Manuscript Oai280-C@1 · Oai280-C.tex · section · 7d44b8c0

One connected two-cell encoding

One connected two-cell encoding · lines 682–716

Oai280-C@1#eq:F, Oai280-C@1#eq:W, Oai280-C@1#eq:W-components

Use two copies of the same tensor. Each cell has one bulk input, four exterior physical legs, and one joining leg. Denote the left bulk by $L$ , the right bulk by $S$ , the exterior legs by

a_{1}, \dots, a_{4}, c_{1}, \dots, c_{4},

and the joining legs by $e, e^{'}$ .

Reshape the block encoder into

F_{a; l, e} = 2 V_{a, e; l}, F : C^{2}_{L} \otimes C^{2}_{e} ⟶ (C^{2})_{a}^{\otimes 4} .

Use the same convention for the right cell. Define

W = (F_{L} \otimes F_{R}) (I_{L S} \otimes ∣ Φ_{2} ⟩_{e e^{'}}) .

In components,

W_{a, c; l, s} = 2 e = 0 \sum^{1} V_{a, e; l} V_{c, e; s} = 22 e = 0 \sum^{1} T_{l, a, e} T_{s, c, e} .

Displayed formulas in this passage

Oai280-C@1#eq:F · line 693
$F_{a; l, e} = 2 V_{a, e; l}, F : C^{2}_{L} \otimes C^{2}_{e} ⟶ (C^{2})_{a}^{\otimes 4} .$
Oai280-C@1#eq:W · line 700
$W = (F_{L} \otimes F_{R}) (I_{L S} \otimes ∣ Φ_{2} ⟩_{e e^{'}}) .$
Oai280-C@1#eq:W-components · line 708
$W_{a, c; l, s} = 2 e = 0 \sum^{1} V_{a, e; l} V_{c, e; s} = 22 e = 0 \sum^{1} T_{l, a, e} T_{s, c, e} .$
unlabeled · line 687
$a_{1}, \dots, a_{4}, c_{1}, \dots, c_{4},$

One connected two-cell encoding

Displayed formulas in this passage

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