18 votes

Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?

A first remark This same phenomenon of 'control' qubits changing states in some circumstances also occurs with controlled-NOT gates; in fact, this is the entire basis of eigenvalue estimation. So not ...
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15 votes
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Why does the "Phase Kickback" mechanism work in the Quantum phase estimation algorithm?

Imagine you have an eigenvector $|u\rangle$ of $U$. If you have a state such as $|1\rangle|u\rangle$ and you apply controlled-$U$ to it, you get out $e^{i\phi}|1\rangle|u\rangle$. The phase isn't ...
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9 votes
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Quantum phase estimation and HHL algorithm - knowledge of eigenvalues required?

You should know a bound on the eigenvalues (both upper and lower). As you say, you can then normalise $A$ by rescaling $t$. Indeed, you should do this to get the most accurate estimate possible, ...
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6 votes
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Are the squared absolute values of the eigenvalues of a unitary matrix always 1?

Good question. The answer turns out to be Yes. You don't even need the vector to be normalized. Watch: Start with the definition of eigenvalues and eigenvectors: $$ \begin{align} U|\psi\rangle &=...
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6 votes
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Quantum Circuit for $e^{iAt}$ Hamiltonian Simulation in HHL algorithm

As requested through the comment by the OP. Given a Hermitian matrix $H$, we can always write it as linear combination of Pauli strings. That is, $$ H = \sum_i \alpha_i P_i \hspace{1 cm} P_i \in \{I,...
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6 votes
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What is the difference between amplitude amplification, amplitude estimation, and phase estimation?

To find the differences between them, you only need to know the aim of the problems. The aim of AA is to find the answers from unstructured data(or more directly, amplify the probability of the right ...
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  • 2,468
6 votes
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Can we use Hadamard test to estimate phases?

So QPE using $\mathcal{O}(1/\epsilon)$ queries to $U$ outputs an estimate of the eigenphase $\theta$ given a corresponding eigenvector with additive error and $\Omega(1)$ probability. The method using ...
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  • 314
5 votes
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Which angle is estimated by the phase estimation algorithm?

Neither. Phase estimation algorithm does not estimate a property of a qubit state (and the angles $\theta$ and $\varphi$ in your question are exactly that - a property of a given qubit state). ...
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5 votes
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Quantum algorithm for linear systems of equations (HHL09): Step 1 - Confusion regarding the usage of phase estimation algorithm

It depends on the papers but I saw 2 approaches: In most of the papers I read about the HHL algorithm and its implementation, the Hamiltonian evolution time $t$ is taken such that this factor ...
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  • 4,497
5 votes

Are the squared absolute values of the eigenvalues of a unitary matrix always 1?

@user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the ...
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5 votes
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Given a state $|\phi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+e^{i\theta}|1\rangle)$, how do I know the angle $\theta$?

Answer to the first question: As mentioned in the comments of the question I assume that we can prepare $|\phi \rangle$ as many as we want. Let's calculate the relative phase for this one qubit pure ...
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5 votes
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How to get the relative phase of an entangled pair of qubits

Apply a CNOT gate with one of the qubits as control and the other as target. You'll get $$\frac{1}{\sqrt{2}}(|0\rangle+e^{i\theta}|1\rangle) \otimes |0\rangle$$ Use the methodology from How to get ...
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5 votes
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Can quantum search be performed without phase estimation?

I'm sure I'm not the first to notice this; does it appear somewhere? Section 6 of the 1996 paper "Tight Bounds on Quantum Searching" by Boyer et al uses this strategy of iteratively trying ...
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5 votes
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Phase estimation using $U_3$ gate

If you were to apply QPE to this unitary, what you will get, assuming you start with a proper eigenvector $|\Lambda\rangle$, is an estimation of $x$, if the associated eigenvalue $\Lambda$ is written ...
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5 votes
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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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4 votes

What is Quantum Phase Estimation in Shor's Algorithm?

Phase estimation is the process by which you are given a controlled-$U$ unitary, and a state that you are promised is an eigenvector of $U$ with eigenvalue $e^{2\pi ix/2^t}$, then you can use a $t$-...
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4 votes
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Why can I use the Sum of Eigenvectors for Phase Estimation in Shor

As you say, we would be able to use phase estimation if we knew the eigenvector $|u_s \rangle$ that depends on the order $r$ and the integer $s$ which is $0 \leq s \leq r - 1$. However, we don't know ...
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4 votes

A question regarding quantum phase estimation algorithm

The objective of all those gates is to put the quantum state in a "nice form" to manipulate. Let me explain. Let's note $U |\psi \rangle = e^{2i\pi\theta}|\psi\rangle$ all the variables we ...
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4 votes
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Simulating QPE + Grover using Low-Rank Stabilizer Decomposition

I agree that the Bravyi et al. paper is not easy to understand and they should have made some reference implementation available. Without going into details, I don't think it is likely to get an ...
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4 votes

Question about the phase kickback in the phase estimation algorithm

The phase is applied to the overall wave function $|\phi\rangle$, therefore you can factor the phase to any individual qubit. For example if we have a wave function as a result of a controlled ...
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4 votes
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Interpreting the phase in QPE

Given your definition $U|\psi\rangle=e^{2\pi i\theta}|\psi\rangle$, then the $U$ that you use can be any value of $\theta$. However, the answer you get out will (with high probability) be the best $t$-...
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3 votes
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How to implement exponentiation of a gate without breaking complexity?

As a general rule, just because you can produce controlled-$U$, it does not mean that you can produce controlled-$U^{2^k}$ with the same complexity. Modular exponentiation is a very special case where ...
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  • 47.2k
3 votes

Given a state $|\phi\rangle=\frac{1}{\sqrt{2}}(|0\rangle+e^{i\theta}|1\rangle)$, how do I know the angle $\theta$?

I would just like to share a code for testing a phase measurement on IBM Q: ...
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3 votes
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How to analyze highly entangled quantum circuits?

The $U$ used in phase estimation is not only a diagonal matrix with the same diagonal elements. Instead, it is an arbitrary unitary matrix. The way that you analyse it, instead, is that the input $|u\...
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3 votes
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SWAP gate(s) in the $R(\lambda^{-1})$ step of the HHL circuit for $4\times 4$ systems

I don't see the need for the swap gate either. Although I also don't think you need the set of swap gates that you're wanting. Remember that the standard implementation of the Fourier transform ...
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  • 47.2k
3 votes

Quantum algorithm for linear systems of equations (HHL09): Step 1 - Confusion regarding the usage of phase estimation algorithm

What am I missing here? Where did the factor of $\frac{t}{2\pi}$ vanish in their algorithm? Remember that in Dirac notation, whatever you write inside the ket is an arbitrary label referring to ...
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  • 47.2k
3 votes

Phase estimation error analysis

Let me augment the discussion by adding some insight into the derivation of the estimate provided. This will give you a good understanding of when the result is an approximation and when it is precise....
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3 votes
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Sketch the quantum logic gates correctly and give a proof for the identity

These two circuits produce the same results - in both cases you'll get $|0\rangle \otimes |\psi\rangle$ with 50% probability or $|1\rangle \otimes U|\psi\rangle$ with 50% probability. But I don't ...
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3 votes
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Finding phase angle in Q#

If this is part of a debugging effort then you can use the DumpRegister function. If you have a qubit q which is in the general ...
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  • 752
3 votes
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Quantum Phase estimation with $2\pi$ replaced with $2e$

No - the key intuition of QPE is that $e^{2 \pi i \theta} $ holds for $\theta \in [0, 1) $, and the reading out the ancilla provides the binary representation of the fraction. If you changed the ...
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