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A preparation with finite mean position and infinite variance

Physical scope

A finite model computation in the spinless massive-particle model exhibiting a legitimate preparation whose position law has a finite mean but infinite second moment: the state is a product of a symmetric profile with a power tail in the first coordinate and an arbitrary normalized transverse factor, its first absolute moment is finite while its second moment diverges, so the mean position exists and the variance of the position law does not. The item asserts no apparatus behaviour, no measured value and no empirical claim; it is a demonstration of the domain hypotheses of the moment statements.

Example

Work in the spinless massive-particle model of P5 with $H = L^{2} (R^{3}; C)$ and position operator $Q_{1}$ (Kinematics and massive-particle model postulate P5: Newtonian frames and one-particle operators, Position and momentum operators on Euclidean L2). Use the numerical SI spatial coordinates of Position and momentum outcome laws and the Born density of the massive particle model: $x_{1}, x_{2}, x_{3}, t$ are read in metres and the assigned observable is $(Q_{1}, m)$ . Thus the cutoff $∣ t ∣ \leq 1$ below means a distance of at most one metre; powers of $t$ in the profile apply to its numerical value in that fixed unit. The unnormalized $g_{0}$ is dimensionless, $Z$ has length units, and $g$ , $φ$ , $ψ$ carry units $m^{- 1 / 2}$ , $m^{- 1}$ , $m^{- 3 / 2}$ respectively, so the mean is read in metres and the second moment and variance in $m^{2}$ . Define the symmetric profile $g_{0} (t) : = {\begin{cases} 1, & ∣ t ∣ \leq 1, \\ ∣ t ∣^{- 3 / 2}, & ∣ t ∣ > 1, \end{cases} Z : = \int_{R} ∣ g_{0} (t) ∣^{2} d t, g : = Z^{- 1 / 2} g_{0},$ so that $g \in L^{2} (R; C)$ has $∥ g ∥_{2} = 1$ ; fix a unit vector $φ \in L^{2} (R^{2}; C)$ , choose a finite Borel representative as in the cited outcome-law theorem, and put $ψ (x) : = g (x_{1}) φ (x_{2}, x_{3})$ . Then $ψ \in H$ is a unit vector, $[ψ]$ is a preparation, and for its position law $μ_{ψ}^{Q_{1}}$ with density $∣ g ∣^{2}$ :

(i) The mean exists and equals $0$ : $\int_{R} ∣ t ∣ ∣ g (t) ∣^{2} d t < \infty, \int_{R} t ∣ g (t) ∣^{2} d t = 0,$ and $ψ \in D (∣ Q_{1} ∣^{1 / 2})$ while $ψ \notin D (Q_{1})$ .

(ii) The second moment diverges, so the variance of the position law does not exist: $\int_{R} t^{2} ∣ g (t) ∣^{2} d t = + \infty, \int_{R} (t - 0)^{2} d μ_{ψ}^{Q_{1}} (t) = + \infty .$

Thus a preparation of the model can have a finite mean position without a finite second moment: the moment identities of the framework are conditional on the domains, and a finite mean does not require membership in $D (Q_{1})$ .

The example is a model computation; it asserts no measured value and no apparatus behaviour.

Facts & Assumptions

Given: The spinless massive model with $H = L^{2} (R^{3}; C)$ (Kinematics and massive-particle model postulate P5: Newtonian frames and one-particle operators, Position and momentum operators on Euclidean L2); the profile $g_{0}$ , the constant $Z$ and the normalized profile $g = Z^{- 1 / 2} g_{0}$ ; a unit vector $φ \in L^{2} (R^{2}; C)$ ; the product state $ψ = g \otimes φ$ ; the outcome-law reading of Assigned sharp observables and the outcome law of the framework; the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice).

[F1]

Power series and harmonic divergence. $\sum_{k \geq 1} k^{- p}$ converges exactly for $p > 1$ (The p-series for a real exponent p converges exactly when p is greater than one), and a nonnegative series converges if and only if its partial sums are bounded (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum), so for $p = 1$ the partial sums of $\sum_{k \geq 1} k^{- 1}$ are unbounded and the series diverges to $+ \infty$ .

[F2]

Block comparison for nonnegative integrals. On $[k, k + 1]$ with $k \geq 1$ one has $t^{- 2} \leq k^{- 2}$ and $t^{- 1} \geq (k + 1)^{- 1}$ ; monotonicity and additivity of the nonnegative Lebesgue integral (The nonnegative Lebesgue integral, Additivity of the nonnegative Lebesgue integral) therefore give $\int_{k}^{k + 1} t^{- 2} d t \leq k^{- 2}$ and $\int_{k}^{k + 1} t^{- 1} d t \geq (k + 1)^{- 1}$ for every $k$ , and monotone convergence (Monotone convergence for the integral) passes the resulting bounds to the improper integrals: $\int_{1}^{\infty} t^{- 2} d t \leq \sum_{k \geq 1} k^{- 2} < \infty$ and $\int_{1}^{\infty} t^{- 1} d t \geq \sum_{k \geq 2} k^{- 1} = \infty$ .

[F3]

Product states and their position law. For the unit product state $ψ = g \otimes φ$ , the law of $Q_{1}$ has density $f_{1} (t) = \int_{R^{2}} ∣ ψ (x_{⊥}, t) ∣^{2} d x_{⊥} = ∣ g (t) ∣^{2} \int ∣ φ ∣^{2} = ∣ g (t) ∣^{2}$ , and $μ_{ψ}^{Q_{1}} ({λ}) = 0$ for every $λ$ (Position and momentum outcome laws and the Born density of the massive particle model clauses (a) and (d), Tonelli's theorem for nonnegative measurable functions on a sigma-finite product,).

[F4]

Moment criterion. The mean of the position law is finite exactly when the state lies in $D (∣ Q_{1} ∣^{1 / 2})$ , the second moment is finite exactly when it lies in $D (Q_{1}) = {h : x_{1} h \in L^{2}}$ , and for a unit vector in $D (Q_{1})$ the mean is $⟨ Q_{1} ψ, ψ ⟩$ and the variance is the second moment minus the square of the mean (Mean, second moment and variance of an assigned observable in a preparation clauses (a)-(c), Mean, second moment and variance of the scalar spectral law clauses (a)-(c)).

[F5]

Evenness. The substitution $t \mapsto - t$ is a $C^{1}$ change of variables with unit Jacobian, so for the even profile $g$ the function $t \mapsto t ∣ g (t) ∣^{2}$ is odd and its integral over $R$ vanishes whenever it is absolutely integrable (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Verification

technique · direct; normalize the profile and evaluate the first and second absolute moments by comparison with the $p$-series

1.1givenF1F2algebra

[The profile is a normalizable $L^{2}$ function] The profile $g_{0}$ is measurable (it is continuous on $∣ t ∣ > 1$ and on $∣ t ∣ < 1$ and bounded near the junction) and nonnegative, and $Z = \int_{- 1}^{1} 1 d t + 2 \int_{1}^{\infty} t^{- 3} d t = 2 + 2 \int_{1}^{\infty} t^{- 3} d t$ , where by the block comparison of [F2] with $h (t) = t^{- 3}$ the tail satisfies $0 < \int_{1}^{\infty} t^{- 3} d t \leq \sum_{k \geq 1} k^{- 3} < \infty$ ; hence $0 < Z < \infty$ and $g = Z^{- 1 / 2} g_{0}$ is a well-defined element of $L^{2} (R; C)$ with $∥ g ∥_{2}^{2} = Z^{- 1} Z = 1$ .

2.1givenstep 1.1F1F2algebra

[The first moment is finite and the second is infinite] Using evenness and the explicit form of $g$ , $\int_{R} ∣ t ∣ ∣ g (t) ∣^{2} d t = Z^{- 1} (\int_{- 1}^{1} ∣ t ∣ d t + 2 \int_{1}^{\infty} t \cdot t^{- 3} d t) = Z^{- 1} (1 + 2 \int_{1}^{\infty} t^{- 2} d t)$ , which is finite and at least $Z^{- 1} > 0$ by the bound $\int_{1}^{\infty} t^{- 2} d t \leq \sum_{k \geq 1} k^{- 2} < \infty$ of [F2]; and $\int_{R} t^{2} ∣ g (t) ∣^{2} d t = Z^{- 1} (\int_{- 1}^{1} t^{2} d t + 2 \int_{1}^{\infty} t^{2} \cdot t^{- 3} d t) = Z^{- 1} (\frac{2}{3} + 2 \int_{1}^{\infty} t^{- 1} d t) = + \infty$ because $\int_{1}^{\infty} t^{- 1} d t \geq \sum_{k \geq 2} k^{- 1} = \infty$ by [F2] and [F1].

3.1givenstep 1.1F3algebra

[The state and its position law] The product $ψ = g \otimes φ$ is an element of $H$ with $∥ ψ ∥^{2} = ∥ g ∥_{2}^{2} ∥ φ ∥^{2} = 1$ by Tonelli and the normalizations of step 1.1; hence $[ψ]$ is a pure preparation. By [F3] the law of $Q_{1}$ in this preparation has density $∣ g ∣^{2}$ on $R$ (independent of $φ$ ), so its first and second absolute moments are exactly the integrals computed in step 2.1, and every exact value has probability zero.

4.1step 2.1step 3.1F4F5algebra∎

[Mean finite, variance infinite] By step 3.1 the position law of the preparation has density $∣ g ∣^{2}$ , so [F5] applies: the mean $\int t d μ_{ψ}^{Q_{1}} = Z^{- 1} \int t ∣ g_{0} (t) ∣^{2} d t$ vanishes because $t ∣ g_{0} (t) ∣^{2}$ is odd and integrable by step 2.1, and the mean is finite; the second moment is $+ \infty$ by steps 2.1-3.1, so the variance, which is the second moment minus the square of the (zero) mean, does not exist as a finite number. By the criterion [F4] this is exactly the statement $ψ \in D (∣ Q_{1} ∣^{1 / 2})$ but $ψ \notin D (Q_{1})$ ; the divergence of $\int t^{2} ∣ g ∣^{2} d t$ is the same as the failure of $x_{1} ψ \in L^{2}$ .

Remarks

The profile is symmetric, so the finite mean is zero; a non-symmetric variant shifts the mean while keeping the second moment infinite, since the tail exponent $3$ is what decides finiteness: $∣ g ∣^{2}$ decays like $t^{- 3}$ , whose first moment $t \cdot t^{- 3} = t^{- 2}$ is integrable while the second is not.
The example shows that the domain hypotheses of Mean, second moment and variance of an assigned observable in a preparation are not automatic: the state here is a perfectly legitimate preparation of the model, yet the variance of the position law is infinite and the statement "the spread $Δ Q_{1}$ " has no finite value in this preparation.

A preparation with finite mean position and infinite variance

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources