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some changes proposed by Andreas

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Daniel Knüttel 2020-03-05 10:48:13 +01:00
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54
thesis/chapters/quantum_computing.tex
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@ -9,7 +9,7 @@
$ \{\ket{\uparrow} \equiv \ket{1}, \ket{\downarrow} \equiv \ket{0}\} $ $ \{\ket{\uparrow} \equiv \ket{1}, \ket{\downarrow} \equiv \ket{0}\} $
with $\braket{\uparrow}{\downarrow} = 0$. In the following this basis will be called with $\braket{\uparrow}{\downarrow} = 0$. In the following this basis will be called
the $Z$ basis in analogy to the conventions used in spin systems ($\sigma_Z$). For some computations the $Z$ basis in analogy to the conventions used in spin systems ($\sigma_Z$). For some computations
it can be useful to have component vectors, $\ket{\uparrow} \equiv \left(\begin{array}{c} 0 \\ 1\end{array} \right)$ and it can be useful to have component vectors; $\ket{\uparrow} \equiv \left(\begin{array}{c} 0 \\ 1\end{array} \right)$ and
$\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ $\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$
are used in these cases. are used in these cases.
\end{definition} \end{definition}
@ -17,7 +17,7 @@
A gate acting on a qbit is a unitary operator $G \in SU(2)$. One can show that A gate acting on a qbit is a unitary operator $G \in SU(2)$. One can show that
$\forall G \in SU(2)$ $G$ can be approximated arbitrarily good as a product of unitary generator matrices $\forall G \in SU(2)$ $G$ can be approximated arbitrarily good as a product of unitary generator matrices
\cite[Chapter 4.3]{kaye_ea2007}\cite[Chapter 2]{marquezino_ea_2019}; \cite[Chapter 4.3]{kaye_ea2007}\cite[Chapter 2]{marquezino_ea_2019};
common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with common choices for the generators are $ X, H, R_{\phi}$ or $Z, H, R_{\phi}$ with
\label{ref:singleqbitgates} \label{ref:singleqbitgates}
\begin{equation} \begin{equation}
@ -26,19 +26,19 @@ common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ wit
& Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) \\ & Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) \\
& H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) \\ & H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) \\
& R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)\\ & R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)\\
& I := \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right) \\ & I := \left(\begin{array}{cc} 1 & 0 \\ 0 & 1\end{array}\right). \\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
Note that $X = HZH$ and $Z = R_{\pi}$, so the set of $H, R_\phi$ is sufficient. Note that $X = HZH$ and $Z = R_{\pi}$, so the set of $H, R_\phi$ is sufficient.
Further note that the basis vectors are chosen s.t. $Z\ket{0} = +\ket{0}$ and $Z\ket{1} = -\ket{1}$; Further note that the basis vectors are chosen s.t. $Z\ket{0} = +\ket{0}$ and $Z\ket{1} = -\ket{1}$;
transforming to the other Pauli eigenstates is done using $H$ and $SH$: transforming to the other Pauli eigenstates is done using $H$ and $SH$
\begin{equation} \begin{equation}
S := R_{\frac{\pi}{2}} = \left(\begin{array}{cc} 1 & 0 \\ 0 & i\end{array}\right) S := R_{\frac{\pi}{2}} = \left(\begin{array}{cc} 1 & 0 \\ 0 & i\end{array}\right)
\end{equation} \end{equation}
\begin{equation} \begin{equation}
S H Z H^\dagger S^\dagger = S X S^\dagger = Y S H Z H^\dagger S^\dagger = S X S^\dagger = Y.
\end{equation} \end{equation}
The following states are the $\pm 1$ eigenstates of the $X,Y,Z$ operators The following states are the $\pm 1$ eigenstates of the $X,Y,Z$ operators
@ -59,12 +59,12 @@ and will be used in some calculations later:
\label{ref:many_qbits} \label{ref:many_qbits}
\begin{postulate} \begin{postulate}
A $n$-qbit quantum mechanical state is the tensor product \cite[Definition 14.3]{wuest1995} of the $n$ one-qbit A $n$-qbit quantum mechanical state is the tensor product of the $n$ one-qbit
states. states \cite[Postulate 4]{nielsen_chuang_2010}.
\end{postulate} \end{postulate}
Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$ Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$
be the basis of the one-qbit systems. Then two-qbit basis states are: be the basis of the one-qbit systems. Then two-qbit basis states are
\begin{equation} \begin{equation}
\ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right) \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)
@ -76,21 +76,21 @@ be the basis of the one-qbit systems. Then two-qbit basis states are:
\ket{2} := \ket{0b10} := \ket{1}_s \otimes \ket{0}_s := \left(\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right) \ket{2} := \ket{0b10} := \ket{1}_s \otimes \ket{0}_s := \left(\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right)
\end{equation} \end{equation}
\begin{equation} \begin{equation}
\ket{3} := \ket{0b11} := \ket{1}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right) \ket{3} := \ket{0b11} := \ket{1}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right).
\end{equation} \end{equation}
The $n$-qbit basis states can be constructed in a similar manner. The $n$-qbit basis states can be constructed in a similar manner.
A general $n$-qbit state can then be written as a superposition of the A general $n$-qbit state is now a superposition of the
integer states: integer states
\begin{equation} \begin{equation}
\label{eq:ci} \label{eq:ci}
\ket{\psi} = \sum\limits_{i = 0}^{2^n - 1} c_i \ket{i} \ket{\psi} = \sum\limits_{i = 0}^{2^n - 1} c_i \ket{i} .
\end{equation} \end{equation}
With the normation condition: with the normation condition
\begin{equation} \begin{equation}
\sum\limits_{i = 0}^{2^n - 1} |c_i|^2 = 1 \sum\limits_{i = 0}^{2^n - 1} |c_i|^2 = 1.
\end{equation} \end{equation}
The states $\ket{i}$ for $i = 0, ..., 2^{n}-1$ are called integer states. Note The states $\ket{i}$ for $i = 0, ..., 2^{n}-1$ are called integer states. Note
@ -133,12 +133,12 @@ is acting on qbit $j$.
where without loss of generality $j < i$; the other case is analogous. where without loss of generality $j < i$; the other case is analogous.
In particular for $X, Z$: In particular for $X, Z$
\begin{equation}\label{eq:CX_pr} \begin{equation}\label{eq:CX_pr}
CX(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes X_i CX(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes X_i
\end{equation} \end{equation}
\begin{equation}\label{eq:CZ_pr} \begin{equation}\label{eq:CZ_pr}
CZ(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes Z_i CZ(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes Z_i .
\end{equation} \end{equation}
\end{definition} \end{definition}
@ -148,13 +148,13 @@ $CU$ applies the gate $U$ to the act-qbit if the control-qbit is in its $\ket{1}
One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough
to generate an arbitrary $n$-qbit gate \cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}. to generate an arbitrary $n$-qbit gate \cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}.
The matrix representation of $CX$ and $CZ$ for two qbits is given by: The matrix representation of $CX$ and $CZ$ for two qbits is given by
\begin{equation} \begin{equation}
CX_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right) CX_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)
\end{equation} \end{equation}
\begin{equation} \begin{equation}
CZ_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right) CZ_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right).
\end{equation} \end{equation}
\subsubsection{Measurements} \subsubsection{Measurements}
@ -181,13 +181,13 @@ $i$ $Z_i$ is measured. The $+1$ eigenvalue of $Z_i$ is $\ket{0}_i$, $\ket{1}_i$
\end{corrolary} \end{corrolary}
\begin{proof} \begin{proof}
The measuerment in not injective: Measuring both The measuerment is not injective: Measuring both
$\ket{0}$ and $\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ (can) map to $\ket{0}$. $\ket{0}$ and $\frac{1}{\sqrt{2}}(\ket{0} + \ket{1})$ (can) map to $\ket{0}$.
Any unitary matrix $U$ has the inverse $U^\dagger \equiv U^{-1}$. Any unitary matrix $U$ has the inverse $U^\dagger \equiv U^{-1}$.
\end{proof} \end{proof}
Because a measurement is not unitary it is not a gate in the sense the definition above. Because a measurement is not unitary it is not a gate in the sense of the definition above.
In the following discussion the term \textit{measurement gate} will be used from time In the following discussion the term \textit{measurement gate} will be used from time
to time as a measurement can be treated similarely while doing numerics. to time as a measurement can be treated similarely while doing numerics.
@ -196,27 +196,27 @@ to time as a measurement can be treated similarely while doing numerics.
As mentioned in \ref{ref:many_qbits} one can approximate an arbitrary $n$-qbit As mentioned in \ref{ref:many_qbits} one can approximate an arbitrary $n$-qbit
gate $U$ as a product of some single-qbit gates and either $CX$ or $CZ$. gate $U$ as a product of some single-qbit gates and either $CX$ or $CZ$.
Writing (possibly huge) products of matrices is quite unpractical and very much Writing (possibly huge) products of matrices is quite unpractical and
unreadable. To address this problem quantum circuits have been introduced. unreadable. To address this problem quantum circuits have been introduced.
These represent the qbits as a horizontal line and a gate acting on a qbit is These represent the qbits as a horizontal line and a gate acting on a qbit is
a box with a name on the respective line. Quantum circuits are read from a box with a name on the respective line. Quantum circuits are read from
left to right. This means that a gate $U_i = Z_i X_i H_i$ has the left to right. This means that a gate $U_i = Z_i X_i H_i$ has the
circuit representation: circuit representation
\[ \[
\Qcircuit @C=1em @R=.7em { \Qcircuit @C=1em @R=.7em {
& \gate{H} & \gate{X} & \gate{Z} &\qw \\ & \gate{H} & \gate{X} & \gate{Z} &\qw \\
} }.
\] \]
The controlled gates (such as $CX$ and $CZ$) have a vertical line from the control-qbit to The controlled gates (such as $CX$ and $CZ$) have a vertical line from the control-qbit to
the gate, for instance the circuit for $CZ_{2, 1}CX_{2,0}$ is: the gate, for instance the circuit for $CZ_{2, 1}CX_{2,0}$ is
\[ \[
\Qcircuit @C=1em @R=.7em { \Qcircuit @C=1em @R=.7em {
& \qw & \ctrl{2} & \qw & \qw &\qw \\ & \qw & \ctrl{2} & \qw & \qw &\qw \\
& \qw & \qw & \qw & \ctrl{1} &\qw \\ & \qw & \qw & \qw & \ctrl{1} &\qw \\
& \qw & \gate{X} & \qw & \gate{Z} &\qw \\ & \qw & \gate{X} & \qw & \gate{Z} &\qw &\rstick{.}\\
} }
\] \]
@ -241,12 +241,12 @@ algorithm that can be used to analyze the spectrum of the transfer matrix:
T = \exp(itH) T = \exp(itH)
\end{equation} \end{equation}
The eigenvalue of $T$ can now be estimated by using the phase estimation circuit: The eigenvalue of $T$ can now be estimated by using the phase estimation circuit
\[ \[
\Qcircuit @C=1em @R=.7em { \Qcircuit @C=1em @R=.7em {
& \lstick{\ket{0}} & {/^N} \qw & \gate{H^{\otimes n}} & \ctrl{1} & \gate{FT^\dagger} & \qw & \meter & \rstick{x}\\ & \lstick{\ket{0}} & {/^N} \qw & \gate{H^{\otimes n}} & \ctrl{1} & \gate{FT^\dagger} & \qw & \meter & \rstick{x}\\
& \lstick{\ket{\varphi}} & {/} \qw & \qw & \gate{T} & \qw & {/} \qw & \qw& \rstick{\ket{\varphi}} \\ & \lstick{\ket{\varphi}} & {/} \qw & \qw & \gate{T} & \qw & {/} \qw & \qw& \rstick{\ket{\varphi}.} \\
} }
\] \]

118
thesis/chapters/stabilizer.tex
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@ -18,10 +18,10 @@ performance has since been improved to $n\log(n)$ time on average \cite{andersbr
\begin{definition} \begin{definition}
\begin{equation} \begin{equation}
P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\} P := \{\pm I, \pm X, \pm Y, \pm Z, \pm iI, \pm iX, \pm iY, \pm iZ\}
\end{equation} \end{equation}
Is called the Pauli group. with the matrix product is called the Pauli group \cite{andersbriegel2005}.
\end{definition} \end{definition}
The group property of $P$ can be verified easily. Note that The group property of $P$ can be verified easily. Note that
@ -34,7 +34,7 @@ the elements of $P$ either commute or anticommute.
P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i \middle| p_i \in P\right\} P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i \middle| p_i \in P\right\}
\end{equation} \end{equation}
is called the multilocal Pauli group on $n$ qbits. is called the multilocal Pauli group on $n$ qbits \cite{andersbriegel2005}.
\end{definition} \end{definition}
The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition
@ -44,13 +44,15 @@ via the tensor product.
\subsubsection{Stabilizers} \subsubsection{Stabilizers}
The discussion below follows the argumentation given in \cite{nielsen_chuang_2010}.
\begin{definition} \begin{definition}
\label{def:stabilizer} \label{def:stabilizer}
An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff
\begin{enumerate} \begin{enumerate}
\item{$\forall i,j = 1, ..., N$: $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute \item{$\forall i,j = 1, ..., N$: $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute
} }
\item{$-I \notin S$} \item{$-I \notin S$ }
\end{enumerate} \end{enumerate}
\end{definition} \end{definition}
@ -61,7 +63,7 @@ via the tensor product.
\begin{enumerate} \begin{enumerate}
\item{$\pm iI \notin S$} \item{$\pm iI \notin S$}
\item{$(S^{(i)})^2 = I$ for all $i$} \item{$(S^{(i)})^2 = I$ for all $i$}
\item{$S^{(i)}$ are hermitian for all $i$} \item{$S^{(i)}$ are hermitian for all $i$ }
\end{enumerate} \end{enumerate}
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
@ -71,7 +73,7 @@ via the tensor product.
\item{From the definition of $S$ ($G_n$ respectively) follows that any \item{From the definition of $S$ ($G_n$ respectively) follows that any
$S^{(i)} \in S$ has the form $\pm i^l \left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ where $S^{(i)} \in S$ has the form $\pm i^l \left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ where
$\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $\left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ $\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $\left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$
is hermitian $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly. is hermitian and unitary $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly.
} }
\item{Following the argumentation above $\left(S^{(i)}\right)^2 = -I \Leftrightarrow l=1$ \item{Following the argumentation above $\left(S^{(i)}\right)^2 = -I \Leftrightarrow l=1$
therefore $\left(S^{(i)}\right)^2 = -I \Leftrightarrow \left(S^{(i)}\right)^\dagger \neq S^{(i)}$.} therefore $\left(S^{(i)}\right)^2 = -I \Leftrightarrow \left(S^{(i)}\right)^\dagger \neq S^{(i)}$.}
@ -113,7 +115,7 @@ can be diagonalized simultaneously. This motivates and justifies the following d
\end{equation} \end{equation}
is called the space of stabilizer states associated with $S$ and one says is called the space of stabilizer states associated with $S$ and one says
$\ket{\psi}$ is stabilized by $S$. $\ket{\psi}$ is stabilized by $S$ \cite{nielsen_chuang_2010}.
\end{definition} \end{definition}
It is clear that to show the stabilization property of It is clear that to show the stabilization property of
@ -121,13 +123,13 @@ $S$ the proof for the generators is sufficient,
as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$. as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$.
The dimension of $V_S$ is not immediately clear. One can however show that The dimension of $V_S$ is not immediately clear. One can however show that
for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension
$dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important $\dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important
result: result:
\begin{theorem} \begin{theorem}
\label{thm:unique_s_state} \label{thm:unique_s_state}
For a $n$ qbit system and stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer For a $n$ qbit system and stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer
space $V_S$ has $dim V_S = 1$, in particular there exists an up to a trivial phase unique space $V_S$ has $\dim V_S = 1$, in particular there exists an up to a trivial phase unique
state $\ket{\psi}$ that is stabilized by $S$. state $\ket{\psi}$ that is stabilized by $S$.
Without proof. Without proof.
@ -147,7 +149,7 @@ and a unitary transformation $U$ that describes the dynamics of the system, i.e.
\end{equation} \end{equation}
It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$ anymore. There are It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$ anymore. There are
however some statements that can still be made: however some statements that can still be made \cite{nielsen_chuang_2010}:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -169,7 +171,7 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers.
C_n := \left\{U \in SU(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\} C_n := \left\{U \in SU(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\}
\end{equation} \end{equation}
is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group. is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group \cite{andersbriegel2005}.
\end{definition} \end{definition}
\begin{theorem} \begin{theorem}
@ -212,10 +214,10 @@ product of $n$ Pauli matrices. This has led to the simulation using stabilizer t
Interestingly also measurements are dynamics covered by the stabilizers. Interestingly also measurements are dynamics covered by the stabilizers.
When an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is measured When an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is measured
one has to consider the projector: one has to consider the projector
\begin{equation} \begin{equation}
P_{g_a,s} = \frac{I + (-1)^s g_a}{2} P_{g_a,s} = \frac{I + (-1)^s g_a}{2}.
\end{equation} \end{equation}
If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$ If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$
@ -230,19 +232,20 @@ and the stabilizers are left unchanged:
\end{aligned} \end{aligned}
\end{equation} \end{equation}
As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$. As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$ \cite{nielsen_chuang_2010}.
If $g_a$ does not commute with all stabilizers the following lemma gives If $g_a$ does not commute with all stabilizers the following lemma gives
the result of the measurement. the result of the measurement.
\begin{lemma} \begin{lemma}
\label{lemma:stab_measurement} \label{lemma:stab_measurement}
Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$. When measuring Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$ and
$J^c := \left\{S^{(i)} \middle| S^{(i)} \notin J \right\}$. When measuring
$\frac{I + (-1)^s g_a}{2} $ $\frac{I + (-1)^s g_a}{2} $
$s=1$ and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing $s=1$ and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing
a $j \in J$ the new state $\ket{\psi'}$ is stabilized by a $j \in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010}
\begin{equation} \begin{equation}
\langle \{(-1)^s g_a\} \cup \left\{S^{(i)} S^{(j)} \middle| S^{(i)} \in J \setminus \{S^{(j)}\} \right\} \cup J^c\rangle \langle \{(-1)^s g_a\} \cup \left\{S^{(i)} S^{(j)} \middle| S^{(i)} \in J \setminus \{S^{(j)}\} \right\} \cup J^c \rangle.
\end{equation} \end{equation}
\end{lemma} \end{lemma}
@ -276,7 +279,7 @@ the result of the measurement.
\end{equation} \end{equation}
The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by
$S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$. $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$ \cite{nielsen_chuang_2010}.
\end{proof} \end{proof}
\subsection{The VOP-free Graph States} \subsection{The VOP-free Graph States}
@ -287,17 +290,17 @@ vertex operator-free will be clear in the following section about graph states.
\begin{definition} \begin{definition}
\label{def:graph} \label{def:graph}
The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$. The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements.
In the following $V = \{0, ..., n-1\}$ will be used. In the following $V = \{0, ..., n-1\}$ will be used.
$E$ is the set of edges $E = \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$. $E$ is the set of edges $E \subset \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$.
For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is called the neighbourhood For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is called the neighbourhood
of $i$. of $i$ \cite{hein_eisert_briegel2008}.
\end{definition} \end{definition}
This definition of a graph is way less general than the definition of a mathematical graph. This definition of a graph is way less general than the definition of a graph in graph theory.
Using this definition will however allow to avoid an extensive list of constraints on the Using this definition will however allow to avoid an extensive list of constraints on the
mathematical graph that are implied in this definition. graph from graph theory that are implied in this definition.
\begin{definition} \begin{definition}
For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are
@ -305,7 +308,7 @@ mathematical graph that are implied in this definition.
K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_j K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_j
\end{equation} \end{equation}
for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by
$\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$. $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$ \cite{hein_eisert_briegel2008}.
\end{definition} \end{definition}
It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute
@ -321,16 +324,19 @@ from the graph.
\begin{lemma} \begin{lemma}
For a graph $G = (V, E)$ the associated state $\ket{\bar{G}}$ is For a graph $G = (V, E)$ the associated state $\ket{\bar{G}}$ is
constructed using constructed using \cite{hein_eisert_briegel2008}
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
\ket{\bar{G}} &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right)\left(\prod\limits_{i \in V} H_i\right) \ket{0} \\ \ket{\bar{G}} &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right)\left(\prod\limits_{i \in V} H_i\right) \ket{0} \\
&= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+} \\ &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+} .\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
FIXME: This
Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before.
Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$. Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$.
In the following discussion the direction $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$ In the following discussion the direction $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$
@ -374,28 +380,28 @@ is done by using the symmetric set difference:
\begin{definition} \begin{definition}
For two finite sets $A,B$ the symmetric set difference $\Delta$ is For two finite sets $A,B$ the symmetric set difference $\Delta$ is
defined as: defined as \cite{hein_eisert_briegel2008}
\begin{equation} \begin{equation}
A \Delta B = (A \cup B) \setminus (A \cap B) A \Delta B = (A \cup B) \setminus (A \cap B).
\end{equation} \end{equation}
\end{definition} \end{definition}
Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$. Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$.
Another transformation on the VOP-free graph states for a vertex $a \in V$ is: Another transformation on the VOP-free graph states for a vertex $a \in V$ is \cite{andersbriegel2005}
\begin{equation} \begin{equation}
M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j} M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}.
\end{equation} \end{equation}
This transformation toggles the neighbourhood of $a$ which is an operation This transformation toggles the neighbourhood of $a$ which is an operation
that will be used later. that will be used later\cite{andersbriegel2005}.
\begin{lemma} \begin{lemma}
\label{lemma:M_a} \label{lemma:M_a}
When applying $M_a$ to a state $\ket{\bar{G}}$ the new state When applying $M_a$ to a state $\ket{\bar{G}}$ the new state
$\ket{\bar{G}'}$ is again a VOP-free graph state and the $\ket{\bar{G}'}$ is again a VOP-free graph state and the
graph is updated according to: graph is updated according to\cite{andersbriegel2005}:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
n_a' &= n_a \\ n_a' &= n_a \\
@ -489,7 +495,7 @@ clear that both $C_L$ and $CZ$ can be applied to a general graph state.
+1 \ket{G} = \left(\prod\limits_{j=1}^no_j\right) K_G^{(i)} \left(\prod\limits_{j=1}^no_j\right)^\dagger \ket{G} +1 \ket{G} = \left(\prod\limits_{j=1}^no_j\right) K_G^{(i)} \left(\prod\limits_{j=1}^no_j\right)^\dagger \ket{G}
\end{equation} \end{equation}
$o_i$ are called the vertex operators of $\ket{G}$. $o_i$ are called the vertex operators of $\ket{G}$ \cite{andersbriegel2005}.
\end{definition} \end{definition}
Recalling the dynamics of stabilizer states the following relation follows immediately: Recalling the dynamics of stabilizer states the following relation follows immediately:
@ -499,16 +505,16 @@ Recalling the dynamics of stabilizer states the following relation follows immed
\end{equation} \end{equation}
The great advantage of this representation of a stabilizer state is its space requirement: The great advantage of this representation of a stabilizer state is its space requirement:
Instead of storing $n^2$ $P_1$ matrices only some vertices (which often are implicit), Instead of storing $n^2$ $P$ matrices only some vertices (which often are implicit),
the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem
will improve this even further: instead of $n$ matrices it is sufficient to store $n$ integers will improve this even further: instead of $n$ matrices it is sufficient to store $n$ integers
representing the vertex operators: representing the vertex operators:
\begin{theorem} \begin{theorem}
$C_L$ has $24$ degrees of freedom. $C_L$ has $24$ degrees of freedom \cite{andersbriegel2005}.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof}
It is clear that $\forall a \in C_L$ a is a group isomorphism $P_1 \circlearrowleft$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. It is clear that $\forall a \in C_L$ a is a group isomorphism $P \rightarrow P$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$.
Therefore $a$ will preserve the (anti-)commutator relations of $P$. Therefore $a$ will preserve the (anti-)commutator relations of $P$.
Further note that $Y = iXZ$, so one has to consider the anti-commutator relations Further note that $Y = iXZ$, so one has to consider the anti-commutator relations
of $X,Z$ only. of $X,Z$ only.
@ -521,7 +527,7 @@ representing the vertex operators:
From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used. From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used.
One can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is One can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is
$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific operation on graph states. $C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific operation on graph states \cite{andersbriegel2005}.
\begin{equation} \begin{equation}
S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right) S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)
@ -541,12 +547,12 @@ basically a stabilizer tableaux that might require less memory than the tableaux
CHP. The true power of this formalism is seen when studying its dynamics. The simplest case CHP. The true power of this formalism is seen when studying its dynamics. The simplest case
is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers of are changed to is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers of are changed to
$\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation $\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation
it is clear that just the vertex operators are changed and the new vertex operators are given by: it is clear that just the vertex operators are changed and the new vertex operators are given by
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
o_i' &= o_i &\mbox{if } i \neq j\\ o_i' &= o_i &\mbox{if } i \neq j\\
o_i' &= c o_i c^\dagger &\mbox{if } i = j\\ o_i' &= c o_i c^\dagger &\mbox{if } i = j.\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -554,6 +560,7 @@ The action of a $CZ$ gate on the state $(V, E, O)$ is in most cases less trivial
Let $i \neq j$ be two qbits, now consider the action of $CZ_{a,b}$ on $(V, E, O)$. Let $i \neq j$ be two qbits, now consider the action of $CZ_{a,b}$ on $(V, E, O)$.
The cases given here follow the implementation of a $CZ$ application in \cite{pyqcs}, The cases given here follow the implementation of a $CZ$ application in \cite{pyqcs},
the respective paragraphs from \cite{andersbriegel2005} are given in italic. the respective paragraphs from \cite{andersbriegel2005} are given in italic.
Most of the discussion follows the one given in \cite{andersbriegel2005} closely.
\textbf{Case 1}(\textit{Case 1})\textbf{:}\\ \textbf{Case 1}(\textit{Case 1})\textbf{:}\\
@ -590,7 +597,7 @@ to the following theorem:
and right-multiplying $M_j^\dagger$ to the vertex operators in the sense and right-multiplying $M_j^\dagger$ to the vertex operators in the sense
that $\sqrt{-iX}^\dagger = \sqrt{iX}$ is right-multiplied to $o_j$ and that $\sqrt{-iX}^\dagger = \sqrt{iX}$ is right-multiplied to $o_j$ and
$\sqrt{iZ}^\dagger = \sqrt{-iZ}$ is right-multiplied to $o_l$ for all $\sqrt{iZ}^\dagger = \sqrt{-iZ}$ is right-multiplied to $o_l$ for all
neighbours $l$ of $j$. neighbours $l$ of $j$ \cite{andersbriegel2005}.
Without proof. Without proof.
\end{theorem} \end{theorem}
@ -627,13 +634,13 @@ clearing $o_b$ one can retry to clear $o_a$.
In any case at least one vertex operator has been cleared. If both vertex operators have been In any case at least one vertex operator has been cleared. If both vertex operators have been
cleared Case 1 will be applied. If there is just one cleared vertex operator it cleared Case 1 will be applied. If there is just one cleared vertex operator it
is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality
assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005}: assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005}
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
\ket{G} &= \left(\prod\limits_{o_i \in O} o_i\right) \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+}_n \\ \ket{G} &= \left(\prod\limits_{o_i \in O} o_i\right) \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+}_n \\
&= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) o_b (CZ_{a,b})^s \ket{+}_n \\ &= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) o_b (CZ_{a,b})^s \ket{+}_n \\
&= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) \ket{+}_{n-2} \otimes \left(o_b (CZ_{a,b})^s \ket{+}_2\right) \\ &= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) \ket{+}_{n-2} \otimes \left(o_b (CZ_{a,b})^s \ket{+}_2\right) .\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -655,11 +662,14 @@ If one wants to do computations using this formalism it is however also necessar
\subsubsection{Measurements on Graph States} \subsubsection{Measurements on Graph States}
This is adapted from \cite{andersbriegel2005}; measurement results and updating the graph after
a measurement is described in \cite{hein_eisert_briegel2008}.
Recalling \ref{ref:meas_stab} it is clear that one has to compute the commutator of Recalling \ref{ref:meas_stab} it is clear that one has to compute the commutator of
the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes
which is a quite expensive computation in theory. It is possible to simplify which is a quite expensive computation in theory. It is possible to simplify
the problem by pulling the observable behind the vertex operators. For this consider the problem by pulling the observable behind the vertex operators. For this consider
the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$: the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -667,32 +677,32 @@ the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$:
&= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right)P_a o_a \ket{\bar{G}} \\ &= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right)P_a o_a \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right) o_a o_a^\dagger P_a o_a \ket{\bar{G}} \\ &= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right) o_a o_a^\dagger P_a o_a \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O} o_i \right) o_a^\dagger P_a o_a \ket{\bar{G}} \\ &= \left(\prod\limits_{o_i \in O} o_i \right) o_a^\dagger P_a o_a \ket{\bar{G}} \\
&= \left(\prod\limits_{o_i \in O} o_i \right) \tilde{P}_{a,s} \ket{\bar{G}} \\ &= \left(\prod\limits_{o_i \in O} o_i \right) \tilde{P}_{a,s} \ket{\bar{G}} .\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
This transformed projector has the important property that it still is a Pauli projector This transformed projector has the important property that it still is a Pauli projector
as $o_a$ is a Clifford operator: as $o_a$ is a Clifford operator
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
\tilde{P}_{a,s} &= o_a^\dagger P_a o_a \\ \tilde{P}_{a,s} &= o_a^\dagger P_a o_a \\
&= o_a^\dagger \frac{I + (-1)^sZ_a}{2} o_a \\ &= o_a^\dagger \frac{I + (-1)^sZ_a}{2} o_a \\
&= \frac{I + (-1)^s o_a^\dagger Z_a o_a}{2} \\ &= \frac{I + (-1)^s o_a^\dagger Z_a o_a}{2} \\
&= \frac{I + (-1)^s \tilde{g}_a}{2} \\ &= \frac{I + (-1)^s \tilde{g}_a}{2} .\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is enough to study the measurements Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is enough to study the measurements
of any Pauli operator on the vertex operator free graph states. The commutators of the observable of any Pauli operator on the vertex operator free graph states. The commutators of the observable
with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute
and it is easier to list the operators that anticommute: and it is easier to list the operators that anticommute
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
A_{\pm X_a} &= \left\{j \middle| \{j, a\} \in E\right\}\\ A_{\pm X_a} &= \left\{j \middle| \{j, a\} \in E\right\}\\
A_{\pm Y_a} &= \left\{j \middle| \{j, a\} \in E\right\} \cup \{a\} \\ A_{\pm Y_a} &= \left\{j \middle| \{j, a\} \in E\right\} \cup \{a\} \\
A_{\pm Z_a} &= \{a\}\\ A_{\pm Z_a} &= \{a\}.\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -704,29 +714,29 @@ and $\tilde{g}_a$ are related by just inverting the result $s$.
The calculations to obtain the transformation on graph and vertex operators are lengthy and follow The calculations to obtain the transformation on graph and vertex operators are lengthy and follow
the scheme of Lemma \ref{lemma:M_a}. \cite[Section IV]{hein_eisert_briegel2008} also contains the scheme of Lemma \ref{lemma:M_a}. \cite[Section IV]{hein_eisert_briegel2008} also contains
the steps required to obtain the following results: the steps required to obtain the following results
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
U_{Z,s} &= \left(\prod\limits_{b \in n_a} Z_b^s\right) X_a^s H_a \\ U_{Z,s} &= \left(\prod\limits_{b \in n_a} Z_b^s\right) X_a^s H_a \\
U_{Y,s} &= \prod\limits_{b \in n_a} \sqrt{(-1)^{1-s} iZ_b} \sqrt{(-1)^{1-s} iZ_a}\\ U_{Y,s} &= \prod\limits_{b \in n_a} \sqrt{(-1)^{1-s} iZ_b} \sqrt{(-1)^{1-s} iZ_a}.\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}. These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}.
The second part makes sure that the qbit $a$ is diagonal in measured observable and has the correct eigenvalue. The second part makes sure that the qbit $a$ is diagonal in measured observable and has the correct eigenvalue.
When comparing with Lemma \ref{lemma:stab_measurement} in both cases $Y,Z$ the anticommuting stabilizer When comparing with Lemma \ref{lemma:stab_measurement} in both cases $Y,Z$ the anticommuting stabilizer
$K_G^{(a)}$ is chosen. The graph is changed according to: $K_G^{(a)}$ is chosen. The graph is changed according to
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
E'_{Z} &= E \setminus \left\{\{i,a\} \middle| i \in V\right\}\\ E'_{Z} &= E \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} \middle| i \in V\right\}\\ E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} \middle| i \in V\right\}.\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are: For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -739,7 +749,7 @@ For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are:
E'_{X} = E &\Delta (n_b \otimes n_a) \\ E'_{X} = E &\Delta (n_b \otimes n_a) \\
& \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\ & \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\
& \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\ & \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\
& \setminus \left\{\{i,a\} \middle| i \in V\right\}\\ & \setminus \left\{\{i,a\} \middle| i \in V\right\}.\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}