# Tag Info

### 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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### 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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### 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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### 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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### $QFT^{-1}$ at the end of Shor's algorithm and $QFT$ at the end of Hidden Subgroup algorithm

$\text{QFT}\big(|0\rangle^{\otimes n}\big) = \text{QFT}^{-1}\big(|0\rangle^{\otimes n}\big) = |{+}\rangle^{\otimes n}$, so a QFT, inverse QFT, or a column of Hadamard gates are all equivalent at the ...
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### $QFT^{-1}$ at the end of Shor's algorithm and $QFT$ at the end of Hidden Subgroup algorithm

Note that $\text{QFT}^2$ is a permutation $|k\rangle \rightarrow |(-k) \bmod 2^n\rangle$. This is a classical operation. It can be applied in the post processing of the measurements, and in fact it ...
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### Requirement of vector 'b' in the definition of Phase Estimation Sampling (PES)

The discussion in question appears to be discussing usage of the Quantum Phase Estimation algorithm when we do not have access to an eigenstate $|\eta_j \rangle$ of the unitary matrix $U$ in question. ...
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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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### 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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### 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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### 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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### 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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### Why are $U$s raised to successive powers of two in quantum phase estimation when we use $n$ register qubits $|0\rangle|0\rangle|0\rangle$?

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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### 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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### 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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### Exponential Quantum Speedup for the Traveling Salesman Problem - where is the catch?

The catch is the parameter $\kappa$, the condition number of the embedding matrix. I can't comment on the correctness of the whole work. This will require a thorough examination of the content. But ...
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### 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....

### 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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### 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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### 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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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 ...