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I want to use the CNOT gate over a qubit $|0+\rangle$, but the definition says that CNOT flips the second qubit if a $1$ is found in the first one (i.e. 0 to 1, and 1 to 0). However, what is is to flip a $|+\rangle$?

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    $\begingroup$ Write $|+\rangle$ in the $|0\rangle$, $|1\rangle$ basis. $\endgroup$
    – noah
    Mar 16, 2021 at 15:36
  • $\begingroup$ Hi, thanks. I put it as 0,1 basis and it is a sum. How do a flip a sum? $\endgroup$
    – Theo Deep
    Mar 16, 2021 at 15:39
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    $\begingroup$ CNOT is an operator so linear $\endgroup$ Mar 16, 2021 at 15:40

3 Answers 3

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Some very basic quantum information:

$$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$

therefore

$$|0+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)$$

and

$$\operatorname{CNOT}|0+\rangle = \frac{1}{\sqrt{2}}(\operatorname{CNOT}|00\rangle + \operatorname{CNOT}|01\rangle) = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)$$ using the convention that the control qubit of the $\operatorname{CNOT}$ is the first one in the state. If you want the control to be the $|+\rangle$ state, we have $$\operatorname{CNOT}|{+}0\rangle = \frac{1}{\sqrt{2}}(\operatorname{CNOT}|00\rangle + \operatorname{CNOT}|10\rangle) =\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$

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Just to add to the answer provided by 'noah' for any Qubit $| \psi \rangle = \alpha|0 \rangle + \beta |1 \rangle $:

$$\text{CNOT}|0 \psi\rangle = \alpha \text{CNOT}|0 0\rangle + \beta \text{CNOT}|0 1\rangle=\alpha |0 0\rangle + \beta |0 1\rangle = |0 \psi\rangle$$

And conversely:

$$\text{CNOT}|1 \psi\rangle = \alpha \text{CNOT}|1 0\rangle + \beta \text{CNOT}|1 1\rangle=\alpha |1 1\rangle + \beta |1 0\rangle = |1 \big(X\psi\big)\rangle$$

You can then use linearity to evaluate any CNOT. Also you could just use the matrix but this is the intuition behind it.

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Note that $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$. $X$ (or NOT) gate interchange 0 and 1 which means that $$ X|+\rangle = \frac{1}{\sqrt{2}}(X|0\rangle + X|1\rangle) =\frac{1}{\sqrt{2}}(|1\rangle + |0\rangle) = |+\rangle, $$ so the state is unchanged.

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