New answers tagged entanglement
2
The set of separable states is closed.
Thus, around any entangled state - not necessarily pure - there is an $\epsilon$-ball which lies entirely within the entangled states.
Or, in the language of your question: All entangled states are "robust".
(As illustrated by DaftWullie's answer, the size of this ball can depend on the state: There are pure ...
1
For a fixed $\epsilon_0$, why not simply consider
$$
|\psi\rangle=\cos\theta|00\rangle+\sin\theta|11\rangle?
$$
Since it's a two-qubit state, entanglement can be determined using the PPT criterion. Hence,
$$
\rho=(1-\epsilon)|\psi\rangle\langle\psi|+\epsilon I/4
$$
is entangled if $\epsilon<2\sin(2\theta)/(1+2\sin(2\theta))$. Any $\epsilon$ you give me, ...
1
You are analysing the case where you know a unitary has definitely been applied on the first qubit. In that case, it should not be surprising that there's no change in entanglement. You can take a couple of perspectives:
single qubit unitaries do not change entanglement. To change entanglement with a unitary requires a two-qubit unitary.
If you know what ...
2
Given a separable bipartite state as $|\psi\rangle\otimes|\phi\rangle$, you "get" the states of the single systems but taking only the corresponding state, e.g. here $|\psi\rangle$ or $|\phi\rangle$.
More generally, you might not know the structure of the state, and you might have entanglement between the different subsystems, in which case the ...
1
I would like to add to keisuke.akira answer.
The Noise Model in which only a Single Qubit Flips is correct. However we can assume a more general Noise Model which may be more realistic and still see the use of Bit Flip Code.
Since Quantum Circuits are analog, hence it is rare that a qubit flips completely. It is more likely that there is a small coherent ...
1
We're trying to build a code to protect against single bit flips. That is, we are assuming the noise model. By assumption, it has the form $\sigma_x \otimes \mathbb{I} \otimes \mathbb{I}$, therefore, it only flips one of them. Of course, in general, the noise does whatever it wants, and therefore, we need to build codes that can protect against more general ...
0
While designing your circuit are you sure 6 Qubits were really
required for X gate . As i am observing in your above plot there are
two possible probabilities hopefully that might be right ,
I would suggest you to start designing the above circuit with 2 Qubits
and observe the outcome.
2
An example of a state which yeilds an entangled state when you trace one qubit out, is the 3-qubit "W" state:
$$
\lvert W_3 \rangle = \tfrac{1}{\sqrt3} \Bigl( \lvert 100 \rangle + \lvert 010 \rangle + \lvert 001 \rangle \Bigr)
$$
Taking the outer product with itself, we obtain
$$
\lvert W_3 \rangle\!\langle W_3 \rvert = \tfrac{1}{3} \Bigl(
\!\...
6
Symmetric Werner states in any dimension $n\geq 2$ provide examples.
Let's take $n=2$ as an example for simplicity. Define $\rho\in\mathrm{D}(\mathbb{C}^2\otimes\mathbb{C}^2)$ as
$$
\rho = \frac{1}{6}\,
\begin{pmatrix}
2 & 0 & 0 & 0\\
0 & 1 & 1 & 0\\
0 & 1 & 1 & 0\\
0 & 0 & 0 & 2
\end{pmatrix},
$$
which is ...
0
To see intuitively why this protocol increases the entanglement after each iteration, we can work out an example where our initial state is say $\lvert 00\rangle+\lvert 11\rangle$. Upon passing through a 50:50 Beam Splitter, we get:
$$|00\rangle+|11\rangle = |00\rangle+ a_{0}^{+}a_{1}^{+}|00\rangle \hspace{3mm}transforms\hspace{3mm}|00\rangle + \frac{1}{2}(...
1
A mixed separable state is written in the form
$$
\rho=\sum_ip_i\sigma^A_i\otimes\sigma^B_i
$$
where the $\sigma_i$ are valid density matrices on a single site.
The example you give, say $\rho=\frac12|00\rangle\langle00|+\frac12|11\rangle\langle 11|$ is exactly of this form. Specifically,
$$
p_0=p_1=\frac12,\qquad \sigma^A_0=\sigma^B_0=|0\rangle\langle 0|,\...
2
Let $\rho_{d}, \sigma_{d}$ be the (simultaneously diagonal) density matrices whose eigenvalues are $\{ p_{j} \}, \{ q_{j} \}$, respectively (represented as probability vectors below). Then, if $\vec{p} \succ \vec{q}$, the following sequence of arguments can be observed:
There exists a bistochastic matrix $M$ such that $M \vec{p} = \vec{q}$ (basic result of ...
0
$\newcommand{\calU}{\mathcal{U}}\newcommand{\ket}[1]{\lvert#1\rangle}$Suppose such a circuit existed, and denote with $\calU$ the unitary describing its overall action.
For this to be a "circuit detecting entanglement", there should be two types of output states, one corresponding to the answer "yes, the input was entangled" and the other ...
1
In your argument, you are missing the relativistic causality. Bob and Alice simply cannot decide who did the measurement first because their space-time measurement events are connected by the space-like curve. (because they do the measurements targeting the faster than light communication.) Depending on the inertial reference frame, Bob can be first or Alice ...
5
Apply a CNOT gate with one of the qubits as control and the other as target. You'll get
$$\frac{1}{\sqrt{2}}(|0\rangle+e^{i\theta}|1\rangle) \otimes |0\rangle$$
Use the methodology from How to get the relative phase of a qubit? for the first qubit :-)
2
No.
For instance, if I either give you $|00\rangle$ or $|11\rangle$ with 50% probability each, or $|00\rangle\pm|11\rangle$ with 50% probability each, there is no way to distinguish these two cases - not even with any whatsoever small probability. The mathematical reason is that those are described by the same density matrix - but you always get some pure ...
2
It sounds like the source of your misunderstanding is the following
Bob is able to harness the information of a qubit which contains more information than regular classical bits
This statement is not true. There is a theorem known as Holevo's theorem which states that the maximum information that can be obtained from a single qubit is one bit.
2
As soon as you impose that $\lambda_x=\lambda_{\bar x}$, your state is of the form that my previous results apply to. This is because you can write
$$
|x\rangle\langle x|+|\bar x\rangle\langle\bar x|=\frac{1}{2}(|x\rangle+|\bar x\rangle)(\langle x|+\langle\bar x|)+\frac{1}{2}(|x\rangle-|\bar x\rangle)(\langle x|-\langle\bar x|),
$$
and so this state is ...
9
First, about teleportation, you say that you think quantum communication takes place in the protocol, but it doesn't. They only share an EPR pair they created together, hence the coordination and after, what Alice sends to Bob when communication takes place are classical bits, she sends the measured bits of the 2 qubits she has, so the only communication we ...
2
Remember that the partial transpose condition is generally good for detecting entanglement, i.e. a bipartite state $\rho$ is certainly entangled if the partial transpose is not non-negative. In other words, if there exists a state $|\psi\rangle$ such that
$$
\langle\psi|I\otimes\text{T}(\rho)|\psi\rangle<0,
$$
then the state is certainly entangled.
If you ...
2
I don't know the details of this paper, although there are much better things that you can say about the GHZ case (in general, properties of GHZ states are much easier to analyse than W states). I'll summarise the key result in this context below, but further details are available in my paper, here.
There are some very simple entanglement criteria that one ...
2
Apparently, the specific question posed here has been answered in the affirmative--at least (first, we point out) through numerical means--by
Arsen Khvedelidze and Ilya Rogojin in Table 2 of their 2018 paper,
"On the Generation of Random Ensembles of Qubits and Qutrits: Computing Separability Probabilities for Fixed Rank States" ArsenIlya
They ...
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