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Let's deal specifically with a two-level system. I understand that ``coherence'' is due to the off-diagonal elements in the density matrix $\rho(t) = \sum_{i}|\psi_i(t)\rangle p_i\langle\psi_i(t)|$ for the system (generally written for mixed states). Entanglement can be measured by finding the eigenvalues $p_i$ of the density matrix at some time $t$ and adding up the terms $p_i\times \log(p_i)$.

What are these values physically? I have seen problems where one would start off with a pure product state of one type and another, and observe as the two states become entangled and coherence is transferred from one to the other and back. What does this actually mean?

I also find mention of correlations. How do those fit into this picture? Can one have entanglement without coherence, or coherence without entanglement?

Edit: "Entanglement can be measured..." actually means that we are quantifying the entanglement spectrum via entropy.

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    $\begingroup$ This is a very confusing question. First, the density matrix $\rho$ isn't generally equal to "psi psi". That would completely miss the point. That's the density matrix for a pure state - a case in which the concept of density matrices is useless. In general, a density matrix is a linear combination of such "psi psi" terms with coefficients $p_i$ whose interpretation is the probability of a given "psi". The off-diagonal entries prove the coherent phase between two basis vectors. Entanglement is nothing else than a correlation written in the most general quantum way. $\endgroup$
    – Luboš Motl
    Commented Mar 24, 2013 at 8:22
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    $\begingroup$ Two subsystems are entangled if the total state vector or density matrix can't be written as a tensor product of the state vectors or density matrices for these subsystems - but only as a linear combination of such tensor products. Finally, entanglement needs some complications on the Hilbert space and it can't exist if the Hilbert space is just two-dimensional because there's no way to factorize 2 to a product of two integers greater than one that would describe (nonempty) information about the subsystems. $\endgroup$
    – Luboš Motl
    Commented Mar 24, 2013 at 8:23
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    $\begingroup$ eqb: "Entanglement can be measured by finding the eigenvalues $p_i$ of the density matrix at some time t and adding up the terms $p_i×\log(p_i)$" This is not quite right. For a pure state $\rho_{AB} = |\psi_{AB}\rangle\langle\psi_{AB}|$ the eigenvalues of the reduced density matrix $\rho_A = \mathrm{Tr}_B \rho_{AB}$ determine the entanglement, via the expression you describe (von Neumann entropy). For mixed states, there is no unique measure of entanglement, so I would stick to the simple case of pure states for now. $\endgroup$
    – Mark Mitchison
    Commented Mar 24, 2013 at 20:14
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    $\begingroup$ The correct condition for unentangled mixed states is that $\rho = \sum_i p_i \rho_{A,i} \otimes \sigma_{B,i}$, where the $p_i$ are probabilities: $p_i \geq 0$, $\sum_i p_i =1$, and the $\rho_i$ and $\sigma_i$ are valid density matrices, i.e. $\rho_{A,i} = \rho_{A,i}^{\dagger}$, $\rho_{A,i} \geq 0$ and Tr($\rho_{A,i})$ = 1. $\endgroup$
    – Mark Mitchison
    Commented Mar 24, 2013 at 20:20
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    $\begingroup$ The conditions for a valid density matrix that I gave simply ensure that the eigenvalues of the density matrix define a valid probability distribution. $\rho \geq 0$ means that all of the eigenvalues $p_i$ of $\rho$ satisfy $p_i \geq 0$, or equivalently that $\langle \psi|\rho |\psi\rangle \geq 0$ $\forall |\psi\rangle$. $\endgroup$
    – Mark Mitchison
    Commented Mar 26, 2013 at 0:48

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This is just the comments formalised into a partial answer. First, it's important to realise that entanglement is a type of quantum correlation between two distinct systems. So, it's not useful to consider a single two-level system, and there is no such thing as entanglement between states. As Lubos Motl points out, you need to consider a system that has a natural decomposition into separate subsystems in order for entanglement to become a meaningful concept. The simplest such example is a pair of two-level systems that are separated in space.

OP wrote:"Entanglement can be measured by finding the eigenvalues $p_i$ of the density matrix at some time $t$ and adding up the terms $p_i\times\log(p_i)$"

This is not quite right. For a pure state $\rho_{AB}=|\psi_{AB}⟩⟨\psi_{AB}|$ the eigenvalues of the reduced density matrix $\rho_A = \mathrm{Tr}_B \rho_{AB}$ determine the entanglement, via the von Neumann entropy $S(\rho_A) = -\mathrm{Tr}(\rho \log \rho)$. The eigenvalues represent the probabilities of finding the subsystem in a given state. Any bipartite pure state can be written using the Schmidt decomposition as $$|\psi_{AB}\rangle = \sum_i \lambda_i |i_A\rangle \otimes |i_B\rangle. $$ Then you can show easily that the eigenvalue of the reduced density matrix $p_i = |\lambda_i|^2$ is the probability of finding system $A$ in the state $|i_A\rangle$. Note that the structure of the state $|\psi_{AB}\rangle$ implies that finding system $A$ in state $i_A$ means that system $B$ will necessarily be in state $i_B$. Therefore two systems in an entangled state are correlated. Also, in order for entanglement to be present, the state must be in a coherent superposition, so that at least two of the $\lambda_i \neq 0$, otherwise you can easily show that $S(\rho_A) = 0$.

For mixed states, there is no unique measure of entanglement, and things get much more complicated. Lubos Motl wrote:"Two subsystems are entangled if the total state vector or density matrix can't be written as a tensor product of the state vectors or density matrices for these subsystems - but only as a linear combination of such tensor products."

This is also not strictly true: it is only true for pure states. Mixed states such as $$\rho_{AB}=\frac{1}{2}(|0_A0_B⟩⟨0_A0_B|+|1_A1_B⟩⟨1_A1_B|)$$ are not entangled. The state above still has classical correlations between the two systems: if you find system $A$ in state $0$ then $B$ will also be in that state. However, there is nothing quantum about this state, it can be prepared simply by flipping coins and talking on the telephone. The correct condition for unentangled mixed states is that $$ \rho = \sum_i p_i \;\rho_{i,A}\otimes \tau_{i,B}, $$ where the probabilities $p_i$ and density operators $\rho_i$, $\tau_i$ must define valid probability distributions.

So in summary, entanglement always implies correlations between two systems, which must also exist in a coherent superposition. However, neither the presence of coherence nor correlations imply that a system is entangled.

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  • $\begingroup$ You need to consider two systems, but they don't need to be separated in space. For instance, you could consider two properties of the same particle. $\endgroup$
    – a06e
    Commented Mar 27, 2013 at 0:27
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    $\begingroup$ @becko Yes, thanks, that has been corrected. The question is what do we mean by "two systems", which is a fairly imprecise term in quantum mechanics. Two properties of a single particle can equally be considered "one system" with a larger Hilbert space. The point is to find a decomposition into subsystems that is natural from the point of view of macroscopic, localised observers. Usually this decomposition results from spatial separation, although you're right that it doesn't have to. The interesting nonlocal aspects only become meaningful with spacelike separation, though. $\endgroup$
    – Mark Mitchison
    Commented Mar 27, 2013 at 0:41
  • $\begingroup$ This is a good explanation of entanglement! $\endgroup$
    – freecharly
    Commented Mar 10, 2018 at 2:05

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