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So when I run through risk, it displayed it had an equal 25% chance to get 00 01 10 11 respectively. I know how the CNOT output looks like when you apply hadamard gate on control part before CNOT, but I'm not quite sure how the outcomes look like when you add a 2nd hadamard gate to it. Here is what I mean:

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Thanks in advance

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  • $\begingroup$ Have you tried to simply do the math? Just recall how $H$ and CNOT act in the computational basis $\endgroup$
    – Markus Heinrich
    Apr 22, 2021 at 7:43

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You want to be able to work through a circuit like this step by step. However, in this particular case, there is a shortcut. Note that acting $H$ on the second qubit produces the state $|+\rangle$ which is a $+1$ eigenstate of the $X$ operator. This actually means that the controlled-$X$ operator does absolutely nothing on this state! (Whether the control qubit is 0 or 1, the output of the second qubit is either $|+\rangle$ or $X|+\rangle=|+\rangle$ respectively.) Hence, the overall output is just $|++\rangle$, as if the controlled-not weren't there.

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