# How does a superposition work with the cNOT quantum gate?

Note: In this question I am going to list the inputs in the order of control and then target.

Using the cNOT gate, what would happen in the following scenarios (i.e., what would the outputs be):

• Input a superposition (all superpostions assumed to be of the form $\alpha |00\rangle + \beta |01\rangle+\gamma|10\rangle+\delta|11\rangle$)
• Input $|1\rangle$ and a superposition
• Input a superposition and a superposition

Guesses:

• I'm not really sure...
• It would still flip the superposition like in the NOT gate (though I'm not quite sure what form that would take with four variables)
• Depends on the first one, but if the target does flip, then my guess for the second one would apply

Any help would be appreciated, as I'm not really sure what the result would be here.

Edit: I don't see why this is being voting to be closed as a homework question; it is asking conceptually what happens and includes my own guesses, so I don't see how it violates homework policy. (For the curious, this is not even from a textbook, though I know that isn't relevant.)

• Way, way out of my league, but just in case it's related: quantiki.org/wiki/basic-concepts-quantum-computation – user108787 Oct 4 '16 at 0:37
• @CountTo10, actually, yes, that was quite related. I'm still reading through it, but so far it's been rather helpful. Obviously, though, I'll still wait for an answer. – heather Oct 4 '16 at 0:48
• A CNOT gate is like any other linear operator, i.e. $T(a|x\rangle + b|y\rangle) = a T |x\rangle + b T |y\rangle$. – DanielSank Nov 7 '16 at 8:13
• @heather Regarding the homework policy: You should include your thoughts, not just guesses without any backup thoughts. Also, I don't see how "I'm not really sure..." qualifies as a thought. This can be appended to any homework question! – Norbert Schuch Nov 7 '16 at 9:08

Quantum operations are linear. So if you know how CNOT acts on the basis vectors then you can work out any case. Specifically in the usual $1/0$ basis, CNOT adds the control qubit to the target qubit modulo 2. For instance $$CNOT |1,1 \rangle = |1, 1\oplus 1 \rangle = |1,0\rangle$$.