Math 2.0

Statistical mechanics, structure of matter Quantum statistics and ideal gases Example

Lieb–Liniger model

Builds on Bethe ansatz

Example

The Lieb–Liniger model describes NN identical bosons on a circle of length LL with a repulsive contact interaction of strength c>0c > 0. In units with ℏ=2m=1\hbar = 2m = 1, its Hamiltonian acts on symmetric wave functions ψ(x1,…,xN)\psi(x_1, \ldots, x_N) with xj∈[0,L)x_j \in [0, L) periodic: H=−∑j=1N∂2∂xj2+2c∑1≤i<j≤Nδ(xi−xj).H = -\sum_{j=1}^N \frac{\partial^2}{\partial x_j^2} + 2c \sum_{1 \le i < j \le N} \delta(x_i - x_j). Elliott Lieb and Werner Liniger solved it in 1963 with the Bethe ansatz. In the region x1<x2<⋯<xNx_1 < x_2 < \cdots < x_N the eigenfunctions are finite sums of plane waves ei(kP1x1+⋯+kPNxN)e^{i(k_{P1} x_1 + \cdots + k_{PN} x_N)} over permutations PP, with real quasi-momenta k1,…,kNk_1, \ldots, k_N. Periodicity forces the Bethe equations eikjL=∏l≠jkj−kl+ickj−kl−ic,j=1,…,N,e^{i k_j L} = \prod_{l \ne j} \frac{k_j - k_l + i c}{k_j - k_l - i c}, \qquad j = 1, \ldots, N, and the energy is E=k12+⋯+kN2E = k_1^2 + \cdots + k_N^2.

Check in two limits. For c=0c = 0 every factor is 11, so eikjL=1e^{i k_j L} = 1 and kj∈(2π/L)ℤk_j \in (2\pi/L)\Z: these are free bosons on the circle. As c→∞c \to \infty every factor tends to −1-1, so eikjL=(−1)N−1e^{i k_j L} = (-1)^{N-1}: the spectrum is that of free fermions, the Tonks–Girardeau gas.

Remarks

In the thermodynamic limit at fixed density, the ground state is described by a density of quasi-momenta that solves the Lieb–Liniger integral equation. The model is one of the basic exactly solvable models of one-dimensional quantum gases, and it has been realized with cold atoms in narrow traps.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Lieb%E2%80%93Liniger_model
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Kindexample
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MSC82B23, 81Q80, 81V73
Instance ofExactly solvable model
UsesBethe ansatz
Same as
Changed3 November 2026

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