Physics Asked on June 24, 2021
I read this question which clarified the concepts of mixed/pure/separable/entangled states, but I can’t see if one of "mixed/pure" implies one of "separable/entangled" and/or vice versa.
Let $mathcal{H}=mathcal{H}_Aotimesmathcal{H}_B$.
and vice versa, for each point.
I think point (3) is true (and so whould be point (4)) but I can’t see why.
Separability and purity refer to different classes of properties of a state. A generic bipartite state $rho$ being separable means that is can be written as a convex combination of product states, that is, it has a decomposition of the form $$rho = sum_i p_i rho_i^A otimes rho_i^B,$$ for some $p_ige0$ with $sum_i p_i=1$ and collections of states $rho_i^A,rho_i^B$ on the two systems. This general definition works for arbitrary (i.e. generally non-pure) states. If a state $rho$ is pure, it means that it is a rank-1 projector, and can be written as $rho_{Psi}=|Psirangle!langle Psi|$ for some ket state $|Psirangle$. A pure, separable state is always a product state, meaning $$|Psirangle = |psirangleotimes|phirangle, qquad rho_Psi = rho_psiotimesrho_phi.$$ For example:
As above, no, entanglement/separability and purity/mixedness measure different types of properties. An entangled state can be both pure and mixed. For example, a two-qubit Bell state is entangled and pure. A Werner state is (for $p<1/3$, or some other range depending on the way it is defined) entangled, but not pure. More explicitly, you can use as an example $$(1-epsilon)|Psi^+rangle!langlePsi^+|+epsilon frac{Iotimes I}{4},$$ where $|Psi^+rangle$ is a(ny) Bell state, and $epsilon>0$ is small enough. The idea is that you are mixing a maximally entangled state with a maximally mixed one. If the mixed component is small enough, the state remains entangled.
One thing that can be said though is that entanglement in pure states is relatively easy to detect and classify, whereas things get way more tricky for general entangled states.
No. For example, the maximally mixed state shown above is separable and its partial states are also maximally mixed one-qubit states, which are not pure.
If the state is pure and entangled, then the partial states are necessarily mixed, and more precisely have rank equal to the Schmidt rank of the full state. More generally, if the partial states are pure, that means that the original state is not only separable, but also a product state, as it produces no correlations whatsoever. Turning this around, whenever the state is not a product state (and thus in particular whenever the state is entangled), the partial states ought to be mixed.
Here's a brief summary of the examples of pure/mixed separable/entangled states:
pure | mixed | |
---|---|---|
product | $|0rangle!langle0|otimes |0rangle!langle0|$ | $frac{1}{4}(Iotimes I)$ |
separable | always also product | $frac{1}{2}(|00rangle!langle00|+|11rangle!langle11|)$ |
entangled | Bell states | (some) Werner states |
Correct answer by glS on June 24, 2021
The concept of pure and entangled quantum states are decoupled from each other. That is to say, there exists states for all the 4 possible combinations.
Analogous to the pure state scenario, bipartite mixed states $rho$ that can be written as $displaystyle rho=sum_{i} p_{i};rho_{A}^{i}otimes rho_{B}^{i}$ are defined to be separable and all other mixed states are said to be entangled.
Answered by Arnab on June 24, 2021
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