Quantum Information Apps
Exercise PLAN-417
primary source nl only. Current content-bound mathematical second pass accepted. No new Lean verification is claimed for this NL view.
Assumptions and conventions
Question
Let rho be a state on A^{tensor n}, and let rhobar = (1/n!) sum_{pi in S_n} Q_pi rho Q_pi^dagger be its symmetrization (Q_pi the permutation operators). Let Pi_n be the projector onto the symmetric subspace Sym^n(H_A). Prove the identity Pi_n rhobar Pi_n = Pi_n rho Pi_n. In your proof establish and use the intertwining relation Pi_n Q_pi = Pi_n = Q_pi Pi_n for every pi (the symmetric projector absorbs permutations), and be explicit about why the 1/n! average collapses.
Hint: mathematical route
Because the symmetric projector absorbs permutations (Pi_n Q_pi = Pi_n), the symmetrization average collapses and Pi_n rhobar Pi_n = Pi_n rho Pi_n, so symmetric-subspace compression cannot distinguish a state from its symmetrization.
- Pi_n absorbs permutations: Pi_n Q_pi = Pi_n = Q_pi Pi_n.
- Each averaged term becomes Pi_n rho Pi_n independent of pi.
- The 1/n! average collapses to a single Pi_n rho Pi_n.
Reference solution
First establish Pi_n Q_pi = Pi_n for all pi. The symmetric subspace Sym^n(H_A) is by definition the set of vectors fixed by every Q_pi, and Pi_n is the orthogonal projector onto it. For any vector |w>, Q_pi |w> and |w> have the same component in Sym^n because Q_pi acts as the identity on Sym^n and permutes the orthogonal complement among itself (the decomposition into isotypic components is Q_pi-invariant, and the trivial/symmetric isotype is pointwise fixed). Concretely, Pi_n = (1/n!) sum_sigma Q_sigma (the projector onto the trivial isotype equals the group average), and Q_sigma Q_pi = Q_{sigma pi} ranges over all of S_n as sigma does, so Pi_n Q_pi = (1/n!) sum_sigma Q_{sigma pi} = (1/n!) sum_tau Q_tau = Pi_n; likewise Q_pi Pi_n = Pi_n and Pi_n is self-adjoint with Q_pi^dagger = Q_{pi^{-1}}, giving Pi_n Q_pi^dagger = Pi_n too. Now compute Pi_n rhobar Pi_n = Pi_n [ (1/n!) sum_pi Q_pi rho Q_pi^dagger ] Pi_n = (1/n!) sum_pi (Pi_n Q_pi) rho (Q_pi^dagger Pi_n). Using Pi_n Q_pi = Pi_n and Q_pi^dagger Pi_n = Pi_n for every pi, each summand becomes Pi_n rho Pi_n, independent of pi. Hence Pi_n rhobar Pi_n = (1/n!) sum_pi Pi_n rho Pi_n = (1/n!)(n!) Pi_n rho Pi_n = Pi_n rho Pi_n. The 1/n! average collapses precisely because the permutations are absorbed by the flanking symmetric projectors, leaving n! identical copies of Pi_n rho Pi_n. This shows that, as far as the symmetric-subspace compression is concerned, a state and its symmetrization are indistinguishable.
Assessment rubric
- Establishes Pi_n Q_pi = Q_pi Pi_n = Pi_n using Pi_n = (1/n!) sum_sigma Q_sigma and the group reshuffle Q_sigma Q_pi = Q_{sigma pi} (1 weight)
- Writes Pi_n rhobar Pi_n = (1/n!) sum_pi (Pi_n Q_pi) rho (Q_pi^dagger Pi_n) (1 weight)
- Reduces each summand to Pi_n rho Pi_n via the absorption identity (including Q_pi^dagger Pi_n = Pi_n) (1 weight)
- Collapses the n! identical terms to conclude Pi_n rhobar Pi_n = Pi_n rho Pi_n (1 weight)
Graph-grounded prerequisites
These are support/source anchors, not an automatically extracted theorem-dependency proof.
- Operator and semigroup theory (frozen curriculum anchor; see graph registry)
- Matrix analysis (frozen curriculum anchor; see graph registry)
Lean view and evidence boundary
No aligned Lean view has been admitted for this native core in this textbook checkpoint. This is not a claim that the underlying mathematics lacks a formal proof. Contributions must preserve this exact target contract and split.
Sources and provenance
- website/research/sources.json
leditzky-2025-repth-qit
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Contract: 5f9b00f9eef7b7b85ea81cf919c3102526b3d01decc38f5077a4a521caee2040
Item digest: e2e13bbb34865744a0e5a9f2ab454f606e78c8144fe0237f4ed4c708ddceebdc
Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a
Authored support steps: 3. This is not measured graph distance or model difficulty.