CSS codes Shyam Sundhar R November 12, 2017 Shyam Sundhar R CSS - - PowerPoint PPT Presentation

CSS codes

Shyam Sundhar R November 12, 2017

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Outline

1

Codes as a group

2

Code Construction

3

Hadamard Transform

4

Error Modelling

5

Error Correction

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Codes as a group

W.K.T sum of any two code-words is also a code-word. closed under addition And 0 is also a codeword. identity element exists So any classical code C can be seen as a sub-group of Z n

2

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Let C1 and C2 be two linear codes. C2 ⊂ C1 C2 is a subgroup of C1 C1 can be partitioned into cosets of C2

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Code construction

C1 and C2 are [n,k,d] and [n, k1, d1] codes respectively. C2 ⊂ C1 C1 and C ⊥

2 are both t-error correcting codes.

C1 = C2 ∪ (c1 + C2) ∪ (c2 + C2) . . . ∪ (cn + C2) where n is the number

• f cosets.

N = 2k

2k1

We associate each basis state with a unique coset. |si = 1 √ 2k Σc′∈C2|ci + c′ Each basis state is in a superposition of all the codewords in a coset. ci ∈ C1 so If c1

i and c2 i belong to the same coset , they define the

same basis codeword.

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Application of Hadamard

Phase flips become bit flips after hadamard application. Hadamard transforms a codes to its dual W.K.T |i H⊗n − − → Σj(−1)(i.j)|j Σc′∈C2|c′ H⊗n − − → ΣjΣc′∈C2(−1)(c

′.j)|j

Σc′∈C2(−1)(c

′.j)

non-zero only if j ∈ C ⊥

2

Σc′∈C2|c′ H⊗n − − → Σj∈C ⊥

2 |j Shyam Sundhar R CSS codes November 12, 2017 6 / 10

|si = 1 √ 2k Σc′∈C2|ci + c′ H⊗n − − → ΣjΣc′∈C2(−1)(ci+c

′).j|j

− → Σj(−1)(ci.j)Σc′∈C2(−1)c

′.j|j

By previous result |si H⊗n − − → Σj∈C ⊥

2 (−1)c ′.j|j = |wi

In |si basis the states are in the superposition of codewords of C1 , bit flips can be corrected In |wi basis the states are in the superposition of codewords of C ⊥

2 ,

phase flips can be corrected

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Error Modelling

We model bit flips as an addition of a vector with same length |ci + c′

|e

− − − − →

bit flip |ci + c′ + e

We model phase flips as the multiplication by (−1)m where m is the number of phase flipped qubits and state as ’1’. |ci + c′

|e

− − − − − →

phase flip (−1)(ci+c′).e|ci + c′

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Error Correction

Let e1 and e2 be 2 vectors of hamming weight atmost ’t’ |ci + c′ error − − → (−1)(ci+c′).e1|ci + c′ + e2 where ci ∈ C1 and c

′ ∈ C2

use C1

− − − − → (−1)(ci+c′).e1|ci + c′ Apply hadamard

H⊗n

− − → ΣjΣc′∈C2(−1)(ci+c

′).(j+e1)|j

j iterates over all possible strings, so replacing j with j+e

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− → ΣjΣc′∈C2(−1)(ci+c

′).(j)|j + e1

− → Σj(−1)ci.jΣc′∈C2(−1)c

′.j|j + e1

Each term is non zero only if j ∈ C ⊥

2

− → Σj∈C ⊥

2 (−1)c ′.j|j + e1

W.K.T C ⊥

2 is t-error correcting useC ⊥

2

− − − − → Σj∈C ⊥

2 (−1)c ′.j|j

apply hadamard again

H⊗n

− − → |ci + c′

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