Deriving or building a Hamiltonian from a Density Matrix Is it possible to create a Hamiltonian if given a Density Matrix.
If you already the the Density Matrix, then is the Partition
Function (Z) even needed?
This Q is not about physics. Its about an application of math 
to poorly defined and dynamic systems such as populations
(any kind) and stock/bond portfolios (just another type of 
population).
I see a connection here.
 A: 
Is it possible to create a Hamiltonian if given a Density Matrix.

No. The hamiltonian describes the system and the density matrix describes the state of the system. Your question's equivalent in classical mechanics would be something like

is it possible to recreate the force's dependence on the position if you're given a single snapshot of the particle's position at a single instant?,

which should be immediately recognizable as ridiculous.

That said, since you mention partition functions, it's reasonably likely that you're asking about thermal states, in which case your question can be rephrased as 

Is it possible to recover the hamiltonian of a system given the density matrix of a thermal state under that hamiltonian?

to which the answer is yes.
The simple answer to that is that the density matrix of a thermal state is given by
$$
\hat \rho = \frac1Z e^{-\beta\hat H},
$$
so therefore the hamitonian can be recovered as
$$
\hat H =  -\frac{\ln(\hat\rho)+\ln(Z)}{\beta }
$$
That might make you understandably nervous, since it includes the logarithm of an operator. This is to be understood as acting on the eigenvalues in any spectral decomposition: if 
$$
\hat\rho = \sum_{n,m} p_n |n,m⟩⟨n,m|
$$
is a spectral decomposition of $\hat\rho$, with $m$ indexing the eigenprojectors within each degenerate subspace, then the hamiltonian can be constructed explicitly as
$$
\hat H =  -\sum_{n,m} \frac{\ln(p_n)+\ln(Z)}{\beta } |n,m⟩⟨n,m|.
$$
Now, if all you know is $\hat\rho$, then you have access to the $p_n$ and the $|n,m⟩⟨n,m|$, but you won't be able to access the partition function, so you will only be able to calculate $\hat H$ up to the first term in
$$
\hat H =  -\sum_{n,m} \frac{1}{\beta}\ln(p_n) |n,m⟩⟨n,m| -  \frac{1}{\beta }\ln(Z) \mathbb I,
$$
i.e. you will only be able to recover $\hat H$ modulo a constant times the identity operator. This is obviously to be expected, since offsetting the hamiltonian by a constant will not affect the thermal states it produces.

As for this comment, on the other hand,

The idea is to jump from "Quantum Mechanics" to any complex system: stock portfolios, funds, population dynamics (of organisms), etc.

if the system is described by the same mathematics as QM, then the mathematics will apply. However, if you don't have a solid basis of modelling work on the other field that tells you that the mathematical formalism is valid, then it's not valid. None of the fields you've mentioned have such an analogy.
A: In both classical and quantum mechanics you have two relatively separate concepts:


*

*State of the system.

*The evolution law of the system.
In the classical mechanics of $N$ point particles, you can think of the first as of particles' positions and momenta, $\{q_1,\ldots,q_N,p_1,\ldots,p_N\}$, while of the second as of the Hamiltonian function $H(q_1,\ldots,q_N,p_1,\ldots,p_N)$.
For example, the (initial) state of a one-dimensional system at time $t_0$ can be given by some numbers $q=q_0, p=p_0$. What will happen to the particle in the next second, depends on the form of $H$, which can be, for example, $H=p^2+q^2$ or $H=p^2+q^4$. At any moment of time, given the values of your $q$'s and $p$'s, you can calculate any observable (which is a simply a function of those). For example, if $q$ stands for the Cartesian coordinate, the kinetic energy at $t=t_0$ is simply
$$E_K(q,p) = \dfrac{p_0^2}{2m}$$
Lastly, we should say that, in case if your description of the quantum system is indeterministic, instead of precise values of $q_0$ and $p_0$, you have to deal with some probability distribution of those, $f_0(p,q)$. In such a case, your best guess for the energy at $t_0$ would be
$$E_K = \int  f_0(q,p) E_K(q,p)dqdp$$
Now, coming back to your original question. We have more or less same thing in quantum mechanics. The state of a 'deterministic' system given by a wave vector (one way to represent which is a wave function). I used quotes because even if you know the state of the system precisely, depending on the particular form of the measurement, you still may not be able to predict its outcome. However, you may also have a quantum-mechanical state which would be an analogue of a classical indeterministic state. And that's where you would have to introduce the density matrix (actually, Landau did that for you). Moral: the state of any quantum-mechanical system is given by a density matrix which, in certain cases, can be reduced to a single wave vector.
However, so far we've been only talking about the states of the system. The system's evolution is defined by its Hamiltonian, which is now not a function but rather an operator acting on the space of wave vectors describing the system's state (it is typically a function of $\hat{q}$ and $\hat{p}$, which are also operators, and no longer define the state of the system.
So, the correspondence goes as follows. State of the system:
$$
\text{Deterministic (pure) states:}\quad\{q,p\}\longleftrightarrow |\psi\rangle \\
\text{Indeterministic (mixed) states:}\quad f(q,p)\longleftrightarrow\hat{\rho}=\int f(q) |q\rangle\langle q| dq\\
$$
Evolution of the system:
$$
H(q,p)\longleftrightarrow\hat{H}(\hat{q},\hat{p})
$$
Bottom line: the density matrix and the Hamiltonian are two different concepts which have to be specified separately in order to give the full description of the system: the former specifies its initial state, the latter — how the it will evolve with time.
A: No.  Suppose you have an admixture of spin-up/spin-down states:
$$
\rho=\left(\begin{array}{cc}
1-\alpha & 0 \\
0 &\alpha\end{array}\right)\, . 
$$
There is no information about the evolution of the system, in the sense that there is no reason to suppose that this density matrix must evolve according to $H=\omega \sigma_z$ or $H=\omega\sigma x$ or the more general $H=\omega \hat n\cdot\vec\sigma$ for that matter.
