Joseph Geipel
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2 answers
5 votes
446 views
N-Toffoli on Cirq
Accepted answer
7 votes

This is actually very easy in Cirq. The controlled_by method can be used to automatically make any given gate controlled by an arbitrary number of control qubits. Here is a simple example for creating ...

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1 answers
6 votes
334 views
Grover's diffusion on subset of input space
6 votes

This can work. There's no reliance on powers of two or anything like that in the basic conception of the algorithm. If $S$ is a subset of computational basis states with $N$ elements and you have a ...

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1 answers
4 votes
240 views
How do I prove that $P_\pm=\frac12(1\pm U)$ if $U^2=I$?
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5 votes

First, we can start with $U = P_+ - P_-$, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If $U^2 = I$, that means $I = (P_+ - P_-)(P_+ - P_-)...

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1 answers
4 votes
489 views
Second reflection in the Grover's algorithm
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5 votes

The Grover diffusion operator, the second phase of the algorithm, is given by: $$D = 2\left|s\right>\left<s\right| - I$$ where $\left|s\right> = \frac{1}{\sqrt{N}}\sum_{x=0}^{N-1}\left|x\...

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3 answers
2 votes
122 views
Why can no pair of single qubits look like $\frac{1}{\sqrt2}(|00\rangle+|11\rangle$)?
4 votes

The tensor product of two vectors $\begin{pmatrix} a \\ b \end{pmatrix}$ and $\begin{pmatrix}c \\ d \end{pmatrix}$ is $\begin{pmatrix} ac \\ ad \\ bc \\ bd \end{pmatrix}$. In this case, determining ...

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1 answers
1 votes
66 views
What does "large-scale universal quantum computation" mean?
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4 votes

"Large-scale" generally means that the quantum computer's ability scales to practical instances of problems. For the case of factoring, RSA keys are usually 1024/2048/4096 bit sized: if a ...

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1 answers
3 votes
293 views
What is the procedure of finding z-y decomposition of unitary matrices?
3 votes

For a 2x2 unitary matrix given by four complex numbers $\begin{pmatrix} a & b\\ c & d \end{pmatrix}$, the aspects of the $z$-$y$ decomposition $e^{i\alpha}R_z(\beta)R_y(\gamma)R_z(\delta)$ can ...

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2 answers
3 votes
183 views
How does the QFT represent the frequency domain?
3 votes

If you apply the $n$-qubit QFT defined as $\frac{1}{\sqrt{N}}\sum_{k = 0}^{N - 1}\sum_{n = 0}^{N - 1}a_n e^{2 \pi i n k/N}\left|k \right>$ acting on a state $\sum_{x = 0}^{N - 1}a_x\left|x\right>...

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2 answers
0 votes
188 views
Explanation of NOT gate in Hadamard basis
3 votes

If $\left|\psi\right> = \alpha \left|0\right> + \beta \left|1\right>$, then $\alpha = \left<0|\psi\right>$ and $\beta = \left<1|\psi\right>$. A 1-qubit unitary operator can be ...

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2 answers
8 votes
742 views
What does it mean to "measure an operator"?
3 votes

In this case, "measuring an operator" is meant to describe measuring the observable associated with the operator. More commonly one would put a "using" or "with" or "...

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1 answers
3 votes
210 views
Implementing 3-Qubit Grover Algorithm in Qiskit
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2 votes

Number of states marked isn't a strict correspondence with number of CZ gates: your new oracle still marks two states, but, rather than $\left|101\right>$ and $\left|110\right>$, it marks $\left|...

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2 answers
-3 votes
179 views
Have you ever seen the preparation of the state $a^{*}|0\rangle+b^*|1\rangle$ and $a|0\rangle+b|1\rangle$ from one initial state?
2 votes

$\left<\psi\right|$ is not a state of a quantum system, it is a linear functional that takes a quantum state and returns a scalar. In terms of basic Linear Algebra, it is a row vector rather than a ...

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3 answers
4 votes
880 views
How can a CNOT gate change the control qbit (e.g. in the Deutsch Oracle problem)?
2 votes

As everyone knows, all functions of quantum computing are inverses of each other. Hence, the 2 H gates cancel out. Quantum gates all have inverses, but the inverse of a gate is not necessarily the ...

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1 answers
2 votes
46 views
Using Quantum Bit String Comparision inside Grover's algorithm
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1 votes

The problem is actually in the oracle. First, $O_1$'s state is 1 if $a$ (the constant which you are checking against) is greater than $b$, so 000 and 001 will be the only marked states: you want $O_2$ ...

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