basically finished the theory on the stabilizer formalism
This commit is contained in:
parent
d6df56a51b
commit
1c4430dd58
|
@ -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}
|
||||||
|
|
||||||
|
|
|
@ -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}
|
||||||
|
}
|
||||||
|
|
Loading…
Reference in New Issue
Block a user