Imrove Step by step circuit explanation #16

Author: Ilia Lazarev

manuscript.tex

@@ -219,7 +219,7 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 	\caption{
 		Quantum circuit for the quantum parallelized Hamming distance calculating  between all pairs of binary vectors from two sets ${X}$ and ${Y}$ encoded \cite{trugenberger2001} in $X$ and $Y$ quantum registers   respectively.
 		First, we encoded information about pairwise different qubits in a quantum state of the $X$-register with applying the CNOT gates. 
-		Second, Hamming distance values are extracted into the amplitudes of superposition with the controled rotation around $z$-axis gate and Hadamard gates. 
+		Second, Hamming distance values are extracted into the amplitudes of superposition with the controled rotation around $z$-axis gate~(\ref{eq:controled_rotation}) and Hadamard gates. 
 		Finally, a quantum state of the $X$-register returned to the initial basis for information retrieval. 	
 	} 
 	\label{fig:qcircuit}
@@ -255,12 +255,12 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 Meanwhile, during the procedure it stores the differences between input vectors and cluster states.
 
 Let us assume we have $k$ input vectors and $l$ cluster states. 
-The $i$th input vector and $j$th cluster vector are respectively denoted as $\left| X_i \right\rangle$, $\left| Y_j \right\rangle$. 
+The $i$th input vector and $j$th cluster vector are respectively denoted as $\left| x_i \right\rangle$, $\left| y_j \right\rangle$. 
 The registers $\left| X \right\rangle$ and $\left| Y \right\rangle$ are initialized to store the input vectors and cluster vectors according to
 %
 \begin{align}
-    \left| X \right\rangle  & = \frac{1}{\sqrt{k}} \sum\limits_{i=1}^{k} \left| X_i \right\rangle,  \\
-    \left| Y \right\rangle&  = \frac{1}{\sqrt{l}} \sum\limits_{j=1}^{l} \left| Y_j \right\rangle .
+    \left| X \right\rangle  & = \frac{1}{\sqrt{k}} \sum\limits_{i=1}^{k} \left| x_i \right\rangle,  \\
+    \left| Y \right\rangle&  = \frac{1}{\sqrt{l}} \sum\limits_{j=1}^{l} \left| y_j \right\rangle .
     \label{eq:encodnig}
 \end{align}
 % 
@@ -276,28 +276,26 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 %
 where $\left| a \right\rangle$ is an auxiliary qubit in the state $\left| 0 \right\rangle$ initially.
 
-Given this initial state we may begin the processing of the problem. We start by applying a CNOT gate between $\left| X \right\rangle$ and $\left| Y \right\rangle$
+Given this initial state we may begin the processing of the problem. We start by applying a CNOT gate on $\left| x^{(\alpha)} y^{(\alpha)} \right\rangle$ and $\alpha = 1 \dots n$
 
-\begin{align}
-    | \psi_1 \rangle & = 
-    \mathrm{CNOT} (Y,X)| \psi_0 \rangle \nonumber \\
-		& =  
+\begin{equation}
+    | \psi_1 \rangle  =  
     \frac{1}{\sqrt{kl}} \sum_{i, j=1}^{k} 
     | d^{(1)}_{ij}, \dots, d^{(n)}_{ij} \rangle 
     | y^{(1)}_j, \dots, y^{(n)}_j \rangle
     | 0 \rangle 
-\end{align}
+\end{equation}
 %
-where $d^\alpha_{ij} = \mathrm{CNOT}(y^\alpha_i, x^\alpha_j)$, $\alpha = 1 \dots n$, and $i,j$ are the qubit indexes in the registers. 
+where $d^{(\alpha)}_{ij} = \mathrm{CNOT}(y^{(\alpha)}_i, x^{(\alpha)}_j)$, and $\alpha = 1 \dots n$  is the qubit index in the register. 
 At this stage of the computation the $\left| X \right\rangle$ no longer stores the input vectors,
 instead it stores the information about pairwise different qubits between the input vector $\{X\}$ and cluster vector $\{Y\}$. 
 Next, for each pair $\{X\}$ and $\{Y\}$, the accumulated information of all the differences is projected onto the amplitude of the superposed state. 
 This is achieved by applying the Hadamard gate on auxiliary qubit, 
 followed by a controlled rotation around $z$-axis gate on $\left| Xa \right\rangle$ defined as
 %
 \begin{equation}
-    \label{eq:control_phase_rotation}
-    R_{(X,a)}(\phi) = 
+    \label{eq:controled_rotation}
+    C_{R_z}(\phi) = 
     \begin{pmatrix}
         1 & 0 & 0 & 0 \\
         0 & e^{-i \frac \phi 2} & 0 & 0 \\
@@ -312,46 +310,57 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 After the first Hadamard on the ancilla qubit the state is
 %
 \begin{equation}
-    \left| \psi_2 \right\rangle = H_a\left| \psi_1 \right\rangle = 
+    \left| \psi_2 \right\rangle = 
     \frac{1}{\sqrt{kl}}\sum\limits_{i, j=1}^{k} 
-    \left| d^{(1)}_{ij}, \dots, d^{(n)}_{ij} \right\rangle 
-    \left| y^{(1)}_j, \dots, y^{(n)}_j \right\rangle
+    \left| d_{ij} \right\rangle 
+    \left| y_j \right\rangle
     \dfrac{(\left| 0 \right\rangle + \left| 1 \right\rangle)}{\sqrt{2}}  .
 \end{equation}
 %
-Applying the controlled rotation around $z$-axis gate the state then becomes
+Applying the controlled rotation around $z$-axis gate on $\left| x^{(\alpha)} a \right\rangle$ where $\alpha = 1\dots n$ and the state then becomes
 %
 \begin{multline}
-    \left| \psi_3 \right\rangle = R_{(X,a)}\left(\dfrac{\pi}{n}\right)\left| \psi_2 \right\rangle
-    \\ = \dfrac{1}{\sqrt{2kl}}
-				\sum\limits_{i, j=1}^{k} 
-				\left| d^{(1)}_{ij}, \dots, d^{(n)}_{ij} \right\rangle 
-        \left| y^{(1)}_j, \dots, y^{(n)}_j \right\rangle 
-        \left| 0 \right\rangle
-        \\ + \dfrac{1}{\sqrt{2kl}}
-				\sum\limits_{i, j=1}^{k}
-        \exp\left(\dfrac{-i \pi}{n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-        \left| d^{(1)}_{ij}, \dots, d^{(n)}_{ij} \right\rangle 
-\\ \times        \left| y^{(1)}_j, \dots, y^{(n)}_j \right\rangle 
-        \left| 1 \right\rangle
+    \left| \psi_3 \right\rangle  
+	= \dfrac{1}{\sqrt{2kl}} \sum\limits_{i, j=1}^{k} 
+		\exp\left(
+		    \dfrac{-i \pi}{2n}
+		    \sum\limits_{\alpha=1}^n d^{(\alpha)}_{ij}
+		\right)
+		\left| d_{ij} \right\rangle 
+		\left| y_j \right\rangle 
+		\left| 0 \right\rangle \\
+    + \dfrac{1}{\sqrt{2kl}} \sum\limits_{i, j=1}^{k}
+		\exp\left(
+		    \dfrac{i \pi}{2n}
+		    \sum\limits_{\alpha=1}^n d^{(\alpha)}_{ij} 
+		\right)
+		\left| d_{ij} \right\rangle  
+		\left| y_j \right\rangle 
+		\left| 1 \right\rangle
 \end{multline}
 %
 Applying another Hadamard on the ancilla qubit we obtain
 %
 \begin{multline}
     \left| \psi_4 \right\rangle = 
     \frac{1}{\sqrt{kl}}\sum\limits_{i, j=1}^{k} 
-    \exp \left(\dfrac{-i \pi}{2n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-		\\ \times
-        \left[ \cos\left(\dfrac{\pi}{2n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-        \left| d^{(1)}_{ij}, \dots, d^{(n)}_{ij} \right\rangle 
-        \left| y^{(1)}_j, \dots, y^{(n)}_j \right\rangle 
-        \left| 0 \right\rangle\right.
-        \\+ 
-        \left. i \sin\left(\dfrac{\pi}{2n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-        \left| d^{(1)}_{ij}, \dots, d^{(n)}_{ij} \right\rangle 
-        \left| y^{(1)}_j, \dots, y^{(n)}_j \right\rangle 
-        \left| 1 \right\rangle\right] .
+        \left[ 
+			\cos\left(
+			    \dfrac{\pi}{2n}
+			    \sum\limits_{\alpha=1}^n d^{(\alpha)}_{ij}
+			\right)
+			\left| d_{ij} \right\rangle 
+			\left| y_j \right\rangle 
+			\left| 0 \right\rangle\right.
+			\\+ 
+			\left. i \sin\left(
+			    \dfrac{\pi}{2n} 
+			    \sum\limits_{\alpha=1}^n d^{(\alpha)}_{ij} 
+			\right)
+			\left| d_{ij} \right\rangle 
+			\left| y_j \right\rangle 
+			\left| 1 \right\rangle
+		\right] .
 \end{multline}
 %
 This completes the step for projecting differences between pairs of $\{X\}$ and $\{Y\}$ onto the amplitude of the auxiliary qubit. 
@@ -363,19 +372,25 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 At this stage, the information regarding the differences between pairs of $\{X\}$ and $\{Y\}$ \hl{encoded in the amplitudes, in order to extract the Hamming distances between the relevant $\left| x_i \right\rangle$, $\left| y_j \right\rangle$ we return to our initial basis} for register $\left| X \right\rangle$ by applying pairwise CNOT gates:
 %
 \begin{multline}
-    \left| \psi_f \right\rangle = 
-    \mathrm{CNOT} (Y,X)\left| \psi_4 \right\rangle \\=  
+    \left| \psi_f \right\rangle =  
     \frac{1}{\sqrt{kl}}\sum\limits_{i, j=1}^{k} 
-    \exp \left(\dfrac{-i \pi}{2n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-				\left[ \cos\left(\dfrac{\pi}{2n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-        \left| X_i \right\rangle 
-        \left| Y_j \right\rangle 
+	\left[ 
+	    \cos\left(
+	        \dfrac{\pi}{2n}
+	        \sum\limits_{\alpha=1}^n d^{(\alpha)}_{ij} 
+	    \right)
+        \left| x_i \right\rangle 
+        \left| y_j \right\rangle 
         \left| 0 \right\rangle\right.
         \\+
-        \left. i \sin\left(\dfrac{\pi}{2n}\sum\limits_{l=1}^n d^{(l)}_{ij} \right)
-        \left| X_i \right\rangle 
-        \left| Y_j \right\rangle 
-        \left| 1 \right\rangle\right] .
+        \left. i \sin\left(
+            \dfrac{\pi}{2n}
+            \sum\limits_{\alpha=1}^n d^{(\alpha)}_{ij} 
+        \right)
+        \left| x_i \right\rangle 
+        \left| y_j \right\rangle 
+        \left| 1 \right\rangle
+    \right] .
 \end{multline}
 %
 This makes $\{X\}$ store the input vectors again, as in the initial step 
@@ -386,11 +401,11 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 In this case, the biggest amplitude of the measurement result coincides with the smallest Hamming distance when the measurement result of the ancilla qubit is 0. 
 If the ancilla qubit is 1, the smallest amplitude of the measurement result coincides with the smallest Hamming distance.  
 
-Measuring the Hamming distance of a particular pair of input vectors $\left| X_i \right\rangle$ and cluster vector $\left| Y_j \right\rangle$ consists of extracting the relevant amplitude from the subspace that those states form, 
+Measuring the Hamming distance of a particular pair of input vectors $\left| x_i \right\rangle$ and cluster vector $\left| y_j \right\rangle$ consists of extracting the relevant amplitude from the subspace that those states form, 
 this can be done using the following projection operator
 %
 \begin{align}
-\Pi_{i,j} = &\left| X_i \rangle\langle X_i \right| \otimes \left| Y_j \rangle\langle Y_j \right| \otimes I .
+\Pi_{i,j} = &\left| x_i \rangle\langle x_i \right| \otimes \left| y_j \rangle\langle y_j \right| \otimes I .
 \end{align} 
 %
 Using the above projection operator, the subspace of the Hilbert space formed by a particular pair of input and cluster vectors can be traced out as
@@ -407,7 +422,7 @@ \subsection{Optimized quantum scheme for Hamming distance calculation}
 \end{align}
 %
 In order to reduce noise we average the measurement results over different states of the ancilla qubit, 
-thus the measured Hamming distance between the input vector $\left| X_i \right\rangle$ and cluster vector $\left| Y_j \right\rangle$ is
+thus the measured Hamming distance between the input vector $\left| x_i \right\rangle$ and cluster vector $\left| y_j \right\rangle$ is
 %
 \begin{align}
     d_{i,j}^H & \propto 1 - \frac{1}{2}(a_0(x_i,y_j) + (1-a_1(x_i,y_j))) .