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9

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 attached to a specific register, it's just an overall multiplicative factor. Now let's use a superposition on the first register: $$(|0\rangle+|1\rangle)|u\rangle\... 7 A first remark This same phenomenon of 'control' qubits changing states in some circumstances also occurs with controlled-NOT gates; in fact, this is the entire basis of eigenvalue estimation. So not only is it possible, it is an important fact about quantum computation that it is possible. It even has a name: a "phase kick", in which the control qubits (or ... 6 You should know a bound on the eigenvalues (both upper and lower). As you say, you can then normalise A by rescaling t. Indeed, you should do this to get the most accurate estimate possible, spreading the values \lambda t over the full 2\pi range. Bounding the eigenvalues is not typically a problem. For example, you're probably requiring your matrix ... 5 Neither. Phase estimation algorithm does not estimate a property of a qubit state (and the angles \theta and \varphi in your question are exactly that - a property of a given qubit state). Rather, it estimates the eigenvalue of a given unitary U that corresponds to the given eigenvector |\psi\rangle - the angle \alpha in the following equation: ... 5 It depends on the papers but I saw 2 approaches: In most of the papers I read about the HHL algorithm and its implementation, the Hamiltonian evolution time t is taken such that this factor disappear, i.e. t = t_0 = 2\pi. The approximate eigenvalue is often written \tilde \lambda. In some paper this notation really means "the approximation of the ... 4 @user1271772's answer is excellent, and absolutely the right answer. I just wanted to add in some additional perspective, given recent questions regarding Hamiltonians. Many physicists start from the Hamiltonian being the underlying thing that determines evolution, and unitaries are derived as a consequence. They start from the Schrödinger equation,$$ i\...

3

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 $|u\rangle$ is specifically chosen to be an eigenvector of $U$. That means $U|u\rangle=e^{i\phi}|u\rangle$. But there are different eigenvectors with different ...

3

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 think this is going to help you with the phase estimation algorithm. In quantum phase estimation application of $U^c$ is followed by inverse Fourier transform ...

3

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. After the algorithm has run, we are left with the following state on the first register: $$\frac{1}{2^{n}}\sum_{x=0}^{2^n - 1} \sum_{k=0}^{2^n - 1}e^{-\frac{... 3 I don't see the need for the swap gate either. Although I also don't think you need the set of swap gates that you're wanting. Remember that the standard implementation of the Fourier transform outputs the binary string j\in\{0,1\}^4 where the eigenvalues are of the form 2\pi j/16 but in reverse order, so the least significant bit is at the top, and the ... 2 What am I missing here? Where did the factor of \frac{t}{2\pi} vanish in their algorithm? Remember that in Dirac notation, whatever you write inside the ket is an arbitrary label referring to something more abstract. So, it is true that you are finding the (approximate) eigenvector to U, which has eigenvalue e^{-i\lambda t} and therefore what you're ... 2$$x^0 \bmod N = 1 \implies x^r \bmod N = 1$$as by definition, r is the order of x \bmod N i.e. r is the least integer that satisfies x^r=pN+1 for some integer p. 2 Was a clarifying comment: If I'm interpreting your confusion correctly. You're thinking you just need to say e+1 \leq l not the l \leq 2^{t-1}  part. After all there is no such coefficient as \alpha_{100000} if t is only 3 for example. All that 2^{t-1} is doing is making sure there are only 2^t coefficients. That is it is a qudit with d=t. Is ... 1 One thing that I noticed. If cu3 gate from q[2] to q[0] is some U, then the cu3 from q[2] to q[0] should be U^2 in the phase estimation algorithm, but the comparisons of operators with the help of numpy.array showed me that it's not true here. I tried to implement by replacing cu3 part of the QASM code with the following: cu3(1.6, -1.12, 2.03) q[... 1 3 things I see from your implementation of inverse QFT: SWAP gates are missing prior to applying Hadamard gates and cu1 gates. The Hadamard gate should come first before cu1 gates. The angles of cu1 gates, how I understand inverse QFT, should be different. Here is inverse QFT that worked for me with not touching other parts of the code: def qft_dagger(... 1 So, are you asking whether it's possible (in the qubit case) to perform an arbitrary map of the form$$ ae^{i\theta}|0\rangle+be^{i\phi}|1\rangle\rightarrow \frac{1}{\sqrt{2}}(e^{i\theta}|0\rangle+e^{i\phi}|1\rangle),  if $a$ and $b$ are unknown real numbers satisfying $a^2+b^2=1$? If this is the case, this is impossible. Consider a simple case where ...

1

First point: in most of the cases, the QPE algorithm cannot output the integer $x$. That is why $w$ is introduced in the algorithm: to represent the closest approximation of $x$ that can be returned by the QPE. About your question, no the QPE is not always 100% successful (in this case successful means that the algorithm returns $w$, the closest ...

1

Numerators and denominators of convergents of continued fractions are always co-prime. If $s$ and $r$ have a common factor, the $r'$ returned by the continued fractions algorithm would be a factor of $r$. For example, $r=60$, provided the phase estimation step gives a good approximate to $40/60$ (i.e. $s=40$), you'll get $2/3$ as a convergent. Performing ...

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