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I need a set of gates, even controlled, which are providing the following unitary matrix with three qubits:

$$ U = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 \\ \end{bmatrix}. $$

The problem is that I do not know any general method to express a unitary matrix in terms of Pauli matrices. In case, I managed to obtained something similar to the one above by applying a Z-gate to the first qubit and a multicontrolled-Z gate (can be obtained with multicontrolled X gate and two Hadamard gates) to any qubit and using the other two as controls (it seems it doesn't matter the application). The matrix in this case reads as:

$$ U = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}. $$ So, the first and the last block are exchanged.The code to obtain it is:

qc.z(0) 
qc.h(0)  # H on target qubit
qc.mcx([1,2], 0, mode='noancilla')
qc.h(0)  # again H on target qubit

In general, I need this for any number of qubits.

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2 Answers 2

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This circuit implements the required unitary:

enter image description here

Code:

from qiskit import QuantumCircuit

circ = QuantumCircuit(3)
circ.cz(0, 1)
circ.cz(0, 2)
circ.ccz(0, 1, 2)

To check:

from qiskit.quantum_info import Operator
from qiskit.visualization import array_to_latex

array_to_latex(Operator(circ))
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  • $\begingroup$ Many thanks! Could you tell me how you got it? Any tip about how to generalize it to n qubits? $\endgroup$
    – francler
    Oct 16, 2023 at 9:39
  • $\begingroup$ The matrix is diagonal with $+1$ and $-1$ entries only. So, we can implement it using $Z, CZ, CCZ$ gates only. Now, the first $CZ$ will give us $diag(1, 1, 1, -1, 1, 1, 1, -1)$. The second $CZ$ will give us $diag(1, 1, 1, 1, 1, -1, 1, -1)$. Multiply them together to get $diag(1, 1, 1, -1, 1, -1, 1, 1)$. So, we need $CCZ$ to provide the remaining $-1$ $\endgroup$ Oct 16, 2023 at 9:52
  • $\begingroup$ Is there a way to generalize it to n qubits? I tried your appproach with 4 qubits but seems not to work. $\endgroup$
    – francler
    Oct 16, 2023 at 10:23
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This can be solved semi-mechanically.

$\newcommand{\ket}[1]{|{#1}\rangle}\newcommand{\bra}[1]{\langle{#1}|}\newcommand{\+}{\oplus}$This matrix $U$ is of form $U\ket{abc}=(-1)^{f(abc)}\ket{abc}$, for boolean variables $a, b, c$. Here, $f(a, b, c)$ is a boolean function.

And we can read the table for the boolean function $f$: $f(abc)=1$ iff $abc=011, 101$, or $111$. (I think it really depends on the ordering we take on the tensor product basis vectors, but, anyway...)

And, if you know the boolean table, then you can write it as a boolean polynomial: $f(abc)=(1\+a)bc\+a(1\+b)c\+abc$, since $f$ is true iff $\neg a\wedge b\wedge c$, or $a\wedge\neg b\wedge c$, or $a\wedge b\wedge c$ (corresponding to $011$, $101$, and $111$), and those 'OR's are in fact exclusive ORs, since these three cannot be true at the same time.

But, $(1\+a)bc\+a(1\+b)c\+abc=bc\+ac\+abc\+abc\+abc=bc\+ac\+abc$.

Then, we have $U\ket{abc}=(-1)^{bc+ac+abc}\ket{abc}=(-1)^{bc}(-1)^{ac}(-1)^{abc}\ket{abc}$.

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