Manuscript Hawk05-B@1 · Hawk05-B.tex · section · f2e5d4f5

The Haar fourth-moment calculation for code decoupling

The Haar fourth-moment calculation for code decoupling · lines 1851–1948

Hawk05-B@1#eq:fourth, Hawk05-B@1#eq:omegapurity, Hawk05-B@1#eq:sigmapurity

Haar invariance permits us to take $J ∣ i ⟩$ to be the first $d$ coordinate vectors. Write $V_{ab, i} = U_{(a, b), i}$ , with $a = 1, \dots, r$ , $b = 1, \dots, b$ .

For $D \geq 2$ , the required Haar fourth moment is

E [U_{x i} U_{z k} U_{y j} U_{w ℓ}] = \frac{δ _{x y} δ _{z w} δ _{ij} δ _{k ℓ} + δ _{x w} δ _{z y} δ _{i ℓ} δ _{k j}}{D ^{2} - 1} - \frac{δ _{x y} δ _{z w} δ _{i ℓ} δ _{k j} + δ _{x w} δ _{z y} δ _{ij} δ _{k ℓ}}{D ( D ^{2} - 1 )} .

This is the order-two Haar integration identity CollinsSniady. It can also be obtained directly: left and right unitary invariance restrict the tensor to the two row pairings and two column pairings; symmetry under exchange gives two coefficients. Contracting with $\sum_{x} U_{x i} U_{x j} = δ_{ij}$ gives the two linear equations whose solution is $1/ (D^{2} - 1)$ and $- 1/ [D (D^{2} - 1)]$ .

For the normalized complementary Choi state,

(ω_{QB})_{ib, j b^{'}} = \frac{1}{d} a \sum V_{ab, i} V_{a b^{'}, j} .

Therefore

Tr ω_{QB}^{2} = \frac{1}{d ^{2}} i, j, a, a^{'}, b, b^{'} \sum V_{ab, i} V_{a b^{'}, j} V_{a^{'} b^{'}, j} V_{a^{'} b, i} .

Substitution of (eq:fourth) gives four index counts:

d r^{2} b, d^{2} r b^{2}, d^{2} r^{2} b, d r b^{2} .

The last two carry the negative coefficient divided by $D$ . Thus

E Tr ω_{QB}^{2} = \frac{r ( D / d - 1 ) + b ( D - 1/ d )}{D ^{2} - 1} .

Similarly,

(σ_{B})_{b, b^{'}} = \frac{1}{d} a, i \sum V_{ab, i} V_{a b^{'}, i},

and the corresponding counts are

d^{2} r^{2} b, d r b^{2}, d r^{2} b, d^{2} r b^{2} .

Hence

E Tr σ_{B}^{2} = \frac{r ( D - 1/ d ) + b ( D / d - 1 )}{D ^{2} - 1} .

Because $Tr_{Q} ω_{QB} = σ_{B}$ ,

∥ ω_{QB} - I_{Q} / d \otimes σ_{B} ∥_{2}^{2} = Tr ω_{QB}^{2} - \frac{1}{d} Tr σ_{B}^{2} .

Subtracting (eq:sigmapurity) from (eq:omegapurity) with this factor gives

\frac{( 1 - d ^{- 2} ) ( b D - r )}{D ^{2} - 1} = \frac{( 1 - d ^{- 2} ) r ( b ^{2} - 1 )}{D ^{2} - 1},

which proves (eq:HS). The expression vanishes for a one-dimensional residual, as it must.

Displayed formulas in this passage

Hawk05-B@1#eq:fourth · line 1860
$E [U_{x i} U_{z k} U_{y j} U_{w ℓ}] = \frac{δ _{x y} δ _{z w} δ _{ij} δ _{k ℓ} + δ _{x w} δ _{z y} δ _{i ℓ} δ _{k j}}{D ^{2} - 1} - \frac{δ _{x y} δ _{z w} δ _{i ℓ} δ _{k j} + δ _{x w} δ _{z y} δ _{ij} δ _{k ℓ}}{D ( D ^{2} - 1 )} .$
Hawk05-B@1#eq:omegapurity · line 1907
$E Tr ω_{QB}^{2} = \frac{r ( D / d - 1 ) + b ( D - 1/ d )}{D ^{2} - 1} .$
Hawk05-B@1#eq:sigmapurity · line 1925
$E Tr σ_{B}^{2} = \frac{r ( D - 1/ d ) + b ( D / d - 1 )}{D ^{2} - 1} .$
unlabeled · line 1887
$(ω_{QB})_{ib, j b^{'}} = \frac{1}{d} a \sum V_{ab, i} V_{a b^{'}, j} .$
unlabeled · line 1893
$Tr ω_{QB}^{2} = \frac{1}{d ^{2}} i, j, a, a^{'}, b, b^{'} \sum V_{ab, i} V_{a b^{'}, j} V_{a^{'} b^{'}, j} V_{a^{'} b, i} .$
unlabeled · line 1902
$d r^{2} b, d^{2} r b^{2}, d^{2} r^{2} b, d r b^{2} .$
unlabeled · line 1915
$(σ_{B})_{b, b^{'}} = \frac{1}{d} a, i \sum V_{ab, i} V_{a b^{'}, i},$
unlabeled · line 1921
$d^{2} r^{2} b, d r b^{2}, d r^{2} b, d^{2} r b^{2} .$
unlabeled · line 1933
$∥ ω_{QB} - I_{Q} / d \otimes σ_{B} ∥_{2}^{2} = Tr ω_{QB}^{2} - \frac{1}{d} Tr σ_{B}^{2} .$
unlabeled · line 1940
$\frac{( 1 - d ^{- 2} ) ( b D - r )}{D ^{2} - 1} = \frac{( 1 - d ^{- 2} ) r ( b ^{2} - 1 )}{D ^{2} - 1},$