A student argues: 'The uncertainty principle says that measuring a particle's position very precisely disturbs its momentum — the measurement kicks the particle and randomizes its momentum.' This account of the Kennard inequality Δx Δp ≥ ℏ/2 is:
ACorrect — measurement disturbance is the source of the position-momentum uncertainty
BPartially correct — disturbance explains most cases, but there are some exceptions
CMisleading — the inequality holds for the quantum state itself before any measurement occurs, as a consequence of the Fourier transform relationship between position and momentum representations
DCorrect for electrons but not for photons
The Kennard inequality Δx Δp ≥ ℏ/2 is a theorem about quantum states, not about measurement procedures. A Gaussian wave packet sitting undisturbed in free space already satisfies it — no measurement has occurred. The uncertainty is a property of the wave function: a narrow (localized) wave packet requires a broad superposition of momentum eigenstates (a wide spread in p-space), by the mathematics of Fourier transforms. Measurement disturbance is a real and separate phenomenon, but it is not the source of the canonical uncertainty relations.
Question 2 Multiple Choice
A Gaussian wave packet is prepared with minimum uncertainty Δx · Δp = ℏ/2. Which statement correctly describes this state?
ABoth Δx and Δp are zero — the state is as classical as possible
BThis is the minimum uncertainty state; narrowing Δx further would require increasing Δp to compensate
CThe uncertainty principle is violated at the minimum — this state is quantum mechanically impossible
DThe uncertainties are only defined after a measurement is performed on the state
Δx · Δp = ℏ/2 is the *minimum* allowed by the Robertson relation — achieved by Gaussian (coherent) states. It is a perfectly valid quantum state, the most 'classical-like' in the sense of simultaneously minimizing both uncertainties. Narrowing the spatial width (decreasing Δx) necessarily broadens the momentum distribution (increasing Δp) to keep the product ≥ ℏ/2. The uncertainties are properties of the state, not of measurements — they are defined as standard deviations over an ensemble of identical preparations.
Question 3 True / False
The Robertson uncertainty relation ΔA ΔB ≥ ½|⟨[Â, B̂]⟩| is a mathematical theorem proven from the Cauchy-Schwarz inequality applied to Hilbert space vectors, not an empirical generalization.
TTrue
FFalse
Answer: True
The derivation is purely mathematical: define two vectors |u⟩ = (Â − ⟨Â⟩)|ψ⟩ and |v⟩ = (B̂ − ⟨B̂⟩)|ψ⟩, apply the Cauchy-Schwarz inequality ‖u‖·‖v‖ ≥ |⟨u|v⟩|, and separate the result into real and imaginary parts. The imaginary part gives ½|⟨[Â, B̂]⟩|. No experiment is needed; the bound follows from the algebra of Hilbert space and the definition of standard deviation. It holds exactly, not approximately.
Question 4 True / False
Two observables that commute ([Â, B̂] = 0) cannot both be measured precisely in the same quantum state — there will typically be some uncertainty in at least one of them.
TTrue
FFalse
Answer: False
If [Â, B̂] = 0, the Robertson bound is ΔA · ΔB ≥ 0, which imposes no constraint. Commuting operators share a complete set of simultaneous eigenstates. If the system is in a simultaneous eigenstate of both, then ΔA = ΔB = 0 — both can be measured with perfect precision. Non-zero uncertainty for commuting observables is a property of specific states, not a universal constraint. The canonical uncertainty between position and momentum is irreducible precisely because [x̂, p̂] = iℏ ≠ 0.
Question 5 Short Answer
Why is a spatially narrow (highly localized) wave packet necessarily associated with a broad spread of momenta?
Think about your answer, then reveal below.
Model answer: Because position-space and momentum-space wave functions are related by the Fourier transform. A spatially narrow wave packet ψ(x) is a sharp spike in x-space, and its Fourier transform φ(p) — the momentum-space wave function — must be broad to reproduce that spike. This is a mathematical fact about Fourier transforms: narrow functions and their transforms cannot both be narrow simultaneously. The standard deviation of |ψ(x)|² (which is Δx) and the standard deviation of |φ(p)|² (which is Δp) satisfy Δx · Δp ≥ ℏ/2 as a consequence of this Fourier relationship.
This is the deepest way to see why the uncertainty principle holds: it is the quantum expression of a universal mathematical constraint on conjugate Fourier pairs. The same relationship appears in signal processing — a time-limited signal cannot also be frequency-limited. In quantum mechanics, position and momentum are conjugate Fourier variables, so the constraint is not a mysterious quantum weirdness but a consequence of representing the state as a wave.