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CX and CZ commutation

$$CX=(I\otimes H) CZ( I\otimes H)$$ So we can express $$CX_{1,3}CZ_{2,3}=(I\otimes I\otimes H) CZ_{1,3}(I\otimes I\otimes H) CZ_{2,3}$$ By the first identity I showed, $CX=(I\otimes H) CZ( I\otimes H)$...
GaussStrife's user avatar
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CX and CZ commutation

$E$ is the map given by $E|i,j\rangle |\psi \rangle = |i,j\rangle Z^j X^i |\psi \rangle$. $E'$ is the map given by $E'|i,j\rangle |\psi \rangle = |i,j\rangle X^i Z^j |\psi \rangle = (-1)^{ij}|i,j\...
john's user avatar
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Creating a uniform superposition of a subset of basis states

Just adding to Quantum_Brains' answer, the algorithm by Alok Shukla et al. has been implemented as UniformSuperPositionGate in Qiskit release 1.2. Therefore, you can directly use it in Qiskit. Here is ...
Quantum Learner's user avatar
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Block encoding of a diagonal matrix with equidistant eigenvalues

$\newcommand{ket}[1]{|#1\rangle}\newcommand{bra}[1]{\langle#1|}$ We can construct a block-encoding with optimal scale factor $\alpha = 2^n-1$ and using only a single ancilla qubit as follows: Let $...
Neel Sandeep Modi's user avatar
4 votes

Finding a unitary transformation to swap the control bit

Unitaries $U$ are diagonalisable. That means there exists a unitary $V$ such that $VUV^\dagger=P$, where $P$ is a phase gate (diagonal matrix). This means that we can think of controlled-$U$ as the ...
DaftWullie's user avatar
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Finding a unitary transformation to swap the control bit

Given a 2 qubit controlled unitary $U$, one generally expresses it as $$|0\rangle\langle0|\otimes I + |1\rangle\langle1|\otimes P,$$ where in this example $P$ can be any unitary operator. In your ...
GaussStrife's user avatar
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3 votes
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Power of Toffoli vs T in quantum logic

I think of the gatesets as equivalent. Clifford+Toffoli can do everything Clifford+T can do, if given access to a single $|T\rangle$ state. Clifford+Toffoli can duplicate the T state, allowing an ...
Craig Gidney's user avatar
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3 votes
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Swapping amplitudes of orthogonal components

Yes, unitaries map orthonormal bases to orthonormal bases. As $|A\rangle$ and $|A^{\perp}\rangle$ are orthogonal they can be extended to an orthonormal basis on the space. Let's call such a basis $\{|\...
Rammus's user avatar
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