) = 1.53 × 10⁻¹⁹ J = 0.95 eV.
- Compare with the exact answer of 9.40 eV: the bound is low by a factor 9.87 = π². It is a floor of the right form and the right scale, not the energy itself — the well's ground state carries σₓ σₚ = 0.568ħ rather than ħ/2, and that shortfall is where the π² went.
Answer⟨T̂⟩ ≥ 0.95 eV, exactly π² below the true 9.40 eV. Holding an electron inside an atom-sized box costs about an electronvolt of kinetic energy no cooling can remove — which is why chemistry happens on the eV scale.
MediumWithout solving the Schrödinger equation, find the lowest ⟨Ĥ⟩ allowed for Ĥ = p̂²/2m + ½mω²x̂², then evaluate it for carbon monoxide, whose stretching mode absorbs at ν̃ = 2143 cm⁻¹ with reduced mass μ = 6.857 u.
- The ground state of a symmetric potential has definite parity, so ⟨x̂⟩ = ⟨p̂⟩ = 0 and the second moments are the variances: ⟨Ĥ⟩ = σₚ²/2m + ½mω²σₓ².
- The relation allows σₚ no smaller than ħ/2σₓ, so the best a state of width σₓ can do is E(σₓ) = ħ²/8mσₓ² + ½mω²σₓ². The first term punishes localisation, the second punishes spreading.
- Minimise: dE/dσₓ = −ħ²/4mσₓ³ + mω²σₓ = 0, so σₓ⁴ = ħ²/4m²ω² and σₓ = √(ħ/2mω).
- Substitute back. Kinetic term: (ħ²/8m)(2mω/ħ) = ħω/4. Potential term: ½mω²(ħ/2mω) = ħω/4. So Eₘᵢₙ = ħω/2, exactly the true ground-state energy — and the two halves are equal, the virial split special to a quadratic potential.
- Numbers: ω = 2πcν̃ = 2π × 2.998 × 10¹⁰ cm s⁻¹ × 2143 cm⁻¹ = 4.037 × 10¹⁴ rad s⁻¹, so E₀ = ħω/2 = 2.13 × 10⁻²⁰ J = 0.133 eV.
- The width comes free: μ = 6.857 u = 1.139 × 10⁻²⁶ kg gives σₓ = √(ħ/2μω) = 3.39 × 10⁻¹² m, about 3% of the 113 pm bond — small enough that the quadratic expansion behind Ĥ is safe.
AnswerE₀ = ħω/2 = 0.133 eV, with σₓ = 3.4 pm. The bound is exact here rather than a floor, because the oscillator's ground state really is a Gaussian and Gaussians saturate σₓ σₚ ≥ ħ/2.
HardBefore 1932 the nucleus was supposed to contain electrons, which β⁻ decay then ejected with kinetic energies of order 1 MeV. Test that picture for a nucleus of radius 5.0 fm, then apply the same box to a proton. Take ħc = 197.3 MeV fm, mₑ c² = 0.511 MeV, mₚ c² = 938 MeV, and a nuclear well about 8 MeV per nucleon deep.
- Along one Cartesian axis the density is confined to a 10 fm span, so σₓ ≤ 5.0 fm and σₚ ≥ ħ/2σₓ ≥ ħ/(10 fm).
- Put it in energy units, so no mass is needed yet: σₚ c ≥ ħc/(10 fm) = 197.3 MeV fm ÷ 10 fm = 19.7 MeV.
- Check the regime before picking a kinetic-energy formula. 19.7 MeV ≫ mₑ c² = 0.511 MeV, so an electron here is ultra-relativistic and E ≈ pc: its energy scale is about 20 MeV, forty times its own rest energy.
- Now the test. A well 8 MeV deep cannot hold a 20 MeV particle, and the electrons actually seen leave with about 1 MeV. Both comparisons fail by more than a factor of ten, so no factor-of-two looseness in the estimate rescues the picture: the β electron is created at the decay, n → p + e⁻ + ν̄ₑ.
- The proton in the identical box gets the identical σₚ c ≥ 19.7 MeV, but now 19.7 MeV ≪ 938 MeV, so use the non-relativistic form: ⟨T̂⟩ ≥ (σₚ c)²/2mₚ c² = 19.7² ÷ (2 × 938) = 389/1876 = 0.21 MeV.
- 0.21 MeV against 8 MeV of binding is comfortable — and it is only a floor; a realistic three-dimensional well puts nucleons at tens of MeV, still bound. Same box and same ħ, so it is the mass in the denominator, not the size, that decides who fits.
AnswerAn electron would need about 20 MeV against the 1 MeV observed and the 8 MeV available, so nuclei hold no electrons; a proton in the same box needs at least 0.21 MeV and binds easily.