2
$\begingroup$

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 $|+\rangle$?

$\endgroup$
3
  • 1
    $\begingroup$ Write $|+\rangle$ in the $|0\rangle$, $|1\rangle$ basis. $\endgroup$ – noah Mar 16 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 at 15:39
  • 4
    $\begingroup$ CNOT is an operator so linear $\endgroup$ – Mathphys meister Mar 16 at 15:40
5
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.