14 votes
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Why can the QFT be replaced by Hadamard gates?

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 ...
DaftWullie's user avatar
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14 votes
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Why isn't $Ry(\pi/2)$ gate equivalent to Hadamard gate?

While Hadamard gate is defined as $$ H= \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, $$ $y$-rotation by $\pi/2$ leads to gate $$ Ry(\pi/2)= \frac{1}{\sqrt{2}} \begin{...
Martin Vesely's user avatar
10 votes

Why 2 $H$ gates in series create a probability of 100% for one value of the qubit and 0% of the second value of the qubit?

The reason for this is because the inverse of Hadamard gate is itself. That is, giving that $$ H = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix}$$ then $$ H^{-1} = \dfrac{...
KAJ226's user avatar
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How do I prove that the Hadamard satisfies $H\equiv e^{i\pi H/2}$?

First of all, note that the statement, as written, is wrong (or rather, it is correct only as long as the "$\equiv$" symbol is taken to mean "equal up to a phase"). An easy way to see it is by ...
glS's user avatar
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9 votes
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How do I apply the Hadamard gate to one qubit in a two-qubit pure state?

First, you should note that the Hadamaard gate is nothing more than a $2 \times 2$ Discrete Fourier Transform matrix (two-point DFT). That is the reason why, $H \bigg( \dfrac{|0\rangle + |1\rangle}{2}\...
KAJ226's user avatar
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Why is a Hadamard gate unitary?

The Hadamard gate is described by this matrix \begin{equation} H=\frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \end{equation} Conjugate transpose of $H$ is again $H$. Hence ...
Martin Vesely's user avatar
8 votes
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Could the Hadamard gate have been constructed differently with similar characteristics?

The Hadamard gate has close ties to the discrete Fourier transform. Consider the DFT for an $N$-level system: $$\vert j \rangle = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} e^{\frac{i2 \pi j k}{N}} \...
Jonathan Trousdale's user avatar
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How are the IBM's and Google's Hadamard gates fabricated and operated?

A Hadamard gate isn't usually a physical object that you pass qubits through. In the case of superconducting qubits, the Hadamard gate is performed by bouncing microwaves off of the qubits. It doesn't ...
Craig Gidney's user avatar
8 votes
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What Hamiltonians generate Hadamard and CNOT matrices?

Let us denote Hadamard with $H$ and the two Hamiltonians as $\mathcal H_H$ and $\mathcal H_{CNOT}$, i.e. $$ H = \exp (-i\mathcal{H}_H) \\ CNOT = \exp (-i\mathcal{H}_{CNOT}). $$ We will make use of the ...
Adam Zalcman's user avatar
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8 votes

How do you represent a Hadamard gate as a product of $R_x$ and $R_y$ gates?

If you're not concerned with global phase then the following works using only two rotation gates: \begin{align} R_y\left(-\frac{\pi}{2}\right) R_x\left(\pi\right) &= \exp \left(i\frac{\pi}{4}Y\...
forky40's user avatar
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7 votes

Is there a gate that puts a qubit into superposition with a not so purely probabalistic (50 50) outcome?

You can use $Ry$ gate to prepare a qubit in superposition with arbitrary probabilities. When you apply the gate on qubit in state $|0\rangle$, you get a qubit in superposition $$ |\psi\rangle = \cos(\...
Martin Vesely's user avatar
7 votes
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How to translate the Hadamard gate matrix into Dirac notation?

First remember that each matrix element can be written as outer products in Dirac notation: $$|0\rangle\langle 0| = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix},|1\rangle\langle 1| = \begin{...
user1271772 No more free time's user avatar
6 votes

How do 2 Hadamard gates act on a single qubit?

If you apply the same gate twice on a state $|\psi\rangle$ you will get out the state $|\psi\rangle$ if the hermitian of the gate is the same as the gate. In your case, you applied the Hadamard gate, $...
auden's user avatar
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6 votes
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What is the intuition of using Hadamard gate in quantum fourier transform?

The intuition, roughly speaking, is that the only way that you're going to get some difference between classical and quantum computing is if you are able to prepare qubits in a superposition. If you ...
DaftWullie's user avatar
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6 votes

How do I prove that the Hadamard satisfies $H\equiv e^{i\pi H/2}$?

For questions like this, the conventional physics notation is easier to work with than the QIT gate notation. Define $\vec \sigma = (\sigma_1,\sigma_2,\sigma_3)$ to represent the three Pauli matrices ...
Jonathan Trousdale's user avatar
6 votes

How to visualize Hadamard gate as $X$-$Z$-$X$ decomposition?

As mentioned in the other answer, the Hadamard gate is a pi rotation (180 degree) around the $X + Z$ axis. That is, it is a 180 degree rotation around the purple axis indicated in the below figure: ...
KAJ226's user avatar
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Initial state preparation for Hadamard test

You could almost do it. In fact, the following circuit would work: You can convince yourself that what you do is in fact applying the unitary $\mathbf{VUV^\dagger}$, which will allow you to evaluate ...
Tristan Nemoz's user avatar
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6 votes
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Can we use Hadamard test to estimate phases?

So QPE using $\mathcal{O}(1/\epsilon)$ queries to $U$ outputs an estimate of the eigenphase $\theta$ given a corresponding eigenvector with additive error and $\Omega(1)$ probability. The method using ...
dylan7's user avatar
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6 votes
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Commutation rules between Pauli $X$ and controlled-Hadamard

There are two cases: $X$ on the controlled qubit and $X$ on the target. $X$ on the controlled qubit For the first case, note that for any controlled unitary $CU$, we have $$ (X_1\otimes I_2)\circ ...
Adam Zalcman's user avatar
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What is the intuition behind uniform Hadamard superposition?

The first Hadamard gates applied to a fiducial all-zero's ket on $n$ qubits serve to prepare your input register into the uniform superposition over all $2^n$ basis states. The last Hadamard gates ...
Mark Spinelli's user avatar
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Hadamard transform on state

This is a little fiddly to get right the first time. Let's start by rewriting eq (3) as $$ \sum_{b\in\{0,1\}}\alpha_b|b\rangle\otimes X^{x_{b,y}}|0\rangle. $$ Now, it probably helps to think of this $...
DaftWullie's user avatar
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5 votes

What are the $|+\rangle$ and $|-\rangle$ states?

The $|+⟩$ and $|-⟩$ are states given by the following decomposition in the Z-basis: \begin{equation} \begin{aligned} |+⟩ &= \frac{1}{\sqrt{2}} \Big(|0⟩ + |1⟩\Big)\\ |-⟩ &= \frac{1}{\sqrt{2}} ...
luigi's user avatar
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5 votes

Could the Hadamard gate have been constructed differently with similar characteristics?

You can easily check that $H^1=XHX$, which is another way to say that $H^1$ is the same as $H$ modulo swapping $|0\rangle$ and $|1\rangle$. This means that it is a different gate. On the other hand, ...
glS's user avatar
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5 votes
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Projecting $\lvert ++ \rangle$ on Bell Basis

The Hadamard gate is: $$\frac{1}{\sqrt 2} \left(|0\rangle \langle 0 | + |0\rangle\langle 1| + |1\rangle \langle 0| - |1\rangle \langle 1|\right)$$ And since $|+\rangle$ is $\frac{1}{\sqrt 2}\left(|0\...
Mahathi Vempati's user avatar
5 votes
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Analysis of the second Hadamard in the Detusch-Jozsa Algorithm

If you look at the formula you want to prove term-by-term, you'll notice that the sum and the $(-1)^f(x)$ part is the same in both formulas; you just need to show that $$H^{\otimes n} |x\rangle = \...
Mariia Mykhailova's user avatar
5 votes
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How should I understand the change of qubit's basis as a rotation?

Hadamard gate can be interpreted as a rotation in 3D Euclidean space (on Bloch sphere) by angle $\pi$ around X+Z axis. The qubit rotation by angle $\theta$ around axis pointed by unit vector $\textbf{...
kludg's user avatar
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Is there a gate that puts a qubit into superposition with a not so purely probabalistic (50 50) outcome?

Welcome to QCSE. You already know that $a^2=b^2=0.5$. For a single qubit gate akin to the Hadamard gate you can achieve any two probabilities you want, as long as they add to $1$. For example one ...
Mark Spinelli's user avatar
5 votes
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Randomness using simple parallel Hadamard circuit

The issue is that you are using noisy hardware with imperfect operations and measurements. In particular, the most likely problem here is that after you prepare a qubit it immediately begins decaying ...
Craig Gidney's user avatar
5 votes
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How do I apply a Hadamard gate on a given qubit, in matrix formalism?

Applying quantum gates to quantum states is indeed represented as matrix multiplication. To multiply two matrices, you need only one dimension to match: $$\frac{1}{\sqrt2}\begin{bmatrix}1 & 1 \\ 1 ...
Mariia Mykhailova's user avatar
5 votes
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A CNOT between two Hadamard gates: why does the CNOT changed the output of the second Hadamard gate?

This is because the CNOT gate created an an entangled state and the system after the CNOT gate can't be written individually. That is, you can't stay that your first qubit is in the state $\dfrac{|0\...
KAJ226's user avatar
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