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Operator support weight is invariant under wire permutation
Statement
Let the source register have ordered factors , and let the target register have factors in the QC02 order . Let be the corresponding wire bijection and set . An original factor labeled is in target position . Thus is supported in exactly when is supported in ; consequently .
Facts & Assumptions
Given: The source and target ordered tensor registers in the Statement, a permutation , its wire bijection , and .
Support and weight of an operator on a finite register: Support means conjugating by and its inverse gives a local operator tensored with identity on the complementary factors; weight is the minimum support size.
Ordered tensor registers, their inner product, and wire permutations: A wire permutation sends a product-basis tensor to the tensor with the same factor labels in the declared output order and is extended linearly.
Finite wire permutations are unitary basis permutations: Every finite wire permutation is a bijective linear isometry, with inverse the wire permutation for the inverse ordering.
Tensor products of finitely many linear maps: A tensor map is the unique linear extension of its factorwise action on elementary tensors.
The product basis is orthonormal and tensor dimensions multiply: Elementary product tensors form a basis and span each finite ordered tensor register.
Proof
Put , the target positions occupied by source factors from . Let reorder those source factors into their target order, and let reorder the complementary factors likewise; these are wire bijections by [F2]. On each elementary tensor, both and place each factor vector in the same target order. Elementary tensors span by [F4], and the maps agree on them by [F1], so .
If is supported in , [L1] gives with . Conjugating by step 1.1 gives . On the right side is , hence equals by [F3]; thus is supported in .
Apply the same implication to and the inverse wire permutation. It gives the converse: if is supported in , then is supported in . Thus bijects the support sets of and , preserving cardinality; their minimum support sizes agree.
Remarks
The QC02 bodies were inspected. This proof uses their declared factor-order action and bijective inverse; it does not identify differently ordered registers without a map.
Depends on
Used by
Nothing in the library uses this result yet.
Cited to discharge well-definedness by Support and weight of an operator on a finite register.
Dependency tree · two levels
18 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.
Sources
- Richard Cleve, Quantum Information Processing course notes, §§36.1, 36.3, and 37.3 (standard reference, not scraped)
- John Preskill, Quantum Computation, Chapter 7, §7.4.1 (standard reference, not scraped)