diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex
index 8763c5d..36aec57 100644
--- a/thesis/chapters/stabilizer.tex
+++ b/thesis/chapters/stabilizer.tex
@@ -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.
 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 projector $P_a = \frac{I + (-1)^sZ_a}{2}$:
+the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$:
 
 \begin{equation}
     \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) 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) \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{equation}
 
@@ -651,7 +651,7 @@ as $o_a$ is a Clifford operator:
 
 \begin{equation}
     \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 \\
                     &= \frac{I + (-1)^s o_a^\dagger Z_a o_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{equation}
 
-This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a$ 
-is measured the result $+1$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.
+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 $s=0(1)$ 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}
 
diff --git a/thesis/main.bib b/thesis/main.bib
index cc3f5b0..afacd94 100644
--- a/thesis/main.bib
+++ b/thesis/main.bib
@@ -139,3 +139,11 @@
     year=2020,
     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}
+}