added missing reverse permutation

content.tex

@@ -497,7 +497,7 @@ A quantum gate is just a unitary operator on a select set of qubits. Because the
   \label{def:gate_nbit_extension}
   Let $\ket{\mathbf{x}} = \ket{x_1, \dots, x_n}$ be the $n$ qubits of a quantum system and let $U$ be a $k$-qubit quantum gate, with $k < n$. The $k$-tuple $I = (i_1, \dots, i_k)$ with $i_l, i_g \in [1,n]$ and $i_l \neq i_g$ for all $l,g$ maps $k$ qubits from $\ket{\mathbf{x}}$ to the inputs of $U$. Then,
   \begin{equation*}
-    G_n(U,I) \coloneqq V_{\pi_I} (U \otimes \idmat_{n-k})
+    G_n(U,I) \coloneqq V_{\pi_I} (U \otimes \idmat_{n-k}) V_{\pi_I}^\dagger
   \end{equation*}
   is the $n$-qubit extension of $U$, with $V_{\pi_I}$ being a permutation matrix that satisfies
   \begin{equation*}
@@ -509,7 +509,7 @@ A quantum gate is just a unitary operator on a select set of qubits. Because the
 \begin{remark}
   Two gates $U_1 \in \C^{2^{k_1}}$ and $U_2 \in \C^{2^{k_2}}$ acting on $I_1 = (i_1^1,\dots,i_{k_1}^1)$ and $I_2 = (i_1^2,\dots,i_{k_2}^2)$ respectively can be computed in parallel iff $I_1 \cap I_2 = \emptyset$ and thus
   \begin{equation*}
-    G_n(U_1,I_1) G_n(U_2,I_2) = G_n(U_1 \otimes U_2, I_1I_2) = V_{\pi_{I_1 I_2}}(U_1 \otimes U_2 \otimes \idmat_{n - (k_1 + k_2)})
+    G_n(U_1,I_1) G_n(U_2,I_2) = G_n(U_1 \otimes U_2, I_1I_2) = V_{\pi_{I_1 I_2}}(U_1 \otimes U_2 \otimes \idmat_{n - (k_1 + k_2)}) V_{\pi_{I_1 I_2}}^\dagger
   \end{equation*}
   with $I_1I_2$ being the concatenation of $I_1$ and $I_2$ and 
   \begin{equation*}
@@ -525,4 +525,4 @@ A quantum gate is just a unitary operator on a select set of qubits. Because the
   The circuit $C$ is said to be maximally parallelized if all neighboring gate extensions $G_n(U_l,I_l)$ and $G_n(U_{l+1}, I_{l+1})$ with $I_l \cap I_{l+1} = \emptyset$ are reduced to $G_n(U_l \otimes U_{l+1}, I_l I_{l+1})$.
 \end{definition}
 
-
+\contentsketch{Circuit example Deutsch's algorithm}