A physicist prepares many identical copies of state |ψ⟩ and measures position on half the copies and momentum on the other half. The results show Δx·Δp > ℏ/2. Their colleague claims the measurement process must have disturbed the state, causing this spread. Is the colleague correct?
AYes — Heisenberg's microscope argument shows that measuring position always disturbs momentum by at least ℏ/(2Δx)
BNot necessarily — the Robertson relation says Δx and Δp reflect the intrinsic spread of |ψ⟩ before any measurement; the bound holds even for ensembles where no single particle is measured twice
CYes — any quantum measurement introduces uncontrollable disturbance, which is the physical source of the uncertainty
DNot necessarily — the uncertainty principle only applies when position and momentum are measured on the same particle
The Robertson relation ΔA·ΔB ≥ ½|⟨[Â,B̂]⟩| is a theorem about the statistical spreads of outcomes over many measurements on identically prepared states — not about disturbance of any individual particle. Because position and momentum are measured on different copies of |ψ⟩, no single particle is disturbed by both measurements. The spread arises from the quantum state itself, not from the act of observation. Heisenberg's original 'microscope' argument (option A) is a heuristic that captures some intuition but does not capture the formal Robertson bound.
Question 2 Multiple Choice
For two observables  and B̂ with commutator [Â, B̂] = iC where C is a positive real constant, the Robertson uncertainty relation gives:
AΔA·ΔB ≥ C²
BΔA·ΔB ≥ C/2
CΔA·ΔB ≥ C
DΔA + ΔB ≥ C/2
The Robertson relation is ΔA·ΔB ≥ ½|⟨[Â,B̂]⟩|. With [Â,B̂] = iC (C real and positive), the expectation value ⟨[Â,B̂]⟩ = iC in every state, so |⟨[Â,B̂]⟩| = C. The bound becomes ΔA·ΔB ≥ C/2. For position and momentum, [X̂,P̂] = iℏ so C = ℏ and the bound is Δx·Δp ≥ ℏ/2. The factor of ½ comes from the proof via the Cauchy-Schwarz inequality; a common error is to forget it, giving ΔA·ΔB ≥ C (option C, off by a factor of 2).
Question 3 True / False
A Gaussian wavefunction saturates the Robertson uncertainty bound, achieving exactly Δx·Δp = ℏ/2.
TTrue
FFalse
Answer: True
The Gaussian wavepacket ψ(x) ∝ exp(−x²/4σ²) is the minimum-uncertainty state. Its Fourier transform is also Gaussian with width 1/(2σ), giving Δx = σ and Δp = ℏ/(2σ), so Δx·Δp = ℏ/2 exactly. Any other normalized wavefunction gives Δx·Δp > ℏ/2. This means Gaussians occupy the smallest possible 'volume' in phase space consistent with quantum mechanics, which has applications in coherent states of the harmonic oscillator and quantum optics.
Question 4 True / False
The Heisenberg uncertainty principle states that measuring the position of a particle precisely disturbs its momentum, and this disturbance is the fundamental source of the uncertainty relation.
TTrue
FFalse
Answer: False
This disturbance picture is Heisenberg's original 1927 heuristic argument (the gamma-ray microscope thought experiment) and is not the formal uncertainty principle. The Robertson relation is a theorem about quantum states — it says any state |ψ⟩ has standard deviations Δx and Δp satisfying Δx·Δp ≥ ℏ/2, regardless of whether any measurement has been made. The uncertainty is intrinsic to the state, not a result of measurement back-action. This is demonstrated by the ensemble argument in the previous question: even if position and momentum are measured on different particles from the same preparation, the bound holds.
Question 5 Short Answer
Why is the formal uncertainty principle (Robertson relation) considered a mathematical theorem rather than an independent physical postulate, and what does this imply about the origin of quantum uncertainty?
Think about your answer, then reveal below.
Model answer: The Robertson relation ΔA·ΔB ≥ ½|⟨[Â,B̂]⟩| follows as a consequence of the Cauchy-Schwarz inequality applied in Hilbert space — a purely mathematical result. It requires no additional physical assumptions beyond the existing structure of quantum mechanics (states as vectors in Hilbert space, observables as self-adjoint operators). This implies that quantum uncertainty is not a limitation of our instruments or a result of unavoidable measurement disturbance, but a geometric feature of Hilbert space: states that are narrow in one observable must be broad in any observable that does not commute with it.
The proof outline: for any two self-adjoint operators Â, B̂ and any state |ψ⟩, consider the vectors |u⟩ = (Â−⟨A⟩)|ψ⟩ and |v⟩ = (B̂−⟨B⟩)|ψ⟩. The Cauchy-Schwarz inequality gives ⟨u|u⟩⟨v|v⟩ ≥ |⟨u|v⟩|². Expanding both sides and using the identity |⟨u|v⟩|² ≥ |Im⟨u|v⟩|² = |⟨[Â,B̂]⟩|²/4 yields the Robertson bound. Every step is a mathematical identity — no physics is added.