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lholmes135
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- TL;DR Summary
- 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.
\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.