CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Finite matrices cannot satisfy exact canonical commutation relations

Statement refuted

For every positive integer $d$ and every $ℏ > 0$ , there exist matrices $Q, P \in M_{d} (C)$ such that $Q P - P Q = i ℏ I_{d}$ .

Facts & Assumptions

Given: A positive integer $d$ , $ℏ > 0$ , arbitrary complex $d \times d$ matrices $Q, P$ , and $tr (M) : = \sum_{j = 1}^{d} M_{j j}$ for $M \in M_{d} (C)$ .

[A1]

The proposed pair, if it existed, would satisfy $Q P - P Q = i ℏ I_{d}$ .

Counterexample

1.1givenalgebra

By the definition of matrix multiplication, $tr (Q P) = \sum_{j = 1}^{d} \sum_{k = 1}^{d} Q_{j k} P_{k j}$ and $tr (P Q) = \sum_{j = 1}^{d} \sum_{k = 1}^{d} P_{j k} Q_{k j}$ . Interchanging the finite indices $j, k$ in the second sum and commuting complex scalars makes the sums equal, so $tr (Q P - P Q) = 0$ .

2.1A1step 1.1algebra∎

Taking the trace in [A1] gives $0 = i ℏ tr (I_{d}) = i ℏ d$ , which is impossible because $d \geq 1$ and $ℏ > 0$ . Therefore the proposed universal existence statement is false.

Remarks

This proof concerns bounded finite matrices. It does not differentiate a Weyl relation for unbounded generators or supply a domain argument for such a differentiation.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Anthony W. Knapp, Advanced Real Analysis, Digital Second Edition, Chapter II, §5, trace-class definitions and cyclicity exercise, pp. 49–52 (the proof here is the explicit finite sum)