Δx Δp ≥ ħ/2 and the cost of confinement

Robertson's theorem gives σ_A σ_B ≥ 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, L̂_y] = iħL̂_z: in the state |l, m = 0⟩ the bound reads σ_Lx σ_Ly ≥ 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.

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.

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.

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.

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. Nor could anything hold it: the strong force ignores electrons, leaving only the protons' Coulomb pull, of order 10 MeV at that radius. That 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, tiny beside the strong-force well, some 50 MeV deep, that holds nucleons. Same box, same ħ; the mass in the denominator decides.

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, σ_H ⋅ σ_A/|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.