some more work on the graph formalism
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@ -187,24 +187,86 @@ Where every $o_i$ acts on the $i$-th qbit.
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One can show that any stabilizer state can be realized as a graph state (for instance in \cite{schlingenmann2001}).
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\subsubsection{The Vertex Operator-Free Graph States}
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In order to understand some essential transformations of graph states it is necessary
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to study the vertex operator-free graph states first, partially because the graph states as used in this paper
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were derived from the vertex operator-free graph states.
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\begin{definition}
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\label{def:vop_free_g_state}
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A $n$ qbit vertex operator-free graph state $\ket{\overline{G}}$ is associated with a graph $(V, E)$
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by the $n$ operators
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\begin{equation}
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K^{(i)}_G := X_i \left(\prod\limits_{\{i, j\} \in E} Z_j\right)
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\end{equation}
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for all $i \in V$ where for some operator $O$ $O_i$ indicates that it acts on the $i$-th qbit.
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A state $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$.
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\end{definition}
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\begin{corrolary}
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All $K^{(i)}_G$ commute and are hermitian. Therefore they have a common set of eigen states
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(in particular definition \ref{def:vop_free_g_state} is well defined).
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In terms of quantum mechanics $K^{(i)}_G$ are observables.
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Further as $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$ which are
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multi-local Pauli operators, $\{K^{(i)}_G | i \in \{0, ..., n-1\}\}$ is the stabilizer
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of $\ket{\overline{G}}$ and $\ket{\overline{G}}$ is a stabilizer state.
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\end{corrolary}
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\begin{proof}
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As $X_i$ and $Z_i$ are hermitian their product is hermitian.
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Consider the case $\{i,j\} \notin E$ first:
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\begin{equation}
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\begin{aligned}
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\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
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\end{aligned}
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\end{equation}
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As operators acting on different qbits commute. The case $\{i,j\} \in E$ is slightly less trivial:
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\begin{equation}
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\begin{aligned}
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\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] \\
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&= \left[X_i Z_j \prod\limits_n Z_n, X_j Z_i \prod\limits_m Z_m\right]\\
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&= \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 \\
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&= \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 \\
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&= \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 \\
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&= 0
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\end{aligned}
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\end{equation}
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as $X$, $Z$ anticommute.
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\end{proof}
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\subsection{Operations on the Graph State}
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\subsubsection{Single Qbit Gates}
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Recalling \eqref{eq:g_state}
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Makes it clear that for any single qbit gate $o \in C_L$ with $o^{(k)}$ being the gate
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Makes it clear that for any single qbit gate $g \in C_L$ with $g_k$ being the gate
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acting on qbit $k$ the state changes according to
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\begin{equation}
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\begin{aligned}
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o^{(k)} \ket{G} &= o^{(k)} \left(\bigotimes\limits_{i=0}^{n-1} o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right) \ket{+} \\
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&= \left(\bigotimes\limits_{i=0}^{n-1} o^{\delta_{i,k}}o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right)\ket{+}
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g_k \ket{G} &= g_k \left(\bigotimes\limits_{i=0}^{n-1} o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right) \ket{+} \\
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&= \left(\bigotimes\limits_{i=0}^{n-1} g_k^{\delta_{i,k}}o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right)\ket{+}
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\end{aligned}
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\end{equation}
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meaning that the graph state $(V, E, O)$ changes to $(V, E, \{o_0, ..., o_{k-1}, oo_k, o_{k+1}, ..., o_{n-1}\})$
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as $C_L$ is almost a group the element $oo_k \in C_L$ up to a global phase that is disregarded. All the results
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meaning that the graph state $(V, E, O)$ changes to $(V, E, \{o_0, ..., o_{k-1}, go_k, o_{k+1}, ..., o_{n-1}\})$
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as $C_L$ is almost a group the element $go_k \in C_L$ up to a global phase that is disregarded. All the results
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of $C_L \times C_L \rightarrow C_L, a,b \mapsto ab$ have been precomputed in a lookup table and the vertex operators
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are updated according to that lookup table.
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