I don't know if the following helps, but for me it was very insightful to visualize the scaling behavior of quantum systems:
The main problem comes from the fact that the Hilbert space of quantum states grows exponentially with the number of particles. This can be seen very easily in discrete systems. Think of a couple of potential wells that are conected to each other, may just two: well 1 and well 2. Now add bosons (e.g., Rubidium 87, just as an example), at first only one. How many possible basis vectors are there?
- basis vector 1: boson in well 1
- basis vector 2: boson in well 2
They can be written like $\left|1,0 \right\rangle$ and $\left|0,1 \right\rangle$
Now suppose the boson can hop (or tunnel) from one well to the other. The Hamiltonian that describes the system can then be written is matrix notation as
$$
\hat{H}=\pmatrix{ \epsilon_{1} & t \\ t & \epsilon_{2}}
$$
where $\epsilon_{1,2}$ are just the energies of the boson in well 1 and 2, respectively, and t the tunneling amplitude.
The complete solution of this system, i.e., the solution containing all the information necessary to compute the system's state at any given point of time (given an initial condition), is given by the eigenstates and eigenvalues. The eigenstates are linear superpositions of the basis vectors (in this case $\left|1,0 \right\rangle$ and $\left|0,1 \right\rangle$).
This problem is so simple that it can be solved by hand.
Now suppose we have more potential wells and more bosons, e.g., in the case of four wells with two bosons there are 10 different possibilities to distribute the bosons among the wells. Then the Hamiltonian would have 10x10=100 elements and 10 eigenstates.
One can quickly see that the number of eigenstates is given by the binomial coefficient:
$$
\text{number of eigenstates}=\pmatrix{\text{number of wells} + \text{number of bosons} - 1 \\ \text{number of bosons}}
$$
So even for "just" ten bosons and ten different potential wells (a very small system), we'd have 92,378 eigenstates. The size of the Hamiltonian is then $92,378^2$ (approximately 8.5 billion elements). In a computer they'd occupy (depending on your system) about 70 gigabytes of RAM and is therefore probably impossible to solve on most computers.
Now let's assume we have a continuous system (i.e. no potential wells, but free space) and 13 water molecules (for simplicity I treat them each of them as a particle). Now in computer we can still model free space using many tiny potential wells (we discretize space... which is ok, as long as the relevant physics takes place on larger length scales then the discretization length). Let's say there are 100 different possible positions for each of the molecules in each of the x, y and z directions. So we end up with 100*100*100 = 1,000,000 little boxes. Then we'd have more than $2.7 \cdot 10^{53}$ basis vectors, the Hamiltonian would have almost $10^{107}$ elements, occupying so much space that we'd need all the particles from 10 million universes like ours just to encode that information.