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I'm reading the paper which introduces a method to characterize the Pauli noise channel. In eq(5) the authors state that the stabilizer state can be written as the following form $$ \left|\phi_e^{\mathrm{S}}\right\rangle\left\langle\phi_e^{\mathrm{S}}\right|=\frac{1}{2^{n-k}} \sum_{s \in \mathrm{S}}(-1)^{\langle s, e\rangle} P_s \tag{1} $$ where $\mathrm{S}$ is a stabilizer group on $n-k$ qubits, i.e. consists of $2^{n-k}$ commuting Pauli operators. $s$ here is the bit string used to describe a Pauli operator, namely(write it as $a$ for a general bit string) $a=a_{x, 1} a_{z, 1} a_{x, 2} a_{z, 2} \cdots a_{x, n} a_{z, n}$ used to stand for $P_a=\otimes_{k=1}^n i^{a_{x, k} a_{z, k}} X^{a_{x, k}} Z^{a_{z, k}}$ and the inner product between two bit strings is in a skew inner product style, namely $$ \langle a, b\rangle=\sum_{k=1}^n\left(a_{x, k} b_{z, k}+a_{z, k} b_{x, k}\right) \quad \bmod 2 $$

My problem is, what does $e$ mean in eq(1)? The author mentioned that

$e \in \mathrm{S}^{\perp}:=\mathbb{Z}_2^{2(n-k)} / \mathrm{S}$ known as the error syndrome.

But I can't understand this statement about the meaning of $e$. Instead, I try to understand eq(1) by the example of a maximally entangled state of two qubits which can be written as $\frac{1}{4}\left( I\otimes I+X\otimes X+Z\otimes Z-Y\otimes Y \right)$. The corresponding bit string for $X\otimes X$ is $1010$ and for $Z\otimes Z$ is $0101$ and for $Y\otimes Y$ is $1111$ and for $I\otimes I$ is $0000$. However, I can't find any bit string $e$ to satisfy eq(1). What does $e$ mean here?

Edit I may wrongly understand the eq(1), I tried another example while I still get a wrong result. Suppose $n-k=1$ and $\mathrm{S}=\left\{ I,X \right\}$ now. Then we have $e$ to be element of $\left\{ Y,Z \right\} $. We can write $X,Y,Z,I$ as bit string as $\left( 1,0 \right) ,\left( 1,1 \right) ,\left( 0,1 \right) ,\left( 0,0 \right) $ repectively. Hence we have the following result $$|\phi _{\left( 1,1 \right)}^{\mathrm{S}}\rangle \langle \phi _{\left( 1,1 \right)}^{\mathrm{S}}|=\frac{1}{2}\left( -1 \right) ^{\langle \left( 1,1 \right) ,\left( 0,0 \right) \rangle}P_{\left( 0,0 \right)}+\frac{1}{2}\left( -1 \right) ^{\langle \left( 1,1 \right) ,\left( 1,0 \right) \rangle}P_{\left( 1,0 \right)}=\frac{1}{2}I-\frac{1}{2}X \\ |\phi _{\left( 0,1 \right)}^{\mathrm{S}}\rangle \langle \phi _{\left( 0,1 \right)}^{\mathrm{S}}|=\frac{1}{2}\left( -1 \right) ^{\langle \left( 0,1 \right) ,\left( 0,0 \right) \rangle}P_{\left( 0,0 \right)}+\frac{1}{2}\left( -1 \right) ^{\langle \left( 0,1 \right) ,\left( 1,0 \right) \rangle}P_{\left( 1,0 \right)}=\frac{1}{2}I-\frac{1}{2}X$$ But I finally get the result $|\phi _{\left( 1,1 \right)}^{\mathrm{S}}\rangle \langle \phi _{\left( 1,1 \right)}^{\mathrm{S}}|=|\phi _{\left( 0,1 \right)}^{\mathrm{S}}\rangle \langle \phi _{\left( 0,1 \right)}^{\mathrm{S}}|$ which is not right since the paper require $|\phi _{e}^{\mathrm{S}}\rangle \langle \phi _{e}^{\mathrm{S}}|$ to be an orthogonal basis for the $n-k$ qubit subsystem.

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  • $\begingroup$ You are misreading Eq (5). It says that given any $e$ (outside $S$) of your choice, the definition of $|\phi_e^S\rangle \langle \phi_e^S|$ is the RHS of Eq (5). $\endgroup$ Commented May 26, 2023 at 3:55
  • $\begingroup$ @AbdullahKhalid Thank you for your comment. Do you mean the notation $/$ in $\mathbb{Z}_2^{2(n-k)} / \mathrm{S}$ stands for deleting the element of $\mathrm{S}$ from the $\mathbb{Z}_2^{2(n-k)}$? $\endgroup$
    – Sherlock
    Commented May 26, 2023 at 5:03
  • $\begingroup$ That is correct. This is set theory notation. $\endgroup$ Commented May 26, 2023 at 17:13
  • $\begingroup$ @AbdullahKhalid But in my new edit, this still seems not right.. $\endgroup$
    – Sherlock
    Commented May 27, 2023 at 0:38
  • $\begingroup$ Your example calculations are correct. I really can't say why the orthogonality result fails. This is not the standard definition of stabilizer states, so I am not sure where the orthogonality claim comes from. $\endgroup$ Commented May 27, 2023 at 21:56

1 Answer 1

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Stabilizer states are states that are eigenstates of a group of operators $S$, known as the stabilizer group. The error syndrome in general refers to some information that can be used to deduce the error. In a classical error correcting code such as Hamming code, syndromes are given by parity check bits and the error is deduced from these. Here, the phrase 'error syndrome' is misleading; $e$ ($P_e$) is the error on the state and the corresponding syndrome would be the eigenvalues of the (generators of the) stabilizer group, and is used here to generate a set of eigenstates of the stabilizers by displacing/causing errors on the stabilizer state $|\phi_0^S\rangle\langle\phi_0^S|$.

In this work, they have utilized the density representation of stabilizer states. The eigenvalue of densities can be found by $a = Tr(A\rho)$ and $a =\pm 1$ for $A\in S$ from the stabilizer group. The notation is confusing, so for the rest of the answer I'm adopting the following notation $|\phi_e^S\rangle\langle\phi_e^S| => \rho_e^S$.

Since $P_s\cdot P_{s'} = P_{s''},\text{ for } s,s', s'' \in S$, the state $\rho_0^S = \frac{1}{2^{n-k}}\sum_{s\in S} P_s$ is invariant when multiplied by $P_s, s\in S$.

\begin{equation} \Rightarrow P_s\rho_0^SP_s = P_s\rho_0^S = \rho_0^S \end{equation}

Thus $Tr(P_s\rho_0^S) = 1, \forall s\in S$.

Notice that the coefficient of $P_s$ in the expression of $\rho_e^S$ given is $\pm 1$ depending on whether they commute/anticommute with $P_e$. This suggests that one can obtain this as $\rho_e^S = P_e\rho_0^S P_e = \frac{1}{2^{n-k}}\sum_{s\in S} P_eP_sP_e = \frac{1}{2^{n-k}}\sum_{s\in S} (-1)^{<s, e>}P_s$.

Now lets check the eigenvalue with respect to the stabilizers \begin{equation} Tr(P_s\rho_e^S) = Tr(P_sP_e\rho_0^SP_e) = Tr((-1)^{<s, e>}P_eP_e\rho_0^SP_e) = (-1)^{<s, e>}Tr(\rho_e^S) = (-1)^{<s, e>} \end{equation}

Based on the eigenvalues, we can determine the 'error' $e$. But $<s, e> = <s, e + s'> \forall s'\in S$ and $\rho_{s+e}^S = \rho_e^S$, i.e., they are equivalent upto an addition of $s' \in S$. $e\in \mathbb{Z}_2^{2(n-k)}/S$ (it's from the quotient set where the equivalence relation is the inner product with the stabilizers, or belong to the same coset of $S$ in $\mathbb{Z}_2^{(n-k)}$).

In your example for the Stabilizer states of the Bell state, the stabilizers are given by $S = \{I\otimes I, X\otimes X, Z\otimes Z\}$. Try $e = (1000)$ or $(0010)$, you will obtain distinct states.

In your edited tests/calculations for $e_1 = (1, 1), e_2 = (0, 1)$ are correct, they belong to the same equivalence class with respect to your chosen Stabilizer group as $(1, 1) = (0, 1) + (1, 0),\text{ where } (1, 0) \equiv X \in S$, i.e., they will give the same stabilizer states.

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  • $\begingroup$ Thank you for your answer! Do you know why in the density representation of stabilizer states, the maximally entangled state can be written as $\frac{1}{4^n}\sum_{u\in \mathbb{Z} _{2}^{2n}}{P_u\otimes P_{u}^{T}}$ where the dimension of the maximally entangled state is $4^n$? $\endgroup$
    – Sherlock
    Commented May 31, 2023 at 7:11
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    $\begingroup$ @Sherlock consider the two qubit Bell state circuit, the initial state in density form is of the form $|00\rangle\langle 00| = (1/4)(I + Z)\otimes (I + Z)$ and when you apply the H[0], CNOT[0-> 1] (for densities, $\rho \Rightarrow U\rho U^\dagger$) gates you'll obtain that sum of Pauli operators. I thought about it and unfortunately I do not have any intuition as to why the form you specified is the case for maximally entangled state apart from the fact that it works. The transpose reminds me of Schmidt decomposition, maybe there's a clue there? $\endgroup$ Commented Jun 1, 2023 at 22:12

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