9 votes
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Worst Bell inequality violation with non-maximally entangled state?

What you're after is Gisin's theorem. He proved that the maximal violation of the CHSH inequality with a fixed state $|\psi\rangle = c_0 |00\rangle + c_1|11\rangle$ is given by $2\sqrt{1+4 |c_0 c_1|^2}...
Mateus Araújo's user avatar
9 votes
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Is it true that if $U$ sends computational basis states to product states, then it sends product states to product states?

TL;DR: The claim is false, i.e. it is not true that if $U|x\rangle$ is a product state for all computational basis states $|x\rangle$, then $U$ sends product states to product states. Counterexamples ...
Adam Zalcman's user avatar
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8 votes
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Why do we need Entanglement?

Almost every quantum state is entangled$^1$. It is therefore unsurprising that quantum algorithms bump into them every so often. In fact, an attempt to navigate the space of all quantum states without ...
Adam Zalcman's user avatar
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8 votes
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How to convert a partially entangled state into maximally entangled using SLOCC

This case is pretty straightforward, if you're already familiar with generalised measurements. Assume $\alpha>\beta$ are real numbers. We can define $$ M_1=\left(\begin{array}{cc} \sqrt{\frac{\beta}...
DaftWullie's user avatar
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7 votes
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What is the complexity of determining if a state is entangled?

At least when the state that you're testing is a mixed state (rather than known to be a pure state), there's been quite a bit of work on this. In 2003, Gurvits showed that the problem is NP hard. ...
DaftWullie's user avatar
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7 votes
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Truthfulness of Entanglement Fidelity

Initial answer The entanglement fidelity isn't intended to measure a change in entanglement. Rather, it measures how coherently a channel preserves a given state, meaning how well it preserves the ...
John Watrous's user avatar
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6 votes
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Why is a Bell state involved in quantum teleportation?

The motivation for introducing the Bell state is simple: it gets the job done! This of course raises the question: what properties of the Bell state make it suitable for use as a resource in quantum ...
Adam Zalcman's user avatar
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5 votes
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When does CNOT entangle?

A necessary condition is that the control state is not an eigenstate of Z, and the target state is not an eigenstate of X. A sufficient condition to have a fully entangled state is that the control ...
Yaron Jarach's user avatar
5 votes

Controlled Z gate entanglement

Showing that the matrix of a gate cannot be written as the tensor product is a tedious approach. It's easier to come up with a product state that the gate entangles. For this, it's helpful to have a ...
Adam Zalcman's user avatar
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5 votes
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Is it possible to produce an entangled state by measuring an unentangled state?

Depends on what you mean by "measuring". A "measure", in and of itself, produces (classical) outcomes, not necessarily states. But you can consider the quantum states that "...
glS's user avatar
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5 votes
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Open neighborhood of an entangled state with non-decreasing Schmidt rank

TL;DR: Yes. The set of pure bipartite quantum states with Schmidt rank at least $r$ is open$^1$. Let $S_{r}$ denote the set of all pure bipartite quantum states in $H_A\otimes H_B$ with Schmidt rank ...
Adam Zalcman's user avatar
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5 votes
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Is possible to write a separable state as a finite or countable infinite sum of product states?

Yes, you can always write an integral of separable states as a (finite) convex combination of product states, owing to the fact that the set of separable states is compact. A quick way to see this is ...
John Watrous's user avatar
  • 5,703
4 votes
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Is it advantageous for a state to have higher Bell CHSH violation?

Yes, the CHSH game/scenario is an example of a robust self-test. What this means is that if you obtain a value that is close to the optimal Bell violation, then the state $\hat{\rho}$ and the ...
Condo's user avatar
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4 votes

Is this state entangled?

$$ \frac{{\sqrt 2 }}{4}\left| {\left. {000} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {001} \right\rangle } \right. + \frac{{\sqrt 2 }}{4}\left| {\left. {010} \right\rangle } \right....
hft's user avatar
  • 868
4 votes

Is it true that if $U$ sends computational basis states to product states, then it sends product states to product states?

No - the CNOT gate is a simple counterexample. It maps computational basis states to computational basis states, but it can be used to create entangled states (e.g. when acting on $|+\rangle|0\rangle$)...
Norbert Schuch's user avatar
4 votes

Transformation of 0 state to superposition of 0 and + state with only using single qubit gates

I believe the following will work... How did I come up with this? I realised that the state you were after, $|0\rangle+|+\rangle$ is a +1 eigenstate of the Hadamard gate. There's a standard circuit ...
DaftWullie's user avatar
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4 votes
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What is a general entangled bipartite system?

I didn't see the question asked about pure states - then yes! A general entangled pure state takes the following form, via the Schmidt decomposition: $$|\Psi\rangle=\sum_i \psi_i |a_i\rangle\otimes |...
Quantum Mechanic's user avatar
4 votes

Why can't we use a two-qubit entangled state to send information faster than light?

The state that you start with can be written as $(|0\rangle+|1\rangle)\otimes (|0\rangle+|1\rangle)$. This means it is separable. Nothing you do on the first qubit will ever change anything on the ...
DaftWullie's user avatar
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4 votes
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How to create a maximally entangled state of two 4-level quantum mechanical systems?

Too long for a comment so writing it here. Say, for simplicity, $$ \begin{align} |0\rangle &:= |\text{red}\rangle \,, \\ |1\rangle &:= |\text{blue}\rangle \,, \\ |2\rangle &:= |\text{green}...
FDGod's user avatar
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4 votes
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What is the relation between maximally entangled and maximally mixed?

What is a maximally mixed state? Intuitively, think of a maximally mixed state as a quantum state where all possible measurement outcomes are equally likely to occur at random upon measurement. If ...
FDGod's user avatar
  • 1,843
4 votes

Equivalence of two unitary transformation with respect to local operations

Equivalent $U_1$ and $U_2$ should have the same spectrum. If eigenvalues of $U_1$ are all different, then there is essentially a unique unitary $V$ (up to a set of phases) which gives the equivalence $...
Danylo Y's user avatar
  • 6,989
4 votes
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Why can't you efficiently simulate quantum computers on classical computers

Superposition isn't what causes problems for simulation, but rather the way multiple qubits are combined (which is what gives rise to the notion of entanglement). Yes, the state of a single qubit is ...
Cody Wang's user avatar
  • 1,138
4 votes

Is it possible to derive a Schmidt decomposition for a mixed state?

Usually, the Schmidt rank of a mixed state $\rho$ is defined as the minimum of the maximal Schmidt rank of an element in any decomposition of $\rho$ into a mixture of pure states (there are many ...
Danylo Y's user avatar
  • 6,989
3 votes
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How to prove that the distillable entanglement satisfies $E_D(|\psi_d\rangle\!\langle\psi_d|)\ge \log d$?

The distillable entanglement of this state is equal $\log d$ by definition. Thus, both $E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\ge \log d$ and $E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\le \...
Norbert Schuch's user avatar
3 votes
Accepted

Can we use purity for separable states?

Yes, purity $p=\mathrm{tr}(\rho^2)$ is well-defined for both separable and inseparable $\rho$. Note that the purity of the state $\rho$ of a system $A$ does not tell us anything about internal ...
Adam Zalcman's user avatar
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3 votes

What is the complexity of determining if a state is entangled?

This question requires a careful modification, so that one would be able to speak about the complexity of a solution. Suppose you have an algorithm that guaranties to run in time $<f(n)$, i.e. ...
Danylo Y's user avatar
  • 6,989
3 votes

pennylane:fidelity calculation after swap test between entagled states. Swap test issue

As the error suggests, you cannot use QubitStateVector twice in the same QNode. Take, for instance, this simpler example that reproduces your error. Here, I want to ...
isaac's user avatar
  • 171
3 votes

How are logical operators performed and measured on the surface code, in a defect-free way?

I am still not sure how to perform [...] $X_L$ or $Z_L$ To perform logical X or logical Z in any code, apply the Pauli operators of the observable. This is always fault tolerant, in any stabilizer ...
Craig Gidney's user avatar
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3 votes
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(When) must the ground state of a frustrated Hamiltonian be entangled?

In the Ising model, I don't think frustration is the key feature. Let's start with an arbitrary Ising model with no transversal field. (I'll let you have arbitrary interactions, and arbitrary ...
DaftWullie's user avatar
  • 57.1k
3 votes
Accepted

Gottesman-Knill simulation and Bell states

See Nielson and Chuang 10.5.3. The state just before measurement is stabilized by $\langle X\otimes X, Z\otimes Z\rangle $. Measuring qubit one in the $Z$ basis, i.e. measuring $Z \otimes I$ works as ...
Abdullah Khalid's user avatar

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