CNOT over a $|+\rangle$ qubit

Question

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 $|+\rangle$?

Answer 1

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 $$\operatorname{CNOT}|{+}0\rangle = \frac{1}{\sqrt{2}}(\operatorname{CNOT}|00\rangle + \operatorname{CNOT}|10\rangle) =\frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$$

Answer 2

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.

Answer 3

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.