# Tag Info

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The state $$|\Psi \rangle = \frac{1}{\sqrt{2}}\left(|\psi_1\rangle +|\psi_2\rangle \right)$$ is a pure state. Meaning, there's not a 50% chance the system is in the state $|\psi_1\rangle$ and a 50% it is in the state $|\psi_2\rangle$. There is a 0% chance that the system is in either of those states, and a 100% chance the system ...

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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 may be nonseparable). The set of mixed states $S(\cal H)$ is a convex body in the complex linear space of trace class operators $B_1(\cal H)$ which is a two-side ...

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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 measurements + unitary time evolution + ancillary systems, all of which is described in your usual QM formalism. The Measurement Postulate Let's start from ...

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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 $\mathcal E$ to $\rho$: Denote the system of $\rho$ by $A$. Add a maximally entangled state $|\omega\rangle=\tfrac{1}{\sqrt{D}}\sum_{i=1}^D|i,i\rangle$ of the ...

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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, the set of separable states has measure zero. (It gets a bit more complicated if you have $\mathcal H_0^{\otimes 4}$ and you're allowed to split it any way ...

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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 space of states. Keep in mind that we are talking about the full state; I haven't decomposed it into basis states, and I will not. This is not what the ...

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Yes, the density matrix reconciles all quantum aspects of the probabilities with the classical aspect of the probabilities so that these two "parts" can no longer be separated in any invariant way. As the OP states in the discussion, the same density matrix may be prepared in numerous ways. One of them may look more "classical" – e.g. the method following ...

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Let $\{|i\rangle\}$ be an orthonormal basis for the Hilbert space of the system. Then the trace of an operator $O$ is given by (See the Addendum below) \begin{align} \mathrm {tr}(O) = \sum_i \langle i|O|i\rangle \end{align} For a given state $|\psi\rangle$, we define an operator $P_\psi$ by \begin{align} P_\psi|\phi\rangle = \langle\psi|\phi\rangle|\psi\...

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Physically, a CP map represents evolution processes even in the presence of entanglement. After all positive maps sends a state to a state This is false when the system is entangled with something else. The classic example is the transpose on the system of interest. (Technically, a transpose, since it's basis-dependent.) Since positivity can be ...

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The sentence of Wikipedia : "For example, there may be a 50% probability that the state vector is $| \psi_1 \rangle$ and a 50% chance that the state vector is $| \psi_2 \rangle$ . This system would be in a mixed state." is false. The difference between pure states and partially or completely mixed states, is only a difference of structure of the ...

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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 you specify which degrees of freedom in the universe are inside and outside of your system. The content of this statement isn't really different, but it ...

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No macroscopic quantum system is described by a pure state. For example, notions like temperature or pressure, which apply to macroscopic systems do not even exist for systems described by a pure state. The description of macroscopic objects (discussed in statistical mechanics) is always in terms of a density operator (or the essentially equivalent notion of ...

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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 corresponds to the case where you send, at random i.e. 50% chance, through either one of the slits. This will result in no interference pattern on the receiving screen. ...

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Yes, subsystems of an entangled state – if this subsystem is entangled with the rest – is always in a mixed state or "statistical mixture" which is used as a synonym in your discussion (or elsewhere). If we're only interested in predictions for a subsystem $A$ in a system composed of $A,B$, then $A$ is described by a density matrix $\rho_A$ calculable by "...

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Let me try to convince you that the density operator is a mathematical convenience and not a fundamental aspect of quantum mechanics by describing a very general setup for states and observables in both classical and quantum mechanics. This may not directly answer your question, but hopefully it will settle whatever motivated this question. Briefly, ...

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Well, the von Neumann equation holds witin Schroedinger picture and it is immediate to prove in quantum physics (differently from Liouville equation which needs to preventively establish the non-trivial Liouville theorem for the symplectic measure on the space of phases). Indeed, as $\rho$ is an incoherent superposition of pure states, one has, $$\rho = \... 13 \rho_\psi, the density matrix, is not an observable/operator evolving in the sense of the Heisenberg equation of motion$$ \mathrm{i}\hbar\frac{\mathrm{d}}{\mathrm{d}t} A = - [H,A]$$since it is defined, as you correctly write, as a projector on states, hence it is time-dependent in the Schrödinger picture (since there the states it projects on are time-... 12 General form, properties A Lindblad form$$\dot \rho = -i[\eta, \rho] + A \rho A^\dagger - \frac 12 A^\dagger A \rho - \frac 12 \rho A^\dagger A$$has three important properties: It is still linear dynamics, in terms of \rho. It is trace-free regardless of the trace of \rho. This means that the total sum of the eigenvalues, which starts out as 1, does ... 12 It means that all the eigenvalues are nonnegative, or equivalently \langle \psi | \rho | \psi \rangle \geq 0 for any vector | \psi \rangle in the Hilbert space. See https://en.wikipedia.org/wiki/Positive-definite_matrix. Whether or not the matrix elements of a finite-dimensional operator are nonnegative depends on the choice of basis, so non-negativity ... 12 This really depends whether you believe in the "church of the larger Hilbert space". If you feel that pure states are more fundamental than mixed states, then you might argue that any mixed state is just a lack of knowledge, and somewhere out there is the missing piece of the system which will give you full information (i.e., a pure state). Even though you ... 11 But what mathematical space do density matrices belong to? It is clear that the expression \hat{ρ}\in\mathcal{H} is incorrect, as a mixed state cannot possibly be described in terms of a state in a Hilbert space. A bounded linear operator on \mathcal{H} is of trace class iff it has finite trace independent of choice of basis (cf. also nuclear operator). ... 11 A unitary operator U can only take a pure state |\psi \rangle to a pure state. Let U | \psi \rangle = | \psi' \rangle. Then acting the unitary U on the pure state | \psi \rangle yields$$ U \left( | \psi \rangle \langle \psi | \right) U^\dagger = | \psi' \rangle \langle \psi' |,$$which is a pure state. Another way to see this is that a (... 11 The statement by yuggib is correct. To put it in perspective, I'll start with a completely general formulation, and then I'll show how vector-states and density operators fit into that picture. I won't try to be mathematically rigorous here, but I'll try to give an overview with enough keywords and references to enable further study. State = normalized ... 10 It is easy. Assume that \psi \in \cal H is normalized to 1. In this case, |\psi\rangle\langle\psi| it is nothing but the orthogonal projector P_\psi onto the one-dimensional linear space generated by the vector \psi. Putting |\psi\rangle and \langle\psi| together simply means to exploit the tensor product. If \phi' \in \cal H' the (... 10 The reduced density matrix can be found by taking the trace over the subspaces of the Hilbert space that represent systems you're not interested in. For the Bell state the density matrix of the whole system is$$\tfrac{1}{2}(|00\rangle+|11\rangle)(\langle 00|+\langle 11|)\\ = \tfrac{1}{2} (|00\rangle\langle 00|+|00\rangle\langle 11|+|11\rangle\langle 00|+|...

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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. Entanglement is (mostly?) nonclassical correlations. Bipartite states are correlated, when the outcome of the measurement on one particle tells us something about ...

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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 for saying this is that there exist approximation methods which work by restricting themselves to a low-entanglement subspace, such as Matrix Product States (...

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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 expressions of this form come up in practice. So, let's go through a brief rundown of some pertinent points. The definition of a density matrix is just an ...

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The reduced matrix is defined as the partial trace of the density matrix. Be $A$, $B$ finite dimensional Hilbert spaces and be $T$ $\in$ $L(A \otimes B)$ (Linear operators on $A \otimes B$), then the partial trace of T is defined as $\rm{Tr}_B [T]$ in $L(A)$ is defined by \langle a | \rm{Tr}_B [T]| b \rangle = \sum_n \langle a | \langle n |...

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