# Tag Info

4

If you know, in advance, that the state you want to deamplify is specifically $|000\rangle$, there are a couple of strategies that you could follow. For example, introduce an ancilla and perform the multi-controlled not, targeting the ancilla, where it is controlled off every qubit in the original state being in the $|0\rangle$ state. So, you'd be doing $$\... 3 Any state |\Psi\rangle can be decomposed using a two-dimensional subspace comprising marked and unmarked states,$$ |\Psi\rangle=\alpha|\Psi_m\rangle+\beta|\Psi_u\rangle. $$Part of the assumption is that you have a marking oracle that acts as$$ |\Psi\rangle|0\rangle\rightarrow\alpha|\Psi_m\rangle|1\rangle+\beta|\Psi_u\rangle|1\rangle. $$This is exactly ... 3 Grover's algorithm is a special case of the amplitude amplification algorithm where the number of good entries G in the N-item database is 1. In a nutshell: In the Grover's algorithm Wiki page that you linked, the keyword is "unique". Given f:\{0, \ldots, N-1\} \to \{0, 1\} such that f(x) = 1 for exactly one x (say \omega), Grover's algorithm ... 3 Only a partial answer, the Deutsch-Jozsa algorithm is an example of an exact algorithm. In my view, the algorithms differ exactly in how the answer is given. Either with probability 1 for exact algorithms, or with a bounded probability for approximate ones. I would say you cannot use amplitude amplification in exact algorithms, as this would imply that ... 2 It seems that there is a missing \mathrm{X} and \mathrm{H} gates on qubit q_2. I used this Grover algoritm shape: Note 1: controlled \mathrm{Z} is replaced by \mathrm{CNOT} and Hadamards on both sides. Note 2: put your Oracle instead of dashed line. Note 3: you do not have to measure q_2. 2 Suppose we have two quantum circuits, the first computes (or at least approximates) the classical \sqrt{\cdot} function$$S|x\rangle|0\rangle=|x\rangle |\sqrt{x}\rangle,$$while the second circuit A computes (again could probably just approximate) the \arccos(\cdot) function$$A|x\rangle|0\rangle=|x\rangle |\arccos(x)\rangle. Lastly, suppose we have ...

2

I asked basically the same question on CS stack exchange before this community was created. The answer is that the class of exact quantum algorithms has a name (EQP) but isn't very natural to study theoretically, because whether or not an exact algorithm can be executed depends entirely on the gate set that you have available, and moreover there's no ...

2

Say you want to factorise a large integer $N$. We know (inefficient) classical algorithms to do this, a naive example being: just check all combinations of smaller numbers until you find one that multiplies to $N$. You can make this into a quantum algorithm by simply converting each operation in your classical algorithm into a reversible one (there are ...

1

It seems that $\mathrm{CNOT}$ gates should not be in your circuit. Here is Grover algorithm for 3 qubits: Put your Oracle instead of dashed line. The Oracle should have three inputs $q_0$, $q_1$ and $q_2$, output should be on qubit $q_3$ after $\mathrm{H}$ gate.

1

Yes. It is, for instance, part of Grover's algorithm and to be precise it is the 'Amplitude Amplification' part. $2| \psi \rangle \langle \psi | - I$, which will increase the amplitudes by their difference from the average

1

The probability amplitudes of a quantum system comprising of several qubits can be changed (amplified / de-amplified) with the application of suitable Rotation Gate. However, the sum of probabilities of collapsing to all possible states will remain unity, so if some states are getting amplified then others will get de-amplified. Is there any practical ...

1

I think your confusion starts at the point where you refer to 'amplitude of the qubit'. The amplitude of a qubit is constant and always 1. When you measure it, you will always get a quantity of "1". Basically, this is the Bloch sphere: the qubit can point to any point on the surface of the sphere, so it has radius (amplitude) of 1. Amplitude amplification ...

Only top voted, non community-wiki answers of a minimum length are eligible