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Results tagged with Search options user 58382
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Applies also to pre-Hilbert spaces, rigged Hilbert spaces, and spaces with negative norm or zero-norm states.

No more than one. It follows from Holevo's theorem that $n$ qubits cannot be used to store more than $n$ bits of information. See for example these notes for an explanation of Holevo's theorem (intere …
answered Jan 29 '18 by glS
The other answer already points out the typo, but just in case it may help someone else stumbling upon this: this formula (the corrected version in the other answer) becomes almost trivial using diagr …
answered Jun 16 '18 by glS
The correct normalization factor is $$N = \frac{1}{\sqrt{2}}.$$ To see this, note that you can write your wave-function in ket notation as $$\psi(x) = \langle x | a \rangle + \langle x | -a \rangle \ … answered Dec 27 '14 by glS The name of the concept you are looking for is probability amplitude. The two states \lvert T\rangle = \frac{1}{\sqrt2}(\lvert 10\rangle + \lvert 01\rangle) and \lvert S\rangle = \frac{1}{\sqrt2}( … answered Nov 6 '16 by glS You need all 16 expectation values to totally reconstruct the density matrix. Knowing these values, the density matrix is simply written as$$\rho=\sum_{i,j=1}^{4} \langle \sigma_i\otimes\sigma_j\ra …
The Schmidt decomposition is nothing but the singular value decomposition (SVD) applied to the coefficients of a bipartite state. Any matrix $A$ can be written, using the SVD, as $A=\sum_k s_k\lvert … answered Oct 24 '18 by glS We know that$\psi_{l,m}$satisfies, for each$l$and$m$, the equations $$L^2\psi_{l,m}(r,\theta,\phi)=l(l+1)\hbar^2\psi_{l,m}(r,\theta,\phi),$$ $$L_z\psi_{l,m}(r,\theta,\phi)=m\hbar \psi_{l,m}(r,\ … answered Feb 6 '16 by glS 1answer Consider a generic single-qubit state$$\rho=\lambda_1\lvert \lambda_1\rangle\!\langle \lambda_1\rvert+\lambda_2\lvert \lambda_2\rangle\!\langle \lambda_2\rvert\in\mathcal H_S.$$I am interested in un … asked Nov 5 '18 by glS Detailed version Let \rho be a density matrix describing a state shared by Alice and Bob. We can generically write it as$$\newcommand{\ketbra}{\lvert#1\rangle\!\langle#2\rvert}\rho=\sum_{ijkl}\ … answered May 22 by glS It is in the sense that given any pair of states$\rho$and$\sigma, you have \begin{align} \operatorname{Tr}_2(\rho\otimes\sigma)&=\rho,\\ \operatorname{Tr}_1(\rho\otimes\sigma)&=\sigma. \end{align} … answered Oct 24 '18 by glS A GHZ inM=2$would be$|00\rangle+|11\rangle$. Sure you could still call it a GHZ, but it is not very useful because this state already has a name: it's a Bell state. A standard reference to unders … answered Apr 13 '18 by glS As was already pointed out in the comments, the "blank state" cannot be chosen as a function of$|\phi\rangle$. The reason is that a cloning operation is a unitary$U$such that for any state$|\phi\ …