New answers tagged linear-algebra
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A physical quantum operation ${\cal E}$ can be described as a map between the set of density operators of the form
$${\cal E}(\rho) =\sum_k E_k \rho E_k^{\dagger}, \qquad \sum_kE_k^{\dagger}E_k \leq {\bf 1},$$
cf. Ref. 1. As Norbert Schuch correctly notes this implies that a physical quantum operation ${\cal E}$ must be a completely positive map. In this ...
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The plane is enough because all the vectors – before and after the operations (which are really simple rotations) – belong to a two-dimensional plane. The Hilbert space has many more dimensions but they're orthogonal to the plane of the picture and the coordinates (amplitudes) in these additional directions don't change during the calculation so we don't ...
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From the definition of an eigenvector, any eigenvector $|\psi\rangle$ of $U$ must satisfy $U |\psi\rangle = \lambda |\psi\rangle$ for some $\lambda$. Then, for any scalar factor $\alpha \ne 0$, we also have
$$ U\Bigl(\alpha|\psi\rangle\Bigr) = \alpha \lambda |\psi\rangle = \lambda\Bigl(\alpha|\psi\rangle\Bigr)$$
so that $\alpha |\psi\rangle$ is an ...
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Q1: That looks like a typo. If it is supposed to be an eigenvalue equation (it probably is), then it is indeed $|\psi_n\rangle$ on both sides.
Q2: A state is described by a vector in Hilbert space. This is an abstract mathematical object, but it is often convenient to choose a basis and use the components in that basis to describe the state. If you have a ...
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Your map fails to be completely positive. If you apply it to half of a maximally entangled state $(|0\rangle|0\rangle+|1\rangle|1\rangle)/\sqrt{2}$, you can easily see that $\phi(\rho)=\rho$ and $\phi(\rho')=\rho'$ imply that $\phi(|0\rangle\langle1|) = \alpha |0\rangle\langle1|$ and $\phi(|1\rangle\langle0|) = \alpha^* |1\rangle\langle0|$ for the resulting ...
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The slave particle approach is based on the assumption of spin-charge separation in the strongly correlated electron systems (typically Mott insulators). It was proposed that the electrons can decay into spinons and chargons (holons/doublons). But to preserve the fermion statistics of the electrons, the spinon-chargon bound state must be fermionic, so the ...
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Ron Maimon is entirely correct when he says that GA is precisely Clifford algebra, as any book or paper using the phrase "Geometric Algebra" is sure to say. But I think he misses both the point of the question and the point of "GA".
The question
I'll paraphrase the question as: Is GA a good, pedagogical way to introduce the mathematical side of physics to ...
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$\Gamma_{ii}$ is the $i$-th entry on the diagonal of $\Gamma = L^+ = (D-A)^+$, $\Gamma_{jj}$ is the $j$-th entry, and $\Gamma_{ij}$ is the entry located at row $i$, column $j$. Thus $\Omega_{ij}$ is a scalar, but you could assemble all such values into a matrix $\Omega$ that gives the resistances between all pairs of vertices.
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