Clifford+T on the Bloch sphere: channel representation, sde, T-count

The Pauli-channel representation Û of a gate U is the matrix of ρ ↦ UρU† in the Pauli basis: its columns are the images of x̂, ŷ, ẑ, i.e. the rotation of the Bloch vector.

The six Pauli directions ±X, ±Y, ±Z are the vertices of an octahedron. Clifford gates (X, Y, Z, H, S) permute these vertices, so Û has entries in {0, ±1}.

T is a rotation by π/4 about z: it sends a vertex to an edge midpoint, and 1/√2 appears in Û.

sde(Û) = the smallest k such that √2^k·Û has entries in ℤ[√2]. Clifford: sde 0. T: sde 1.

Apply a gate (or press the key)

X, Y, Z: rotation by π about the axis of the same name. H: rotation by π about (x+z)/√2. S: π/2 about z. T: π/4 about z.

Track

the frame (images of x̂, ŷ, ẑ)

a state

Speed

Gates applied (in time order)

Channel matrix Û (columns = images of x̂, ŷ, ẑ)

sde(Û) = 0 · T gates applied = 0

image of x̂ image of ŷ image of ẑ
Yellow entries are non-integers. Drag to orbit, scroll to zoom.

At sde = k every Bloch coordinate reads (a + b√2)/√2^k with a, b ∈ ℤ. This looks like a grid of pitch 1/√2^k, but it is not one: ℤ[√2] is dense in ℝ (e.g. 99 − 70√2 ≈ 0.005).

What makes each level finite is unitarity, Σu_i² = 2^k, read in both real embeddings √2 ↦ ±√2. Both are sums of real squares, so |a| ≤ √2^k and |b| ≤ √2^k−1.

Result: exactly 6·2^k new directions at each level, dense in the limit. One T = one bit. Toggle the naive grid to see that points do not sit on it.

T-count (← → keys)

k = 2resolution ≈ 2^−k = 0.25

sde ≤ k sde = k only

Display

axis ticks at the coordinate values reached

naive grid of pitch 1/√2^k on the equatorial plane

auto-rotate

Counts

directions with sde = k: · = 6·2^k
directions shown: ·
distinct values of x / y / z: ·
naive grid pitch 1/√2^k = ·

Colour = sde of the direction

Octahedron vertices at k = 0, edge midpoints at k = 1, then (±1, ±1, ±√2)/2 at k = 2. Drag to orbit, scroll to zoom.