Lieb–Liniger model

In physics, the Lieb–Liniger model describes a gas of particles moving in one dimension and satisfying Bose–Einstein statistics. More specifically, it describes a one dimensional Bose gas with Dirac delta interactions. It is named after Elliott H. Lieb and Werner Liniger [de] who introduced the model in 1963.^[1] The model was developed to compare and test Nikolay Bogolyubov's theory of a weakly interaction Bose gas.^[2]

Definition

Given ${\displaystyle N}$ bosons moving in one-dimension on the ${\displaystyle x}$-axis defined from ${\displaystyle [0,L]}$ with periodic boundary conditions, a state of the N-body system must be described by a many-body wave function ${\displaystyle \psi (x_{1},x_{2},\dots ,x_{j},\dots ,x_{N})}$. The Hamiltonian, of this model is introduced as

${\displaystyle H=-\sum _{i=1}^{N}{\frac {\partial ^{2}}{\partial x_{i}^{2}}}+2c\sum _{i=1}^{N}\sum _{j>i}^{N}\delta (x_{i}-x_{j})\ ,}$

where ${\displaystyle \delta }$ is the Dirac delta function. The constant ${\displaystyle c}$ denotes the strength of the interaction, ${\displaystyle c>0}$ represents a repulsive interaction and ${\displaystyle c<0}$ an attractive interaction.^[3] The hard core limit ${\displaystyle c\to \infty }$ is known as the Tonks–Girardeau gas.^[3]

For a collection of bosons, the wave function is unchanged under permutation of any two particles (permutation symmetry), i.e., ${\displaystyle \psi (\dots ,x_{i},\dots ,x_{j},\dots )=\psi (\dots ,x_{j},\dots ,x_{i},\dots )}$ for all ${\displaystyle i\neq j}$ and ${\displaystyle \psi }$ satisfies ${\displaystyle \psi (\dots ,x_{j}=0,\dots )=\psi (\dots ,x_{j}=L,\dots )}$ for all ${\displaystyle j}$.

The delta function in the Hamiltonian gives rise to a boundary condition when two coordinates, say ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ are equal. The condition is that as ${\displaystyle x_{2}}$ approaches ${\displaystyle x_{1}}$ from above (${\displaystyle x_{2}\searrow x_{1}}$), the derivative satisfies

${\displaystyle \left.\left({\frac {\partial }{\partial x_{2}}}-{\frac {\partial }{\partial x_{1}}}\right)\psi (x_{1},x_{2})\right|_{x_{2}=x_{1}+}=c\psi (x_{1}=x_{2})}$.

Solution

Fig. 1: The ground state energy (per particle) ${\displaystyle e}$ as a function of the interaction strength per density ${\displaystyle \gamma =Lc/N}$, from.^[1]

The time-independent Schrödinger equation ${\displaystyle H\psi =E\psi }$, is solved by explicit construction of ${\displaystyle \psi }$. Since ${\displaystyle \psi }$ is symmetric it is completely determined by its values in the simplex ${\displaystyle {\mathcal {R}}}$, defined by the condition that ${\displaystyle 0\leq x_{1}\leq x_{2}\leq \dots ,\leq x_{N}\leq L}$.

The solution can be written in the form of a Bethe ansatz as^[2]

${\displaystyle \psi (x_{1},\dots ,x_{N})=\sum _{P}a(P)\exp \left(i\sum _{j=1}^{N}k_{Pj}x_{j}\right)}$,

with wave vectors ${\displaystyle 0\leq k_{1}\leq k_{2}\leq \dots ,\leq k_{N}}$, where the sum is over all ${\displaystyle N!}$ permutations, ${\displaystyle P}$, of the integers ${\displaystyle 1,2,\dots ,N}$, and ${\displaystyle P}$ maps ${\displaystyle 1,2,\dots ,N}$ to ${\displaystyle P_{1},P_{2},\dots ,P_{N}}$. The coefficients ${\displaystyle a(P)}$, as well as the ${\displaystyle k}$'s are determined by the condition ${\displaystyle H\psi =E\psi }$, and this leads to a total energy

${\displaystyle E=\sum _{j=1}^{N}\,k_{j}^{2}}$,

with the amplitudes given by

${\displaystyle a(P)=\prod _{1\leq i<j\leq N}\left(1+{\frac {ic}{k_{Pi}-k_{Pj}}}\right)\,.}$^[4]

These equations determine ${\displaystyle \psi }$ in terms of the ${\displaystyle k}$'s. These lead to ${\displaystyle N}$ equations:^[2]

${\displaystyle L\,k_{j}=2\pi I_{j}\ -2\sum _{i=1}^{N}\arctan \left({\frac {k_{j}-k_{i}}{c}}\right)\qquad \qquad {\text{for }}j=1,\,\dots ,\,N\ ,}$

where ${\displaystyle I_{1}<I_{2}<\cdots <I_{N}}$ are integers when ${\displaystyle N}$ is odd and, when ${\displaystyle N}$ is even, they take values ${\displaystyle \pm {\frac {1}{2}},\pm {\frac {3}{2}},\dots }$ . For the ground state the ${\displaystyle I}$'s satisfy

${\displaystyle I_{j+1}-I_{j}=1,\quad {\rm {for}}\ 1\leq j<N\qquad {\text{and }}I_{1}=-I_{N}.}$

Thermodynamic limit