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 Alsop Apr 19 '18 at 11:58

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    $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Apr 13 '18 at 11:16
  • $\begingroup$ what is $|\bar{A}\rangle$? $\endgroup$ – glS Apr 13 '18 at 21:28
  • $\begingroup$ @glS It's NOT operation, so that $|\overline 0\rangle = |1\rangle$ $\endgroup$ – Laurynas Tamulevičius Apr 13 '18 at 21:32

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.

  • $\begingroup$ 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. $\endgroup$ – Laurynas Tamulevičius Apr 19 '18 at 18:17

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