ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Two-Mode CAR Signs and a Matrix Unit

Example

Order the two-mode occupation basis as $e_{\emptyset}, e_{{1}}, e_{{2}}, e_{{1, 2}}$ . The operators defined by the finite occupation formulas are

$a_{1} = (\begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{array}), a_{1}^{†} = (\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array}),$

$a_{2} = (\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & - 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}), a_{2}^{†} = (\begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & - 1 & 0 & 0 \end{array}) .$

The minus signs occur when mode $2$ is toggled while mode $1$ is occupied.

Facts & Assumptions

Given: The two-mode occupation basis and the operators in Finite CAR Occupation Operators Give All Matrix Units.

[F1]

For two modes, $N_{j} = a_{j}^{†} a_{j}$ is the projection onto basis states containing $j$ , and $P_{\emptyset} = (I - N_{1}) (I - N_{2})$ projects onto $e_{\emptyset}$ (Finite CAR Occupation Operators Give All Matrix Units).

[F2]

The universal two-mode CAR algebra is $M_{4} (C)$ (The Finite-Mode CAR Algebra Is a Full Matrix Algebra).

Verification

1.1givenF1algebra

Reading columns of the displayed matrices gives $a_{1} e_{{1, 2}} = e_{{2}}$ and $a_{2} e_{{1, 2}} = - e_{{1}}$ . Applying the operators in the reverse order gives $a_{2} a_{1} e_{{1, 2}} = e_{\emptyset}$ and $a_{1} a_{2} e_{{1, 2}} = - e_{\emptyset}$ , while both products vanish on the other three basis vectors. Hence ${a_{1}, a_{2}} = 0$ . The adjoint matrices give the corresponding creation relations, and their same-mode products satisfy $a_{j} e_{S} = 0$ when $j \notin S$ , $a_{j}^{†} e_{S} = 0$ when $j \in S$ , and ${a_{j}, a_{j}^{†}} = I$ on each of the four basis vectors.

2.1givenstep 1.1algebra

For the mixed distinct-mode relation, on $e_{{1}}$ one has $a_{1} a_{2}^{†} e_{{1}} = - e_{{2}}$ and $a_{2}^{†} a_{1} e_{{1}} = e_{{2}}$ ; on $e_{\emptyset}, e_{{2}}, e_{{1, 2}}$ both terms sum to zero by the displayed matrices. Thus ${a_{1}, a_{2}^{†}} = 0$ . This confirms the sign convention for modes in the order $1 < 2$ .

3.1givenF1F2step 2.1algebra∎

By [F1], $P_{\emptyset} = diag (1, 0, 0, 0)$ . The operator $a_{1}^{†} P_{\emptyset} a_{2}$ sends $e_{{2}}$ to $e_{{1}}$ and kills the other three basis vectors, so it is the matrix unit $E_{{1}, {2}}$ . Together with the general formula in [F1], this is one explicit off-diagonal matrix unit inside the algebra $M_{4} (C)$ of [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.