Take the 2-minute tour ×

Let's say that I have a system of $n$ particles $p_1,\ldots,p_n\in\mathbb{R}^3$ (where $n$ here is on the order of 10,000). Furthermore, suppose we have a graph $G=(V,E)$ describing some network, where the set of vertices $V$ is the set of particles and the set of edges $E$ satisfies $|E|\ll n^2$. Each edge in $G$ will represent a spring between the corresponding pair of particles holding them together. I will also give each particle $p_i$ its own scalar potential $V_i$.

The wavefunction of this system will be of very large dimension: $\Psi(x_1,\ldots,x_n;t)$. We can write the Hamiltonian as $$H=\sum_{i=1}^n\left(-\frac{h^2}{2m} \nabla_i^2+V_i(x_i)\right)+\sum_{(p_i,p_j)\in E} c\|x_i-x_j\|^2$$

Obviously I cannot step the Schrodinger equation for this huge system forward in time using any standard discrete simulation method, nor can I even write down a discretized version of $\Psi$ due to its large dimensionality.

Are there techniques for computing approximate low-energy states of this system? What approximations are reasonable here? Can one find likely positions of each particle $p_i$?

share|improve this question
It depends what $V_i$ is. If $V_i=0$ or bilinear function of $x$, then the system is solvable - it's just another multi-dimensional harmonic oscillator. For other $V$, you want to find some good zeroth approximation and solve it by perturbation theory. –  Luboš Motl Oct 25 '12 at 17:09
add comment

1 Answer

The way you wrote it is probably not what you intended to write down--- you probably want $||x_i - x_j - C_{ij}||^2$ instead of $||x_i-x_j||^2$, so that the springs have a nonzero stationary length, so that the masses have a shape in the absence of external forces. If you don't do this, the classical least-energy solution is all the particles are exactly on top of each other.

But I will solve it as you wrote it (it's not much harder the other way, but it's more annoying to figure out the stationary configuration). In this case, you write the spring term as

$$ \sum A_{ij} x_i x_j $$

and you diagonalize A by a rotation (this is possible since A is symmetric).

$$ y_i = R_{ij} x_j $$

Where the sum on j is implicit, and R is a rotation matrix. You can compute the R using any diagonalization algorithm, they are all easy in the symmetric case. Then in terms of y, you have decoupled oscillators

$$ \sum \lambda_i y_i^2 $$

Where the $\lambda_i$ are the eigenvalues of A, and each one is a normal harmonic oscillator that you solve independently. To find the expected values of the x's, you write the x's in terms of the y's.

When you have a symmetry, like an abstract lattice of x's where there are only finitely many classes of points not related by lattice symmetry to each other, you can write the explicit solution for even an infinite lattice, using Fourier theory. This is the starting point for quantum field theory.

share|improve this answer
In the end I'll actually be doing this computation on a manifold rather than $\mathbb{R}^n$, so zero-length springs are alright. But, such factorizations won't be possible. Are there more generic tricks? –  Justin Solomon Oct 26 '12 at 18:14
@JustinSolomon: Are you restricting the x's to lie on a manifold? This is a completely different problem. If you have constraints, the kinetic terms become nonlinear, and then the springs linearity is not at all the complicated thing, it's the other nonlinearity in the kinetic term. –  Ron Maimon Oct 26 '12 at 19:05
Ah yes, I am restricting them to be on a manifold. Since I'm doing simulation, the manifold is a triangulated surface, and I can write down its Laplacian operator if that's useful. –  Justin Solomon Oct 27 '12 at 20:08
I don't think this is what you mean, a triangulated surface. This is not the right way to do restriction, you should find a polynomial approximation to the restricting function--- triangulated surfaces are usually no good for doing Newton's laws on. –  Ron Maimon Oct 27 '12 at 21:12
@JustinSolomon: Can you ask the question using your setup then--- the main problem in doing newton's laws on a simplicial complex is that the complex is not a differentiable manifold already at the intersections of the intersections of the simplices, where there is infinite curvature. Your main question should be "how do you do Newon's laws on a triangulated manifold in such a way that it reproduces smooth Newton's laws?" The problem you are asking in your question is trivial. Please, ask another question, because there is little relation between what you asked and what you want. –  Ron Maimon Oct 29 '12 at 1:57
show 1 more comment

Your Answer


