# CNOT over a $|+\rangle$ qubit

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 $$|+\rangle$$?

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

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

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.

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.