basically finished the theory on the stabilizer formalism

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Daniel Knüttel 2020-02-18 13:27:37 +01:00
parent d6df56a51b
commit 1c4430dd58
2 changed files with 58 additions and 6 deletions

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@ -634,15 +634,15 @@ Recalling \ref{ref:meas_stab} it is clear that one has to compute the commutator
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.
This is a quite expensive computation in theory however it is possible to simplify This is a quite expensive computation in theory however 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 $P_a = \frac{I + (-1)^sZ_a}{2}$: the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
P_a \ket{\psi} &= P_a \left(\prod\limits_{o_i \in O} o_i \right) \ket{\bar{G}} \\ P_{a,s} \ket{\psi} &= P_{a,s} \left(\prod\limits_{o_i \in O} o_i \right) \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)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 \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}
@ -651,7 +651,7 @@ as $o_a$ is a Clifford operator:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
\tilde{P}_a &= 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} \\
@ -671,6 +671,50 @@ so it is easier to list the operators that anticommute:
\end{aligned} \end{aligned}
\end{equation} \end{equation}
This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a$ This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$
is measured the result $+1$ is obtained with probability $1$ and $(V, E, O)$ is unchanged. is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.
In any other case the results $s=1$ and $s=0$ both have probability $\frac{1}{2}$ and both
graph and vertex operators have to be updated. Further it is clear that measurement of $-\tilde{g}_a$
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 scheme of Lemma \ref{lemma:M_a}. \cite[Section IV]{hein_eisert_briegel2008} also contains
the steps required to obtain the following results:
\begin{equation}
\begin{aligned}
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}\\
\end{aligned}
\end{equation}
These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}
and the second part makes sure that the qbit $a$ is diagonal in the correct state of the measured state.
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
\begin{equation}
\begin{aligned}
E'_{Z} &= E \setminus \left\{\{i,a\} | i \in V\right\}\\
E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} | i \in V\right\}\\
\end{aligned}
\end{equation}
For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are
\begin{equation}
\begin{aligned}
U_{X,0} &= \sqrt{iY_b} \prod\limits_{c \in n_a \setminus n_b \setminus \{b\}} Z_c \\
U_{X,1} &= \sqrt{-iY_b} \prod\limits_{c \in n_b \setminus n_a \setminus \{a\}} Z_c \\
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
E'_{X} = E &\Delta (n_b \otimes n_a) \\
& \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\
& \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\
& \setminus \left\{\{i,a\} | i \in V\right\}\\
\end{aligned}
\end{equation}

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@ -139,3 +139,11 @@
year=2020, year=2020,
note={https://github.com/daknuett/pyqcs}, note={https://github.com/daknuett/pyqcs},
} }
@article{
hein_eisert_briegel2008,
title={Multi-party entanglement in graph states},
year=2008,
author={M. Hein, J. Eisert, H.J. Briegel},
note={https://arxiv.org/abs/quant-ph/0307130v7}
}