bachelor_thesis/notes/chapters/stabilizer.tex

\section{The Stabilizer Formalism and VOP-Free Graph States}

\subsection{Stabilizers and Stabilizer States}

This chapter discusses the stabilizer formalism that was introduced by Gottesman\cite{gottesman1997}
for quantum error correction but soon proved to be a useful tool to describe a subset of states:
the stabilizer states which can be simulated in polynomial time \cite{gottesman2008}.



\begin{definition}
    \begin{equation}
        p \in P_n \Rightarrow p = \bigotimes\limits_{i=0}^n p_i \\
 \forall i: p_i \in P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\}
    \end{equation}

Where $X = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right)$,
    $Y = \left(\begin{array}{cc} 0 & i \\ -i & 0\end{array}\right)$ and
    $Z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right)$ are the Pauli matrices and 
    $I$ is the identity. 

    $p \in P_n$ is called a multi-local Pauli operator.
\end{definition}

\begin{definition}
    For a group $G$, $g_1, ..., g_n$ are called the generators of $G$ iff $\forall g \in G: g = \prod\limits_{i \in J} g_i$ for a
    subsed $J$ of $I = \{1, ..., n\}$. We write $G = \langle g_i \rangle_i$ if G is generated by the $g_i$. The generators $g_i$ are chosen 
    to be the smallest set of generators of $G$. 
\end{definition}

The notation $\langle g_i \rangle_i \equiv \langle g_i \rangle_{i \in I}$ is used used as a shorthand
notation for $\langle \{g_i\}_{i \in I} \rangle$.

\begin{definition}
    \label{def:stabilizer}
    For a $n$ qbit state $\ket{\psi}$ $\langle S^{(i)} \rangle_{i}$ is called the stabilizer of $\ket{\psi}$ if

    \begin{enumerate}
        \item{$\forall i = 1, ..., n$ $S^{(i)} \in P_n$}
        \item{$\forall i,j = 1, ..., n$ $[S^{(i)}, S^{(j)}] = 0$ $S^{(i)}$ and $S^{(j)}$ commute}
        \item{$\forall i = 1, ..., n$ $S^{(i)}\ket{\psi} = +1 \ket{\psi}$}
    \end{enumerate}
\end{definition}

\begin{lemma}
    For every $\langle S^{(i)} \rangle_i$ fulfilling the first two conditions in definition \ref{def:stabilizer} there exists
    a (up to a global phase) unique state $\ket{\psi}$ fulfilling the third condition. This state is called stabilizer state.
\end{lemma}

\begin{proof}
    All multi-local Pauli operators are hermitian (observables in terms of quantum mechanics), as the $S^{(i)}$
    commute they have a common set of eigenstates. Because each $S^{(i)}$ has eigenvalues $+1, -1$, there 
    exist $2^n$ eigenstates, one state $\ket{\psi}$ with eigenvalue $+1$ for all $S^{(i)}$. As the dimension of $n$ qbits is $2^n$
    the state $\ket{psi}$ is unique up to a global phase.
\end{proof}

One can study the dynamics of stabilizer states using only the stabilizers\cite{nielsen_chuang_2010}. Two important
cases are the unitary transformation of a state and the measurement of a qbit. When applying a unitary gate to a stabilizer
state $\ket{\psi}$ the resulting state will in general be no stabilizer state anymore, however there exists a group of
transformations that map stabilizers to other stabilizers: the Clifford group.

\begin{definition}
    \begin{equation}
        C_n := \{U \in SU(2) | UpU^\dagger \in P_n \forall p \in P_n\}
    \end{equation}
    is called the Clifford group on $n$ qbits.
    $C_1$ is called the local Clifford group.
\end{definition}

The properties of this group will be discussed later, for the time being is existence is enough. 

\begin{lemma}
    Let $\ket{\psi}$ be stabilized by $\langle S^{(i)} \rangle_i$, then $U\ket{\psi}$ is stabilized
    by $\langle US^{(i)}U^\dagger \rangle_i$.
\end{lemma}

\begin{proof}
$$ U\ket{\psi} = US^{(i)}\ket{\psi} = US^{(i)}U^\dagger U\ket{\psi}$$
So $U\ket{\psi}$ is a $+1$ eigenstate of $US^{(i)}U^\dagger$.
\end{proof}

This is an important insight that is used for simulations\cite{gottesman_aaronson2008}, as
updating the $n$ stabilizers that are a tensor product of $n$ Pauli matrices scales with roughly
$\mathcal{O}(n^2)$ instead of $\mathcal{O}(2^n)$ for the state vector approach.

Every qbit can be measured in the $X, Y$ or  $Z$ basis which is a projection using

\begin{equation}
    \frac{I + (-1)^s g_a}{2}
\end{equation}

Where $g_a \in \{Z_a, Y_a, X_a\}$ and $s \in \{0, 1\}$. How the stabilizers change when
measuring a qbit is given by the following lemma:

\begin{lemma}
    \label{lemma:stab_measurement}
    Let $J := \{ S^{(i)} | [g_a, S^{(i)}] \neq 0\}$. Then

    \begin{enumerate}
        \item{If $J = \{\}$, one value is measured with probability $1$ and the stabilizers are unchanged.}
        \item{If $J \neq \{\}$, $1$ and $0$ are measured with probability $\frac{1}{2}$ and after choosing
            a $j \in J$ the new state $\ket{\psi'}$ is stabilized by 
            \begin{equation}
                \langle \{(-1)^s g_a\} \cup \{S^{(i)} S^{(j)} | S^{(i)} \in J \setminus \{S^{(j)}\} \} \cup J^c\rangle
            \end{equation}}
    \end{enumerate}
\end{lemma}

\begin{proof}
    \begin{enumerate}
        \item{As $g_a$ commutes with all stabilizers, $\ket{\psi}$ is an eigenstate of $g_a$,
            so the result of the measurement is deterministic and $\ket{\psi}$ is left unchanged.}
        \item{As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators,
            $S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
            
            \begin{equation}
            \begin{aligned}
                P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\
                        &= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\
                        &= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
                        &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\
                        &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
                        &= P(s=-1)
            \end{aligned}
            \notag
            \end{equation}

            With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$.
            Further for $S^{(i)},S^{(j)} \in J$

            \begin{equation}
            \begin{aligned}
                \frac{I + (-1)^sg_a}{2}\ket{\psi} &= \frac{I + (-1)^sg_a}{2}S^{(j)}S^{(i)} \ket{\psi} \\
                    &= S^{(j)}\frac{I + (-1)^{s + 1}g_a}{2}S^{(i)} \ket{\psi} \\
                    &=  S^{(j)}S^{(i)}\frac{I + (-1)^{s + 2}g_a}{2}\ket{\psi} \\
                    &=  S^{(j)}S^{(i)}\frac{I + (-1)^sg_a}{2}\ket{\psi}
            \end{aligned}
            \notag
            \end{equation}

            the state after measurement is stabilized by $S^{(j)}S^{(i)} i,j \in J$, and by 
            $S^{(i)} \in J^c\setminus\{a\}$. $g_a$ trivially stabilizes $\ket{\psi'}$.
            }
    \end{enumerate}
\end{proof}

\subsection{The Vertex Operator-Free Graph States}

In order to understand some essential transformations of graph states it is necessary
to study the vertex operator-free graph states first, partially because the graph states as used in this paper
were derived from the vertex operator-free graph states.

\begin{definition}
    \label{def:vop_free_g_state}
    A $n$ qbit vertex operator-free graph state $\ket{\overline{G}}$ is associated with a graph $\bar{G} = (V, E)$
    by the $n$ operators

    \begin{equation}
        K^{(i)}_G := X_i \left(\prod\limits_{\{i, j\} \in E} Z_j\right) 
    \end{equation}

    for all $i \in V$ where for some operator $O$ $O_i$ indicates that it acts on the $i$-th qbit.

    $\ket{\overline{G}}$ is the $+1$ eigenstate of all $n$ $K^{(i)}_G$.
\end{definition}

\begin{corrolary}
    All $K^{(i)}_G$ commute and are hermitian. Therefore they have a common set of eigenstates
    (in particular definition \ref{def:vop_free_g_state} is well defined).
    In terms of quantum mechanics $K^{(i)}_G$ are observables.

    Further as $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$ which are
    multi-local Pauli operators, $\{K^{(i)}_G | i \in \{0, ..., n-1\}\}$ is the stabilizer
    of $\ket{\overline{G}}$ and $\ket{\overline{G}}$ is a stabilizer state.
\end{corrolary}

\begin{proof}
    As $X_i$ and $Z_i$ are hermitian their product is hermitian.

    Consider the case $\{i,j\} \notin E$ first:
    \begin{equation}
    \begin{aligned}
        \left[K^{(i)}_G, K^{(j)}_G\right] = \left[X_i \prod\limits_{\{i, n\} \in E} Z_n, X_j \prod\limits_{\{j, m\} \in E} Z_m\right] = 0
    \end{aligned}
    \end{equation}

    As operators acting on different qbits commute. The case $\{i,j\} \in E$ is slightly less trivial:
    \begin{equation}
    \begin{aligned}
        \left[K^{(i)}_G, K^{(j)}_G\right] &= \left[X_i \left(\prod\limits_{\{i, n\} \in E, n \neq j} Z_n\right) Z_j, X_j \left(\prod\limits_{\{j, m\} \in E, m \neq i} Z_m\right) Z_i\right] \\
        &= \left[X_i Z_j \prod\limits_n Z_n, X_j Z_i \prod\limits_m Z_m\right]\\
        &= \left(X_i Z_j X_j Z_i - X_j Z_i X_i Z_j\right)  \prod\limits_n Z_n \prod\limits_m Z_m \\
        &= \left(Z_j X_j X_i Z_i - X_j Z_j Z_i X_i\right)  \prod\limits_n Z_n \prod\limits_m Z_m \\
        &= \left((-1)^2X_j Z_j Z_i X_i - X_j Z_j Z_i X_i\right)  \prod\limits_n Z_n \prod\limits_m Z_m \\
        &= 0
    \end{aligned}
    \end{equation}

    as $X$, $Z$ anticommute.
\end{proof}

\begin{lemma}
    \begin{equation}
        \ket{\overline{G}} = \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right) \left(\prod\limits_{l \in V} H_l\right) \ket{0}
    \end{equation}

    In particular definitions \ref{def:vop_free_g_state} and \ref{def:graph_state} are consistent, when there are no 
    vertex operators on the graph state $\ket{G}$.
\end{lemma}
\begin{proof}
    Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$ $X_i \ket{+} = +1 \ket{+}$.
    Set $\ket{\tilde{G}} := \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right)\ket{+}$.

    \begin{equation}
        \begin{aligned}
            K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,j\} \in E} CZ_{l,j} \right) \ket{+} \\
            & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\
            & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) X_i \ket{+} \\
            & = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{2\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\
			& = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\	
		    & = +1 \ket{\tilde{G}}	
        \end{aligned}
    \end{equation}

    as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$.
\end{proof}

\begin{definition}
    Let $\bar{G} = (V, E)$ be a graph as in definition \ref{def:vop_free_g_state}.
    For a vertex $i \in V$ $n_i := \{j \in V | \{i,j\} \in E\}$ is called the neighbourhood
    of $i$.
\end{definition}

\begin{lemma}
    Let $\bar{G}$, $\ket{\bar{G}}$, $K_G^{(i)}$ be as in definition \ref{def:vop_free_g_state} and for
    a vertex $a \in V$ set

    \begin{equation}
        M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}
    \end{equation}

    Then the graph $\bar{G}'$ associated with $\ket{\bar{G}'} = M_a\ket{\bar{G}}$ is changed according to 
    the following equations:

    \begin{equation}
        \begin{aligned}
            n_a' &= n_a \\
            n_j' &= n_j, \hbox{ if } j \notin n_a\\
            n_j' &= (n_j \cup n_a) \setminus (n_j \cap n_a), \hbox{ if } j \in n_a
        \end{aligned}
    \end{equation}

    I.e. the neighbourhood of $a$ is toggled.
\end{lemma}

\begin{proof}
    $\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger \rangle_i$, so it is sufficient
    to study how the $ K_G^{(i)}$ change under $M_a$.
    At first note that $[K_G^{(a)}, M_a] = 0$ and $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$,
    so the first two equations follow trivially. For $j \in n_a$ set 

    \begin{equation}
        \begin{aligned}
            S^{(j)} &= M_a K_G^{(j)} M_a^\dagger \\
                &= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right)
                                 \sqrt{iZ_j}  K_G^{(j)} \sqrt{iZ_j}^\dagger 
                                 \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right)
                                 \sqrt{-iX_a}^\dagger \\
                &= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right)\sqrt{iZ_j}
                                X_j \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) Z_a
                                \sqrt{iZ_j}^\dagger
                                \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right)
                                \sqrt{-iX_a}^\dagger \\
                &= \sqrt{iZ_j} X_j\sqrt{iZ_j}^\dagger\left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) 
                                \sqrt{-iX_a} Z_a \sqrt{-iX_a}^\dagger \\
                &= (-1)^2 Y_j Y_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)\\
                &= (-1)i^2 Z_j X_a X_j Z_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)
        \end{aligned}
    \end{equation}

    One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenvalue
    of the new $K_{G'}^{(i)}$. It is clear that $\forall j \notin n_a$  $K_{G'}^{(j)} = K_G^{(j)}$.
    To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a = \{j\} \cup I$ and $n_j = \{a\} \cup J$.
    Then

    \begin{equation}
        \begin{aligned}
            S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in J} Z_l\\
                    &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in J} Z_l\right)
                                        \left(\prod\limits_{l \in I}Z_l\right)
                                        \left(\prod\limits_{l \in I}Z_l\right) \\
                    &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in ((I\cup J) \setminus (I\cap J))} Z_L\right)
                                        \left(\prod\limits_{l \in I}Z_l\right) \\
                    &= K_{G'}^{(a)} K_{G'}^{(j)} \\
                    &= K_{G}^{(a)} K_{G'}^{(j)}
        \end{aligned}
    \end{equation}

    Using this ine can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$:

    \begin{equation}
        \ket{\bar{G}'} = S^{(j)}\ket{\bar{G}'} = K_{G}^{(a)} K_{G'}^{(j)}\ket{\bar{G}'} 
                        = K_{G'}^{(j)}K_{G}^{(a)}\ket{\bar{G}'} = K_{G'}^{(j)}\ket{\bar{G}'}
    \end{equation}

    Because $\{K_G^{(i)} | i \notin n_a\} \cup \{S^{(i)} | i\in n_a\}$ and 
    $\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}$ are both $n$ commuting
    multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$
    and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$ 
    $\langle\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}\rangle$
    are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given
    in the third equation.

\end{proof}

These insights can be used to understand how measurement works on the vop-free graph state \cite{nielsen_chuang_2010}:
Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a Pauli operator $g \in \{Z_a, Y_a, X_a\}$ 
that is to be measured.
Recalling lemma \ref{lemma:stab_measurement} the following relations follow immideately:

\begin{enumerate}
    \item{For $g = X_a$ if $a$ is an isolated qbit $+1$ is measured with probability $1$
        and the state $\ket{\bar{G}}$ is unchanged.}
    \item{For $g = X_a$ and $a$ is non-isolated, choose $b \in n_a$ and the new stabilizers are
        \begin{equation}
            \langle \{(-1)^sX_a\} 
                \cup \{K_G^{(b)}K_G^{(i)} | i \in n_a \setminus \{b\}\} 
                \cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
        \end{equation}}
    \item{For $g = Z_a$ the new stabilizers are
        \begin{equation}
            \langle \{(-1)^sZ_a\} 
                \cup \{K_G^{(i)} | i \in V \setminus n_a\} \rangle
        \end{equation}}
    \item{For $g = Y_a$ the new stabilizers are
        \begin{equation}
            \langle \{(-1)^sY_a\} 
                \cup \{K_G^{(a)}K_G^{(i)} | i \in n_a\} 
                \cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle
        \end{equation}}
\end{enumerate}

The states after the measurement are in general no vop-free graph states anymore,
the following discussion will allow to construct new vop-free graph states and
Clifford transformations from the vop-free graph state to the resulting state.
In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged.

\begin{lemma}
    \label{lemma:Z_measurement}
    \begin{enumerate}
        \item{For a result $+Z_a$ the new state is
                $\ket{+_Z}_a \otimes \ket{\bar{G}'}$
                with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for
                $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$.
            }
        \item{For a result $-Z_a$ the new state is $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$
            with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$
            and $U = \prod\limits_{i \in n_a}Z_i$.}
    \end{enumerate}
\end{lemma}
\begin{proof}
    \begin{enumerate}
        \item{
                It is trivial that $Z_a$ and $K_G^{(i)} i \neq a$ stabilize $\ket{+_Z}_a \otimes \ket{\bar{G}'}$.
                $Z_aK_G^{(i)}$ for $i \in n_a$ do not act on $a$, so $\ket{\bar{G}'}$ is well-defined.
            }
        \item{
                The state $\ket{-_Z}_a \otimes \ket{\bar{G}'}$ is the $-1$ eigenstate of $K_G^{(i)}$ for all
                $i \in n_a$. This can be corrected by transforming $K_G^{(i)}$ to $-K_G^{(i)} = Z_i K_G^{(i)} Z_i^\dagger$
                which stabilizes the state $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$.
            }
    \end{enumerate}

\end{proof}

\begin{lemma}
    When $Y_a$ is measured with a result $s \in \{0, 1\}$ the state after the measurement is

    \begin{equation}
        \ket{(-1)^s_Y}_a \otimes U_{Y,s} \ket{\bar{G}'}
    \end{equation}
    \begin{equation}
        U_{Y,s} = \prod\limits_{l \in n_a} \sqrt{(-1)^s iZ_l}
    \end{equation}

    Where $\ket{\bar{G}'}$ is associated with the graph $G' = (V \setminus \{a\}, E')$
    with $E'$ being changed according to $\forall i \in n_a$ $n_i' = (n_i \cup n_a) \setminus (n_i \cap n_a) \setminus \{a\}$.
\end{lemma}

\begin{proof}
    It is known that $(-1)^s Y_a$ has to stabilize the state after measurement,
    further $\forall j \in n_a$ $S^{(j)} = K_G^{(a)} K_G^{(j)}$, are the stabilizers of the new
    state around the measured qbit and the stabilizers $K_G^{(j)}, j \notin n_a$ are unchanged.
    Let $j \in n_a$ and $n_a =: \{j\} \cup I$, $n_j =: \{a\} \cup J$.

    \begin{equation}
        \begin{aligned}
            S^{(i)} &= K_G^{(a)} K_G^{(j)} \\
            &= X_a \left(\prod\limits_{l \in n_a} Z_l\right)X_j\left(\prod\limits_{l\in n_j} Z_l\right) \\
            &= X_a Z_a Z_j X_j \left(\prod\limits_{l \in I} Z_l\right)\left(\prod\limits_{l \in J} Z_l\right)\\
            &= X_a Z_a  Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)} Z_l\right) \\
            &= iY_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J) \setminus \{a\}} Z_l\right) \\
            &= (-1)^si S^{(a)} Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)\setminus\{a\}} Z_l\right) \\
            &= S^{(a)} \sqrt{(-1)^siZ_j} K_{G'}^{(j)} \sqrt{(-1)^siZ_j}^\dagger
        \end{aligned}
    \end{equation}
     with $G'$ as above, the rest of the argument is analogous to lemma \ref{lemma:Z_measurement}.
\end{proof}