New answers tagged

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 ...
amihart's user avatar
  • 41
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, ...
Abdullah Khalid's user avatar
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 ...
Adam Zalcman's user avatar
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 ...
Adam Zalcman's user avatar
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 ...
Craig Gidney's user avatar
  • 32.5k
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}\...
Mariia Mykhailova's user avatar
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 ...
DaftWullie's user avatar
  • 54.5k
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 ...
DaftWullie's user avatar
  • 54.5k
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\}^...
Dudu Ponar's user avatar
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 ...
Mark Spinelli's user avatar

Top 50 recent answers are included