New answers tagged circuit-construction
0
Just to emphasise the point on the accepted answer...
The two gates, when applied to a specific state, give the same output. That does not mean they are the same, because the action on another state may be different (and I don't just mean a global phase).
For example, if you take the state $|+\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$ then
$$
H|+\rangle=|0\...
7
While Hadamard gate is defined as
$$
H= \frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & 1 \\
1 & -1
\end{pmatrix},
$$
$y$-rotation by $\pi/2$ leads to gate
$$
Ry(\pi/2)= \frac{1}{\sqrt{2}}
\begin{pmatrix}
1 & -1 \\
1 & 1
\end{pmatrix}.
$$
So, there is a difference in position of -1 in the second column. Application of the $X$ gate returns the -1 in $...
6
If you are trying to implement a fault-tolerant quantum computation, you need to implement unitary gates that act on logical qubits. You typically have a finite set of these gates available, and what you really care about is making your operations in such a way as to keep the fault-tolerant threshold as small as possible.
If you calculate a fault-tolerant ...
1
Leaving this here. Recently published, might be useful for this question.
Efficient Evaluation of Exponential and Gaussian Functions on a Quantum Computer
https://arxiv.org/abs/2110.05653
1
There currently isn't a controlled Pauli product gate in Stim. You have to decompose it into a series of CX, CY, and CZ gates.
# Apply X1*Y2*Z3 controlled by qubit 0
CX 0 1
CY 0 2
CZ 0 3
# Apply X1*Y2*Z3 if latest measurement result was True
CX rec[-1] 1
CY rec[-1] 2
CZ rec[-1] 3
0
5.1 Implementation of oracles P_A and P_b (p. 13)
https://arxiv.org/pdf/1910.10649.pdf
5
The matrix
$$
M = \frac{1}{\sqrt{2}}\begin{bmatrix}-i & 1\\-1 & i\end{bmatrix}
$$
resembles
$$
X/2 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & -i\\-i & 1\end{bmatrix}\tag1
$$
where we follow the notation $\pm X/2$ for the $\pm\frac{\pi}{2}$ rotation around the $X$ axis as used in the table B.6 on page 101 in Julian Kelly's PhD thesis. We can make ...
1
Qiskit defines the $RX$ gate as follows:
$$
RX(\theta) = \exp\left(-i \frac{\theta}{2} X\right) =
\begin{pmatrix}
\cos{\frac{\theta}{2}} & -i\sin{\frac{\theta}{2}} \\
-i\sin{\frac{\theta}{2}} & \cos{\frac{\theta}{2}}
\end{pmatrix}
$$
Thus, setting $\theta = \pi$, would give us:
$$
RX(\pi) = \...
1
Retrieving an arbitrary qubit, even using an ancilla qubit, would violate the no-cloning theorem. This can be seen in the simplest case where the array stores a single qubit and is indexed by a single qubit. So the array is of the form
$$ \left|\psi\right> = a \left|00\right> + b \left|01\right> + c \left|10\right> + d\left|11\right>$$
where $|...
3
The decomposition they give is the following:
$$
R_z(\theta) = R_x\left(\pi / 2\right) R_y(\theta) R_x\left(-\pi / 2\right)
$$
Therefore, the Qiskit code would look like:
from qiskit import QuantumCircuit
from qiskit.quantum_info import Operator
import numpy as np
theta = np.pi / 4
qc = QuantumCircuit(1)
qc.rx(-np.pi / 2, 0)
qc.ry(theta, 0)
qc.rx(np.pi / 2,...
3
You can decompose the T gates themselves to create a Toffoli Gate. Here is one way of doing this:-
You can refer to this Qiskit chapter if you are interested and want to understand gate decomposition: https://qiskit.org/textbook/ch-gates/more-circuit-identities.html
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