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TL;DR

Let $U_1, U_2, U$ be arbitrary 1-qubit quantum gates. Can 2-qubit gates of the form $U_1\oplus U_2$ always be decomposed into a combination of controlled gates ($I\oplus U$) and single qubit gates?

Context

The paper Optimal local implementation of non-local quantum gates presents the following quantum circuit to teleport a controlled gate.

gate teleportation circuit

This circuit does not only work for controlled gates of the form $\left[\begin{matrix}I & 0 \\ 0 & U\end{matrix}\right]$, but also for 2-qubit gates of the from $\left[\begin{matrix}U_1 & 0 \\ 0 & U_2\end{matrix}\right]$.

I'm not sure if that observation allows for the teleportation of other 2-qubit gates than achievable with only 1 controlled gate plus single qubit operations. Hence my question above.

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4 Answers 4

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$A \oplus B$ translates to:

if q1:
    apply B to q2
else:
    apply A to q2

which is equivalent to

if q1:
    apply A to q2
    apply A† to q2
    apply B to q2
else:
    apply A to q2

which is equivalent to:

apply A to q2
if q1:
    apply A† to q2
    apply B to q2

which is equivalent to:

apply A to q2
if q1:
    apply (B·A†) to q2

which is a single qubit gate followed by a controlled gate, which is of the form you asked for.

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  • $\begingroup$ This is very interesting! I had never thought about it like this! $\endgroup$
    – FDGod
    Mar 27 at 1:55
  • $\begingroup$ Your gates are applied from right to left? $\endgroup$ Mar 27 at 3:56
  • $\begingroup$ @NorbertSchuch That's unfortunately the standard ordering of matrix multiplication, not my choice. $\endgroup$ Mar 27 at 3:58
  • $\begingroup$ I'd say the problem is that we use column vectors, not matrix multiplication ;) ... but either way, in pseudocode one could just as well apply gates from left to right (indeed, this is way I assumed until the last line). Then again, just today I saw Jeongwan Haah draw a circuit from right to left ... :-o $\endgroup$ Mar 27 at 4:19
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Any two-qubit gate can be decomposed in terms of one-qubit gates and CNOTs (or some other two-qubit gate). Thus, this is in particular true for the special subclass you ask about.

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$$\begin{pmatrix} U_1\\& U_2 \end{pmatrix} = \underbrace{\begin{pmatrix} U_1\\& U_1 \end{pmatrix}}_{I \otimes U_1} \underbrace{\begin{pmatrix} I\\& U_1^\dagger U_2 \end{pmatrix}}_{I\oplus (U_1^\dagger U_2)} $$

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The main mathematical observation is that you can decompose direct sums as sums of tensor products as $$A\oplus B = \mathbb{P}_0 \otimes A + \mathbb{P}_1\otimes B, \qquad \mathbb{P}_i\equiv |i\rangle\!\langle i|.$$ The rest is a simple algebraic manipulation: $$A\oplus B = (I\otimes A)(\mathbb{P}_0\otimes I + \mathbb{P}_1\otimes (A^\dagger B)) = (I\otimes A)[I\oplus (A^\dagger B)] \\ = [I\oplus (BA^\dagger)](I\otimes A),$$ and the observation that $(I\otimes A)$ is a one-qubit gate and $I\oplus (A^\dagger B)$ is the controlled gate.

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  • $\begingroup$ What is the added value of the decomposition into projectors, rather than going right away from the 1st to the 3rs step in the equation? $\endgroup$ Mar 27 at 17:00
  • $\begingroup$ @NorbertSchuch I guess to see more clearly why $A\oplus B$ becomes $I\otimes A$ etc? It's just another way to write the direct sum decomposition without having to write the matrices $\endgroup$
    – glS
    Mar 27 at 17:47

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