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?

  • $\begingroup$ $\frac1{sqrt 2}(|0\rangle+|1\rangle)$ is not entangled, because it is single qubit state. It is a superposition. Please clarify your question... $\endgroup$
    – draks ...
    Commented May 4, 2020 at 6:32
  • $\begingroup$ sorry, i just edit my question. $\endgroup$ Commented May 4, 2020 at 6:37
  • $\begingroup$ Just note that two qubit states you described are so-called Bell states. $\endgroup$ Commented May 4, 2020 at 21:35

2 Answers 2


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.

  • $\begingroup$ yes, the details steps are as what you said. but in quantum circuit, the X gate can only pass the single qubit, is that right? so you mean the situation will never happen?or the actual quantum circuit will resolve this situation into X*I as you said? $\endgroup$ Commented May 4, 2020 at 6:47
  • $\begingroup$ @Henry_Fordham if we are applying $X$ gate on the first qubit, mathematically it will correspond to $X \otimes I$ operator, if we have a two-qubit state. We can always apply one qubit gate, no matter is it in an entangled state or not, because the qubit exists in some fixed place and we always can point our laser (or other physical things that can help to implement one qubit gate) to that particular qubit. $\endgroup$ Commented May 4, 2020 at 6:59
  • $\begingroup$ Ah,Thx, i got it, and then i edit my question and put another confusion on it, can u help me for that? Thx $\endgroup$ Commented May 4, 2020 at 7:21
  • $\begingroup$ @Henry_Fordham, please can you write a separate question? I think it will be better that way :) $\endgroup$ Commented May 4, 2020 at 7:26
  • 1
    $\begingroup$ OK, have done that $\endgroup$ Commented May 4, 2020 at 7:52

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.


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