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Quasiprobability decomposition of the CZ-gate

Just for completeness I'll add how to apply this CZ decomposition. One must apply each summand individually to the state and add them afterwards. Let's assume we want to apply the decomposed CZ to an ...
upe's user avatar
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3 votes

$U_1\oplus U_2$ decomposable into $I\oplus U$ and 1-qubit gates?

$$\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 (...
Norbert Schuch's user avatar
2 votes

$U_1\oplus U_2$ decomposable into $I\oplus U$ and 1-qubit gates?

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\!...
glS's user avatar
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3 votes

$U_1\oplus U_2$ decomposable into $I\oplus U$ and 1-qubit gates?

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.
Norbert Schuch's user avatar
6 votes
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$U_1\oplus U_2$ decomposable into $I\oplus U$ and 1-qubit gates?

$A \oplus B$ translates to: if q1: apply B to q2 else: apply A to q2 which is equivalent to ...
Craig Gidney's user avatar
2 votes
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How many gates are necessary to implement an arbitrary n-qubit permutation unitary?

Using the idea of parameter counting, suggested by @NorbertSchuch, I was able to find more information on that topic and work out a proof that the number of required gates is indeed exponential. The ...
QNA's user avatar
  • 171
1 vote

How to retrieve a phase gate from a circuit made out of $CX$ and $T$

If you've got a bit with value $x_1$ and you apply a $T$ gate to it, you're going to get a phase of angle $x_1\pi/4$. Do another $T$ gate on a bit with value $x_2$ and you're up to a total phase of $(...
DaftWullie's user avatar
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