39 votes

How do I show that a two-qubit state is an entangled state?

A two qudit pure state is separable if and only if it can be written in the form $$|\Psi\rangle=|\psi\rangle|\phi\rangle$$ for arbitrary single qudit states $|\psi\rangle$ and $|\phi\rangle$. ...
DaftWullie's user avatar
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19 votes
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What does it mean for two qubits to be entangled?

For a simple example suppose you have two qubits in definite states $|0\rangle$ and $|0\rangle$. The combined state of the system is $|0\rangle\otimes |0\rangle$ or $|00\rangle$ in shorthand. Then if ...
snulty's user avatar
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18 votes
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Why isn't there a contradiction between the existence of CNOT gate/entanglement and the no-cloning theorem?

The cloning theorem requires that the result of the cloning is two independent copies of the starting qubit, i.e., the state of the system in the end should be $\big(\alpha |0\rangle + \beta |1\rangle ...
Mariia Mykhailova's user avatar
16 votes
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How does evolving a two-qubit state through a CNOT gate entangle them?

You are correct that none of the states in the truth table are entangled. Not all states become entangled when acted on by the CNOT. The entangled state in your question would result if the control ...
James Wootton's user avatar
15 votes

Are entangled and Bell states the same thing?

A bipartite state $\rho_{AB}$ is called entangled if it cannot be written as $$ \rho_{AB} = \sum_i \lambda_i \sigma_A^i \otimes \sigma_B^i, $$ for some $\lambda_i\geq 0$, $\sum_i \lambda_i = 1$ and $\...
Rammus's user avatar
  • 5,773
15 votes
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What is a "maximally mixed state"?

The maximally mixed state is a quantum state whose density matrix is proportional to the identity matrix. Physically, it may be interpreted as a uniform mixture of states in an orthonormal basis. The ...
Adam Zalcman's user avatar
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14 votes
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What are min and max overlaps of a maximally entangled state with a separable state?

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 ...
Adam Zalcman's user avatar
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13 votes
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Is entanglement transitive?

TL;DR: It depends on how you choose to measure entanglement on a pair of qubits. If you trace out the extra qubits, then "No". If you measure the qubits (with the freedom to choose the optimal ...
DaftWullie's user avatar
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13 votes
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Is entanglement necessary for quantum computation?

Short answer: yes One has to be a little bit more careful setting up the question. Thinking about a circuit as being composed of state preparation, unitaries, and measurements, it is always in ...
DaftWullie's user avatar
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13 votes
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What does "bipartite" mean?

What does a bipartite system mean? I'll summarize the main definitions below (adapted from Quantiki). Bipartite system and states: If Alice's subsystem is described by the Hilbert space $\mathcal{H}...
Sanchayan Dutta's user avatar
12 votes

How do I show that a two-qubit state is an entangled state?

Actually an even simpler way is as follows (reusing @nbro's notations). We have: \begin{align} |\Phi^{+}\rangle &= |a\rangle \otimes |b\rangle \\ &= \left( \alpha |0\rangle + \beta |1\rangle ...
codester's user avatar
  • 137
12 votes
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Unentangling a qubit from a system: can we convert $\alpha|000\rangle+\beta|111\rangle$ into $\alpha|00\rangle+\beta|11\rangle$?

Yes, it is possible. To obtain the state $$\vert \phi \rangle = \alpha \vert 00 \rangle + \beta \vert 11 \rangle$$ from $$\vert \psi \rangle = \alpha \vert 000 \rangle + \beta \vert 111 \rangle,$$ ...
M. Stern's user avatar
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12 votes
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Does quantum computing already possess the level of abstraction to be explicable even without knowledge of physics?

I don't think you need to know quantum physics to understand quantum computing - similarly to how you don't think about the hardware implementation of the classical computers when you write high-level ...
Mariia Mykhailova's user avatar
11 votes
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What happens if two separately entangled qubits are passed through a C-NOT gate?

$\newcommand{\bra}[1]{\left<#1\right|}\newcommand{\ket}[1]{\left|#1\right>}\newcommand{\bk}[2]{\left<#1\middle|#2\right>}\newcommand{\bke}[3]{\left<#1\middle|#2\middle|#3\right>} % $...
DaftWullie's user avatar
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11 votes
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How to implement the 4 Bell states on the IBM Q (composer)?

Remember that the Hadamard (H) gate maps $|0\rangle$ to $\frac{|0\rangle+|1\rangle}{\sqrt{2}}$ and $|1\rangle$ to $\frac{|0\rangle-|1\rangle}{\sqrt{2}}$, while the CNOT gate has the following ...
Sanchayan Dutta's user avatar
11 votes

How to implement the 4 Bell states on the IBM Q (composer)?

One of the possible solutions: $|\Phi^+\rangle = \textrm{CNOT} \cdot H_1 |00 \rangle$ $|\Phi^-\rangle = Z_1 |\Phi^+\rangle = Z_1 \cdot \textrm{CNOT} \cdot H_1 |00 \rangle$ $|\Psi^+\rangle = X_2 |\...
user1271772 No more free time's user avatar
11 votes
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General construction of $W_n$-state

Yes, there are several implementations of this algorithm in the Superposition quantum kata (tasks 14 and 15): For $n = 2^k$, you can use a recursive algorithm: create a W state on the first $2^{k-1}$ ...
Mariia Mykhailova's user avatar
11 votes

Does entanglement allow enhanced communication efficiency?

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 ...
Lena's user avatar
  • 2,597
11 votes
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Qiskit CNOT-gate matrix mixup?

Qiskit uses "little endian" bit ordering. That means, if A and B are $2 \times 2$ unitary matrices then $B \otimes A$ (note the order) is equivalent to applying $A$ to first qubit and $B$ to ...
Egretta.Thula's user avatar
10 votes

Why is an entangled qubit shown at the origin of a Bloch sphere?

Let $(x,y,z)$ be a point in the unit ball with $x^2+y^2+z^2 \leq 1$. The state associated with this point is \begin{eqnarray*} \rho &=& \frac{1}{2} (I_2 + x \sigma_x + y \sigma_y + z \sigma_z)\...
AHusain's user avatar
  • 3,633
10 votes
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Does Controlled-U gate entangle qubits?

For any controlled-$U$, if the input state is $|+\rangle|\phi\rangle$ where $|\phi\rangle$ is not an eigenstate of $U$, then the output state is entangled. This immediately deals with trivial cases ...
DaftWullie's user avatar
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10 votes
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What are the possible non-entangling two-qubit gates?

There are no other non-entangling gates in $SU(d^2)$ in any dimension $d=2,3,\dots$. Note that the global phase is irrelevant to the problem, so we lose no generality by considering non-entangling ...
Adam Zalcman's user avatar
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9 votes

Is it true to say that one qubit in an entangled state can instantaneously affect all others?

If Alice and Bob have an entangled pair of qubits and Alice locally measures her qubit, it does not affect local state of the Bob's qubit in any way. Mathematically, if Alice measures but does not ...
kludg's user avatar
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9 votes
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When was the first use of the word Entanglement?

I managed to get access to the paper mentioned in the question. Schrödinger in 1935 (the same year the original EPR paper was published) wrote in English: "By the interaction the two representatives (...
user1271772 No more free time's user avatar
9 votes

Is entanglement transitive?

This isn't an answer, but instead just some background facts that are important to know about in order to avoid "not even wrong" territory on these types of questions. "Entanglement" is not all-or-...
Craig Gidney's user avatar
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9 votes
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How to show whether a bipartite high-dimensional system is entangled?

Determining whether a given state is entangled or not is NP hard. So if you include all possible types on entanglement, including mixed states and multipartite entanglement, there is never going to be ...
James Wootton's user avatar
9 votes

Maximally mixed states for more than 1 qubit

For two probability distributions, there is a clear notion how to say which one is more mixed: $\vec p$ is more mixed than $\vec q$ if it can be obtained from $\vec p$ by a mixing process, this is, a ...
Norbert Schuch's user avatar
9 votes

Why is an entangled qubit shown at the origin of a Bloch sphere?

The Bloch sphere only represents the state of a single qubit. What you’re talking about is taking a multi-qubit state, and representing the state of just one of those qubits on the Bloch sphere. If ...
DaftWullie's user avatar
  • 58.1k
9 votes

Why do we have to uncompute rather than simply set registers to zero?

While it is definitely possible to reset registers after using them (namely, by simply measuring them) the issue is that doing so can potentially destroy interference that one might want to achieve by ...
MartinQuantum's user avatar
9 votes
Accepted

Why does measuring one qubit after the other in this entangled system alter the result?

If the simulator is saying that state 00 occurs 75% of the time then the simulator has a bug. Reordering measurements can't make certain outcomes more likely in that way. It would violate the no ...
Craig Gidney's user avatar
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