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Here are the gates: \begin{equation} H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\1&-1 \end{pmatrix} \qquad R_y(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\\sin(\theta/2) & \cos(\t …

In addition to Martin Vesely's answer: Here I want to share an alternative way of implementing ccu1(t, q[0], q[1], q[2]) gate, where $t$ is some phase. It can be shown that ccu1 gate can be decompose …

If $U_1 = U_2$, then $U_1 U_2^{\dagger} = I$. So let's add to the first circuit the inverse/dagger of the second one: This whole thing should be an identity. Note that, for this, we should prove th …

Because $CNOT = I\otimes H \cdot CZ \cdot I\otimes H$ as was mentioned here, and because $CZ(q_1, q_2) = CZ(q_2, q_1)$, we can rewrite the circuit in this way (by adding Hadamards as needed): The lin …

An approach for Hamiltonian simulation: Any Hermitian (Hamiltonian) matrix $H$ can be decomposed by the sum of Pauli products with real coefficients (see this thread). An example of 3 qubit Hamiltonia …

Controlled version of $e^{iHt}$: Often in the algorithms (e.g. in HHL or PEA), we want to construct not the circuit for Hamiltonian simulation $e^{iHt}$, but the controlled version of it. For this, we …

Here are the actions for the given transformation on the computational basis states: $$|000\rangle \rightarrow |000\rangle \qquad |001\rangle \rightarrow |010\rangle \qquad |010\rangle \rightarrow |0 …

By using similar ideas from this answer I have found this circuit: Thought process: The unitary is a permutation matrix that doesn't change bitstrings except $U |100\rangle \rightarrow |011\rangle$ a …

A brute force solution :). You can also obtain CCH via qiskit's basic gates with help of get_controlled_circuit method. from qiskit import * from qiskit.aqua.utils.controlled_circuit import get_contr …

We can't implement $e^{iZ_1 \otimes Z_2 \otimes Z_3 \theta}$ with three separate rotations. In other words: $$e^{iZ_1 \otimes Z_2 \otimes Z_3 \theta} \ne e^{i Z_1 \theta} \otimes e^{i Z_2 \theta} \oti …

Here is the CNOT gate: $$CNOT = |0\rangle \langle 0|\otimes I + |1\rangle \langle 1| \otimes X$$ So: $$(I \otimes H) CNOT (I \otimes H) = |0\rangle \langle 0|\otimes HH + |1\rangle \langle 1| \otim …