New answers tagged phase-estimation
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How can I understand the notation for the final state of QPE algorithm?
Given a unitary operator $U$ the algorithm estimates $\theta$ in $U|\psi \rangle = e^{2\pi i \theta} |\psi\rangle$. So $\theta$ is an angle that determines a complex eigenvalue $e^{2\pi i \theta}$ ...
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How can I understand the notation for the final state of QPE algorithm?
There are "2" cases for the final result:
The outcome imply that $2^t\theta=Integer$, in that case, yes, exactly, it will be a state with the binary encoding of $2^t\theta$
In case $2^t\...
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Does phase kickback require the system to be in the eigenstate?
The first circuit equality fails when $|\psi_k\rangle$ is not an eigenstate of $U$.
A simple way to see this is to set the control qubit to $|1\rangle$. In this case, the RHS circuit is equivalent to $...
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Does phase kickback require the system to be in the eigenstate?
Disclaimer: please do not consider this to be a definite answer as it is not fully rigorous. I am just giving you some evidence or clues from one practical example.
You can write the state $|\psi_k\...
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Understand $U|\psi\rangle = e^{2\pi i\phi}|\psi\rangle$ in phase estimation algorithms
The goal of QPE is to estimate an eigenvalue of $U$. Being $U$ unitary, its eigenvalues have unit modulus. Any complex number with unit modulus can be written as $e^{i\phi}$ with $\phi\in[0,2\pi]$, or ...
glS♦
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Understand $U|\psi\rangle = e^{2\pi i\phi}|\psi\rangle$ in phase estimation algorithms
You are right. The main purpose of using QPE is to examine the eigenstate of Hamiltonian (energies) when applied to some system.
$e^{−iE_1t}$ should be set, just as physics tells you, and qubits do ...
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Applying QPE on a large matrix on amazon-braket
This error occurs because there's a mismatch between the dimensions of the unitary matrix and the qubits it's acting on (for example, a 16 x 16 matrix acting on 5 qubits). I'm suspecting there's a bug ...
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Writing a Phase Estimation function in Cirq
You are passing estimate_phi the an operator: CZPowGate(exponent=2*phi). Then in ...
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