Single qubit Hadamard gate transforms standard basis states (zero and one states) to their superpositions (plus and minus states)

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### A CNOT between two Hadamard gates: why does the CNOT changed the output of the second Hadamard gate?

Applying the Hadamard gate twice in a row, it restores the original input: https://algassert.com/quirk#circuit={%22cols%22:[[%22H%22],[%22H%22]]} However, if a CNOT control is added between the two ...
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### How does the given state collapse to the two given possibilities after fourier sampling?

I'm working through the exercises for Professor Vazirani's course on QM & QC, and I don't understand the given solution to one of the problems (assignment 4, problem 10.a). I think I'm missing ...
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### Generate a random bit sequence of 512 bits [duplicate]

How can I generate a random bit sequence of 512 bits on IBM Q experience using 3 or 5 qubits? Putting a hadammard gate and measuring would only give me smaller bit sequence due to limitation in number ...
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### What is the matrix representation of the Hadamard gate in the computational basis?

I read about Hadamard gate H and found it's matrix representation as follows: $$H_1=\frac{1}{\sqrt 2}\begin{pmatrix}1 & 1 \\1 & -1\end{pmatrix}$$ I wanted to know what will be the matrix ...
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### Hadamard gate for three qubits; inconsistency between IBM and Matlab

I am trying to build a large and quite complex three qubit quantum circuit on IBMs quantum computer. I have a specific unitary which I am trying to implement and I am building a circuit following the ...
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### What is the output of applying the Hadamard matrix to $\sum_{y\in\{0,1\}^n} (-1)^{xy}|y\rangle$?

If, for some $x$, I have the $n$-qubit state $$\sum_{y\in\{0,1\}^n} (-1)^{xy}|y\rangle,$$ and I would like to apply to that the $n$-qubit Hadamard transform, with the aim of calculating the final ...
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### Hadamard Test to calculate imaginary part

I am trying to understand the Hadamard Test by finding the average value of $U_1$, which is a diagonal matrix with $1$ everywhere except on the first element. I performed the regular Hadamard Test as ...
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Say we had a Hadamard-like gate with the -1 in the first entry instead of the last. Let's call it $H^1$. $$H = \begin{bmatrix}1&1\\1&-1\end{bmatrix}$$ $$H^1 = \begin{bmatrix}-1&1\\1&... 3answers 177 views ### Can everything in QM be described with degrees instead of matrices and vectors? I found this explanation. "The Hadamard gate can also be expressed as a 90º rotation around the Y-axis, followed by a 180º rotation around the X-axis. So H=XY^{1/2}H = X Y^{1/2}H=XY^{1/2}." Can ... 1answer 53 views ### Analysis of the second Hadamard in the Detusch-Josza Algorithm Consider the Deutsch-Josza, algorithm, which first initializes the state |0 \rangle^{\otimes n} | 1 \rangle, creates a superposition using the the Hadamard gate and the U_f to get into the state: ... 2answers 98 views ### Prove by induction H^{\otimes n} \left| 0 \right>^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{i=0}^{2^n -1} \left| i \right> Let H be the Hadamard operator.$$ H = (\left| 0 \right> \left< 0 \right| + \left| 0 \right> \left< 1 \right| + \left| 1 \right> \left< 0 \right| -\left| 1 \right> \left< 1 \...
When I perform $2$ Hadamard $H$ gates on a single qubit, why is the probability of getting $0$ as the outcome 100%? Why is it not 50% 0 and 50% 1 instead? Why is the second $H$ gate not putting the ...