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If the CNOT gate applies a NOT on the target qubit if the control qubit is a 1 and the control qubit was in superposition (for example in the $|+\rangle$ state), wouldn't this break the superposition since the CNOT gate is checking if the control qubit is a 1 or not and hence measuring it?

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Interestingly, the CNOT will not only keep the superposition, but in your example it will actually entangle the qubits! That is because, as DaftWullie says, you apply the operator linearly.

Let us take the truth table - i.e. all the possibilities on a basis - of the CNOT gate, where here I have taken the first qubit to be the control and the second to be the target:

$CNOT|00\rangle = |00\rangle$
$CNOT|01\rangle = |01\rangle$
$CNOT|10\rangle = |11\rangle$
$CNOT|11\rangle = |10\rangle$

So what happens if I act a CNOT on a state that has the the $|+\rangle$ state as its control? Let's take as an example the $|+\rangle|0\rangle$ state. Given our truth table above, it is not possible to know what happens if you try and directly apply the CNOT to this state as-is. But you can change basis - i.e. write it in all $0$s and $1$s - and then it becomes apparent. Here is how it looks like step by step:

$CNOT|+\rangle|0\rangle$
$=CNOT\frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)|0\rangle$ (here I have changed to the $0/1$ basis by using $|+\rangle=\frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)$)
$=CNOT\frac{1}{\sqrt{2}}\left(|0\rangle|0\rangle+|1\rangle\right|0\rangle)$ (here I have just multiplied out the $|0\rangle$ into the brackets)
$=\frac{1}{\sqrt{2}}\left(CNOT|0\rangle|0\rangle+CNOT|1\rangle\right|0\rangle)$ (here I have used the fact that the operator is linear, so I can basically multiply it out as well)
$=\frac{1}{\sqrt{2}}\left(|0\rangle|0\rangle+|1\rangle\right|1\rangle)$
$=\frac{1}{\sqrt{2}}\left(|00\rangle+|11\rangle\right)$

And just FYI, this is the famous Bell State :) hope this helps!

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This is a classical description of what controlled-not does which helps you think through the logic. However, controlled-not does not measure the control qubit, so it does not learn what state the control is in, and therefore does not destroy the superposition. Instead, it just applies the unitary operation $$ U=|0\rangle\langle 0|\otimes I+|1\rangle\langle 1|\otimes X. $$ Hopefully you can see, reading that, how it corresponds to the classical description. But the important thing is that it's a linear combination of the two statements, so if something is in superposition, it preserves the superposition.

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