Hot answers tagged quantum-information
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This was too long for a comment.
There are two papers that have a nice characterization of positivity in terms of the Bloch vector: quant-ph/0301152 and quant-ph/0302024. You might be able to use dynamic programming to speed up a check of positivity by using the conditions in these papers. Basically, they reduce the positivity constraint to a constraint on ...
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You are right that there's not much you can do with a single qubit. However it doesn't take many before you can do things that can't be done with classical bits. Here are a few illustrative examples.
Communication
Imagine I make this offer to you and a friend of yours: I will give you each a classical bit, and then you will each give me a classical bit. If ...
3
Because the very act of viewing the code changes it. Maybe not uncrackable but any sniffing will be evident because it will be changed. From RSA Laboratories:
Quantum cryptography has a special defense against eavesdropping: If an enemy measures the photons during transmission, he will use the wrong basis half the time, and thus will change some of the ...
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Computation of the $S_z$ probability distribution for each of the
manifolds of equal entanglement:
Remark: Notations and references from Kuś and Žyczkowski are used.
Case 1: The separable case: The state vector is parametrized as (equation: 24)
$w = \begin{bmatrix} \cos \alpha \cos \beta e^{i \chi_1},& \cos \alpha \sin\beta e^{i \chi_2}, &\sin ...
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Edited to add the second part
Edited again, for part 3 and 4
$\newcommand\ket[1]{\left|#1\right>} \newcommand\bra[1]{\left<#1\right|} $
1. Absence of Quantum Loophole
You can easily see that there is no "quantum loophole" in your argument by writing explicitly any pure separable state. With your notations, we have :
$$
...
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If you read the wikipedia article about the Bloch sphere, you will see that any pure state has the form
$|\psi\rangle = \cos\left(\tfrac{\theta}{2}\right) |0 \rangle \, + \, ( \cos \phi + i \sin \phi) \, \sin\left(\tfrac{\theta}{2}\right) |1 \rangle$
with $0 \leq \theta \leq \pi$ and $0 \leq \phi < 2 \pi$. Notice that if you have a complex coefficient ...
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Quantum cryptography is pretty technical, but here is an elementary argument.
The Heisenberg uncertainty principle implies the no-cloning theorem for quantum-mechanical systems. If you could make an exact copy of a quantum-mechanical system, you could use one copy to measure its position, and the other to measure its momentum. This would violate the ...
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You are halfway there, because you already wrote down the Hadamard gates. The remaining part, $I-2\left|0\right>\left<0\right|$ (I negated it from what you have, since this gives a simpler solution), is diagonal in the computational basis. Write down these diagonal entries and say out loud to yourself what this operator "does".
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You can obtain the coordinates in 3-space corresponding to the Bloch sphere, by taking expectation values of the Pauli spin operators:
$$\begin{align*} x &= \langle\psi | \sigma_x |\psi\rangle \\ y &= \langle\psi | \sigma_y |\psi\rangle \\ z &= \langle\psi | \sigma_z |\psi\rangle \end {align*}$$
and in the case of the state you describe, we have ...
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Here's a calculation to get you started.$\def\ket#1{\lvert#1\rangle}$
Define $\ket{h_j} = H\ket{j}$:
$$\begin{align*}
\ket{h_0} &= \tfrac1{\sqrt 2}\Bigl(\ket0 + \ket1\Bigr) \\
\ket{h_1} &= \tfrac1{\sqrt 2}\Bigl(\ket0 - \ket1\Bigr)
\end{align*}$$
Compute (for example) $ \ket{h_0}\ket{h_1}$ by distributing the tensor product over the addition and ...
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Such statements emerge when trying to capture quantum physics in classical physics terms. In other words, statements like "P or not P is not a valid proposition in a quantum world" are typically made by philosophers who don't really understand quantum physics and stubbornly attempt to map the quantum onto their classical intuition.
A much more insightful ...
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