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Applies also to pre-Hilbert spaces, rigged Hilbert spaces, and spaces with negative norm or zero-norm states.

7
votes
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
0
votes
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
2
votes
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 $$\psi(x) = \langle x | a \rangle + \langle x | -a \rangle \ …
answered Dec 27 '14 by glS
2
votes
The name of the concept you are looking for is probability amplitude. The two states $\lvert T\rangle = \frac{1}{\sqrt2}(\lvert 10\rangle + \lvert 01\rangle)$ and $\lvert S\rangle = \frac{1}{\sqrt2}( …
answered Nov 6 '16 by glS
0
votes
You need all $16$ expectation values to totally reconstruct the density matrix. Knowing these values, the density matrix is simply written as $$\rho=\sum_{i,j=1}^{4} \langle \sigma_i\otimes\sigma_j\ra …
answered Feb 1 by glS
0
votes
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
1
vote
We know that $\psi_{l,m}$ satisfies, for each $l$ and $m$, the equations $$L^2\psi_{l,m}(r,\theta,\phi)=l(l+1)\hbar^2\psi_{l,m}(r,\theta,\phi),$$ $$L_z\psi_{l,m}(r,\theta,\phi)=m\hbar \psi_{l,m}(r,\ …
answered Feb 6 '16 by glS
2
votes
1answer
Consider a generic single-qubit state $$\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.$$ I am interested in un …
asked Nov 5 '18 by glS
2
votes
Detailed version Let $\rho$ be a density matrix describing a state shared by Alice and Bob. We can generically write it as $$\newcommand{\ketbra}[2]{\lvert#1\rangle\!\langle#2\rvert}\rho=\sum_{ijkl}\ …
answered May 22 by glS
2
votes
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
5
votes
A GHZ in $M=2$ 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
3
votes
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\ …
answered Jun 19 '18 by glS