Coherence, Entanglement, Correlation 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.
 A: 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. 
