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 Incoming links: Heisenberg-Weyl operator

Heisenberg-Weyl twirling channel 2026-10-05

Averaging conjugation by all Heisenberg-Weyl operators completely depolarizes a system:

\frac{1}{d ^{2}} \sum_{k, m} W_{k, m} A W_{k, m}^{†} = (Tr A) \frac{I}{d} .

Summing over phases

Z^{m}

kills off-diagonal matrix elements, and averaging shifts

X^{k}

makes all diagonal elements equal. Acting on subsystem

B

of a bipartite operator gives

(Tr_{B} A) \otimes I_{B} / d

. This identity turns joint convexity of quantum relative entropy into monotonicity under partial trace.

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Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 51 / 1 / b / Solution Created 2026-10-03 Updated 2026-10-14

Label an orthonormal basis of the

n

-dimensional Hilbert space by

j = 0, \dots, n - 1

, with indices modulo

n

. Let

ω = e^{2 πi / n}, X ∣ j ⟩ = ∣ j + 1 ⟩, Z ∣ j ⟩ = ω^{j} ∣ j ⟩ .

These are the shift and phase Heisenberg-Weyl operators. Alice holds the input

S

A

of the shared maximally entangled state

∣ Φ_{n} ⟩_{A B} = \frac{1}{n} \sum_{j = 0}^{n - 1} ∣ j ⟩_{A} ∣ j ⟩_{B} .

S A

in the generalized Bell basis

∣ B_{pq} ⟩_{S A} = \frac{1}{n} \sum_{j = 0}^{n - 1} ω^{p j} ∣ j ⟩_{S} ∣ j + q ⟩_{A}, p, q = 0, \dots, n - 1.

Its inner products are

⟨ B_{pq} ∣ B_{p^{'} q^{'}} ⟩ = δ_{q q^{'}} \frac{1}{n} \sum_{j = 0}^{n - 1} ω^{(p^{'} - p) j} = δ_{p p^{'}} δ_{q q^{'}} .

There are

n^{2}

orthonormal vectors in an

n^{2}

-dimensional space, so this is a complete projective measurement.

For an unknown normalized state

∣ u ⟩ = \sum_{j} u_{j} ∣ j ⟩

, its contraction with measurement outcome

(p, q)

gives Bob's unnormalized state

_{S A} ⟨ B_{pq} ∣ (∣ u ⟩_{S} ∣ Φ_{n} ⟩_{A B}) = \frac{1}{n} j \sum u_{j} ω^{- p j} ∣ j + q ⟩_{B} = \frac{1}{n} X^{q} Z^{- p} ∣ u ⟩_{B} .

Each outcome has probability

1/ n^{2}

, independently of the input. Alice communicates the two

n

-valued labels, and Bob applies the ordered inverse

U_{pq} = Z^{p} X^{- q}, U_{pq} (\frac{1}{n} X^{q} Z^{- p} ∣ u ⟩) = \frac{1}{n} ∣ u ⟩ .

The order matters because shift and phase operators do not generally commute. Normalizing the branch recovers the unknown state exactly, and all outcomes are corrected, so success is deterministic. The quantum resource is one maximally entangled pair of local dimension

n

. The classical resource is an

n^{2}

-valued message, with information content

2 lo g_{2} n

⌈ lo g_{2} (n^{2})⌉

ordinary bits suffice in fixed-length binary encoding.

This is faithful qudit teleportation even for an input entangled with a reference

R

. Expand a joint pure state as

\sum_{j} ∣ j ⟩_{S} ∣ r_{j} ⟩_{R}

, without requiring the reference vectors to be orthogonal. The same contraction and correction act on each system basis vector, producing

n^{- 1} \sum_{j} ∣ j ⟩_{B} ∣ r_{j} ⟩_{R}

. Thus the whole joint state, not just a reduced state, is preserved. Linearity extends the conclusion to mixed inputs. This is teleportation as an identity channel on a reference.

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