# Tag Info

## New answers tagged hadamard

2

I'm not sure whether it's completely intuitive if it still has some formulas in it, but here's a try. (I prefer term "reflection", since it makes the geometrical interpretation a bit simpler, but I think they are used interchangeably.) First, let's convince ourselves that Controlled Z does inversion about the $|1...1\rangle$ state. Controlled Z flips the ...

2

Answer to EDIT in question: It is possible to prepeare a state $$|\psi\rangle = \frac{1}{\sqrt{m}}\sum_{i=0}^{m-1}|i\rangle,$$ where $m = 2^n$ and $n$ is a number of qubits, by application of operator $\otimes H ^{n} = H_{q_0} \otimes H_{q_1} \otimes \dots \otimes H_{q_{n-1}}$, i.e. Hadamard gate is applied on each qubit involved. As a results, you will ...

3

The circuit you have described is formed of 2 qubits, both in an equal superposition. This can be achieved by applying the H gate to both qubits, as this puts each qubit into superposition and takes us up to 4 possible states. In the IBM Quantum Experience, this circuit would look like

4

Generally, quantum states are determined up to a phase - i.e. up to multiplication by scalar. So $\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}},\frac{-\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}$ are essentially the same state. It is a good question how the statevector simulator chooses to normalize its statevector representation.

9

While the QFT and Hadamard transforms are different, their action on the input state $|00\ldots 0\rangle$ is identical; both produce the uniform superposition of all states. So, if you've got a choice of which to use, you shoulduse the one that is the easiest to implement: the Hadamard transform. Hadamard Transform: $$H|0\rangle=\frac{|0\rangle+|1\rangle}{\... 2 The H gates do not replace a Quantum Fourier Transform. The Quantum Phase Estimation algorithm is defined as shown in the picture you linked in your question: Hadamard gates on all the "ancillary" qubits Controlled unitary matrices (controlled circuits) Inverse QFT. Having an inverse QFT at the end of the circuit does not necessarily mean that there ... 1 You can take the definition of the function f(x) acting on any normal matrix to be such that if$$ M=\sum_i\lambda_i|\lambda_i\rangle\langle\lambda_i|, $$then$$ f(M):=\sum_if(\lambda_i)|\lambda_i\rangle\langle\lambda_i|.  So for any $M$ such that $M^2=I$, then $M=P_+-P_-$ is described using projectors onto its $\pm1$ eigenspaces, and $P_++P_-=I$. We ...

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