# Questions tagged [superposition]

Quantum superposition is a fundamental principle of quantum mechanics. It states that, much like waves in classical physics, any two (or more) quantum states can be added together ("superposed") and the result will be another valid quantum state; and conversely, that every quantum state can be represented as a sum of two or more other distinct states. Mathematically, it refers to a property of solutions to the Schrödinger equation. (Wikipedia)

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### Is it possible to collapse a superposition into a preset state such that all entangled qubits collapse to the same state?

Ideally, I'd be able to apply an operation to qubit 0 that would collapse the superposition to a set state (say 1) on qubit 0 & 1. The operation applied to qubit 0 would be a one qubit operation (...
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### Applying Hadamard gate to $\sqrt{3/4}|0\rangle + \sqrt{1/4}|1\rangle$

[I am just transferring this from Stack Overflow. It might need editing.] ———— [The reader can skip to “It all sounds fine…”, before the spreadsheet representation.] I am trying to figure out quantum ...
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### Derivation for the result of performing the Hadamard transform on $|0\rangle^{\otimes n}$ being $2^{-n/2}\sum_x|x\rangle$

It's said that the result of performing the Hadamard transform on n qubits initially in the all |0> state is $$\frac{1}{\sqrt{2^n}}\sum_x|x\rangle$$ where the sum is over all possible values of x....
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### Understanding the association rule between classical to quantum data $|x\rangle=\frac{1}{|\vec x|_2}\sum_{i=1}^d x_i|i\rangle$

I've been reading the paper on Quantum Hopfield Networks by Rebentrost et al. and I'm not sure to quite understand the association rule they mention on page 2. Here's what they say : Consider any $d$-...
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### What is the role of entanglement in quantum-computational speed-up?

The way I see it, there are three main quantum properties utilized in quantum computing - superposition, quantum interference, and quantum entanglement. I'm looking to understand which one is ...
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### Do any two distinct pure states form a basis?

$\newcommand{\bra}{\left<#1\right|}\newcommand{\ket}{\left|#1\right>}\newcommand{\bk}{\left<#1\middle|#2\right>}\newcommand{\bke}{\left<#1\middle|#2\middle|#3\right>}$ In ...
$\newcommand{\bra}{\left<#1\right|}\newcommand{\ket}{\left|#1\right>}\newcommand{\bk}{\left<#1\middle|#2\right>}\newcommand{\bke}{\left<#1\middle|#2\middle|#3\right>}$ If ...