How does entangle qubits pass single qubit gate?

Question

How does entangle qubits pass single qubit gate? For example, I initialize two qubits $|0\rangle\otimes|0\rangle$, then first qubit passes $H$ gate to make it an superposition state $\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ and then applied CNOT gate. The state finally become $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$

Then I put the first qubit into $X$ gate. What will happen? It will become $\frac{1}{\sqrt{2}}(|10\rangle+|01\rangle)$? If so, how could it is possible the entangled two quibits pass the single-qubit-gate? If not, what will happen in this situation?

Answer 1

Here are the steps in details:

$$H\otimes I |0\rangle \otimes |0\rangle = H |0\rangle \otimes I|0\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle + |1\rangle \right) \otimes |0\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |10\rangle \right)$$

$$CNOT \frac{1}{\sqrt{2}}\left(|00\rangle + |10\rangle \right) = \frac{1}{\sqrt{2}}\left(CNOT|00\rangle + CNOT|10\rangle \right) = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle \right)$$

$$X \otimes I \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle \right) = \frac{1}{\sqrt{2}}\left(X |0\rangle \otimes I|0\rangle + X |1\rangle \otimes I|1\rangle \right) = \frac{1}{\sqrt{2}}\left(|10\rangle + |01\rangle \right)$$

So, we don't apply to the two-qubit state an $X$ operator, instead, we actually apply $X \otimes I$ operator or $I \otimes X$ depending on what qubit $X$ is applied.

Answer 2

Generally, if you have no gate on qubit, it means that there is an identity operator $I$. So if you have $n$ qubits, a gate $U$ applied on $i$ th qubit and no gate on others, effectively you have a gate $$ I_1 \otimes \dots \otimes I_{i-1} \otimes U \otimes I_{i+1} \otimes I_n, $$

which is $n$ qubits gate. So in your case two qubits go through two qubits gate as Davit shown in his answer.