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The mean position of a normalized state need not exist

Physical scope

A refutation, on an explicit normalized state in the one-dimensional specialization of the P1 state assignment and P2 position-probability rule, of the claim that every normalized preparation has a finite position mean. The displayed state has unit norm but neither a finite second moment nor a finite first absolute moment. The item defines the mean as the Lebesgue expectation and states its absolute-integrability condition; it asserts no measured value and no empirical claim.

Statement refuted

Specialize P1 to a one-dimensional system with state space H=L2(R;C) and specialize P2 to the position observable Qψ=xψ, with outcome coordinate measured in metres. Its spectral projection for a Borel set B⊆R is multiplication by 1B, so a pure preparation [ψ] has position law μψQ(B)=∫B∣ψ(x)∣2 dx. Define its mean position to be the Lebesgue expectation EμψQ[X]=∫Rx dμψQ(x) when it is a finite real number; this is so exactly when x∈L1(μψQ), equivalently ∫R∣x∣ ∣ψ(x)∣2 dx<∞. The refuted claim is that every normalized ψ∈L2(R;C) has such a finite mean.

Facts & Assumptions

Given: Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice (ACω), The Axiom of Choice). The one-dimensional instance of P1 with H=L2(R;C) and a normalized vector ψ; the P2 position observable Q as multiplication by x, with the spectral PVM and law of [L6]; and the quantity conventions of Dimensions, SI unit conventions and quantity types of the framework.

[L1]

Let L0>0 be a length and define ψL0(x)=12L0 for ∣x∣≤L0 and ψL0(x)=L02∣x∣ for ∣x∣≥L0. The formulas agree at ∣x∣=L0, and [ψL0]=ℓ−1/2. It is measurable and represents an almost-everywhere class; its L2 membership is established in step 1.1 using [L5] and monotone convergence. Its squared tail is L0/(4x2), whose improper Riemann integral converges by the rational p-test (Improper integrals over unbounded intervals, The improper p-test for rational exponents, The space Lp(μ) as the quotient by null functions).

[L3]

Monotone convergence for the truncated absolute moments: ∫−NN∣ψ∣2→∫R∣ψ∣2 and ∫−NN∣x∣∣ψ∣2→∫R∣x∣∣ψ∣2∈[0,+∞] along the increasing sequence of sets [−N,N], and similarly for the second moments (Monotone convergence for the integral).

[L4]

Second moments and Cauchy-Schwarz: ψ∈D(Q) means ∫Rx2∣ψ(x)∣2 dx<∞, and for a probability density with finite second moment the first absolute moment is finite by the Cauchy-Schwarz inequality applied to the probability measure ∣ψ(x)∣2dx (Position and momentum operators on Euclidean L2, Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors).

[L5]

For each N>L0, the three global integrands ∣ψL0∣2, ∣x∣∣ψL0∣2 and x2∣ψL0∣2 are real and continuous on [−N,N], including at x=±L0, because their defining formulas agree there. The tail functions x−2, x−1 and 1 of [L2] are real and continuous on [L0,N], which excludes zero. Each of these restrictions is therefore bounded and Riemann integrable by A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion. Under the stated Countable Choice hypothesis, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral identifies its finite-interval Riemann and Lebesgue integrals. Thus [L2] evaluates the truncated Lebesgue integrals used by [L3]; the tail notation ∫L0∞ below denotes the monotone limit of these nonnegative Lebesgue integrals, including +∞.

[L6]

Choice is carried from P2 and is an explicit hypothesis of Spectral measure of a real L2 multiplication operator clauses (a)-(c). On the positive, σ-finite Lebesgue measure space R, the measurable finite multiplier m(x)=x satisfies that lemma's hypotheses. Under the stated Choice assumption (The Axiom of Choice), it identifies the self-adjoint position operator Q on D(Q)={f:xf∈L2} and its spectral PVM EQ(B)=M1B, with pure-state law μψQ(B)=∥EQ(B)ψ∥2=∫B∣ψ∣2dx for every unit ψ. The explicit moment calculations require no additional arbitrary choice.

Counterexample

technique · direct; compute the norm, the second moment and the first absolute moment of the displayed state
1.1givenL1L2L3L5algebra

[Norm one] The squared modulus is 1/(4L0) on [−L0,L0] and L0/(4x2) for ∣x∣≥L0. Its integral is 2L0/(4L0)+2∫L0∞L0/(4x2) dx=12+L02lim⁡N→∞(L0−1−N−1)=1 by [L2] and [L3], so ψL0 is a normalized state.

1.2givenL1L2L3L5algebra

[Second moment infinite] The second moment is ∫Rx2∣ψL0∣2 dx=L026+2∫L0∞x2⋅L04x2 dx=L026+L02lim⁡N→∞(N−L0)=+∞ by [L2] and [L3]; equivalently xψL0∉L2 and ψL0∉D(Q), although it is a normalized state.

1.3givenL1L2L3L5algebra

[First absolute moment infinite] The first absolute moment is ∫R∣x∣ ∣ψL0∣2 dx=L04+2∫L0∞x⋅L04x2 dx=L04+L02lim⁡N→∞ln⁡(N/L0)=+∞ by [L2] and [L3]. Both one-sided integrals diverge, so x is not integrable for the probability measure μψL0Q and its Lebesgue expectation is not a finite real mean.

2.1step 1.1step 1.2step 1.3L4L6∎

By steps 1.1-1.3 the P1 preparation [ψL0] is normalized and its P2 position law has infinite first absolute moment, so the refuted universal claim fails. By the definition in the Statement its mean exists exactly when ∫R∣x∣ dμψL0Q(x)=∫R∣x∣ ∣ψL0(x)∣2 dx<∞; and by [L4], the second-moment condition ψL0∈D(Q) implies this integrability by Cauchy-Schwarz.

Remarks

  • The example concerns the existence of a moment, not the definition of the preparation: the displayed ψ is a legitimate normalized state, and statements about its position distribution are computed from ∣ψ∣2dx. What fails is only the claim that the first moment is automatically finite.
  • Moments are not part of the probability postulate of the page; each mean, variance or higher moment of an outcome law is asserted only where its integrability has been established, exactly as the quantity conventions of the framework require.

Depends on

Dependency tree · 2 levels

Sources