# Questions tagged [pauli-gates]

For questions about Pauli matrices in general or Pauli gates in particular, as relevant to quantum computing and/or quantum information theory. The Pauli matrices are a set of three 2 × 2 complex matrices which are Hermitian and unitary. The three Pauli gates are: Pauli-X gate, Pauli-Y gate & Pauli-Z gate. X = {{0,1},{1,0}}; Y = {{0,-i},{i,0}}; Z = {{1,0},{0,-1}}.

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### Regarding the inductive proof that any Clifford gate can be made of Hadamard, phase and c-not

In Exercise 10.40 of Nielsen and Chunang's textbook, the reader is supposed to construct an inductive proof of Theorem 10.6 that any Clifford gate can be made of Hadamard, phase and c-not. There it is ...
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### What does it mean that a qubit is a triple $(H,X,Z)$ with $H$ Hilbert space and $X,Z$ Pauli operators?

In this paper, http://users.cms.caltech.edu/~vidick/teaching/fsmp/fsmp.pdf, it gives the definition of a qubit as follows: A qubit is a triple $(H, X, Z)$ consisting of a separable Hilbert space H and ...
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### How does Spin Measurement correspond to quantum NOT gate?

Newbie in quantum computing (and stack overflow) here. I am confused regarding the relation between spin measurement in quantum mechanics and the quantum NOT gate. I have a Bloch sphere picture of a ...
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### Efficient quantum algorithms to decompose Hessian matrices into sums of unitaries

Are there efficient quantum algorithms that given a d-sparse hessian $H \in \mathbb{C}^{N \times N}$ decompose it into a sum of unitaries (e.g. Pauli matrices)? $$H = \sum_i^q a_i U_i$$ If an ...
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### Pauli decomposed Hamiltonian as Diagonal U gate

While trying to implement a quantum circuit, I had to apply Hadamard gates to all qubits to achieve equal superposition. Done. The next operation is decomposing the Hamiltonian into a sum of tensor ...
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### Equivalence check between rotational gates and Pauli gates

My question is highly related to this one. I am trying to understand the relationship between rotational gates $R_P(\theta)$, where $P \in \{X,Y,Z\}$. As stated here, $\exp(iPx)=\cos(x)I+i\sin(x)P$. ...