# What is the outcome when you apply 2 hadamard gates on CNOT

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:

• Have you tried to simply do the math? Just recall how $H$ and CNOT act in the computational basis Apr 22 at 7:43
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.