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The angle of each rotation $\theta$ obeys $0\leq \theta \leq \frac{\pi}{2}$ so that $0\leq \frac{\theta}{2} \leq \frac{\pi}{4}$. The Grover Iterations as we know, rotates our initial state vector $|\psi\big>=cos(\theta/2) |\alpha\big> + sin(\theta/2)|\beta\big>$ by the angle $\theta$ a few times, such that eventually $|\psi \big>$ will be ...


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The Grover diffusion operator, the second phase of the algorithm, is given by: $$D = 2\left|s\right>\left<s\right| - I$$ where $\left|s\right> = \frac{1}{\sqrt{N}}\sum_{x=0}^{N-1}\left|x\right>$, which is also known as the uniform superposition, or, in the case of the question, the "original state". To see why this inverts over the mean, first ...


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In the last equation in the question above the left hand, and right hand - the $\Delta \theta$ terms cancel, and the denomimators $sin^{2}(\theta)$ term whose root is as you mentioned inverted(as its in the denominator) to give the bound as the first equation in the question. See related question for the floor operation,


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ControlledOnInt is a library function which applies a specified operation (in this case SetRegisterToInt) to the target register only if the control register is in a state that encodes the given integer. Internally it does the following: convert the given integer to an array of bits (in little-endian format I think); applies X gate to each qubit that ...


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On one hand, the oracle can use an arbitrary subset of the register qubits to decide whether it marks the state as the solution to the encoded problem. For example, if you're looking for any 4-qubit state with the first 2 qubits in state $|11\rangle$, you can do a CCNOT with those qubits as controls and the marked qubit as target to implement this condition -...


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Assuming you have n qubits then the domain of $x$ is all bit-strings $n$. ($2^n$ values) To encode the data, in many cases, we start with the maximum superposition state. Which means each qubit is in an equal superposition of |0> and |1> so the entire system of $n$ qubits is in a massive superposition of all possible states (the entire domain of $x$). Then ...


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Why not uncover the quantum computer—open the box—to reveal the mechanism? Well, we can’t. If we “watch” the computation happen, we expose the quantum computer to an environment and this will break the computation. The kind of things a quantum computer needs to do requires complete isolation from the environment. Just like a magician’s trick, if we reveal ...


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