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Is there any non-trivial many-body system for which the exact solution to Schrödinger's equation is known? (By non-trivial, I mean a system with particle-particle interactions.) Perhaps something like positronium, or two electrons in a box.

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Are you looking for physics or mathematics? It is possible to make long-range nonphysical force models where the asymptotic ground state wavefunction is described exactly by a self-consistent field. For physics, I can only point to the Bethe-Ansatz solutions in 1d, where you can do the nonlinear Schrodinger equation, and the Heisenberg antiferromagnet, and related systems. The sector of fixed particle number in the NLSE is nontrivial SE, and it is exactly solvable. Outside of 1d, I don't know anything. –  Ron Maimon May 8 '12 at 3:29
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Right, similarly, there is e.g. Laughlin wavefunction which is very crisp and accurate except that once again, we don't exactly know the equation it solves, see en.wikipedia.org/wiki/Laughlin_wavefunction . Also solids, with Bloch waves for particles etc. make things "somewhat more solvable" but once again, it's a limit where the interactions effectively become zero in some alternative description. The very fact that the system may be solved means that we may interpret it as a system of non-interacting particles in a proper basis... –  Luboš Motl May 8 '12 at 3:48

4 Answers 4

One of my favorite non-trivial, exact many-body ground states is the solution of a very specific spin-1 magnetic insulator in 1D, with a hamiltonian

$$H_{AKLT}=\sum_{\langle ij\rangle}\vec{S}_i \cdot \vec{S}_j + \frac{1}{3}(\vec{S}_i \cdot \vec{S}_j)^2$$

It turns out that you can construct the ground state by looking at the spin-1 operators as a projection onto the triplet subspace of two spin-1/2 operators, where the spin-1/2 objects form nearest-neighbor singlet bonds in a very special way. (More details can be found http://en.wikipedia.org/wiki/AKLT_Model)

This exact ground state informs our understanding of the spin-1 Heisenberg model (i.e., without the biquadratic interaction), and the "fractionalized" spin-1/2 "edge states" that this state predicts for a magnet with open boundary conditions have been observed in experiments (see again the wiki article and its references)

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For a Hamiltonian, for which Laughlins wavefunction is the exact ground state, see F. D. M. Haldane , Fractional Quantization of the Hall Effect: A Hierarchy of Incompressible Quantum Fluid States, Phys. Rev. Lett. 51, 605–608 (1983), http://prl.aps.org/abstract/PRL/v51/i7/p605_1 .

Like the Hamiltonian in the answer of wsc, this Hamiltonian is a sum of projections, which represent the interactions. And in both cases, the ground state is the state, which is annihilated by all these projections.

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The exact ground state for N structureless bosons interacting with contact interactions $V(x_1-x_2) = g \delta(x_1-x_2)$ is known. In free space (also with infinite width periodic boundary conditions) for $g<0$ this is

$$ \psi_{\rm ground} \propto \exp\left(\frac{m g}{2 \hbar^2} \sum_{1 \le j < k \le N} |x_j-x_k| \right)$$

Which is a state that is localised with pair correlations but has a center-of-mass which is free (described by a plane wave).

See Bethe Ansatz for more detail.

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up vote 0 down vote accepted

The most elegant example I have found is Hooke's atom, also called harmonium. It consists of two electrons that are trapped in a harmonic well:

$$ H=-\nabla^2_1-\nabla^2_2+\frac{1}{\left|\mathbf{r}_1-\mathbf{r}_2\right|}+\frac{1}{2}k\left(\mathbf{r}_1^2+\mathbf{r}_2^2\right) $$

For certain values of the spring constant k, this Hamiltonian can be solved exactly. For example, when k = ¼, the ground state is: $$ \Psi\left(\mathbf{r}_1,\mathbf{r}_2\right)=\frac{1}{2\sqrt{8\pi^{5/2}+5\pi^3}}\left(1+\frac{1}{2}\left|\mathbf{r}_1-\mathbf{r}_2\right|\right)\textrm{exp}\left(-\frac{1}{4}\left(r_1^2+r_2^2\right)\right) $$

Source: Wikipedia

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