Lean, informalized

Nielsen & Chuang · Chapter 4, Quantum circuits · p. 191

Exercise 4.38 - A unitary on dd levels that takes at least d−1d-1 two-level unitaries to build

A unitary with no short two-level decomposition.

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The Exercise

Take a quantum system with dd levels, its state space ℂd\mathbb{C}^d and its computational basis |0⟩,…,|d−1⟩|0\rangle, \dots, |d-1\rangle; a unitary on it is a unitary d×dd \times d complex matrix. The simplest unitaries are the two-level ones. The exercise asks how many of them a product needs in order to equal a given unitary, and bounds that number from below: in every dimension dd there is a unitary UU that is not a product of fewer than d−1d - 1 two-level unitaries.✦

Two-level Unitaries

A two-level unitary VV on ℂd\mathbb{C}^d is a unitary d×dd \times d matrix for which there are indices pp and qq such that VV agrees with the identity matrix outside the block where the rows pp and qq meet the columns pp and qq: Vij=δijwhenever i∉{p,q} or j∉{p,q}.V_{ij} = \delta_{ij} \qquad \text{whenever } i \notin \{p, q\} \text{ or } j \notin \{p, q\}.(1)✦ On the span of |p⟩|p\rangle and |q⟩|q\rangle such a VV acts as a 2×22 \times 2 unitary. The indices need not differ, so a phase on a single basis state is a two-level unitary as well.

The Claim

Theorem 1. Let dd be a natural number. There is a unitary d×dd \times d matrix UU such that, whenever V1,…,VkV_1, \dots, V_k are two-level unitaries on ℂd\mathbb{C}^d with U=V1V2⋯VkU = V_1 V_2 \cdots V_k, k≥d−1.k \geq d - 1.(2)✦

What Is Checked

The claim is stated over the unitary d×dd \times d complex matrices and over the two-level unitaries of (1), multiplied in the order written.