48 votes
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What's the intuition behind the Choi-Jamiolkowski isomorphism?

The intuition Let us consider a channel $\mathcal E$, which we want to apply to a state $\rho$. (This could equally well be part of a larger system.) Now consider the following protocol for applying ...
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47 votes
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What is the difference between general measurement and projective measurement?

Note: There is a short summary at the bottom. This is actually also described in Nielsen&Chuang: You don't learn about general measurements, because they are completely equivalent to projective ...
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  • 14.9k
40 votes
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Density matrix formalism

A mixed state is mathematically represented by a bounded, positive trace-class operator with unit trace $\rho : \cal H \to \cal H$. Here $\cal H$ denotes the complex Hilbert space of the system (it ...
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28 votes
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Are there more entangled states or non-entangled ones?

I'm assuming that you have a finite-dimensional base Hilbert space $\mathcal H_0$ and that you're building your full Hilbert space as $\mathcal H=\mathcal H_0\otimes \mathcal H_0$. In these conditions,...
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27 votes
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What is the actual meaning of the density operator?

Firstly, what is a state? A state gives you the complete description of a system. Let's label the state of a system $\lvert \psi \rangle$. This is a normalised state vector which belongs in the vector ...
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  • 5,268
21 votes

How is a quantum superposition different from a mixed state?

Apart from the already mathematically detailed answers given above, perhaps it would be useful to have a physical picture in mind -- the double slit experiment. The classical 50:50 picture ...
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17 votes

Schrödinger's cat and the difficulty of macroscopic superposition state

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ $\renewcommand{\bra}[1]{\left \langle #1 \right \rvert}$ We can see how decoherence really works, why it messes up superposition states, and ...
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  • 22.9k
17 votes

What is the entropy of a pure state?

I think it is a mistake, in this case, to think of entropy as "a description of our ignorance." Rather, I would suggest that you think of entropy as a well-defined, objective property provided that ...
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  • 7,002
17 votes
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Why does the density matrix $\rho$ obey a wrong-signed Heisenberg equation of motion?

$\rho_\psi$, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion $$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$ since it is ...
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  • 107k
16 votes
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Why can't every quantum state be expressed as a density matrix/operator?

The statement by yuggib is correct. To put it in perspective, I'll start with a completely general formulation, and then I'll show how vector-states and density operators fit into that picture. I won'...
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15 votes
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Liouville-von Neumann equation can be directly derived from Heisenberg picture?

Well, the von Neumann equation holds within Schroedinger's picture and it is immediate to prove in quantum physics (differently from the Liouville equation which needs to preventively establish the ...
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15 votes
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Is there a clear and intuitive meaning to the eigenvectors and eigenvalues of a density matrix?

In general, the density matrix of a given system can always be written in the form $$ \rho = \sum_i p_i |\phi_i\rangle\langle\phi_i|, \tag 1 $$ representing among other things a probabilistic mixture ...
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15 votes
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Examples of density operators $\rho=\sum\limits_n p_n|\phi_n\rangle\langle\phi_n|$ in which the states $\{|\phi_n\rangle\}$ are not orthogonal

This thread has seen a ton of incorrect statements coming from a number of sides, so it's probably a good idea to set the record straight in a bit more detail, and to provide some more examples of how ...
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14 votes

What is the physical meaning of the Lindblad operator?

General form, properties A Lindblad form $$\dot \rho = -i[\eta, \rho] + A \rho A^\dagger - \frac 12 A^\dagger A \rho - \frac 12 \rho A^\dagger A$$ has three important properties: It is still linear ...
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  • 35k
14 votes
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Is purification physicaly meaningful?

This really depends whether you believe in the "church of the larger Hilbert space". If you feel that pure states are more fundamental than mixed states, then you might argue that any mixed state is ...
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14 votes

Proof that you can't disentangle two parties if you only operate on one

If there was some unitary operator factorized as $\mathbb I_A \otimes U_B$ that would send the entangled state $|\psi\rangle$ to a factorized state $|\phi_A\rangle\otimes |\phi_B\rangle$, ie : $$|\...
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  • 4,026
13 votes
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Finding the matrix representation of a superoperator

If you want to write a super-operator representing left- or right-multiplication, there is a distinct method which is simpler and more elegant. Let us define the left-multiplication superoperator by $$...
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13 votes
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Does correlation of the measurement outcomes imply that a state is entangled?

The problem is that entanglement doesn't mean that a measurement on the first particle determines the outcome of the second. Although this is often perpetuated, it's not the gist of entanglement. ...
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  • 14.9k
13 votes

Does Unitary operator take a pure state to a pure state or can it take a pure state to a mixed state?

A unitary operator $U$ can only take a pure state $|\psi \rangle$ to a pure state. Let $U | \psi \rangle = | \psi' \rangle$. Then acting the unitary $U$ on the pure state $| \psi \rangle$ yields $$ ...
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  • 41.5k
12 votes

What does it mean for a density operator to be non-negative?

It means that all the eigenvalues are nonnegative, or equivalently $\langle \psi | \rho | \psi \rangle \geq 0$ for any vector $| \psi \rangle$ in the Hilbert space. See https://en.wikipedia.org/wiki/...
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  • 41.5k
11 votes

Density matrix formalism

But what mathematical space do density matrices belong to? It is clear that the expression $\hat{ρ}\in\mathcal{H}$ is incorrect, as a mixed state cannot possibly be described in terms of a state in a ...
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  • 7,366
11 votes

Are there more entangled states or non-entangled ones?

No, in fact, "most states" are entangled. (This is meant to be a heuristic; I freely admit this is probably a sticky thing to get into formally as far as randomly picking a state.) My intuition/reason ...
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  • 9,623
11 votes
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Why can't the off-diagonal terms in density operators of pure states be always removed?

On the other hand, the basis kets themselves are pure states, and the probability of observing $| \psi_i \rangle$ is $|c_i|^2$, so we should be able to express the density operator in terms of these ...
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  • 2,757
11 votes

Examples of density operators $\rho=\sum\limits_n p_n|\phi_n\rangle\langle\phi_n|$ in which the states $\{|\phi_n\rangle\}$ are not orthogonal

Just consider a two-level system and take the three states $|0\rangle$, $|1\rangle$, and $|+\rangle = (|0\rangle+|1\rangle)/\sqrt{2}$. Then, the mixed state $$ \rho = \tfrac13 |0\rangle\langle0| + \...
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10 votes

Why does the density matrix $\rho$ obey a wrong-signed Heisenberg equation of motion?

Your question apparently stems from a lack of understanding of the different pictures in quantum mechanics, that are Schrödinger picture, Heisenberg picture and Interaction picture. In the ...
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  • 479
10 votes
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Lindblad equation for heisenberg operators?

$\def\dd{{\rm d}} \def\LL{\mathcal{L}} \def\ii{{\rm i}} \def\ee{{\rm e}}$ The trick here is very simple and physically motivated. You simply demand that the expectation value of an operator $A$ is the ...
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10 votes
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What's the physical meaning of the kernel of density matrix?

Suppose the system is in some mixed state described by the density operator: $$\hat { \rho}=\sum_{k=1}^N \rho_k |\psi_k\rangle\langle\psi_k|$$ where $N$ is the dimensionality of the Hilbert space. Now ...
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10 votes
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Space of density matrix

Is this phrase true? $|\psi\rangle \langle\psi| \in V \otimes V_\mathrm{dual}$ Yes, this is a correct way to see things, but the more usual view is to note that $$ V \otimes V_\mathrm{dual} \cong \...
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10 votes

Why is Schrödinger's cat in a superposition and not a mixture if you model decay with Fermi's golden rule?

I think that using Fermi's golden rule here is going a bit too far. If you look at a standard derivation of the rule, you will see that, roughly speaking, they calculate a transition amplitude $\...
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10 votes
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Measurement on mixed states

As mentioned by the OP both versions are the same. For an observable $A$ of the form $$A = \sum\limits_k a_k \, P_k \quad , $$ with the projections $P_k^2 =P_k = P_k^\dagger$ on the eigenspace ...
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