# Introduction ^{▲}

The Bose-Hubbard model describes bosonic particles moving on a lattice structure.
In this system, the Bosons can interact with each other, or tunnel to adjecent lattice sites.
Due to the structure of the Hamiltonian there are two competing groundstates for the system:
**Mott insulator**- particles are bound to one lattice site**Bose-Einstein condensate (BEC)**- particles delocalised over the entire lattice

# Bose-Hubbard Hamiltonian ^{▲}

The Bose-Hubbard Hamiltonian describes a system of $N$ bosnic particles on $M$ lattice sites. There are three contributions to this Hamiltonian. Firstly, the kinetic energy which is determined by the tunnelling element $J$. It describes the hopping of a particle from lattice site $j$ to an adjacent lattice site $i$. If two atoms occupy the same lattice site, these particles interact. This leads to the potential energy term proportional to the on-site interaction energy $U$. Furthermore, it is experimentally necessary to trap the atoms. This external trapping potential gives the last contribution to the Hamiltonian. $$ H=\underbrace{-J\sum_{\langle i,j\rangle}b_i^{\dagger} b_j}_{\text{kin. energy}}+\underbrace{\frac{U}{2}\sum_ib_i^{\dagger 2}b_i^2}_{\text{pot. energy}}+\underbrace{\sum_i\epsilon_i b_i^{\dagger}b_i}_{\text{trap. potential}} $$

## A simple Model ^{▲}

A good way of getting familiar with the dynamics of this model is to look at a system with two atoms on two lattice sites. We can choose the following states as a basis to represent the Hamiltonian: $$ b_1^{\dagger}b_2^{\dagger}|vac\rangle=|1,1\rangle \quad b_1^{\dagger 2}|vac\rangle=\sqrt{2}|2,0\rangle \quad b_2^{\dagger 2} |vac\rangle=\sqrt{2}|0,2\rangle $$ The Hamiltonian can thus be written as: $$ H = -J(b_1^{\dagger} b_2 + b_2^{\dagger} b_1) + \frac{1}{2}U(b_1^{\dagger 2}b_1^2 + b_2^{\dagger 2}b_2^2)$$ This eigenvalue problem can be solved analytically and yields the following results: $$ E_1=U \qquad \text{with} \qquad |\phi_1\rangle=\frac{1}{\sqrt{2}}(|0,2\rangle-|2,0\rangle)$$ $$ E_2= \frac{1}{2}(U - \sqrt{16 J^2 + U^2})\qquad \text{with} \qquad |\phi_2\rangle=\sqrt{\frac{2J^2}{4J^2+E_3^2}}(|2,0\rangle+|0,2\rangle+\frac{E_3}{\sqrt{2}J}|1,1\rangle)$$ $$ E_3= \frac{1}{2}(U + \sqrt{16 J^2 + U^2})\qquad \text{with} \qquad |\phi_3\rangle=\sqrt{\frac{2J^2}{4J^2+E_2^2}}(|2,0\rangle+|0,2\rangle+\frac{E_2}{\sqrt{2}J}|1,1\rangle) $$ In Figure 1 the eigenenergies are plotted as a function of $U/J$.

If $U$ is set to zero the wavefunction of the groundstate has the form: $$|\phi_2\rangle \propto |2,0\rangle+|0,2\rangle+\sqrt{2}|1,1\rangle $$ In this case, both atoms delocalise over both lattice sites. This is called a superfluid phase. In the limit $U \gg J$, the groundstate takes on a form, where both atoms are localised at a lattice site: $$|\phi_2\rangle \propto |1,1\rangle$$ If the system is in this state it is called Mott insulator. In the general case of $N$ particles on $M$ lattice sites, both of these limiting cases are equally present. One difference to this simple model is that the grounstate energy has a non-analytic dependence on the ratio $U/J$. This non-analyticity can be associated with an abrupt change from the superfluid to the Mott phase.

## Limiting Cases ^{▲}

In this section we look at lattice with $M$ lattice sites occupied by $N$ particles. As mentioned before the kinetic and potential energy contributions of the Hamiltonian lead to very different behaviour of the groundstate wavefunction. If we look at the limit with a dominant kinetic energy the atoms can easily hopp to the next lattice site. Thus, the system energy is minimised when the atoms delocalise over the entire lattice.

- groundstate is a superposition:$$|BEC\rangle\propto(\sum_{i=1}^M b_i^{\dagger})^N|vac\rangle$$
- not localised
- long-range order: $\psi=const.$
- no gap in energy spectrum
- interference pattern

On the other hand, if the tunnelling is highly suppressed, the atoms prefer to occupy individual lattice sites, because moving to an already occupied site requires an amount of energy equal to $U$. This very stable phase is called Mott insulator.

- groundstate is a fockstate: $$|MOTT\rangle \propto \prod_i^M (b_i^{\dagger})^n|vac\rangle$$
- localised
- no long-range order: $\psi=0.$
- no gap in energy spectrum
- no interference pattern

This implies that the wavefunction changes its behaviour at a certain value of $U/J$. This reordering of the groundstate is called a quantum phase transition.

## Quantum Phase Transition ^{▲}

A quantum phase transition describes a sudden change in the ground state of the system which is caused by quantum fluctuations. Unlike a classical phase transition, this phenomenon occurs at zero temperature. As shown above, the quantum phase transition can be induced by varying the parameter $U/J$. The exact solution to the Schrödinger equation can only be found in the two limiting cases. Thus, the transition point from superfluid to Mott insulator can only be approximated. We will be looking at the Gutzwiller ansatz for the wavefunction to solve this problem: $$ |\Psi\rangle=\prod_i|\phi_i\rangle=\prod_i \sum_{n=0}^{\infty} f_n^{(i)}|n\rangle_i$$ Minimising the energy with respect to the coefficients $f_n^{(i)}$ yields the ground state. $$\langle\Psi| H |\Psi\rangle-\mu \langle\Psi| \sum_i b_i^{\dagger}b_i |\Psi\rangle \rightarrow \text{min}$$ The part containing the chemical potential $\mu$, shows the grandcanonical nature of the problem. The particle number is only conserved in the mean. Additionally, we assume a homogeneous case, where $\phi_i=\phi$ and $f_n^{(i)}=f_n$. Applying the variation in terms of the coefficients $f_n^{(i)}$, leads to a mean field Hamiltonian. This effectively reduces the many-body problem to a one particle problem described by self-consistency equations: $$ (h-\mu n)|\phi\rangle=\epsilon|\phi\rangle$$ $$\psi=\langle\phi|b|\phi\rangle$$ Where $h$ is the mean field Hamiltonian: $$h=-Jz(b^{\dagger}\psi+\psi^{*} b)+\frac{1}{2}Ub^{\dagger 2}b^2$$ In this equation $z$ denotes the number of neighbouring lattice sites. Using the self-consistency equations, the ordering parameter $\psi$ can be evaluated for every value of $U/J$ and $\mu/J$. The ordering parameter $\psi$ determines the phase of the system, because it takes on different values for a Mott insulator and a superfluid. In the Mott phase the parameter is always zero, whereas in the superfluid phase it is constant. Thus, these calculations lead to the phase diagram in Figure 2.

**Figure 2:**Phase-diagram calculated with the Gutzwiller ansatz for a system of $N=15$ atoms in a 3-dimensional lattice.

# Cold Atoms in an optical Lattice ^{▲}

The physics of cold atoms in an optical lattice can be described by a Bose-Hubbard model. A bosonic gas can be cooled down to form a BEC. Experimentally, this can be achieved by evaporation cooling in a magnetic trap. A lattice potential is induced by counter-propagating laser beams. To show that the Bose-Hubbard Hamiltonian describes this system, we start with the many-body Hamiltonian of an interacting bosonic gas: $$ H = \int\mathrm{dx}^3\, \hat{\psi}^{\dagger}(\vec{x})\left(-\frac{\hbar^2}{2m}\nabla^2+V(\vec{x})\right)\hat{\psi}(\vec{x}) +\frac{g}{2}\int\mathrm{dx}^3\, \hat{\psi}^{\dagger}(\vec{x})\hat{\psi}^{\dagger}(\vec{x})\hat{\psi}(\vec{x})\hat{\psi}(\vec{x}) $$ The field operators can then be expanded in terms of the Wannier basis. Using this expansion, we can derive the Bose-Hubbard Hamiltonian with the tunnelling element: $$ J=\int\mathrm{dx}^3 \, w^*(\vec{x}-\vec{x}_i)\left(-\frac{\hbar^2}{2m}\nabla^2+V(\vec{x})\right)w(\vec{x}-\vec{x}_j)\, $$ and the interaction energy: $$ U=g\int \mathrm{dx}^3\,|{w(\vec{x})}|^4 $$ With increasing laser intensity, the contribution of $J$ reduces and the on-site interaction $U$ increases. The quantum phase transition can thus be induced by varying the laser intensity, which is fortunate for an experimental implementation. For this theory to apply certain conditions have to be met:

**Lamb-Dicke**regime: the extension of the groundstate wavefunction $a_0\ll \lambda/2$ (laser wavelength)- On-site
**interaction has to be smaller than excitation energy**needed to excite an atom to the next band - Gas needs to be dilute enough so that
**only two body interactions**are relevant - The temperature has to be low enough so that
**only s-wave scattering**takes place

These requirements have been fulfilled by current experiments with cold atoms in an optical lattice (see Figure 3).

**Figure 3:**Absorption images of the atomic gas interference pattern for different values of $V_0$ in multiples of the recoil energy $E_\text{R}$.

^{1}a) $0\,E_\text{R}$ b) $3\,E_\text{R}$ c) $7\,E_\text{R}$ d) $10\,E_\text{R}$ e) $13\,E_\text{R}$ f) $14\,E_\text{R}$ g) $16\,E_\text{R}$ h) $20\,E_\text{R}$. The images show the transition from Mott insulator to superfluid phase. Diagram taken from [2].

# Summary ^{▲}

The Bose-Hubbard model is based on a simply structured Hamiltonian that enables a thorough investigation of cold bosonic gases. This problem is not yet fully solved, but the Gutzwiller approximation made it possible to predict the behaviour of these bosnic systems. It can be found that a dominance in kinetic energy leads to a superfluid groundstate. On the other hand, stronger interaction energy produces a Mott insulator state. This phase transition can be experimentally observed in a gas of cold atoms in an optical lattice. The advantage of this setup is that the experimentally valid parameters can be easily controlled via the laser intensity. This makes it very attractive for the implementation of quantum algorithms.

### References ^{▲}

- [1] D. Jaksch. C. Bruder, J. I. Cirac, C. W. Gardiner, P. Zoller; Cold Bosonic Atoms in Optical Lattices; Phys. Rev. Lett. 81, p. 3108, 1998
- [2] M. Greiner, O. Mandel, T. Esslinger, T. Hänsch, I. Bloch; Quantum phase transition from a superfluid to a Mott insulator in a gas of ultracold atoms; Nature 415, p. 39-44, 2001
- [3] H. C. Nägerl;
*Vorlesung Grundkonzepte der Quantenmechanik*; 2014 - [4] P. Zoller;
*Vorlesung Quantenmechanik 2*; 2015 - [5] S. Diehl;
*Lattice systems: Physics of the Bose-Hubbard Model*; 2010