Entanglement-assisted stabilizer formalism explained

\Pi ^{n} over n qubits.The sender can make clever use of her sharedebits so that the global stabilizer is Abelian and thus forms a validquantum error-correcting code.

Definition

l{S}\subset\Pi ^{n} of size n - k =2 c + s .Application of the fundamental theorem of symplectic geometry (Lemma 1 in the first external reference)states that there exists a minimal set of independent generators\left\{ \bar{Z} _{1} ,\ldots,\bar{Z} _{s+c} ,\bar{X} _{s+1} ,\ldots,\bar{X} _{s+c} \right\} for l{S} with the following commutation relations:\left[ \bar{Z} _{i} ,\bar{Z} _{j} \right] = 0 \forall i,j, \left[ \bar{X} _{i} ,\bar{X} _{j} \right] = 0 \forall i,j, \left[ \bar{X} _{i} ,\bar{Z} _{j} \right] = 0 \forall i ≠ j, \left\{ \bar{X} _{i} ,\bar{Z} _{i} \right\} = 0 \forall i. The decomposition of l{S} into the above minimal generating setdetermines that the code requires s ancilla qubits and c ebits. The coderequires an ebit for every anticommuting pair in the minimal generating set.The simple reason for this requirement is that an ebit is a simultaneous+1 -eigenstate of the Pauli operators \left\{ XX,ZZ\right\} . The second qubitin the ebit transforms the anticommuting pair \left\{ X,Z\right\} into acommuting pair \left\{ XX,ZZ\right\} . The above decomposition alsominimizes the number of ebits required for the code---it is an optimal decomposition.l{S} into two subgroups: theisotropic subgroup l{S} _{I} and the entanglement subgroupl{S} _{E} . The isotropic subgroup l{S} _{I} is a commutingsubgroup of l{S} and thus corresponds to ancillaqubits:l{S} _{I} = \left\{ \bar{Z} _{1} ,\ldots,\bar{Z} _{s} \right\} .

The elements of the entanglement subgroup

l{S} _{E} come inanticommuting pairs and thus correspond to ebits:l{S} _{E} = \left\{ \bar{Z} _{s+1} ,\ldots,\bar{Z} _{s+c} ,\bar{X} _{s+1} ,\ldots,\bar{X} _{s+c} \right\} .

Entanglement-assisted stabilizer code error correction conditions

The two subgroups

l{S} _{I} and l{S} _{E} play a role in theerror-correcting conditions for the entanglement-assisted stabilizerformalism. An entanglement-assisted code corrects errors in a setl{E}\subset\Pi ^{n} if for all E _{1} ,E _{2} \in l{E} ,E ^{\dagger} _{1} E _{2} \in l{S} _{I} \cup\left( \Pi ^{n} - l{Z}\left(\left\langle l{S} _{I} , l{S} _{E} \right\rangle \right) \right) .

Operation

The operation of an entanglement-assisted code is as follows. The senderperforms an encoding unitary on her unprotected qubits, ancilla qubits, andher half of the ebits. The unencoded state is a simultaneous +1-eigenstate ofthe following Pauli operators:

\left\{ Z _{1} ,\ldots,Z _{s} , Z _{s+1} |Z _{1} ,\ldots,Z _{s+c} |Z _{c} , X _{s+1} |X _{1} ,\ldots,X _{s+c} |X _{c} \right\} .

The Pauli operators to the right of the vertical bars indicate the receiver's halfof the shared ebits. The encoding unitary transforms the unencoded Pauli operatorsto the following encoded Pauli operators:

\left\{ \bar{Z} _{1} ,\ldots,\bar{Z} _{s} , \bar{Z} _{s+1} |Z _{1} ,\ldots,\bar{Z} _{s+c} |Z _{c} , \bar{X} _{s+1} |X _{1} ,\ldots,\bar{X} _{s+c} |X _{c} \right\} . The sender transmits all of her qubits over the noisy quantum channel. Thereceiver then possesses the transmitted qubits and his half of the ebits. Hemeasures the above encoded operators to diagnose the error. The last step isto correct the error.

Rate of an entanglement-assisted code

We can interpret the rate of an entanglement-assisted codein three different ways (Wilde and Brun 2007b).Suppose that an entanglement-assisted quantum code encodes

k informationqubits into n physical qubits with the help of c ebits.

• The entanglement-assisted rate assumes that entanglement shared between sender and receiver is free. Bennett et al. make this assumption when deriving the entanglement assisted capacity of a quantum channel for sending quantum information. The entanglement-assisted rate is

k/n for a code with the above parameters.

• The trade-off rate assumes that entanglement is not free and a rate pair determines performance. The first number in the pair is the number of noiseless qubits generated per channel use, and the second number in the pair is the number of ebits consumed per channel use. The rate pair is

\left( k/n,c/n\right) for a code with the above parameters. Quantum information theorists have computed asymptotic trade-off curves that bound the rate region in which achievable rate pairs lie. The construction for an entanglement-assisted quantum block code minimizes the number c of ebits given a fixed number k and n of respective information qubits and physical qubits.

• The catalytic rate assumes that bits of entanglement are built up at the expense of transmitted qubits. A noiseless quantum channel or the encoded use of noisy quantum channel are two different ways to build up entanglement between a sender and receiver. The catalytic rate of an

\left[ n,k;c\right] code is \left( k - c\right) /n .

Which interpretation is most reasonable depends on the context in which we usethe code. In any case, the parameters

n , k , and c ultimately governperformance, regardless of which definition of the rate we use to interpretthat performance.

Example of an entanglement-assisted code

We present an example of an entanglement-assisted codethat corrects an arbitrary single-qubit error (Brun et al. 2006). Supposethe sender wants to use the quantum error-correcting properties of thefollowing nonabelian subgroup of

\Pi ^{4} :\begin{array} [c]{cccc} Z & X & Z & I\\ Z & Z & I & Z\\ X & Y & X & I\\ X & X & I & X \end{array} The first two generators anticommute. We obtain a modified third generator bymultiplying the third generator by the second. We then multiply the lastgenerator by the first, second, and modified third generators. Theerror-correcting properties of the generators are invariant under theseoperations. The modified generators are as follows:\begin{array} [c]{cccccc} g _{1} & = & Z & X & Z & I\\ g _{2} & = & Z & Z & I & Z\\ g _{3} & = & Y & X & X & Z\\ g _{4} & = & Z & Y & Y & X \end{array} The above set of generators have the commutation relations given by thefundamental theorem of symplectic geometry:\left\{ g _{1} ,g _{2} \right\} = \left[ g _{1} ,g _{3} \right] = \left[ g _{1} ,g _{4} \right] = \left[ g _{2} ,g _{3} \right] = \left[ g _{2} ,g _{4} \right] = \left[ g _{3} ,g _{4} \right] =0 . The above set of generators is unitarily equivalent to the following canonicalgenerators:\begin{array} [c]{cccc} X & I & I & I\\ Z & I & I & I\\ I & Z & I & I\\ I & I & Z & I \end{array} We can add one ebit to resolve the anticommutativity of the first twogenerators and obtain the canonical stabilizer:\begin{array} [c]{c} X\\ Z\\ I\\ I \end{array} \left\vert \begin{array} [c]{cccc} X & I & I & I\\ Z & I & I & I\\ I & Z & I & I\\ I & I & Z & I \end{array} \right. The receiver Bob possesses the qubit on the left and the sender Alicepossesses the four qubits on the right. The following state is an eigenstateof the above stabilizer\left\vert \Phi ^{+} \right\rangle ^{BA} \left\vert 00 \right\rangle ^{A} \left\vert \psi\right\rangle ^{A} . where \left\vert \psi\right\rangle ^{A} is a qubit that the sender wants toencode. The encoding unitary then rotates the canonical stabilizer to the following set of globally commutinggenerators:\begin{array} [c]{c} X\\ Z\\ I\\ I \end{array} \left\vert \begin{array} [c]{cccc} Z & X & Z & I\\ Z & Z & I & Z\\ Y & X & X & Z\\ Z & Y & Y & X \end{array} \right. The receiver measures the above generators upon receipt of all qubits todetect and correct errors.

Encoding algorithm

We continue with the previous example. Wedetail an algorithm for determining an encoding circuit and the optimal numberof ebits for the entanglement-assisted code---this algorithm first appeared in the appendix of (Wilde and Brun 2007a) and later in the appendix of (Shaw et al. 2008). The operators inthe above example have the following representation as a binarymatrix (See the stabilizer code article):

H = \left[ \left. \begin{array} [c]{cccc} 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array} \right] . Call the matrix to the left of the vertical bar the "Z matrix" and the matrix to the right of the vertical bar the"X matrix."

The algorithm consists of row and column operations on the above matrix. Rowoperations do not affect the error-correcting properties of the code but arecrucial for arriving at the optimal decomposition from the fundamental theoremof symplectic geometry. The operations available for manipulating columns ofthe above matrix are Clifford operations. Cliffordoperations preserve the Pauli group

\Pi ^{n} under conjugation. TheCNOT gate, the Hadamard gate, and the Phase gate generate the Clifford group.A CNOT gate from qubit i to qubit j adds column i to column j in theX matrix and adds column j to column i in the Z matrix. A Hadamardgate on qubit i swaps column i in the Z matrix with column i in theX matrix and vice versa. A phase gate on qubit i adds column i in theX matrix to column i in the Z matrix. Three CNOT gates implement aqubit swap operation. The effect of a swap on qubitsi and j is to swap columns i and j in both the X and Z matrix.

The algorithm begins by computing the symplectic product between the first rowand all other rows. We emphasize that the symplectic product here is thestandard symplectic product. Leave the matrix as it is if the first row is notsymplectically orthogonal to the second row or if the first row issymplectically orthogonal to all other rows. Otherwise, swap the second rowwith the first available row that is not symplectically orthogonal to thefirst row. In our example, the first row is not symplectically orthogonal tothe second so we leave all rows as they are.

Arrange the first row so that the top left entry in the

X matrix is one. ACNOT, swap, Hadamard, or combinations of these operations can achieve thisresult. We can have this result in our example by swapping qubits one and two.The matrix becomes\left[ \left. \begin{array} [c]{cccc} 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right\vert \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array} \right] . Perform CNOTs to clear the entries in the X matrix in the top row to theright of the leftmost entry. These entries are already zero in this example sowe need not do anything. Proceed to the clear the entries in the first row ofthe Z matrix. Perform a phase gate to clear the leftmost entry in the firstrow of the Z matrix if it is equal to one. It is equal to zero in this caseso we need not do anything. We then use Hadamards and CNOTs to clear the otherentries in the first row of the Z matrix.

We perform the above operations for our example. Perform a Hadamard on qubitstwo and three. The matrix becomes

\left[ \left. \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right\vert \begin{array} [c]{cccc} 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{array} \right] . Perform a CNOT from qubit one to qubit two and from qubit one to qubit three.The matrix becomes\left[ \left. \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 \end{array} \right\vert \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 \end{array} \right] . The first row is complete. We now proceed to clear the entries in the secondrow. Perform a Hadamard on qubits one and four. The matrix becomes\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 0 \end{array} \right] . Perform a CNOT from qubit one to qubit two and from qubit one to qubit four.The matrix becomes\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 1 \end{array} \right] . The first two rows are now complete. They need one ebit to compensate fortheir anticommutativity or their nonorthogonality with respect to thesymplectic product.

Now we perform a "Gram-Schmidtorthogonalization" with respect to the symplectic product.Add row one to any other row that has one as the leftmost entry in its

Z matrix. Add row two to any other row that has one as the leftmost entry in itsX matrix. For our example, we add row one to row four and we add row two torows three and four. The matrix becomes\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 \end{array} \right] . The first two rows are now symplectically orthogonal to all other rows per thefundamental theorem of symplectic geometry.We proceed with the same algorithm on the next two rows. The next two rows aresymplectically orthogonal to each other so we can deal with them individually.Perform a Hadamard on qubit two. The matrix becomes\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \end{array} \right] . Perform a CNOT from qubit two to qubit three and from qubit two to qubitfour. The matrix becomes\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right] . Perform a phase gate on qubit two:\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array} \right] . Perform a Hadamard on qubit three followed by a CNOT from qubit two to qubitthree:\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \end{array} \right] . Add row three to row four and perform a Hadamard on qubit two:\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right] . Perform a Hadamard on qubit four followed by a CNOT from qubit three to qubitfour. End by performing a Hadamard on qubit three:\left[ \left. \begin{array} [c]{cccc} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{array} \right\vert \begin{array} [c]{cccc} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right] . The above matrix now corresponds to the canonical Pauli operators. Adding one half of an ebit to the receiver's sidegives the canonical stabilizer whosesimultaneous +1-eigenstate is the above state.The above operations in reverse ordertake the canonical stabilizer to the encodedstabilizer.

References

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