The general uncertainty relation from the Schwarz inequality
One inequality, any commutator, a bound set by the state.
Every uncertainty relation in this course is one inequality with a different commutator fed in. Derive it once, in five lines, then do the part that matters in a laboratory: read it as a limit on two histograms taken from the same preparation, and recognise the states in which it has nothing to say.
Build the model
Two spreads from one preparation, bounded by a number the state itself sets.
1The big idea
Prepare many copies of one state ψ, send some to a detector for A and the rest to a detector for B, and measure each copy once. Two histograms come back, with spreads σA and σB, and the general uncertainty relation limits their product: σA σB ≥ ½|⟨[Â, B̂]⟩|. Nothing in the proof mentions a detector. The deviations f = (Â − ⟨A⟩)ψ and g = (B̂ − ⟨B⟩)ψ are functions whose squared lengths are the two variances, the Schwarz inequality caps their overlap ∫ f* g dx by the product of those lengths, and the imaginary part of that overlap is half the mean commutator divided by i. So the bound is fixed by the preparation, and it moves when the preparation moves. Only when the commutator is a constant, like the iħ of x̂ and p̂, is it the same in every state. Otherwise the right side is an expectation value, and wherever it averages to zero the relation reads σA σB ≥ 0 and says nothing, which is not the same as saying both can be sharp. The inequality is rarely tight, because two things were given up on the way: the real part of the overlap, which measures a correlation between A and B, and the slack in the Schwarz step itself. And it borrowed Hermiticity to move operators across integrals, so it fails where (Â − ⟨A⟩)ψ leaves the domain of B̂: on a ring, a momentum eigenstate has σₚ = 0 and a finite position spread.
2In one sentence
For any normalised state ψ, the spreads of two observables, each measured on separate copies of ψ, obey σA σB ≥ ½|⟨[Â, B̂]⟩|, where the right side is an expectation value computed in that same state.
Example. An electron in a 1.0 nm well prepared as (ψ₁ + iψ₂)/√2 has σE = 0.564 eV and σₓ = 0.227 nm, so σₓσE = 0.128 eV nm, above the bound (ħ/2m)|⟨p⟩| = 0.102 eV nm.
3Two histograms from one preparation
Read the relation as an instruction for an experiment before reading it as algebra. Prepare many copies of ψ. Send some to a detector for A and record one result from each; send the rest, never touched by the first detector, to a detector for B. σA and σB are the standard deviations of those two histograms, computed in UQMI-03.08 as σA² = ⟨A²⟩ − ⟨A⟩². So the relation compares the widths of two experiments on one preparation. It does not describe A then B on the same copy, where the state left by the first result (UQMI-04.04) replaces ψ, nor detector resolution, which UQMI-04.09 separates out. Take an electron in a 1.0 nm infinite well, with the eigenfunctions ψₙ = √(2/a) sin(nπx/a) that Unit 3 borrowed, prepared as (ψ₁ + iψ₂)/√2. Every energy result is 0.376 eV or 1.504 eV, each half the time, so σE = 0.564 eV however good the spectrometer, and the position results scatter with σₓ = 0.227 nm. The relation says those two widths cannot both be small, and how small their product may get.
4Five lines from the Schwarz inequality
Two results from Unit 3 are quoted, not re-proved: a Hermitian operator can act on either side of an overlap, ∫ f*(Âg) dx = ∫ (Âf)* g dx (UQMI-03.05), and σA² = ∫ ψ*(Â − ⟨A⟩)²ψ dx (UQMI-03.08). Put f = (Â − ⟨A⟩)ψ and g = (B̂ − ⟨B⟩)ψ. Moving one factor across gives σA² = ∫ |f|² dx and σB² = ∫ |g|² dx: each variance is the squared length of a function. The Schwarz inequality then gives σA²σB² ≥ |z|², with z = ∫ f* g dx. Any complex number obeys |z|² ≥ (Im z)², and 2i Im z = z − z* = ∫ f* g dx − ∫ g* f dx. Moving the operators across once more, z − z* = ∫ ψ*[(Â − ⟨A⟩)(B̂ − ⟨B⟩) − (B̂ − ⟨B⟩)(Â − ⟨A⟩)]ψ dx, where the numbers ⟨A⟩ and ⟨B⟩ cancel, leaving ⟨[Â, B̂]⟩. Hence σA²σB² ≥ |⟨[Â, B̂]⟩|²/4, and a square root finishes it. The commutator of two Hermitian operators is i times a Hermitian one, so its mean is purely imaginary and the right side is a real, non-negative number in units of A × B, as the left side is.
5The right side belongs to the state
For x̂ and p̂ the commutator is the constant iħ, and the bound is ħ/2 in every state; UQMI-04.08 takes that case. Almost every other pair gives an operator whose mean changes with ψ. Position and energy come first: [x̂, p̂²] = p̂[x̂, p̂] + [x̂, p̂]p̂ = 2iħp̂, and V(x) commutes with x̂, so [x̂, Ĥ] = iħp̂/m and σₓσE ≥ (ħ/2m)|⟨p⟩|. With Ehrenfest's ⟨p⟩ = m d⟨x⟩/dt this reads σₓσE ≥ (ħ/2)|d⟨x⟩/dt|: a position histogram whose centre drifts needs a state with a spread of energies. In the well state √(1 − w)ψ₁ + √w e(iφ)ψ₂, with φ falling at ω₂₁ = 3E₁/ħ, the bound is (8/3)√(w(1 − w))|sin φ| in units of ħ²/(ma); at w = ½ it rises to 1.33 and falls to zero twice in each 3.7 fs period for an electron in a 1.0 nm well. Two limits check it. A stationary state, w = 0 or 1, has σE = 0 and ⟨p⟩ = 0 together. And under a uniform force, V = −Fx, the pair p̂ and V̂ gives [V̂, p̂] = −iħF, a constant; with σV = |F|σₓ the bound collapses onto σₓσₚ ≥ ħ/2.
6When the bound goes empty
Wherever ⟨[Â, B̂]⟩ = 0 the relation reads σA σB ≥ 0, which any two spreads satisfy. It has stopped giving information; it has not declared both quantities sharp. Put φ = 0 in the well state at w = ½: ψ is real, ⟨p⟩ = 0 and the bound is exactly zero, yet σE is still 1.5E₁ and σₓ = 0.139a, a product of 1.03 ħ²/(ma). Every eigenstate of  empties the bound on its own: if Âψ = aψ, then ⟨ÂB̂⟩ = a⟨B⟩ by moving  onto ψ*, which needs B̂ψ in the domain of Â, and ⟨B̂Â⟩ = a⟨B⟩ directly, so ⟨[Â, B̂]⟩ = 0. That is how the relation lets the well's ψ₂, with σE = 0, keep a finite σₓ = 0.266a. The relation forces σB towards infinity as σA shrinks whenever ⟨[Â, B̂]⟩ stays away from zero along the way. For x̂ and p̂ that holds in every state, since their constant commutator rules out every normalisable eigenfunction; for most other pairs it depends on the preparation, as for a packet whose mean position keeps moving. Whether A and B can be sharp together across a whole basis of states is a question about the operators, answered by compatibility in UQMI-04.06; an empty bound in one state settles nothing about it.
7What the proof threw away
Two inequalities were used, and each leaves slack. The step |z|² ≥ (Im z)² discarded Re z = ½⟨ÂB̂ + B̂Â⟩ − ⟨A⟩⟨B⟩, the covariance of A and B in ψ; keeping it gives Schrödinger's stronger form σA²σB² ≥ (Re z)² + ¼|⟨[Â, B̂]⟩|². For x and E in the well state, Re z = −(8/3)√(w(1 − w))(1 − 2w)cos φ in units of ħ²/(ma), zero at w = ½ but not elsewhere: at w = 0.25 and φ = 0 the Robertson bound is empty, yet the stronger form still demands σₓσE ≥ 0.577, against an actual 0.856. Leaving aside eigenstates of Â, where f = 0 and both sides vanish, the Schwarz step is tight only when g = λf, one deviation a constant multiple of the other, and reaching Robertson's equality needs λ purely imaginary as well, so that Re z = 0. At w = ½ and φ = 90° the covariance is zero, so the whole gap between σₓσE = 1.68 and the bound 1.33, a factor of 1.26, is Schwarz slack: (Ĥ − ⟨E⟩)ψ is not a multiple of (x̂ − ⟨x⟩)ψ. For x̂ and p̂ the condition g = λf becomes a first-order differential equation, which UQMI-04.08 solves.
8The domain the proof borrowed
Every move of an operator across an integral used Hermiticity on a particular pair of functions, and UQMI-03.05 showed that the move is legal only when both lie in the operator's domain. Here that means (Â − ⟨A⟩)ψ must be an allowed function for B̂, and the other way round. A ring of circumference L = 1.0 nm shows the failure. Measure arc position s from a cut at s = 0 and take the momentum eigenstate ψ = e(2πis/L)/√L: σₚ = 0 exactly, while |ψ|² is uniform and σₛ = L/√12 = 0.289 nm. The formula [ŝ, p̂] = iħ holds on every smooth function, yet the naive bound σₛ σₚ ≥ ħ/2 would demand σₚ ≥ 1.8 × 10⁻²⁵ kg m s⁻¹. The flaw is that sψ jumps by Lψ(0) across the cut, so it is not periodic and p̂ cannot be moved onto it; the boundary term −iħ[ψ* sψ] taken round the ring equals −iħL|ψ(0)|². Keep it, and the relation that does hold is σₛ σₚ ≥ (ħ/2)|1 − L|ψ(0)|²|, which is zero here because L|ψ(0)|² = 1. A bound is only as good as the domain its proof ran in.
Key equations
What each relation says, and the units it is written in.
The general uncertainty relation
σA σB ≥ ½ |⟨[Â, B̂]⟩|
Bounds the product of two spreads, each from its own run on the same preparation. The right side is computed from ψ, never from the apparatus.
σA and σB carry the units of A and B; ⟨[Â, B̂]⟩ is purely imaginary, in units of A × B. Any normalised ψ.
A variance as a squared length
σA² = ∫ |(Â − ⟨A⟩)ψ|² dx
Turns the width of a histogram into the length of a function, which is what lets a geometric inequality act on it. σA = 0 only when Âψ = ⟨A⟩ψ.
Hermitian  (UQMI-03.05) moves one factor of  − ⟨A⟩ onto ψ*; σA² carries the units of A².
Schwarz inequality
|∫ f* g dx|² ≤ ∫ |f|² dx × ∫ |g|² dx
An overlap can never beat the product of the two lengths. It follows from ∫ |f − λg|² dx ≥ 0 with λ chosen to make that integral smallest.
Any square-integrable f and g. Equality holds only when one function is a constant multiple of the other: g = λf for a single constant λ, or f = 0.
The imaginary part of the overlap
∫ f* g dx − ∫ g* f dx = ⟨[Â, B̂]⟩
The left side is 2i Im z for the overlap z = ∫ f* g dx, so |z|² ≥ (Im z)² turns the Schwarz inequality into the uncertainty relation.
f = (Â − ⟨A⟩)ψ and g = (B̂ − ⟨B⟩)ψ; the numbers ⟨A⟩ and ⟨B⟩ cancel inside the commutator.
Position against energy
σₓ σE ≥ (ħ/2m) |⟨p⟩|
Since ⟨p⟩ = m d⟨x⟩/dt, a mean position that moves demands an energy spread. In a stationary state both sides are zero.
From [x̂, Ĥ] = iħp̂/m. σₓ in m, σE in J, ⟨p⟩ in kg m s⁻¹, so both sides are in J m.
Keeping the real part
σA² σB² ≥ (½⟨ÂB̂ + B̂Â⟩ − ⟨A⟩⟨B⟩)² + ¼ |⟨[Â, B̂]⟩|²
Robertson's form above drops it. When A and B are correlated in the state, this stronger bound can stay nonzero where ⟨[Â, B̂]⟩ = 0.
The first bracket is Re z, the covariance of A and B in ψ, in units of A × B.
Interactive lab
Change one variable. Predict first, then read the result.
Controls
Try this. Hold w = 0.5 and sweep φ, which is time running: the bound rises and falls with |sin φ| and empties at 0° and 180°, where ⟨x⟩ turns round, yet σₓσE never drops below 1.0. Then set w = 0.25 and φ = 0°: the dashed line stays up at 0.58.
Live results
- SPREAD σₓ
- 0.227 a
- SPREAD σE
- 7.40 ħ²/(ma²)
- PRODUCT σₓσE
- 1.683 ħ²/(ma)
- BOUND (ħ/2m)|⟨p⟩|
- 1.333 ħ²/(ma)
Catch the common trap
The mistake this lesson is built to prevent.
The usual error is to carry the x–p reading to every pair:  and B̂ do not commute, so they can never both be sharp, and if A is sharp then σB must be infinite. That treats ½|⟨[Â, B̂]⟩| as a number belonging to the two operators, like ħ/2. It is an expectation value in the prepared state, and any eigenstate of  makes it zero. In the well's energy eigenstate ψ₂, σE = 0 while σₓ = 0.266a, perfectly finite, because ⟨[x̂, Ĥ]⟩ = (iħ/m)⟨p⟩ and ⟨p⟩ = 0 there. A spread forced to run away in every state belongs to pairs such as x̂ and p̂, whose commutator is a nonzero constant, so that no normalisable state can make its mean vanish. For any other pair, recompute the right side in the state you actually prepared.
Test it: what changes?
- Bound follows 1.33|sin φ| in ħ²/(ma)
- Bound empties at φ = 0° and 180°
- σE stays at 1.5E₁ throughout
- σₓσE never drops below 1.0 ħ²/(ma)
- Robertson bound (ħ/2m)|⟨p⟩| stays at zero
- Covariance-keeping bound rises from 0 to 0.58 ħ²/(ma)
- σE falls below 1.5E₁
- σₓσE falls from 1.03 to 0.856 ħ²/(ma)
Practice & worked examples
Three tiers. Try each before opening the solution.
EasyExample 1
An electron wave packet in a long wire has σₓ = 2.0 nm, and its mean position moves at d⟨x⟩/dt = 1.5 × 10⁵ m s⁻¹. What is the smallest standard deviation that energy measurements on copies of this state can show?
View solution →
- Choose the pair and find its commutator: [x̂, p̂²] = p̂[x̂, p̂] + [x̂, p̂]p̂ = 2iħp̂, and V(x) commutes with x̂, so [x̂, Ĥ] = iħp̂/m and ½|⟨[x̂, Ĥ]⟩| = (ħ/2m)|⟨p⟩|.
- Ehrenfest's theorem gives ⟨p⟩ = m d⟨x⟩/dt, so the mass cancels and σₓσE ≥ (ħ/2)|d⟨x⟩/dt|.
- σE ≥ ħ|d⟨x⟩/dt|/(2σₓ) = (1.055 × 10⁻³⁴ J s × 1.5 × 10⁵ m s⁻¹)/(2 × 2.0 × 10⁻⁹ m) = 3.96 × 10⁻²¹ J = 0.0247 eV.
- Checks: J s × m s⁻¹ ÷ m = J, an energy. If the packet stops drifting the bound empties, as it must for a stationary state. As ħ → 0 it vanishes, and a classical particle can move with a sharp position and a sharp energy.
Answer. σE ≥ 4.0 × 10⁻²¹ J, about 0.025 eV. That is a floor: the relation does not say the actual spread equals it.
MediumExample 2
An electron in an infinite well of width a = 1.0 nm is prepared in ψ = (√3 ψ₁ + iψ₂)/2. Given ∫ ψ₁(p̂ψ₂) dx = 8iħ/(3a) and ⟨x²⟩ = a²(1/3 − 1/(2n²π²)) in ψₙ, predict σE for an energy run and σₓ for a position run on this preparation, and test them against σₓσE ≥ (ħ/2m)|⟨p⟩|.
View solution →
- Weights: |c₁|² = 3/4 and |c₂|² = 1/4, which sum to 1. With E₁ = π²ħ²/(2ma²) = 0.376 eV and E₂ = 4E₁ = 1.504 eV, ⟨E⟩ = 0.658 eV and σE = (E₂ − E₁)√(¾ × ¼) = 1.128 eV × 0.433 = 0.488 eV.
- Mean momentum: the diagonal terms vanish because each ψₙ is real, so ⟨p⟩ = c₁* c₂ (8iħ/(3a)) + c₂* c₁ (−8iħ/(3a)). With c₁ = √3/2 and c₂ = i/2 this is −(4√3/3)ħ/a = −2.31ħ/a = −2.44 × 10⁻²⁵ kg m s⁻¹.
- Bound: (ħ/2m)|⟨p⟩| = (2√3/3)ħ²/(ma) = 1.155 × 0.0762 eV nm = 0.0880 eV nm, using ħ²/(ma²) = 0.0762 eV for a = 1.0 nm.
- Position: every cross term is proportional to Re(c₁* c₂) = 0, so ⟨x⟩ = a/2 and ⟨x²⟩ = ¾a²(1/3 − 1/(2π²)) + ¼a²(1/3 − 1/(8π²)) = 0.2922a². Then σₓ = √(0.2922 − 0.25) a = 0.205 nm.
- Product: σₓσE = 0.205 nm × 0.488 eV = 0.100 eV nm, 1.14 times the bound. The covariance is zero at this phase, so the 14% excess is Schwarz slack.
- Checks: both sides in eV nm. As |c₂|² → 0, σE and the bound both shrink as √(|c₂|²(1 − |c₂|²)), reaching zero together in the stationary state ψ₁.
Answer. σE = 0.488 eV and σₓ = 0.205 nm, so σₓσE = 0.100 eV nm, 1.14 times the bound (ħ/2m)|⟨p⟩| = 0.0880 eV nm.
HardExample 3
A particle in the potential V = ½κx², with κ = 2.0 × 10⁻⁵ N m⁻¹, is prepared in the real Gaussian ψ = (2πs²)(−1/4) e(−(x − x₀)²/(4s²)), with s = 4.0 nm and x₀ = 12 nm. Derive the bound on σₚσV from the commutator, compute both sides, and say what happens when the packet is centred and when it is displaced far from the centre.
View solution →
- Commutator on a test function f: [V̂, p̂]f = V(−iħf′) + iħ(Vf)′ = iħV′f, so [V̂, p̂] = iħκx̂ and σVσₚ ≥ (ħκ/2)|⟨x⟩| = (ħκ/2)|x₀|.
- Momentum: ψ is real, so ⟨p⟩ = 0, and ⟨p²⟩ = ħ²∫ (ψ′)² dx = ħ²⟨(x − x₀)²⟩/(4s⁴) = ħ²/(4s²). So σₚ = ħ/(2s) = 1.32 × 10⁻²⁶ kg m s⁻¹.
- Potential energy: |ψ|² is a normal distribution of mean x₀ and variance s², so ⟨x²⟩ = x₀² + s² and ⟨x⁴⟩ = x₀⁴ + 6x₀²s² + 3s⁴. Then σV = ½κ√(⟨x⁴⟩ − ⟨x²⟩²) = κs√(x₀² + s²/2) = 9.86 × 10⁻²² J, or 6.16 meV.
- Compare: σₚσV = (ħκ/2)√(x₀² + s²/2) = 1.30 × 10⁻⁴⁷ J kg m s⁻¹ against (ħκ/2)x₀ = 1.27 × 10⁻⁴⁷ J kg m s⁻¹, a ratio √(1 + s²/(2x₀²)) = 1.027. Both sides are in J kg m s⁻¹. With ψ real, f = (V − ⟨V⟩)ψ is real and g = −iħψ′ imaginary, so Re z = 0 and the 2.7% is Schwarz slack, not a dropped covariance.
- Centre the packet, x₀ → 0: ⟨x⟩ = 0 empties the bound, while σₚσV → (ħκ/2)(s/√2) = 2.98 × 10⁻⁴⁸ J kg m s⁻¹ stays finite.
- Displace it far, x₀ ≫ s: the ratio tends to 1. Across a narrow packet the parabola is a straight line of slope κx₀, the force is uniform, σV → κx₀σₓ, and the pair behaves like x̂ and p̂, the classical-correspondence limit.
Answer. σₚσV = 1.30 × 10⁻⁴⁷ J kg m s⁻¹ against a bound of 1.27 × 10⁻⁴⁷ J kg m s⁻¹, a ratio of 1.027. The bound empties for a centred packet and tightens as x₀/s grows.
Quick check
Choose an answer, then check it. Explain to yourself why the others fail.
Key takeaway
σA σB ≥ ½|⟨[Â, B̂]⟩|, evaluated in the prepared state.
Compute the commutator, take its mean in the prepared state, and halve its modulus. The bound limits the product of spreads from separate runs, not one measurement disturbing another. Where ⟨[Â, B̂]⟩ = 0 it goes empty without making both sharp, and every operator move in the proof must stay inside the domain.