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A bounded nonconstant potential with no nontrivial translational symmetry

Physical scope

A counterexample inside the declared massive model with a bounded measurable potential energy: a positive energy scale times the indicator of the ball of radius one metre has no nontrivial translational symmetry, so no nonzero translation commutes with its bounded-potential energy operator, while the potential itself is bounded and measurable and its model is self-adjoint on W^{2,2}. The item refutes a universal claim about translations of bounded nonconstant potentials; it asserts nothing about the physics of a laboratory potential and records no measurement.

Statement refuted

Let $V : R^{3} \to R$ be any real, bounded, measurable, nonconstant potential energy, expressed in joules. Then for every nonzero $a \in R^{3} ∖ {0}$ the translation $T_{a}$ is a symmetry of the declared bounded-potential dynamics with energy $H_{V}$ .

Facts & Assumptions

Given: The postulates P1, P2, P4, P5, P6 in the readings of The massive spin-s particle model of the framework: state space, operators, units and frame readings, Symmetries of the declared dynamics, unitary symmetry groups and conserved sharp observables and the postulate items of those readings; the Axioms of Countable Choice and of Choice and Dependent Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice, The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain); the declared particle of mass $m > 0$ and spin label $s$ with state space $H_{s} = L^{2} (R^{3}; C^{r})$ , $r = 2 s + 1$ , and translations $T_{a}$ (Componentwise position and momentum operators on the component-valued wave-function space clause (c)); the positive model energy scale $E_{*} > 0$ in J and length $R : = 1 m$ ; the potential $V_{0} : = E_{*} 1_{B (0, R)}$ and its shift $V_{0, a} (x) : = V_{0} (x - a)$ ; a vector $a \in R^{3} ∖ {0}$ .

[F1]

Model. For a real bounded measurable $V$ the operator $H_{V} = \frac{1}{2 m} \sum_{k} P_{k}^{2} + M_{V}$ is self-adjoint with domain $D (H_{V}) = W^{2, 2} (R^{3}; C^{r})$ , and $T_{a}$ is a symmetry of $H_{V}$ in the sense of Symmetries of the declared dynamics, unitary symmetry groups and conserved sharp observables clause (b) if and only if $V_{a} = V$ almost everywhere, where $V_{a} (x) = V (x - a)$ ; in fact $T_{a} H_{V} T_{a}^{- 1} = H_{V_{a}}$ (Real multiplication operators and free plus bounded-potential Schrödinger operators clause (c), Translation invariance of bounded potentials and conservation of momentum clause (a)).

[F2]

The indicator potential energy. $B (0, R) = {x \in R^{3} : ∣ x ∣ < R}$ is the Euclidean ball of radius $R$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Open ball, closed ball and sphere in a metric space), with spatial coordinates and $R$ expressed in metres. The function $V_{0} = E_{*} 1_{B (0, R)}$ is real and measurable, has energy unit J and $0 \leq V_{0} \leq E_{*}$ , and is nonconstant since $V_{0} (0) = E_{*} > 0$ and $V_{0} (2 R e_{1}) = 0$ . For every $x$ , $V_{0, a} (x) = E_{*} ⟺ x - a \in B (0, R) ⟺ ∣ x - a ∣ < R,$ so ${x : V_{0} (x) \neq V_{0, a} (x)} = {x : ∣ x ∣ < R \leq ∣ x - a ∣} \cup {x : ∣ x - a ∣ < R \leq ∣ x ∣}$ .

[F3]

Norm geometry. The Euclidean norm satisfies the triangle inequality $∣ y - a ∣ \geq ∣ x - a ∣ - ∣ y - x ∣$ and $∣ y ∣ \leq ∣ x ∣ + ∣ y - x ∣$ for all $x, y, a \in R^{3}$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ clause 1, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

Lebesgue measure. A Euclidean ball $B (x, r)$ with $r > 0$ is Lebesgue measurable with $0 < λ (B (x, r)) < \infty$ (Euclidean balls have positive finite Lebesgue measure); Lebesgue measure is the restriction of Lebesgue outer measure to the measurable sets, and $λ^{*}$ is monotone (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Assuming countable choice, Lebesgue outer measure is an outer measure that restricts to elementary volume, Measure-null sets and almost-everywhere statements relative to a measure); a property holds almost everywhere exactly when its exceptional set is contained in a measurable set of measure zero (Measure-null sets and almost-everywhere statements relative to a measure).

Counterexample

Proof technique: direct; take a positive energy scale times the indicator of the ball of radius $R$ , exhibit a point where the shifted potential differs, show that a whole ball lies in the difference set, and use the positivity of Lebesgue measure of balls to conclude that the potentials are not equal almost everywhere, so that the symmetry criterion fails.

Given: The data of the refuted statement and [F1]-[F4].

1.1F2F3givenalgebra

[A point where the shifted potential differs] Put $u : = a / ∣ a ∣$ , a unit vector, and $δ : = \min {∣ a ∣ / 2, R / 2} > 0$ , and set $x_{0} : = - (R - δ) u$ . Then $∣ x_{0} ∣ = R - δ < R$ , so $V_{0} (x_{0}) = E_{*}$ ; and $x_{0} - a = - (R - δ + ∣ a ∣) u$ , so $∣ x_{0} - a ∣ = R - δ + ∣ a ∣ > R$ because $∣ a ∣ > δ$ for $δ \leq ∣ a ∣ / 2$ , whence $V_{0, a} (x_{0}) = 0$ . Thus $x_{0}$ lies in the difference set ${V_{0} \neq V_{0, a}}$ of [F2].

2.1step 1.1F2F3givenalgebra

[A ball inside the difference set] Let $y \in R^{3}$ with $∣ y - x_{0} ∣ < δ / 2$ and put $r : = δ / 2$ . Then $∣ y ∣ \leq ∣ x_{0} ∣ + ∣ y - x_{0} ∣ < R - δ + δ / 2 < R$ , so $V_{0} (y) = E_{*}$ ; and $∣ y - a ∣ \geq ∣ x_{0} - a ∣ - ∣ y - x_{0} ∣ > R - δ + ∣ a ∣ - δ / 2 = R + ∣ a ∣ - 3 δ / 2 > R$ , the last inequality because $δ \leq ∣ a ∣ / 2$ gives $3 δ / 2 \leq 3 ∣ a ∣ / 4 < ∣ a ∣$ ; hence $V_{0, a} (y) = 0$ . Therefore $B (x_{0}, r) \subseteq {V_{0} \neq V_{0, a}}$ .

3.1step 2.1F4givenalgebra

[The difference set has positive measure] By [F4] the ball $B (x_{0}, r)$ is measurable with $λ (B (x_{0}, r)) > 0$ . If $V_{0, a} = V_{0}$ almost everywhere, then the difference set ${V_{0} \neq V_{0, a}}$ is contained in a measurable null set $N$ , and step 2.1 gives $B (x_{0}, r) \subseteq N$ ; monotonicity of the outer measure [F4] then yields $λ (B (x_{0}, r)) = λ^{*} (B (x_{0}, r)) \leq λ^{*} (N) = λ (N) = 0$ , contradicting $λ (B (x_{0}, r)) > 0$ . Hence $V_{0, a} = V_{0}$ does not hold almost everywhere: their difference set has positive measure.

4.1step 3.1F1F2given∎

[The refutation] By step 3.1 and the symmetry criterion of [F1], $T_{a}$ is not a symmetry of $H_{V_{0}}$ for the displayed nonzero $a$ . Since $a \in R^{3} ∖ {0}$ was arbitrary and $V_{0}$ is real, bounded, measurable and nonconstant by [F2], the universally quantified statement is refuted: a bounded nonconstant potential need not have any nontrivial translational symmetry.

Remarks

Where the failure comes from. The criterion [F1] is an almost-everywhere identity of the shifted potential classes, and the example realizes the failure on a set of positive measure. Boundedness and measurability suffice for the model to be well posed; they do not produce any translation symmetry.
What is not claimed. The counterexample refutes one universal statement about bounded nonconstant potentials; it does not claim that every bounded nonconstant potential has no nontrivial translational symmetry (potentials constant along a direction do have the corresponding translations as symmetries), and it makes no statement about any laboratory potential or measurement.

A bounded nonconstant potential with no nontrivial translational symmetry

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources