remark about shorthand form of circuits in bra-ket notation

content.tex

@@ -490,6 +490,7 @@ The best theoretical model is of no value if it relies on some kind of magic tha
 This wraps up the framework of quantum computing. Fortunately, \cref{def:quantum_state_space,def:unitary_operator,def:measurment_quantum} correspond to the fundamental postulates of quantum mechanics, meaning that quantum computing with all its seemingly strange properties is in fact a physically realizable computational model!
 
 \subsection{Quantum Circuits}
+\label{sec:quantum_circuits}
 In theory, the framework introduced above should be enough to do quantum computation. All that needs to be done in order to develop a new algorithm is to come up with its unitary matrix. At second glance, however, one notices quickly the impracticality of this approach. It implies finding a $2^n \times 2^n$ complex matrix with all required and desired properties needed to solve a $n$-qubit problem. The standard praxis in quantum algorithm design is to formulate the computation as a quantum circuit. The classical pendant to this is the relationship between boolean functions $f : \parensc{0,1}^n \to \parensc{0,1}^m$ and boolean circuits. Where $f$ is realized by a circuit of cascading boolean primitive gates. A similar situation arises in quantum computing, already illustrated by the description of Deutsch's algorithm in \cref{sec:deutschs_algorithm}. Here the complete algorithm is composed of just Hadamard gates, function application gate $U_f$ and one measurement. This makes it way easyer to understand than just one bit unitary matrix. A quantum circuit just allows though processes of the likes of: What operation performed when on which qubit? Which is much more natural to the kind of algorithmic thinking people are usually used to.
 
 A quantum gate is just a unitary operator on a select set of qubits. Because they are unitary, quantum gates always have the same number of input and output qubits, unlike their boolean counterparts. Conceptual wires are used to connect the right qubits to the inputs and outputs of each gate. Let $U$ be a $k$-qubit gate in a $n$-qubit system, with $k < n$. Then $U$ can be extended to a $n$-qubit gate by calculating the tensor product of $U$ and a $(n-k)$-qubit identity matrix.
@@ -519,12 +520,17 @@ A quantum gate is just a unitary operator on a select set of qubits. Because the
 
 \begin{definition}
   A $n$-qubit quantum circuit is formally described by a sequence of gates and input mappings $C = (U_l, I_l)_{l=1}^m$ with $n = \abs{\bigcup_{l=1}^m I_l}$. Using \cref{def:gate_nbit_extension}, the circuit description equals a unitary matrix according to \cref{def:unitary_operator}:
-  \begin{equation*}
+  \begin{equation}
+    \label{eq:unitary_matrix_of_circuit}
     U_C = G_n(U_m,I_m) \cdots G_n(U_1,I_1)
-  \end{equation*}
+  \end{equation}
   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}
 
+\begin{remark}
+  If a quantum circuit $C = (U_l,I_l)_{l=1}^m$ is used in combination with bra-ket notation sometimes out of convenience $C \ket{\psi}$ instead of $U_C \ket{\psi}$ will be written. With the latter being the formally correct application of $C$s unitary matrix representation $U_C$ according to \cref{eq:unitary_matrix_of_circuit}.
+\end{remark}
+
 \begin{definition}
   \label{def:circuit_size_depth}
   Given a quantum circuit $C=(U_l,I_l)_{l=1}^m$ then