New answers tagged entanglement
2
There are lots of different situations one can talk about from a cryptography perspective. But here's one that has a huge practical relevance:
There are physical realisations of quantum computers which, individually, are limited in the number of qubits they can use. For example, ion traps. For the sake of argument, assume you can hold 10 qubits in each ion ...
1
To set a context for my understanding of your question let's start with a single qubit. Then, the state can be described by two complex numbers $|\psi\rangle=a|0\rangle+b|1\rangle$ subject to a constraint $|a|^2+|b|^2=1$ and up to an equivalence $|\psi\rangle \sim e^{i\phi} |\psi\rangle$. Absolute values of $a$ and $b$ determine probabilities to obtain ...
7
Yes. Intuitively, the set of pure product states has lower dimension than the set of all pure states. Therefore, almost all pure two-qubit states are entangled.
Let $\mathcal{F}$ denote the set of all pure states of two qubits and $\mathcal{S}$ denote the set of all pure product states of two qubits. Note that $\mathcal{S}$ can be thought of as the Cartesian ...
1
In the specific $|\psi \rangle$ there is symmetry on the order of bits, so we must only show that one bit is entangled and it then follows that they all are.
Notice we may write:
$$|\psi \rangle = \frac{1}{2}\Big(|0 \rangle \big(|001 \rangle +|010 \rangle +|100 \rangle \big) + |1 \rangle \big(|000 \rangle \big)\Big)=\frac{\sqrt3 |0\rangle |A \rangle + |1\...
5
The need for quantum fields
There are two seemingly unrelated conceptual steps between quantum mechanics (QM) and quantum field theory (QFT). One step reconciles QM with special relativity (SR) and the other replaces a fixed finite number of particles with fields that have infinite degrees of freedom. These two steps appear unrelated, but in order to arrive ...
1
Before talking about upper bounds for $\mathsf{QMA(k)}$, let us first concentrate on upper bounds for $\mathsf{QMA(k)}$.
Recall that the maximum acceptance probability of a $\mathsf{QMA(k)}$ verifier $V_x$ is $\max_{|\psi\rangle} \||1\rangle\langle 1|_{out} V_x |\psi\rangle |\bar{0}\rangle\|_2^2$ where $|\bar{0}\rangle$ are ancillary qubits. It is a ...
2
Entanglement is analogous to correlation.
Correlation over a property simply exists where simultaneous determination is not equivalent to individual determination of a property. What does it mean?
Say you have 2 coins - one black and one white. Let's say if you have a black coin you get a $+1$ colour value whereas with white you get a $-1$ colour value. If ...
4
It depends on what you want to take as your definition of maximally entangled. But, here's a good one:
Given a Bell state, I can convert it into any other two-qubit
entangled state using only local operations and classical
communication
Given that local operations and classical communication cannot increase entanglement, the possibility of performing $|\...
2
Error correcting codes work the same on entangled qubits as any other qubit. All Alice and Bob have to do is separately encode their qubit into a Shor code. They each run an encoding circuit, apply noise, then run a decoding circuit and apply the corrections it inferred.
4
Notice that it is somewhat a coincidence of that particular Bell state and choice of basis. The states $|0\rangle$ and $|1\rangle$ are in the $z$ axis of the Bloch sphere and $|+\rangle$,$|-\rangle$ are in the $x$-axis.
The state you chose is a sum of products of single states that are the same, and it turns out that the same is true when you convert it to ...
2
One strong element of the intuition is related to the fact that it is maximally entangled. One definition of a pure state $|\psi\rangle$ being maximally entangled is that the individual systems have density matrices
$$
\rho_A=\text{Tr}_B(|\psi\rangle\langle\psi|)=\frac{I}{d}
$$
where $d$ is the dimension of $A$'s Hilbert space.
Now, one thing that the ...
14
The minimum overlap is zero and the maximum overlap is $\frac{1}{d}$.
The overlap is a linear function of $\rho$ and the set $S$ of separable states is convex, so the overlap is both minimized and maximized at extreme points. Extreme points of $S$ are the states of the form$^1$ $\rho = \overline\sigma\otimes\tau$. The reason we choose to define $\sigma$ as ...
2
Yet another derivation
Applying a local unitary $U^A$ on the first subsystem of a bipartite maximally entangled state $|\psi^{AB}\rangle$ is equivalent to applying a possibly different unitary $V^B$ on the second subsystem
$$
(U^A\otimes I)|\psi^{AB}\rangle = (I\otimes V^B)|\psi^{AB}\rangle\tag1.
$$
In the specific case of the Bell state $(|00\rangle+|11\...
3
The definition of information is context dependent and has different interpretations. See How to better define information forum (Physics Insights)
As you did not provide a paper, I will assume a very standard context. In general when people talk about information loss in quantum mechanics due to collapse, it is related to irreversibility of collapse.
If you ...
4
If you measure a state $|\psi\rangle = a|0\rangle + b|1\rangle$, the post-measurement state will be either $|0\rangle$ or $|1\rangle$ with $|a|^2$ and $|b|^2$ probability, respectively. The important part here is that the values for $a$ and $b$ are lost in the post-measurement state and you end up with a basis state $\{|0\rangle, |1\rangle\}$ with an ...
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