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Let's say we have a circuit with $2$ Hadamard gates:

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Let's take the $|00\rangle$ state as input. The vector representation of $|00\rangle$ state is $[1 \ 0 \ 0 \ 0]$, but this is the representation of $2$ qubits and H accepts just $1$ qubit, so should we apply the first H gate to $[1 \ 0]$ and the second H gate to $[0 \ 0]$? Or should we input $[1 \ 0]$ in each H gate, because we are applying H gates to just one qubit of state $|0\rangle$ each time?

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Or should we input $[1 \ 0]$ in each H gate, because we are applying H gates to just qubit of state $|0\rangle$ each time?

Yes, when you have a two-qubit state (say you label the two qubits as $A$ and $B$ respectively), you need to apply the two Hadamard gates separately on each qubit's state. The final state will be the tensor product of the two "transformed" single-qubit states.

If your input is $|0\rangle_A\otimes|0\rangle_B$, the output will simply be $$\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)_A\otimes\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)_B$$


Alternative:

If the two input qubits are entangled, the above method won't work since you won't be able to represent the input state as a tensor product of the states of the two qubits. So, I'm outlining a more general method here.

When two gates are in parallel, like in your case, you can consider the tensor product of the two gates and apply that on the 2-qubit state vector. You'll end up with the same result.

$\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\\ \end{bmatrix} \otimes \frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\\ \end{bmatrix} = \frac{1}{2}\begin{bmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1 \end{bmatrix}$

Now, on applying this matrix on the 2-qubit state $\begin{bmatrix}1\\0\\0\\0\end{bmatrix}$ you get:

$$\frac{1}{2}\begin{bmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1 \end{bmatrix} \begin{bmatrix}1\\0\\0\\0\end{bmatrix}=\begin{bmatrix}1/2\\1/2\\1/2\\1/2\end{bmatrix}$$

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

Justification

Tensor product of linear maps:

The tensor product also operates on linear maps between vector spaces. Specifically, given two linear maps $S : V \to X$ and $T : W \to Y$ between vector spaces, the tensor product of the two linear maps $S$ and $T$ is a linear map $(S\otimes T)(v\otimes w) = S(v) \otimes T(w)$ defined by $(S\otimes T)(v\otimes w) = S(v) \otimes T(w)$.

Thus, $$(\mathbf H|0\rangle_A) \otimes (\mathbf H|0\rangle_B) = (\mathbf H\otimes \mathbf H)(|0\rangle_A \otimes |0\rangle_B)$$

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  • $\begingroup$ Thanks. Consider we wanted to do the operation again, by taking the output ((|0⟩+|1⟩2√)A⊗(|0⟩+|1⟩2√)B) as input , we should input (|0⟩+|1⟩2√) and (|0⟩+|1⟩2√) , correct? $\endgroup$ – Archil Zhvania Jun 8 '18 at 12:46
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    $\begingroup$ Yes, correct! By the way, please use MathJax while typing mathematical expressions. Here's a quick tutorial. $\endgroup$ – Sanchayan Dutta Jun 8 '18 at 12:48
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    $\begingroup$ @ArchilZhvania I've updated the answer a bit. It isn't really true that you necessarily have to deal with the two qubit states separately, just if you're having that misconception. $\endgroup$ – Sanchayan Dutta Jun 8 '18 at 13:23
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It's the second option. So, you would apply both Hardmard gates to the state $|0\rangle$, to obtain two $\frac{1}{\sqrt{2}} \left (|0\rangle + |1\rangle \right)$. Therefore, the final two-qubit state would be

\begin{align} \frac{1}{\sqrt{2}} \left (|0\rangle + |1\rangle \right) \otimes \frac{1}{\sqrt{2}} \left (|0\rangle + |1\rangle \right) &= \frac{1}{2} \left( |00\rangle + |01\rangle + |10\rangle + |11\rangle\right) \end{align}

We can easily verify that this is a valid quantum state by checking the normalization condition.

\begin{align} \left| \frac{1}{2} \right|^2 + \left| \frac{1}{2} \right|^2 + \left| \frac{1}{2} \right|^2 + \left| \frac{1}{2} \right|^2 &= \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \\ &= 1 \end{align}

In general, in this context, it's more intuitive if you use Dirac's bra-ket notation (i.e. use $|00 \rangle$ instead of the column vector $(1, 0, 0, 0)^T$). Then, if you have to apply a gate to a subset of the qubits, you can proceed in an analogous way as I did above.

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