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  Quantum gases

Ideal Bose Gas

A Bose–Einstein condensate (BEC) is a defined as the occurrence of the macroscopic occupation of one-particle states. When a dilute gas of bosons cooled to temperatures very close to 0K (usually ~100nK in experiments), a large fraction of bosons occupies the lowest quantum state, at which point macroscopic quantum phenomena become apparent.

In this section, we summarize some basic and useful thermodynamic results and properties for Bose-Einstein condensations. F

Phase Space Density

The fundamental difference between a BEC and a classical gas is the occupancy of a single-particle state. In a classical gas, the mean occupation number for a single quantum state satisfies the Boltzmann distribution $\langle n_{\nu }\rangle =e^{-(\epsilon _{\nu }-\mu )/kT}$ which is much less than unity. This feature is qualitatively captured by the ${\textit {Phase-spaceDensity}}$ defined as (3D, homogeneous gas)

$\rho _{D}=n\lambda _{T}^{3}\,,$

where $\lambda _{T}=(2\pi \hbar ^{2}/mkT)^{1/2}$ is the thermal de Broglie wavelength.

Some typical parameters for

• Classical thermal gas
  • Atom density $n\sim 10^{25}{\text{m}}^{-3}$
  • Interatomic distance $(1/n)^{1/3}\sim 3{\text{nm}}$
  • Thermal de Broglie wavelength Failed to parse (syntax error): {\displaystyle \lambda_T \sim 10^{-2} \text{nm}（T = 300K) }
  • $\rho _{D}\sim 10^{-8}$
• BEC in dilute gas
  • Atom density $n\sim 10^{20}{\text{m}}^{-3}$
  • Interatomic distance $(1/n)^{1/3}\sim 10^{2}{\text{nm}}$
  • Thermal de Broglie wavelength Failed to parse (syntax error): {\displaystyle \lambda_T \sim 10^{3} \text{nm}（T = 100nK) }
  • $\rho _{D}\sim 10^{2}$

The Bose-Einstein Distribution

For non-interacting bosons in thermodynamic equilibrium, the mean occupation number of the single-particle state $\nu$ is

$f(\epsilon ,\mu ,T)={\frac {1}{e^{(\epsilon _{\nu }-\mu )/kT}-1}}={\frac {1}{z^{-1}e^{\epsilon _{\nu }/kT}-1}}\,.$

$z=exp(\beta \mu )$ is defined as fugacity. At high temperature, the chemical potential lies below $\epsilon _{min}$. As temperature is lower, the chemical potential rises until it reaches $\epsilon _{min}$ and the mean occupation numbers increase.

Thermodynamics in Semi-classical Limits

We focused on the semi-classical case where $kT\gg \delta E$. Here $\delta E$ is the scale for enery level spacing in the trapping potential (for example, $\delta E=\hbar \omega$ for 3D harmonic trapping. This is usually valid in a real experiment where $kT\approx 2000Hz$ and $\delta E\approx 100Hz$. In this case, the system can be treated as a continuous excited energy spectrum plus a separated ground state. It seems to be contradictory to the nature of BEC when most of the population is found in the single ground state, but this description is a good enough approximation in many situations. The fully quantum description is necessary for some cases as we will see in the Pedagogical Example in end.

Transition Temperature

When $\nu =\epsilon _{min}$ , the occupation number on the ground state can be arbitrarily large, indicating the emergence of a condensate. The corresponding temperature is the transition temperature $T_{c}$. $T_{c}$ can be calculated with the criteria that the maximum number of particles can be held in the excited states is equal to the total particle number $N$. In the semi-classical limit where the sum over all states is replaced by an integral and simple assumption that $\epsilon _{min}=0$ we have

$N=N_{0}+N_{ex}\,.$

where we define

$N_{ex}:=\int \limits _{0}^{\infty }f(\epsilon ,\mu =0,T_{c})g(\epsilon )\mathrm {d} \epsilon \,.$

Here $g(\epsilon )$ is the density of states. $N_{0}$ is the number of atoms in the ground state. Notice that the chemical potential $\mu =0$ is set to 0 (or in fact $\epsilon _{min}$ without any justifications. In fact, by setting $\mu =0$ what we are calculating here is the maximum possible number of atoms that can be accommodated by the "excited" states. If the total number of atoms is larger tan that, the rest must go to the ground state. A more rigorous calculation without this assumption can be found in the section Finite number effects.

The form of the transition temperature $T_{c}$ and therefore the condensate atom number depends strongly on the form of $g(\epsilon )$ which is affected by the dimension, trapping potential and the dispersion of the system. Under the most general assumption that $g(\epsilon )\propto \epsilon ^{\alpha -1}$, we reach

$N\propto C_{\alpha }(kT_{c})^{\alpha }\int \limits _{0}^{\infty }\mathrm {d} x{\frac {x^{\alpha -1}}{e^{x}-1}}\,.$

where $x=\epsilon /(kT_{c})$. Straightforwardly we have a simple scaling function

$kT_{c}\propto N^{\frac {1}{\alpha }}\,.$

Some common cases are summarized below

cases:	3D box	2D box	3D Harmonic	2D Harmonic
$\alpha$	$3/2$	$1,diverge$	$3$	$2$

We see that the semi-classical picture is already good enough to capture some basic condensate physics. As a quick exaple, for the parameters in a typical AMO experiment (3D, harmonic trapping)

• $N\sim 10^{6}$
• harmonic trapping frequencies $\omega _{i}\sim 2\pi \times 100Hz$

We have $T_{c}\approx 450nK$

Thermodynamic Properties

The thermodynamic properties $H$can be readily calculated from the Bose distributions and sum over all the states.

$H=\int \limits _{0}^{\infty }H(\epsilon )f(\epsilon ,\mu ,T)g(\epsilon )\mathrm {d} \epsilon \,.$

For example, the total energy

$E=\int \limits _{0}^{\infty }\epsilon f(\epsilon ,\mu ,T)g(\epsilon )\mathrm {d} \epsilon \propto (kT)^{\alpha }\int \limits _{0}^{\infty }{\frac {x^{\alpha }}{e^{x}-1}}\mathrm {d} x\,.$

We therefore can obtain a scaling law for all the importnat thermodynamic quantities as listed below assuming that $g(\epsilon )\propto \epsilon ^{\alpha -1}$ and $H\propto T^{\delta }$

Thermodynamic Property:	E	$C_{V}=\partial E/\partial T$	Entropy S
$\delta$	$\alpha +1$	$\alpha$	$\alpha$

It is also useful to express the relationship with dimensionless parameter $\gamma =T/T_{c}$ considering $N\propto T_{c}^{\alpha }$ we therefore obtain

$E\propto NkT\gamma ^{\alpha },C_{V}\propto Nk\gamma ^{\alpha },S\propto Nk\gamma ^{\alpha }\,.$

Beyond Semi-classical Limits

A more quantum way to deal with the system is by treating the energy levels as discrete and replace the integral with summation and also consider the constraint

$\sum _{i=0}^{\infty }f(\epsilon _{i},z,T)=N$

Notice that this time the ground state is not separated from the summation. Given the energy spectrum $\epsilon _{i}$ which is often determined by the trapping potential, and temperature $T$, the constraint allows the calculation of $z(N,T)$ at least numerically. We then immediately get the ground state occupation $N_{0}={\frac {z}{1-z}}$.

Superfluidity and Coherence

Superfluidity

Strictly speaking, Bose-Einstein condensation is not necessary for superfluidity, but also not sufficient. For example, The ideal Bose gas can undergo Bose-Einstein condensation, but it does not show superfluid behavior since its critical velocity is zero. Superfluidity requires interactions. The opposite case (superfluidity without BEC) occurs in lower dimensions. In 1D at T = 0 and in 2D at finite temperature, superfluidity occurs [277], but the condensate is destroyed by phase fluctuations. In 3D, condensation and superfluidity occur together. In a macroscopic bulk system, superfluidity shows itself with several features: absence of viscosity, reduced the momentum of inertia, collective excitations(second sound for example), quantized vortices etc. These effects will be discussed in section "vortices in BEC".

Off-Diagonal Long-Range Order

Another feature of a BEC is the global phase coherence. It is related to the occurrence of the off-diagonal long-range order. We shall see that macroscopic occupation of a single particle state produces such order.

Mathematically, we characterise th between atoms at position ${\vec {r}}$ and ${\vec {r}}'$ with

$\rho ^{(1)}({\vec {r}},{\vec {r}}')=\langle {\hat {\psi }}^{\dagger }({\vec {r}}){\hat {\psi }}({\vec {r}}')\rangle \,.$

The term $\rho ^{(1)}({\vec {r}},{\vec {r}}')$ is sometimes also called real space first-order correlation function. We notice that $\rho ^{(1)}({\vec {r}},{\vec {r}})$ is nothing but the atomic density at potision ${\vec {r}}$.

In a homogeneous system, $\rho ^{(1)}({\vec {r}},{\vec {r}}')$ depends only on the relative coordinate ${\vec {s}}={\vec {r}}-{\vec {r}}'$. The correlation function then reveals the distribution in the momenutm space.

$\rho ^{(1)}({\vec {s}})={\frac {1}{V}}\int \mathrm {d} {\vec {p}}\ \rho ({\vec {p}})e^{i{\vec {p}}\cdot {\vec {s}}}\,.$

• In a thermal gas, we have $\rho ^{(1)}({\vec {s}})\rightarrow 0$ as $|{\vec {s}}|\rightarrow \infty$.
• In a BEC, we have macrosopically populated state ( for example in a box we populate $|{\vec {p}}=0\rangle$), we then have $\rho ^{(1)}({\vec {s}})\rightarrow \sim {\frac {N_{0}}{V}}$ as $|{\vec {s}}|\rightarrow \infty$.

File:Interference bec.jpg

Interference of two BECs. Andrew, Townsend "et. al." Science 275, 637

Direct experimental evidence of the phase coherence was found by observing the interference pattern between two BECs created by splitting a single piece (see Andrew, Townsend "et. al." Science 275, 637).

The periodicity of the fringes is related to the initial splitting and the flight time and is on the order of $\sim 10um$.

Some Remarks

Back to: Quantum gases

Weakly Interacting Homogeneous Bose Gas

Typical introductory statistical mechanics courses examine BECs by assuming that they consist of many non-interacting atoms. That simple model does an excellent job of predicting the condensation temperature and fraction of atoms in the condensate, however it quantitatively and even qualitatively misses some of the properties of real BECs. The issue is that real atoms interact with each other and these interactions can alter many properties of a BEC. Fortunately, a simple mean-field treatment of the interactions can create an excellent model that captures much of the behavior seen in real BECs as will be shown in the following sections.

Weakly Interacting Bose Gas near $T=0$

We can start to account for atom-atom interactions by adding a collisional term to the hamiltonian. We can consider a collision as a process that annihilates a particle with momentum $p$ and a particle with momentum $k$, then creates two particles with momenta $p^{\prime }$ and $k^{\prime }$. By momentum conservation we may write $p^{\prime }=p+q$ and $k^{\prime }=k-q$. We let $U_{q}$ be the matrix element for this process, and so we can write collisional hamiltonian as the sum of all possible collisions (with a factor of 2 to avoid double-counting input states)

$H_{I}={\frac {1}{2V}}\sum _{k,p,q}U_{q}a_{p+q}^{\dagger }a_{k-q}^{\dagger }a_{k}a_{p}\,.$

This hamiltonian is far too complicated to solve in the general case, so we must make some approximations. First, for typical BEC parameters, the spacing between atoms is much larger than the collisional scattering length of the atoms. Therefore the complicated atomic interaction potential can be well approximated by replacing it with a delta function potential. In particular, if the atomic separation is $r$, then we may write the potential as $U(r)=U_{0}\delta (r)$ where $U_{0}=4\pi \hbar ^{2}a/m$ and $a$ is the $s$-wave scattering length. Now $U_{q}$ is the Fourier transform of $U(r)$, and since the Fourier transform of a delta function is a constant function, we have that $U_{q}=U_{0}$. So we may write

$H_{I}\approx {\frac {U_{0}}{2V}}\sum _{k,p,q}a_{p+q}^{\dagger }a_{k-q}^{\dagger }a_{k}a_{p}\,.$

Unfortunately this Hamiltonian is still too complicated to solve. The reason is that it is extremely difficult to diagonalize hamiltonians that are a product of four operators. Therefore we need some way to simplify things down to two operators. We do this with the Bogoliubov approximation, which says that when there are a large number $N_{0}$ of atoms in the condenstate, we may approximate $N_{0}+1\approx N_{0}$. And since $a_{0}^{\dagger }|N_{0}\rangle ={\sqrt {N_{0}+1}}|N_{0}+1\rangle$ and $a_{0}|N_{0}\rangle ={\sqrt {N_{0}}}|N_{0}-1\rangle$, we may then approximate those operators with c-numbers $a_{0}\approx a_{0}^{\dagger }\approx {\sqrt {N_{0}}}$. Furthermore, since $N_{0}$ is large, we see that the terms in the hamiltonian that will dominate are the ones in which there are two or more occurrences of $a_{0}$ and/or $a_{0}^{\dagger }$. We may therefore approximate the hamiltonian as

$H_{I}\approx {\frac {U_{0}N_{0}^{2}}{2V}}+{\frac {U_{0}N_{0}}{2V}}\sum _{k\neq 0}a_{k}^{\dagger }a_{-k}^{\dagger }+a_{k}a_{-k}+2a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\,.$

The full hamiltonian is simply this plus the kinetic energy term

$H\approx {\frac {U_{0}N_{0}^{2}}{2V}}+\sum _{k}\epsilon _{k}a_{k}^{\dagger }a_{k}+{\frac {U_{0}N_{0}}{2V}}\sum _{k\neq 0}a_{k}^{\dagger }a_{-k}^{\dagger }+a_{k}a_{-k}+2a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\,.$

At this point we have made enough approximations to arrive at a solvable hamiltonian, since it only involves quadratic products of operators.

To show that it is possible to diagonalize this hamiltonian, we will outline the method. We start by replacing $N_{0}$ with $N$ using the relation

$N=N_{0}+{\frac {1}{2}}\sum _{k\neq 0}a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\,.$

to give

$H\approx {\frac {U_{0}N^{2}}{2V}}+{\frac {1}{2}}\sum _{k\neq 0}\left(\epsilon _{k}+{\frac {NU_{0}}{V}}\right)\left(a_{k}^{\dagger }a_{k}+a_{-k}^{\dagger }a_{-k}\right)+{\frac {NU_{0}}{V}}\left(a_{k}^{\dagger }a_{-k}^{\dagger }+a_{k}a_{-k}\right)\,.$

Each term in the sum has the form

$H_{k}=E_{0,k}\left(a^{\dagger }a+b^{\dagger }b\right)+E_{1}\left(a^{\dagger }b^{\dagger }+ab\right)\,.$

where we've define $E_{0,k}=\left(\epsilon _{k}+{\frac {NU_{0}}{V}}\right)$, $E_{1}={\frac {NU_{0}}{V}}$, $a=a_{k}$ and $b=a_{-k}$. Now the first term in $H_{k}$ looks nice, it's just the hamiltonian for harmonic oscillators. The second term however is inconvenient, so we'd like to get rid of it. This can be done by performing the Bogoliubov transformations, which rewrites the hamiltonian in terms of different operators $\alpha$ and $\beta$, which are superpositions of $a$ and $b$. In particular we write

Failed to parse (syntax error): {\displaystyle a=u\alpha -v\beta^\dagger \\ b=u\beta -v\alpha^\dagger }

where we have the freedom to choose $u$ and $v$. The first requirement we impose is to make sure $[\alpha ,\alpha ^{\dagger }]=[\beta ,\beta ^{\dagger }]=1$ so that $\alpha$ and $\beta$ obey the bosonic commutation relations. This forces $u^{2}-v^{2}=1$, but still leaves one degree of freedom. When plugging $\alpha$ and $\beta$ into $H_{k}$, we see we'll get something of the form

$H_{k}=(\ldots )\left(\alpha ^{\dagger }\alpha +\beta ^{\dagger }\beta \right)+(\ldots )\left(\alpha ^{\dagger }\beta ^{\dagger }+\alpha \beta \right)$

Therefore, we use our other degree of freedom to choose $u$ and $v$ such that the second prefactor is zero, thereby getting rid of the troublesome terms. We are just left with

$H_{k}=(\ldots )\left(\alpha ^{\dagger }\alpha +\beta ^{\dagger }\beta \right)$

for which we know the solutions are just the harmonic oscillator eigenstates.

Sound propagation in Bose-Einstein condensates

From the previous section, we see that we may write the hamiltonian under the Bogoliubov approximation as

$H=\sum _{k}E_{k}a_{k}^{\dagger }a_{k}+{\text{const.}}$

where we've changed notation a bit so that $a$ now represents what we had called $\alpha$, the annihilation operator for a quasiparticle (not an atom). We've also used

$E_{k}={\sqrt {\left({\frac {\hbar ^{2}k^{2}}{2m}}\right)^{2}+(\hbar ck)^{2}}}\,.$

and $c={\sqrt {NU_{0}/Vm}}$ represents the speed of sound in the BEC. This dispersion relation shows us that the low lying excitaitons are phonons. At $k\rightarrow 0$, $E_{k}=\hbar ck$ that of sound, while at $k\rightarrow \infty$, $E_{k}=\hbar ^{2}k^{2}/2m$, a free particle. Free particles start with a quadratic dispersion relation, while phonons and other Bose systems start with a linear dispersion relation.

The dispersion relation can be experimentally measured using Bragg spectroscopy. In this procedure, two lasers are shined towards the BEC, giving the condensate a chance to absorb a photon from one beam and perform stimulated emission into the other. This process is only allowed if the the momentum and energy change of the photon as it changes beams can be absorbed by the BEC by creating a quasiparticle excitation. By varying the detuning and angle between the beams, one can map out which combinations of energy and momenta create quasiparticles, thereby revealing the dispersion relation.

File:Superfluid to Mott insulator transition-bec3-sound1

File:Superfluid to Mott insulator transition-bec3-sound2

The Bogolubov solution has a great deal of physics in it. It gives the elementary excitation, and the ground state energy. In the simple model that we have a mean field, the ground state energy is

$E_{0}={\frac {U_{0}h}{2}}\left(1+{\frac {128}{15}}{\sqrt {na^{3}/\pi }}\right)\,.$

The extra correction term on the right is a small term, recently observed by the Innsbruck group, due to collective effects. The Bogolubov solution also gives the ground state wavefunction,

$|\psi _{0}\rangle =|0{\rangle }^{\otimes N}+\epsilon \,,$

where $\epsilon$ is the quantum depletion term, which makes the wavefunction satisfy

$\langle n_{k}\rangle ={\frac {v_{k}^{2}}{1-v_{k}^{2}}}$

and

$\langle n_{0}\rangle =N-\sum \langle n_{k}\rangle =N\left[{1-{\frac {8}{3}}{\sqrt {na^{3}/\pi }}}\right]\,.$

The quantum depletion term, which arises from the fact that the gas is weakly interacting, has now been experimentally observed. Recall that in the Bogolubov approximation, the original interaction

$H'=U_{0}\sum a_{p}^{\dagger }a_{q}^{\dagger }a_{r}a_{s}$

is approximated by

$H'=U_{0}a_{0}a_{0}\sum a_{p}^{\dagger }a_{-p}^{\dagger }$

The quantum depletion this leads to is very small. The effect can be more readily experimentally observed by increasing the mass of the particle, and this can be done by placing the particles in a lattice. Plotting the quantum depletion which can be obtained as a function of lattice depth, in such an experiment, one gets:

File:Superfluid to Mott insulator transition-bec3-qdepletion

Beyond a quantum depletion fraction of $0.6$, the Bogolubov approximation breaks down, as the condensate goes through a superfluid to Mott-insulator transition.

Back to: Quantum gases

The inhomogeneous Bose gas

File:Superfluid to Mott insulator transition-bec3-first-expt

The physics of a BEC happens not just in momentum space, but also in position space, and it is useful to analyze it accordingly. With a trapping potential applied, the Hamiltonian is

$H=\int d^{3}r\psi ^{\dagger }(r)[{\frac {-\hbar ^{2}}{2m}}+V_{trap}]\psi (r)+{\frac {1}{2}}\int d^{3}r\int d^{3}r'+....$

This must be approximated, in the spirit of Bogolubov's momentum space approximation, to obtain a useful solution. We thus replace

${\hat {\psi }}(r,t)=\psi (r,t)+{\tilde {\psi }}(r,t)\,,$

where $\psi (r,t)$ is an expectation, and ${\tilde {\psi }}(r,t)$ captures the quantum (+ thermal) fluctuations. The resulting equation is a nonlinear Schr\"odinger equation (also known as a Gross-Pitaevskii equation):

$i\hbar {\frac {\partial \psi }{\partial t}}=\left[{-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V_{trap}+U_{0}N|\psi (r,t)|^{2}}\right]\psi (r,t)$

The $|\psi (r,t)|^{2}$ term captures a potential proportional to the density. In the mean field approximation, it is determined by the trapping potential. This equation can now be solved. In the Thomas-Fermi approximation, with positive (repulsive) interactions, there is a characteristic length $\xi$ which arises, known as the healing length,

$\xi =(8\pi an)^{-1/2}\,,$

arising from

${\frac {\hbar ^{2}}{2m\xi ^{2}}}=U_{0}h\,.$

File:Superfluid to Mott insulator transition-bec3-healing

If the interactions are really strong, the kinetic energy term can be neglected, because the interactions will keep the density constant in its spatial distribution. Such an approximation is the Thomas-Fermi approximation, giving an equation for the wavefunction,

$[V_{ext}+U_{0}|\psi (r,t)|^{2}-\mu ]\psi (r)=0\,,$

giving the solution

$|\psi (r,t)|^{2}={\frac {1}{U_{0}}}(\mu -V(r)\,.$

The wavefunction is essentially just the potential filled up to the chemical potential level, inverted. For a quadaratic potential, $V(r)=m\omega ^{2}r^{2}/2$, the chemical potential is

$\mu ={\frac {\hbar \omega }{2}}(15N_{a}/a_{osc})^{2/5}\,,$

where $x=15N_{a}/a_{osc}$ is a common term worth identifying, and $a_{osc}={\sqrt {\hbar ^{2}/m\omega }}$ is a characteristic length scale of the oscillator, its zero point motion. Defining $R=a_{osc}x^{1/5}$, we may find $\mu =m\omega ^{2}R^{2}/2$. This explains the profile of the condensate data obtained in experiments:

File:Superfluid to Mott insulator transition-bec3-bec-peak

\noindent Note that the size of the ground stat BEC is much larger than the zero-point motion of the harmonic oscillator. This is due to the pressure of the repulsive interactions. The Gross-Pitaevskii interaction gives not only the ground state wavefunction, but also the dynamics of the system. For example, it predicts soliton formation: stable wavefunctions with a size scale determined by a balance of the kinetic energy and the internal interactions. This requires, however, an attractive potential. Such soliton formation can nevertheless be seen in BEC's, with tight traps (see recent Paris experiments).

Length and energy scales in BEC

\begin{itemize}

• Size of atom: $a=3$ nm
• Separation between atoms $n^{1/3}=200$ nm
• Matter wavelength $\lambda _{dB}=1$ $\mu$m
• Size of confinement $a_{osc}=30$ $\mu$m

Note that

$a\ll n^{1/3}\leq \lambda _{dB}<2\pi \xi <a_{osc}\,.$

For a gas, $a\ll n^{1/3}$. For a BEC, $\lambda _{dB}\geq n^{1/3}$ in addition. The corresponding energy scales are also useful to identify. Let $\hbar \omega =a_{osc}$. Then:

$k_{B}T_{s-wave}\gg k_{B}T_{c}\geq k_{B}T>U_{int}>\hbar \omega$

The interaction energy scale $U_{int}\sim (h^{2}/m)na$, corresponding to the healing length.

Vortices in a BEC

The Gross-Pitaevskii equations also predict the formation of topological defects in the condensate, such as vortices. These form with a chacteristic vortex diameter of $\xi$, the healing length.

File:Superfluid to Mott insulator transition-bec3-vortex-creation

File:Superfluid to Mott insulator transition-bec3-vortex-movie

Question: do you get the best votrices by stirring gently, or vigorously?

Hydrodynamics

We may transform the GPE into a hydronamic equation for a superfluid,

${\frac {\partial |\psi |^{2}}{\partial t}}+\nabla {\frac {\hbar }{2mi}}\left({\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*}}\right)\,,$

by introducing flow, from current $j$,

$v={\frac {j}{n}}={\frac {\psi ^{*}\nabla \psi -\psi \nabla \psi ^{*}}{2mi|\psi |^{2}}}\,.$

This gives the continuity equation

${\frac {\partial n}{\partial t}}+\nabla (nv)=0\,.$

Writing $\psi =fe^{i\phi }$, and noting that the gradient of the phase gives us the velocity field, we get equations of motion for $f$ and $\phi$,

$-\hbar {\frac {\partial \phi }{\partial t}}=-frac{\hbar ^{2}}{2mf}\nabla ^{2}f+{\frac {1}{2}}mv^{2}+V(r)+U_{0}f^{2}\,.$

This reduces to

${\begin{array}{rcl}m{\frac {\partial v}{\partial t}}&=&-\nabla (\delta \mu +{\frac {1}{2}}mv^{2}\\\delta \mu &=&v+U_{0}n{\frac {\hbar ^{2}}{2m{\sqrt {n}}}}\nabla ^{2}{\sqrt {n}}-\mu _{0}\,.\end{array}}$

The Thomas-Fermi approximation is now applied, neglecting $\nabla f$, but keeping $\nabla \phi$, giving

$m{\frac {\partial ^{2}\delta n}{\partial t^{2}}}=U_{0}\nabla (n_{0}\nabla (\delta n))\,,$

a wave equation for the density. For $n_{0}$ constant, $U_{0}$ is the speed of sound squared, $c={\sqrt {U_{0}/m}}$. The Thomas-Fermi solution for $n_{0}$ gives collective modes of the condensate. A droplet of condensate can have shape resonances, waves, and many other physical behaviors, captured by these solutions.

Superfluid to Mott-Insulator transition

A condenstate in a shallow standing wave potential is a BEC, well dessribed by a Bogolubov approximate solution. As the potential gets deeper, though, eventually the system transitions into a state of localized atoms, with no long-range coherence, known as a Mott insulator. These physics are important in a wide range of condensed matter systems, and can be explored deeply with BECs. The ultracold atoms are trapped in a periodic potential,

$V(x,y,z)=V_{0}\left({\sin ^{2}kx+\sin ^{2}ky+\sin ^{2}kz}\right[\,,$

and we would like to know what happens when neutral atoms are in this potential. Traditionally, this is studied in condensed matter physics in the context of atoms in a periodic potential, but many anologies exist for the neutral atom system. The Hamiltonian is

$H={\frac {\hbar ^{2}}{2m}}\nabla ^{2}-V_{0}\sin ^{2}kz\,,$

in one dimension, giving the Bloch ansatz wavefunction solution

$\psi _{q,n}(x)=e^{iqx/\hbar }U_{q}(x)\,,$

where $q$ is a quasi-momentum labeling the eigenstate, and $U_{q}(x)$ is a periodic function. For a given $q$ there are several solutions, labeled by $n$, the band index. To solve this equaiton, a Fourier expansion of $U_{q}(x)$ is used, with elements $c_{l}^{a,n}e^{i2lkx}$. The potential is also expanded as $V_{r}e^{i2lkx}$. The solutions to the Schrodinger equation,

$H\psi _{q,n}=E_{a,n}\psi _{a,n}\,,$

give

$\sum _{l'}e^{i(q+2l'kx)}\left[{\left({{\frac {(q+2l'\hbar k)^{2}}{2m}}-E_{q,n}}\right)c_{l}^{q,n}+\sum _{v}c_{l',r}^{q,n}}\right]=0\,.$

Graphically, these solutions are as follows. If energies are plotted as a function of $\hbar k$, we get:

File:Superfluid to Mott insulator transition-bec3-smott-dispersion

With no interations, we get a free particle dispersion diagram. But with the periodic potential, a Brilloiuin zone appears, forbidding momentum beyond $\pm \hbar k$, giving bands. The potential couples the bands; for example, at $V\sim 3E_{r}$ (where $E_{r}=k^{2}/2m$), the band degnericies are lifted to become:

File:Superfluid to Mott insulator transition-bec3-3er

\noindent At higher potentials, the bands become even flatter. We will be interestd in the limit in which $V_{r}/E_{r}$ is large, and we shall focus on the physics of the lowest, $n=1$ band, in which in the limit of tight binding, the energy is

$E_{1}(q)={\frac {3}{2}}\hbar \omega _{0}\,,$

becoming constant with respect to quasi-momentum. The derivation of the energy as a function of quasi-momentum gives

$-2J(\cos q_{x}a+\cos q_{y}a+\cos q_{z}a)\,,$

where $a={\frac {\lambda }{2}}=\pi /k$, and the paramter $J$ tells us how wide the band is and how large the dispersion region is.

BEC to BCS crossover transition

File:Superfluid to Mott insulator transition-bec3-

File:Superfluid to Mott insulator transition-bec3-

File:Superfluid to Mott insulator transition-bec3-

File:Superfluid to Mott insulator transition-bec3-

File:Superfluid to Mott insulator transition-bec3-

File:Superfluid to Mott insulator transition-bec3-