New answers tagged circuit-construction
0
@ahelwer has provided a smart construction, and I will provide a more brute-force method as a complementary for beginners like me. It actually starts out by listing all the possible 16 moves. Let x,y=0,1,2,3, then all the possible moves are
$
x=0, \quad y=(0123)\rightarrow(0123) \\
x=1, \quad y=(0123)\rightarrow(1230) \\
x=2, \quad y=(0123)\rightarrow(2301) ...
1
Constructing a circuit for $C^1(U)$ for rotation operators with TWO single qubit gates and CNOT gate
The question comes in two parts. Firstly, for the top circuit, can you find the gates $A$, $B$ and $C$ such that $ABC=I$ and $CXBXA=R_x(\theta)$ or $R_y(\theta)$. Secondly, can you reduce either of these to the lower circuit with $AB=I$ and $XBXA=R_x(\theta)$ or $R_Y(\theta)$.
The first of these is detailed in the bit of Nielsen and Chuang just before the ...
2
Note: $I_k$ is unit matrix of order $k$ in the following text.
First step of the algorithm is $H \otimes H \otimes I_2 \otimes I_2$ as you mentioned.
A controlled gate $U$ with $n$ qubits between the control qubit and the target qubit can expressed as a matrix
$$
CU_{n} =
\begin{pmatrix}
I_{\frac{N}{2}} & O_{\frac{N}{2}} \\
O_{\frac{N}{2}} & I_{\...
0
You can use $Ry$ (y-rotation gate). Its general matrix is
$$
Ry(\theta) =
\begin{pmatrix}
\cos(\theta/2) & -\sin(\theta/2) \\
\sin(\theta/2) & \cos(\theta/2)
\end{pmatrix}
$$
Applying the gate on $|0\rangle$ state, you get a state
$$
|\psi\rangle = \cos(\theta/2)|0\rangle + \sin(\theta/2)|1\rangle
$$
Hence $\cos(\theta/2) = \frac{1}{\sqrt{3}}$ ...
0
Here is an implementation of a circuit producing state $|\psi\rangle = \frac{1}{\sqrt{3}}(|00\rangle + |01\rangle + |10\rangle)$ on IBM Q:
Note that $\theta = 1.2310$ for $\mathrm{Ry}$ on $q_0$. $\theta = \frac{\pi}{4}$ and $\theta = -\frac{\pi}{4}$ for first and second $\mathrm{Ry}$ on $q_1$.
The $\mathrm{Ry}$ on $q_0$ prepares qubit in superposition $|...
3
Consider some simpler cases, $(j,j+1)$ for general $j$. Then you can do you want with plugging in $j=i$, $j=i+1$, $j=i+3$ and $j=i+4$ and concatenating the circuits appropriately and then simplifying.
So how to do $(j,j+1)$? That is conjugate to $(0,1)$, so just consider that for now.
$(0,1)$ would be NOT but controlled on making sure all the higher places ...
2
A brute force solution :). You can also obtain CCH via qiskit's basic gates with help of get_controlled_circuit method.
from qiskit import *
from qiskit.aqua.utils.controlled_circuit import get_controlled_circuit
q_reg = QuantumRegister(3, 'q')
qc_h = QuantumCircuit(q_reg)
qc_ch = QuantumCircuit(q_reg)
qc_cch = QuantumCircuit(q_reg)
qc_h.h(q_reg[0])
...
2
Summarization based on discussion with user met927:
Transpiled circuit form depends on used backend - it is different for simulator and real quantum processor:
On simulator, the $\mathrm{CH}$ gate is transpiled to the circuit shown above
On real quantum processor, the gate is implemented with two $\mathrm{U2}$ gates and $\mathrm{CNOT}$ (i.e. like in the ...
4
Assuming you've got Toffoli and single-qubit rotations, you can implement the following:
This basically works because if either of the controls is not $|1\rangle$, the Toffoli does nothing and the two single-qubit unitaries cancel each other. Whereas, if both controls are $|1\rangle$, then the net gate on the target qubit is
$$
(\cos\frac{\pi}{8}I+i\sin\...
4
Nielsen and Chuang book may be confusing; I will try to explain.
Any classical algorithm can be presented as a circuit consisting of $NOT$ and $AND$ gates; this means that if we can make quantum gates computing $NOT$ and $AND$, and make fanout, we can run any classical algorithm on quantum computer. $NOT$ is reversible and we have quantum $NOT$ gate; many ...
1
Based on paper Five Two-Bit Quantum Gates are Sucient to Implement the
Quantum Fredkin Gate provided by Norbert Schuch, I realized that there is a more efficient implementation in terms of number of gates. Here is a result:
Matrix of CNOT acting on $|q_1\rangle$ controlled by $|q_2\rangle$ is
\begin{equation}
CNOT_{2}=
\begin{pmatrix}
1 & 0 & 0 &...
7
It's not possible to implement a Toffoli using only Fredkin gates, because Fredkin gates preserve the number of 1s in the state while Toffolis do not.
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