New answers tagged circuit-construction
3
votes
How to simulate a $CNOT$ only using a single qubit?
Quantum operations have to be reversible and the operation you described is not reversible. Both 0 and 1 map to 0, so it is a many-to-one function and thus if you are just given 0 you have no ...
2
votes
Accepted
Encoding circuit for $[\![6, 4, 2]\!]$ code
This can easily be done using stac. More explanations of the process for generating the encoding circuit is explained in a previous answer. But, since, this question is specifically asking for $[[2m, ...
2
votes
Map $n$ qubit state with complex amplitudes to $n+1$ qubit state with real amplitudes
TL;DR: The proposed map fails to be insensitive to the global phase. For example, it can tell apart $|0\rangle\equiv[1,0]^T$ from $|0\rangle\equiv[i,0]^T$ even though these two different mathematical ...
2
votes
Encoding circuit for $[\![6, 4, 2]\!]$ code
We can obtain an encoding circuit for the $[\![m,m-2,2]\!]$ code for any even $m$ by generalizing the circuit displayed in the question. More precisely, for every new qubit we prepend a CNOT gate with ...
4
votes
Map $n$ qubit state with complex amplitudes to $n+1$ qubit state with real amplitudes
Call the operation you want to construct $D$ and call the qubit that ends up storing the real/imaginary distinction $q$.
If I gave you $D$, you could apply $D$ then $Z_q$ then $D^{-1}$. The overall ...
3
votes
Map $n$ qubit state with complex amplitudes to $n+1$ qubit state with real amplitudes
This will violate unitarity of the transformation $U$. Consider states
$$\begin{bmatrix}\frac1{\sqrt2} \\ \frac1{\sqrt2}\end{bmatrix} (b=d=0)$$
and $$\begin{bmatrix}i\frac1{\sqrt2} \\ i\frac1{\sqrt2}\...
3
votes
Superposition on a subset of integers
Perhaps the easiest way is to find $n=\lceil\log_2(k+1)\rceil$. Take $n$ qubits in the state $|0\rangle$ and apply Hadamard to each of them.
If $k+1$ was a power of 2, you're done! If not, get an ...
4
votes
Accepted
Can the oracle for $f:\{0,1\}^n \rightarrow \{0,1\}^m$ be implemented with only $n+m$ qubits?
Your argument is correct, but needs to be made with a little care. The way that a universality proof often goes is that you decompose the target unitary into a series of Givens rotations. Each such ...
5
votes
Accepted
Given $f: \{0, 1\}^n\to\{0, 1\}^m$, how many qubits are needed to implement the oracle $\mathcal U|x,0\rangle^{\otimes m}=|x,f(x)\rangle$?
Usually, saying that there is access to an oracle compute $f$ is equivalence to saying that you assume a model in which it is given that $f$ can be computed for free, Namely, In your case $f :\{0,1\}^...
2
votes
Given $f: \{0, 1\}^n\to\{0, 1\}^m$, how many qubits are needed to implement the oracle $\mathcal U|x,0\rangle^{\otimes m}=|x,f(x)\rangle$?
We usually define the BQP complexity class to limit the number of gates, and hence the number of ancilla qubits, to be polynomial in the number of input qubits $n$ (which is also polynomial in the ...
Top 50 recent answers are included
Related Tags
circuit-construction × 660quantum-gate × 241
qiskit × 144
quantum-circuit × 92
programming × 86
algorithm × 86
quantum-state × 83
gate-synthesis × 39
ibm-q-experience × 31
hamiltonian-simulation × 26
grovers-algorithm × 25
measurement × 24
entanglement × 22
matrix-representation × 21
cirq × 21
universal-gates × 19
textbook-and-exercises × 18
quantum-fourier-transform × 16
mathematics × 15
simulation × 15
hadamard × 15
complexity-theory × 14
unitarity × 14
optimization × 14
oracles × 14