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Does the $\text{CNOT}$ gate activate if the control qubit is $-|1⟩?$

Yes, it would. It is sometimes better to think in the vector-matrix representation than in the dirac notations. $$|1\rangle = \begin{bmatrix} 0\\ 1 \end{bmatrix}\,.$$ Now, $$Z|1\rangle = -|1\rangle =...
FDGod's user avatar
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Does the $\text{CNOT}$ gate activate if the control qubit is $-|1⟩?$

Yes. Think of the $-|1\rangle$ state as a superposition of basis states $|0\rangle$ and $|1\rangle$ (albeit a weird one). In this case, you can apply the general rule for the CNOT gate acting on ...
Mariia Mykhailova's user avatar
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general way to decompose a CZ gate on $n$-qubit system

Yes absolutely it is possible! This algorithm may not use the optimal number of gates, but it gets the job done. You may have heard of the SWAP gate which effectively switches the positions of two ...
River's user avatar
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1 vote
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Problem with the mathematical definition of the eigenvalue algorithm on a specific exercise

Calculate the wavefunction step by step by applying each gate in sequence $$ \begin{align} |000\rangle&\xrightarrow{H\otimes H}\frac12\sum_{a,b=0}^1|a\rangle|b\rangle|0\rangle\tag1\\ &\...
Adam Zalcman's user avatar
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3 votes
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Decomposition of a $4 \times 4$ unitary matrix

That paper appears to do their rotations in a very strange order. The method you're interested in is how to use Givens rotations to perform a QR decomposition (see, e.g. https://en.wikipedia.org/wiki/...
DaftWullie's user avatar
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Decomposition of a $4 \times 4$ unitary matrix

Note the following: if we have a vector with 2 entries $\begin{bmatrix} a \\ b \end{bmatrix}$, then we can use a unitary matrix to change the second entry to $0$. Namely, set $r = \sqrt{|a|^2 + |b|^2}$...
Vladimir Lysikov's user avatar
2 votes
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How much complexity is required to implement $\text{C$_\Pi$NOT}$ gate?

I suspect that depends on $\Pi$. There will be easily categorised cases where you can. For example: $\Pi$ projects onto a subset of basis states for which there exists an efficient classical ...
DaftWullie's user avatar
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1 vote

What role does Landauer's principle play in quantum reversibility?

Actually, classical physics is also reversible, whether you're considering classical dynamics or considering it as the limit of quantum physics (which is reversible). The means for neither classical ...
Sam Jaques's user avatar
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Equivalence checking of quantum circuits up to error

Sam's link to the posting on tcs answers the OP's question - but here's some other comments that were hinted at here and there. The exact classical problem is coNP-complete as this reduces to ...
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Frequency of qubits in qiskit simulators

The qubit frequency is not encoded into the noise model. You can however attach noise to the delay gate. See here.
Yael Ben-Haim's user avatar
1 vote

Netlist of the transpiled circuit in Qiskit

It very much depends on what do you mean by a netlist. If you are searching for a human-readable serialization format of a non-dynamic circuit, I would recommend you OpenQASM 2 ...
luciano's user avatar
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5 votes
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When is the square root of a Clifford gate a Clifford gate?

TL;DR A simple sufficient condition is: if the order $r$ of a given Clifford $U$ and the root's degree $n$ are relatively prime, then $U$ has an$^1$ $n$th root. Let $r$ be the order of $U$, i.e. the ...
Adam Zalcman's user avatar
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Phase accumulation for multiple Rabi driving pulses

Good question! Your first equation mixes a couple of abstractions that we typically sweep under the rug, so let's start by taking a step back. When we apply a control pulse to a qubit, it's usually ...
Chris E's user avatar
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Simulator able to run SwitchCase in qiskit?

The latest version of Qiskit Aer (version 0.13) supports SwitchCaseOp. So, if it's OK to run your circuit locally, you can use Aer simulators: ...
Egretta.Thula's user avatar
1 vote
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Conditioned gates with multiple classical bits

You can use bit_xor function from the newly added classical expressions module ...
Egretta.Thula's user avatar
1 vote

Conditioned gates with multiple classical bits

This can be done quite easily. For instance, take the case where you have two classical bits $c_1, c_2$ and would like to implement $X^{c_1+c_2}$ on a qubit $q$. Clearly, applying $X^{c_1}$ followed ...
Tharrmashastha V's user avatar
2 votes
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If_test() - Dynamic Circuits in qiskit

It appears there's a misunderstanding regarding the use of if_test and _else constructs in your code. In Qiskit, these ...
banercat's user avatar
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2 votes
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Constructing a two 3-qubit state involving either X, Y or Z rotation gate

One way to achieve this is the following way. Initialise 1 qubit in the $|0\rangle$-state and rotate along the $Y$-axis to \begin{equation} \frac{1}{2}|0\rangle+\frac{\sqrt{3}}{2}|1\rangle. \end{...
JoJo P's user avatar
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How to fix two flip-bit errors in a 3 qubit input

I was able to figure it out now via taking 3 ancilla qubits instead of 2 and figuring out the invariant in case 2 bits are flipped.
codeit's user avatar
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Regarding the inductive proof that any Clifford gate can be made of Hadamard, phase and c-not

This question is about the second part of the cited exercise. The first part is the single qubit case. In the second part, one is basically supposed to prove by induction that with the given ...
AYS's user avatar
  • 13
2 votes

Time evolution of Hamiltonian

I found out, that one can do that using the $Z$-rotations instead of $X$-rotations because the $Z$-rotation is already diagonal. that makes it way easier to see what $e^{\sigma^{z}}$ should be. So one ...
Ruebli's user avatar
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