Equivalence of these quantum circuits

[lholmes135]

TL;DR

C-NOT gate equivalent circuits

In the attached image, there are two quantum circuits that are equivalent. I am trying to understand how. Let's call the top qubit $q_1$ and the bottom one $q_2$, and the outputs $q_1'$ and $q_2'$. From what I understand, the C-NOT gate doesn't affect the control qubit. Because Hadamard gates are reversible, it looks to me like $q_1'=q_1$. In the second circuit though, $q_1'=q_1 \oplus q_2$. I also tried doing this through calculation. We'll make $q_1=\begin{bmatrix}c_0\\c_1\end{bmatrix}$ and $q_2=\begin{bmatrix}d_0\\d_1\end{bmatrix}$. Looking at the left circuit, say the input state to the C-NOT gate is $|\psi>$ and the output state $|\psi'>$.

\begin{equation*}
|\psi>=\frac{1}{2}\begin{bmatrix}(c_0+c_1)(d_0+d_1)\\(c_0+c_1)(d_0-d_1)\\(c_0-c_1)(d_0+d_1)\\(c_0-c_1)(d_0-d_1)\end{bmatrix}
\end{equation*}

C_NOT gate should flip third and fourth elements:

\begin{equation*}
|\psi'>=\frac{1}{2}\begin{bmatrix}(c_0+c_1)(d_0+d_1)\\(c_0+c_1)(d_0-d_1)\\(c_0-c_1)(d_0-d_1)\\(c_0-c_1)(d_0+d_1)\end{bmatrix}
\end{equation*}

The problem is that the output state is entangled, and I'm not sure how to calculate the two H gates on the right if I can't separate the states. This way turned out to be a dead end for me.

 

[DrClaude]

You can find the matrix form of the two Hadamard gates acting on the two qubits. You can start by writing the 4x4 version of the Hadamard gate acting on a single qubit, $H_1 \otimes I_2$.

 

[lholmes135]

This is what I was looking for, thanks.

 

[JD_PM]

I have a really basic understanding of quantum circuits but let me try to help

In the attached image, there are two quantum circuits that are equivalent. I am trying to understand how.

There are two ways to show equivalence between circuits (that I am aware of).

$1)$ By means of the computational basis $\alpha$ i.e.

\begin{equation*}
\alpha = \{ |00\rangle, |11\rangle, |10\rangle, |01\rangle\}
\end{equation*}

Where

\begin{equation*}
|0\rangle = \begin{pmatrix}
1 \\
0
\end{pmatrix}, \ \ \ \
|1\rangle = \begin{pmatrix}
0 \\
1
\end{pmatrix}
\end{equation*}

$2)$ By means of the matrix representation of the involved gates with respect to the computational basis.

Let me go for $1)$. I will let you show it via $2)$.

The Hadamard gate acts on a single qubit as follows

\begin{equation*}
H |0 \rangle = \frac{1}{\sqrt{2}}\left( |0 \rangle + |1 \rangle\right) := |+ \rangle
\end{equation*}

\begin{equation*}
H |1 \rangle = \frac{1}{\sqrt{2}}\left( |0 \rangle - |1 \rangle\right) := |- \rangle
\end{equation*}

Or more compacted

\begin{equation*}
H |x \rangle = \frac{1}{\sqrt{2}}\left( |0 \rangle + (-1)^x|1 \rangle\right), \ \ \ \ \text{where} \ x \in \{ 0,1\}
\end{equation*}

The CNOT gate involves two qubits: one is left unchanged while the other is bitwise-added to the first one i.e.

OK, let us focus on the given circuit. Reading from left to right (where $x,y \in \{ 0,1\}$) we get

\begin{align}
&\left( H \otimes H \right) \text{CNOT} \left( H \otimes H \right) |x\rangle \otimes |y\rangle \nonumber \\
&= \frac 1 2 \text{CNOT} \left( |0\rangle + (-1)^x |1\rangle \right) \otimes \left( |0\rangle + (-1)^y |1\rangle \right) \nonumber \\
&= \frac 1 2 \left( H \otimes H \right) \text{CNOT} \left( |00\rangle + (-1)^{x+y}|11\rangle + (-1)^{y}|01\rangle + (-1)^{x}|10\rangle \right) \nonumber \\
&= \frac 1 2 \left( H \otimes H \right) \left( |00\rangle + (-1)^{x+y}|10\rangle + (-1)^{y}|01\rangle + (-1)^{x}|11\rangle \right) \nonumber \\
&= \frac 1 2 \left( |++\rangle + (-1)^{x+y}|-+\rangle + (-1)^{y}|+-\rangle + (-1)^{x}|--\rangle \right) \nonumber \\
&= \frac 1 2 \Big[ |+\rangle \otimes \left( |+\rangle + (-1)^{y}|-\rangle \right) + |-\rangle \otimes \left( (-1)^{x+y}|+\rangle + (-1)^{x}|-\rangle \right)\Big] \nonumber \\
&= \frac 1 2 \Big[ |+\rangle \otimes \left( |+\rangle + (-1)^{y}|-\rangle \right) + (-1)^{x+y} |-\rangle \otimes \left( |+\rangle + (-1)^{-y}|-\rangle \right)\Big] \tag{1}\\
&= \frac 1 2 \Big[ |+\rangle \otimes \left( |+\rangle + (-1)^{y}|-\rangle \right) + (-1)^{x+y} |-\rangle \otimes \left( |+\rangle + (-1)^{y}|-\rangle \right)\Big] \tag{2}\\
&= \frac 1 2 \Big[\left( |+\rangle + (-1)^{x+y}|-\rangle \right) \otimes \left( |+\rangle + (-1)^{y}|-\rangle \right) \Big] \nonumber \\
&= \frac 1 4 \left( (1+ (-1)^{x+y})|0\rangle + (1- (-1)^{x+y})|1\rangle \right) \otimes \left( (1+ (-1)^{y})|0\rangle + (1- (-1)^{y})|1\rangle \right) \nonumber \\
&= |x \oplus y \rangle |y \rangle \nonumber
\end{align}

Where we notice that for $(1) = (2)$ to hold we need $(-1)^{x} = (-1)^{2y + x}$. We have two cases; Case A $x=1$: We note that $y$ is free to be either $0$ or $1$. For any of those two values $2y + x$ will yield an odd number so $(-1)^{x} = (-1)^{2y + x}$ holds. Case B $x=0$: We note that $y$ is again free to be either $0$ or $1$. For any of those two values $2y + x$ will yield an even number so $(-1)^{x} = (-1)^{2y + x}$ again holds.

To show equivalence via $2)$ you will need to use the CNOT matrix representation as well as the $4\times4$ Hadamard matrix.