All Questions
Tagged with quantum-entanglement hilbert-space
163
questions
1
vote
1
answer
248
views
Reduced density matrices and relation to entanglement
I've read that if a state is a product state, the reduced density matrices are pure and if the state is entangled, the reduced density matrices are both mixed.
What would it mean if you had a system ...
6
votes
3
answers
742
views
Bell's inequality for angles 120°
In 1964, John Bell first derived the original Bell inequality, $|E(a,b)-E(a,c)|\leq1+E(b,c)$. Here $a,b,c$ are three different possible spin measurement directions, and $E$ is the measured ...
-4
votes
1
answer
117
views
How to know which states are entangled from a state vector? [closed]
consider the following state vector of three qubits
$$(1/2)|000⟩+(1/2)|011⟩+(1/2)|101⟩+(1/2)|110⟩.$$
how to know which qubits are entangled with respect to their basis states, in other words, how do ...
-2
votes
1
answer
117
views
How can we figure out what fraction of pure states in a Hilbert space are entangled? [duplicate]
The full Hilbert space of a quantum system will generally contain entangled states, and thus when entanglement is lost through decoherence, parts of Hilbert space become inaccessible. Is there a ...
3
votes
1
answer
139
views
Is there a relation between some kind of distance and the Schmidt basis?
Consider two bipartite quantum states $|\phi\rangle^{AB}$ and $|\psi\rangle^{AB}$ (in a finite dimensional Hilbert space $\mathcal H_A\otimes \mathcal H_B$), such that
$$\| |\phi\rangle\langle\phi|^{...
1
vote
1
answer
56
views
Relative "volume" of entangled vs product states [duplicate]
A system containing $n$ qubits is described by a $2^n-$dimensional Hilbert space. Some of these states can be decomposed as product states, but not all of them. The remaining ones are called entangled ...
0
votes
1
answer
63
views
Are there non-trivial two-party stabilizers in bipartite entanglement for product states?
In this recent paper where the authors discuss finite classification of entanglement types, on pg. 29 in appendix A, it is claimed that in bipartite entanglement for product state $|00\rangle$ there ...
3
votes
0
answers
212
views
Confusion about the tensor product structure of a multi-fermion Hilbert space
I often see people study entanglement in fermionic systems. The setup is often like this. Suppose we have a 1d lattice of $2L $ sites, which is divided into a left part and a right part, each with $L ...
0
votes
2
answers
61
views
Are multimode states a product state of single mode states?
Books such as 'Quantum Theory of Light by Rodney Loudon (page 140)' and 'Quantum Optics for Beginners by Ficek and Rizda (page 43)' claim that the multimode state is nothing but a tensor product of ...
-2
votes
1
answer
144
views
What is meant by " a basis is diagonal"?
I am trying to understand Schmidt decomposition. I am stuck in one sentence here. See the example picture.
Here, I can understand everything except the line "For both
HA and HB the Schmidt basis ...
3
votes
1
answer
88
views
Why is the entanglement of formation upper bounded by the Schmidt number?
I have read many times in several articles (such as https://arxiv.org/abs/1609.05033) that the entanglement of formation EoF puts a lower bound on entanglement dimensionality $d$ (i.e., the Schmidt ...
1
vote
0
answers
46
views
What happens when a maximally entangled state passes through a $k$-extendible channel?
In the context of Phys. Rev. A 104, 022401 (arXiv:1803.10710), Figure 3
What happens when a maximally entangled state( not $k$ extendible even for $k=2$) passes through a $k$-extendible channel? We ...
0
votes
1
answer
57
views
Can an entangled state in general be created by destructive interference in some subspace?
For instance, if we have a general two-qubit state $$|\psi\rangle=\frac{1}{2}(|0\rangle+e^{i\varphi_a}|1\rangle)\otimes(|0\rangle+e^{i\varphi_b}|1\rangle)=\frac{1}{2}(|00\rangle+e^{i\varphi_b}|01\...
2
votes
1
answer
137
views
Does spin entanglement imply position entanglement?
My question is whether two electrons can be entangled only with respect to their spins but not with respect to some other observable, such as position.
I initially believed that spin-entanglement ...
2
votes
2
answers
94
views
Why does the fact that all quantum systems are open mean that no quantum state can be pure
I am teaching myself about open quantum systems and I am confused by the following statement on the wikipedia page about open quantum systems:
"The fact that every quantum system has some degree ...