Manuscript Psy04-A@1 · Psy04-A-v1.tex · section · d5800e2a

Encoding and complete readout

A sequential one-qubit construction › Encoding and complete readout · lines 510–544

Psy04-A@1#eq:encoding

Let

σ_{x} = (0110), σ_{y} = (0 i - i 0), σ_{z} = (10 0 - 1) .

For $x = (x_{1}, x_{2})$ , prepare

ρ_{x} = \frac{1}{2} [I + \frac{( - 1 ) ^{x_{1}} σ _{x} + ( - 1 ) ^{x_{2}} σ _{z}}{2}] .

Its Bloch vector has length one, so it is a pure density operator.

For query $j$ , use projectors

P_{b ∣ 1} = \frac{I + ( - 1 ) ^{b} σ _{x}}{2}, P_{b ∣ 2} = \frac{I + ( - 1 ) ^{b} σ _{z}}{2} .

The complete projective instrument is

I_{b ∣ j} (A) = P_{b ∣ j} A P_{b ∣ j} .

Its branches are completely positive and their sum preserves trace.

Displayed formulas in this passage

Psy04-A@1#eq:encoding · line 522
$ρ_{x} = \frac{1}{2} [I + \frac{( - 1 ) ^{x_{1}} σ _{x} + ( - 1 ) ^{x_{2}} σ _{z}}{2}] .$
unlabeled · line 513
$σ_{x} = (0110), σ_{y} = (0 i - i 0), σ_{z} = (10 0 - 1) .$
unlabeled · line 534
$P_{b ∣ 1} = \frac{I + ( - 1 ) ^{b} σ _{x}}{2}, P_{b ∣ 2} = \frac{I + ( - 1 ) ^{b} σ _{z}}{2} .$
unlabeled · line 539
$I_{b ∣ j} (A) = P_{b ∣ j} A P_{b ∣ j} .$