# Should C-NOT gate affect "input" qubit or not?

In MS presentation, they describe the process CNOT gate working with state machine diagram. At first they drew CNOT gate in black bock as following: This implies, that Input qbit is not affected, since it is always $$|x\rangle$$. This also corresponds with what they drew earlier for any irreversible operation made reversible.

Nevertheless, on next slide they drew, that "Input" qubit (left circle on slide, green state change arrow) is affected Why?

I checked myself of applying CNOT matrix to depicted state and found that input state changes, while output does not.

So this was just error on slides, implying "Input" is not changes?

This is known as "Phase Kickback" in quantum computing. It is a trick that uses in a many quantum algorithms.

The phase of the target qubit can kickback to the control-qubit and changes the phase of it.

Take a look here: Phase Kickback Short answer: both can be correct, depending on the state to which the CNOT gate is applied.

At the first slide the presentation talks about the effect of CNOT gate on basis states. And indeed, if the input to the CNOT gate is a basis state, it only changes the state of the target qubit.

At the second slide, the CNOT gate is applied to a superposition of states. If the input state is $$|-\rangle \otimes |-\rangle$$, as the circles seem to imply, the effect of the CNOT gate on it will be:

$$|-\rangle \otimes |-\rangle = \frac12(|00\rangle - |10\rangle - |01\rangle + |11\rangle) \rightarrow \\ \rightarrow \frac12(|00\rangle - |1\color{red}1\rangle - |01\rangle + |1\color{red}0\rangle) = \frac12 (|0\rangle + |1\rangle) \otimes (|0\rangle - |1\rangle) = |\color{red}+\rangle \otimes |-\rangle$$

You see that while the effect on each of the basis states doesn't change the input qubit, the aggregate effect changes the input qubit and doesn't change the output qubit. This is known as phase kickback effect.

It is also possible to have the CNOT gate change the state of both input and output qubits. For example, $$\textrm{CNOT}|0\rangle \frac{1}{\sqrt2}(|0\rangle + |1\rangle) = \textrm{CNOT}\frac{1}{\sqrt2}(|00\rangle + |11\rangle)$$ - now the input and output qubits don't have individual states of their own, but rather are entangled.