Allowed 2-qubit gates [closed]

Question

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?

Answer 1

Trying to not solve the homework but rather give insights into the right thinking direction:

I think case 4 which right now puzzles you is closely related to the Hadamard quantum gate. The link to the wikipedia definition contains an expression of the Hadamard gate from which one could derive the one you need to show that indeed this is a unitary transformation.

However, considering the "we can simply remove the sign at the front" you mention concerning case 5, you may need to rethink how you are understanding reversibility. For the transformation to be reversible (which I agree it is), you need to remove the sign at the front also reversibly, i.e. not deleting any information.