Information contained in a quantum simulation  
The system in the picture has a box filled with $N$ classical particles which can either be on the left or the right side. Let this system have a Hamiltonian $$H= \frac{\mathbf p^2}{\sum 2m_i} + V(x, t) $$ where V becomes infinite at $\pm \frac{L}{2} $. Let it be represented by a point in $2N$-dimensional phase space $$ (x_1, p_1, ... , x_N, p_N) $$ If we say that a positive x coordinate corresponds to 1, and negative to 0, then this system will encode a string of bits $N$ digits long.
Suppose I want a quantum computer to track the information in the string of bits above. The computer will contain $M$ qubits where $M = \log N$. Assuming this is a perfect quantum computer with no errors, the general state of the computer will be a density matrix in $2^M$-dimensional Hilbert space. $$ \rho = \begin{bmatrix} c_{1,1} & c_{1,2} & \dots & c_{1,N} \\ c_{2,1} & c_{2,2}  \\ \vdots & & \ddots \\ c_{N,1} & & & c_{N,N} \end{bmatrix}$$
Now we will quantify the amount of information in both systems.
In order to quantify how much information the box of particles contains, we can use the Shannon Entropy. For a random variable $P_i$ representing the probability of finding a one of a zero in the string at position $i$, the information entropy is$$ S = - \sum_i P_i \log P_i $$ If we assume the particles are sufficiently mixed up in the box, then every $P_i$ will have a probability $\frac{1}{2}$, and then total entropy will be $\frac{N}{2} \log 2$
Next we turn the quantum computer, where the measure of the amount of information is Von Neumann Entropy. Given a density matrix $\rho$ for the quantum computer, the amount of information is given by $$ S = - Tr(\rho \log \rho) $$
Question 1: Does there exist a density matrix that contains the same amount of information as the classical system?
Question 2: Does there exist a Hamiltonian to control the evolution of the quantum computer so that the information contained in both systems are always equal?
 A: What do you mean by "information"? 
There is classical information, which lives in binary strings, and there is quantum information, which lives in Hilbert spaces. They aren't the same thing. 
Holevo's theorem says that the most classical information you can extract from a system of $t$ qubits is $t$ classical bits. However, if you want to approximately describe an arbitrary quantum state on $t$ qubits, you need $c^t$ classical bits to do so ($c$ is a constant that depends on how good you want the approximation to be). 
So for your question, the amount of information you need to prepare the system is the same: $N$ bits, and the amount of information you can extract from the quantum system is only $\log N$ bits. Which of these is the real, bona fide information? That depends on your definition of information.
The standard definition of information in quantum information theory is von Neumann information, which for a system with density matrix $\rho$ is $-\mathrm{Tr} \,\rho \log \rho$. When you extend the theorems of classical information theory to quantum mechanics, this is generally the correct definition. By this measure, the quantum system has $\log N$ bits of information. So by the standard definition, they're different. 
But as for how much "information" there really is in the system — this seems like a philosophy question to me.
A: In general, the entropy of a system is written
$$
S_\text{tot}=S_{EE}+S_\text{thermal}
$$
where EE denotes the entanglement entropy. In a thermal classical system, you have no ignorance associated with the possibility of states being entangled. However, a quantum computer simulates the classical system using some quantum circuit that has non-zero entanglement entropy so clearly the information content of the two systems are different.
Ignoring the quantum computer itself, the information pertaining to the classical simulation the quantum computer performs contains the same amount of information as the classical system being simulated.
