# Allowed 2-qubit gates [closed]

I was working on some Quantum Information problems regarding allowed 2-qubit gates and got stuck. These are the proposed transformations:

1. $|A\rangle |B\rangle \rightarrow |B\rangle|\overline A\rangle$
2. $|A\rangle |B\rangle \rightarrow |\overline A\rangle| A\rangle$
3. $|A\rangle |B\rangle \rightarrow |A\oplus B\rangle| A\rangle$
4. $|A\rangle |B\rangle \rightarrow \frac{1}{\sqrt 2}(|A\rangle|B\rangle + |\overline A\rangle|\overline B\rangle)$
5. $|A\rangle |B\rangle \rightarrow (-1)^{A+B}|A\rangle|B\rangle$

So I think:

1. Possible as unitary is reversible (we preserve $A$ and $B$ at the output)
2. Not possible as we lose information about $B$ and hence cannot recover that state (i.e. the gate would be irreversible)
3. Possible as we know how XOR ($\oplus$) operates and hence can figure $B$ out from $A \oplus B$.
4. Here I get stuck. It seems that we preserve the $A$ and $B$ states but can we revert the output of the gate?
5. I think possible, as we can simply remove the sign at the front and get the original input states.

Is there anything else besides reversibility I should keep in mind when it comes to validation of unitary operations?

## closed as off-topic by Emilio Pisanty, Kyle Kanos, ZeroTheHero, John Duffield, Rory AlsopApr 19 '18 at 11:58

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• Please take a minute to read our guidelines for homework and exercise questions as well as check-my-work questions. We intend our questions to be potentially useful to a broader set of users than just the one asking, and we prefer conceptual questions over those just asking for a specific computation. – Emilio Pisanty Apr 13 '18 at 11:16
• what is $|\bar{A}\rangle$? – glS Apr 13 '18 at 21:28
• @glS It's NOT operation, so that $|\overline 0\rangle = |1\rangle$ – Laurynas Tamulevičius Apr 13 '18 at 21:32

• It was simpler than I thought. I simply calculated the truth tables for every operation in $|0\rangle$ and $|1\rangle$ basis and tested whether the operator was unitary. In the end, I figured that only 2 and 4 were not valid gates. – Laurynas Tamulevičius Apr 19 '18 at 18:17