How to apply single and two qubit gates to 2 qubits multiple times?

Question

Let's say we have the following quantum circuit:

Let's say we input the state $|00\rangle$ . Both of the $H$ gates produce the output $1/\sqrt{2}$, but which one of the following $2$ vectors is the input of $\operatorname{CNOT}$ gate: $1: [1/\sqrt{2}, \ 1/\sqrt{2}, \ 1/\sqrt{2}, \ 1/\sqrt{2}]$ or $2: [1/2, \ 1/2, \ 1/2, \ 1/2]$? Also, after applying a $2$ qubit gate (in this case $\operatorname{CNOT}$) how do we find out what to input in the following single qubit gates (in this case two $H$ gates)? Note that I gave this example for simplicity, a general answer will also be accepted.

Answer 1

Step 1 (application of two hadamard gates):

$$|0\rangle_A \otimes |0\rangle_B \to \left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B$$

This is equivalent to the state vector: $\begin{bmatrix}1/2\\1/2\\1/2\\1/2\end{bmatrix}$, which will act as the input for your CNOT gate.

Step 2: (application of CNOT)

Let's remind ourselves what the CNOT gate does:

So clearly, $$\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)$$ $$\to \frac{1}{2}(|00\rangle + |01\rangle + |11\rangle + |10\rangle) = \left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B$$

Step 3: (re-application of two Hadamard gates)

$$\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle + |1\rangle}{\sqrt{2}}\right)_B \to |0\rangle_A \otimes |0\rangle_B$$

Here, we used that fact that the Hadamard gate maps the state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ to the state $|0\rangle$, for each qubit. You basically need to input the vector $\begin{bmatrix}1/\sqrt{2}\\1/\sqrt{2}\end{bmatrix}$ into the two Hadamard gates at the end. Find the state transformation on each qubit. From there you can construct the state vector for the final 2-qubit state. The final answer will be $\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}$. Clearly, we end up with the same 2-qubit state which we started with.

Exercise: Carry out steps $2$ and $3$ using the matrix notation. Also, check this related answer.