Yes, that notation means the Hadamard on the second qubit depends on the first qubit and the Hadamard on the third qubit depends on the first qubit. The gates aren't connected to each other in any way....

So, let the system be $\rho$, and the environment $|0\rangle \langle 0|$. The given operation (which you can check is unitary, and incidentally happens to be the CNOT operation), is applied on $\rho \... View answer 5 votes The controlled dot doesn't do anything: it merely observes the bottom qubit in order to decide whether to apply the 𝑋 gate onto the top qubit. In the answer below, the qubit that appears first is ... View answer Accepted answer 5 votes The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since$|+\rangle$is$\frac{1}{\sqrt 2}\left(|0\...

As given in the documentation, if your operation is unitary, you can add the statement adjoint auto; within the operation after the body block. This will generate the adjoint (which is the inverse for ...

Here is the solution. The trick is to use "the only connectivity matters" rule. The swap rule of 4.9 helps us reorder the inputs, which then makes it topologically equivalent to the next diagram (...

Posting an answer because I realised what my issue was: What I didn't realise then: When a density matrix is written in any basis, the diagonal elements correspond to the probabilities of the density ...

Now here is where it gets deep an 𝑛 Qubit Array can represent $2^𝑛$ possible array elements (consult anywhere online for an explanation of that or drop a comment). And similarly an 𝑛 Qubit quantum ...

Since this question seems to be in the context of Grover's search, I will explain using what happens in Grover's search, however, this is way more general. The oracle function $f$ itself can be ...

So probability of the second qubit being in state $|1\rangle$ is the probability of the 5 qubit system being in a state that has $|1\rangle$ as the second qubit. So among all the 32 states, find the ...
Assuming your control is $|0\rangle$ to begin with. Then after application of Hadamard, the control is: $$\frac{|0\rangle + |1\rangle}{\sqrt 2}$$. Now using this as control and applying $X$ gate to ...