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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

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 written in terms of bosonic field operators ${\hat {\psi }},{\hat {\psi }}^{\dagger }$ (obeying $[{\hat {\psi }}(r),{\hat {\psi }}^{\dagger }(r')]=\delta (r-r'),[{\hat {\psi }}(r)^{(\cdot )/\dagger },{\hat {\psi }}^{(\cdot )/\dagger }(r')]=0$)

${\hat {H}}=\int d^{3}r\,{\hat {\psi }}^{\dagger }(r){\Big [}{\frac {-\hbar ^{2}}{2m}}\nabla ^{2}+V_{trap}{\Big ]}{\hat {\psi }}(r)+{\frac {1}{2}}\int d^{3}r\int d^{3}r'{\hat {\psi }}^{\dagger }(r){\hat {\psi }}^{\dagger }(r^{\prime })U(r-r^{\prime }){\hat {\psi }}(r^{\prime }){\hat {\psi }}(r)$

Note: this is a general field-quantized expression for a Hamiltonian with two-body interaction. If you are not familiar with second quantization, consult the first chapter of "Quantum Theory of Many-Particle Systems" by Fetter and Walecka, which reviews the mapping between first quantization and second quantization in detail.

We are interested in the time-evolution of the operator ${\hat {\psi }}$ (i.e. we work in the Heisenberg picture). The equation of motion is given by the Heisenberg equation

$i\hbar {\frac {\partial }{\partial t}}{\hat {\psi }}=[{\hat {\psi }},{\hat {H}}]$

Using the field operator commutation relation, we can write the right-hand side of the equation as

$-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}{\hat {\psi }}(r,t)+V_{trap}{\hat {\psi }}(r,t)+{\frac {1}{2}}{\Big (}\int d^{3}r'\,{\hat {\psi }}^{\dagger }(r',t)U(r-r'){\hat {\psi }}(r',t)+\int d^{3}r'\,{\hat {\psi }}^{\dagger }(r',t)U(r'-r){\hat {\psi }}(r',t){\Big )}{\hat {\psi }}(r,t)$

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

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

where the complex number $\psi (r,t)$ is an expectation (mean field), and the operator ${\hat {\delta \psi }}(r,t)$ captures the quantum (+ thermal) fluctuations. We do not consider terms proportional to ${\hat {\delta \psi }}$. We further assume that the interaction potential is a delta function, $U(r-r')=U_{0}\delta (r-r')$ (which is valid for s-wave scattering at short range). With these assumptions, we obtain a nonlinear Schrodinger equation:

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

This is known as the Gross-Pitaevskii equation.

The $|\psi (r,t)|^{2}$ term adds an energy proportional to the density due to the interactions. That density-dependence implies that increasing the density comes with an energy cost and decreasing density lowers the energy. Since the density is multiplied by the wave function, there is a non-linear energy dependence on the density. Therefore, lowering the density in one region and raising it in another costs energy and so this term makes the condensate try to have a uniform density distribution (the total number of particles is fixed so the integrated density cannot change).

The density-dependent term has an interesting interplay with the other two terms. Consider its effect on a condensate in a box potential for instance. If it were not for the interactions, the BEC density distribution would simply be set by the ground state of the box potential. However, the central region of that distribution is very dense, so the density-dependent term will try to flatten out the distribution and push atoms out towards the wings. It can flatten out the distribution near the center very well, but the trap potential makes it impossible to push atoms out too far, so the atom distribution stays roughly flat over a finite region. At the edge of the flat region of the density distribution are the wings which must go to zero at the edges of the box. The density energy cost makes the cloud want to make this crossover region as small as possible. However, making it very small would take a lot of curvature in the wave function, causing the kinetic energy term to increase. Therefore the distribution at the edges reaches a compromise between interaction energy and kinetic energy curvature. The length scale of the crossover region set by this compromise (assuming repulsive interactions) is known as the healing length $\xi$,

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

arising from

${\frac {\hbar ^{2}}{2m\xi ^{2}}}=U_{0}n={\frac {4\pi \hbar ^{2}}{m}}a_{s}n\,.$

where $a_{s}$ is the s-wave scattering length.

We remarked that the Gross-Pitaevskii equation looks like a nonlinear Schrodinger equation. Then what should its time-independent version look like? In other words, what is the $E$ that appears in $\psi (t)=\psi \exp(-iEt/\hbar )$? For ordinary time-independent Schrodinger equation applied to wavefunctions, that would be the energy of an eigenstate. But the notion of an eigenfunction for a nonlinear equation is not clear. The answer is the chemical potential $\mu$, and it can be reasoned as follows (from "Bose-Einstein Condensation and Superfluidity" by Pitaevskii and Stringari). Our assumption ${\hat {\psi }}\sim \langle {\hat {\psi }}\rangle \neq 0$ requires that the bra-state and the ket-state cannot have the same number of particles - the condensate acts like a reservoir. When the number of particles is large and the interaction is not too strong, adding one particle only makes correction to the state on the order of $1/N_{0}$, where $N_{0}$ is the average number of particles. So in this limit, it is valid to interpret $\langle {\hat {\psi }}\rangle$ so that the bra-state has one number of particle less than the ket state. Now if we take an average over stationary states, whose time dependence is governed by $\exp(-iE(N)t/\hbar )$, the time dependence of the mean value ${\hat {\psi }}=\psi$ is given by $\exp(-i[E(N)-E(N-1)]t/\hbar )\approx \exp(-i(\partial E/\partial N)t/\hbar )=\exp(-i\mu t/\hbar )$. The quantity $\langle {\hat {\psi }}\rangle$ is known as the superfluid order parameter in statistical physics. The non-zero, complex value of the superfluid parameter distinguishes the superfluid phase from the normal phase. The term order parameter is used to describe a phase with broken symmetry, which we will explain later in the context of BEC.

In summary, the time-independent version of the GPE reads

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

It is worth emphasizing that the superfluid order parameter is not equivalent to an ordinary Schrodinger wavefunction (from Pitaevskii and Stringari). First of all, we already saw that the order parameter evolves in time with chemical potential, not energy. Also, because the GPE is non-linear, two solutions $\psi _{a}$ and $\psi _{b}$ with different chemical potentials are not necessarily orthogonal ($N^{-1}\int d^{3}r\,\psi _{a}^{*}\psi _{b}\neq 0$). But quantum mechanics is linear and hence different eigenstates are orthogonal. How should these two facts be reconciled? In the weak-interacting limit, so that we can ignore correlations between particles (which the mean-field approximation implies), we can write the many-body wavefunction $\Phi$ corresponding to the field $\psi$ in a factorized form (Hartree-Fock approximation):

$\Phi (r_{1},\ldots ,r_{N})=\prod _{i=1}^{N}{\frac {1}{\sqrt {N}}}\psi (r_{i})$

The inner product of two different many-body wavefunctions corresponding to $\psi _{a}$ and $\psi _{b}$ is $(N^{-1}\int d^{3}r\psi _{a}^{*}\psi _{b})^{N}$, which goes to zero unless $\psi _{a}=\psi _{b}$ when $N\to \infty$. This also emphasizes that the GPE and the mean-field picture is only valid in the limit of large $N$.

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. This is a particularly good approximation in the flat center of the distribution where there is little wave function curvature. 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

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

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

$\mu ={\frac {\hbar \omega }{2}}x^{2/5}\,,$

where $x=15Na/a_{osc}$ 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

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

• 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.

Back to: Quantum gases

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 characteristic vortex diameter of $\xi$, the healing length. Because the Thomas-Fermi approximation breaks down at this length scale, the classical hydrodynamic equations introduced at the end of "Superfluid Hydrodynamics" do not work and we must include the full effects of quantum pressure.

In developing the hydrodynamic equations for the superfluid, it was shown that superfluid flow is irrotational ($\nabla \times {\vec {v}}_{s}=\nabla \times {\frac {\hbar }{m}}\nabla S=0$). This already tells us that rotating superfluids behave differently from rotating normal fluids. For normal fluids with rotational velocity ${\vec {\Omega }}$, the velocity field is given by ${\vec {v}}={\vec {\Omega }}\times {\vec {r}}$, and its curl is not zero. So how can a superfluid rotate (i.e. have a non-zero tangential velocity)? It can rotate as long as the curl of the velocity field becomes nonzero where the density goes to zero.

Let's consider a superfluid confined in a cylindrical vessel of radius $R$ and assume the order parameter is described as $\psi (r)={\sqrt {n(r)}}e^{is\phi }$, where $\phi$ is the angular cylindrical coordinate and $s$ is an integer (so the phase of the order parameter is single-valued). We will show that this is a solution to the Gross-Pitaevskii equation that corresponds to a rotating superfluid with a vortex line through the axis of the cylinder (= z-axis).

First of all, the angular part of the order parameter is an eigenfunction of the rotation operator around the z-axis, with eigenvalue $s\hbar$, so the superfluid carries an angular momentum of $L_{z}=s\hbar$. The tangential velocity is ${\vec {v}}_{s}={\frac {\hbar }{m}}{\frac {1}{r}}{\frac {\partial }{\partial \phi }}(s\phi ){\hat {\phi }}={\frac {\hbar }{m}}{\frac {s}{r}}{\hat {\phi }}$. The curl of this velocity field is equal to ${\frac {1}{r}}{\frac {\partial }{\partial r}}(rv_{s}){\hat {z}}=2\pi s{\frac {\hbar }{m}}\delta ^{(2)}(r){\hat {z}}$ where $\delta ^{(2)}$ is a two-dimensional delta function in the x,y plane (this can be also reasoned by applying Green's theorem to $\oint {\vec {v}}_{s}\cdot {\vec {dl}}=2\pi s{\frac {\hbar }{m}}$). We see that irrotationality is satisfied except on the vortex line. Now we show from the GPE that the density approaches zero as we approach the vortex line.

If we substitute $\psi ={\sqrt {n(r)}}e^{is\phi }$ to the time-independent GPE, by using the expression for the Laplacian in cylindrical coordinates we obtain the following equation for $|\psi |$:

$-{\frac {\hbar ^{2}}{2m}}{\frac {1}{r}}{\frac {d}{dr}}{\Big (}r{\frac {d}{dr}}|\psi |{\Big )}+{\frac {\hbar ^{2}s^{2}}{2mr^{2}}}|\psi |+U_{0}|\psi |^{3}-\mu |\psi |=0$

Far away from the vortex line (distance much longer than the healing length) we expect the density $n(r)$ to approach a uniform value $n_{0}=\mu /U_{0}$. So let's express the modulus of the order parameter in terms of a dimensionless function $f(r/\xi )$ , where $\xi$ is the healing length $\hbar /{\sqrt {2mU_{0}n_{0}}}$

${\sqrt {n(r)}}={\sqrt {n_{0}}}f(r/\xi )\,(f(\infty )=1)$

It is easy to verify that $f(r/\xi )=f(\eta )$ obeys the equation

${\frac {1}{\eta }}{\frac {d}{d\eta }}{\Big (}\eta {\frac {df}{d\eta }}{\Big )}+{\Big (}1-{\frac {s^{2}}{\eta ^{2}}}{\Big )}f-f^{3}=0$

Near the vortex core region ($\eta \to 0$), the solution asymptotically approaches $f\sim \eta ^{|s|}$. One can go further can calculate the energy required to create a vortex line and then the critical angular velocity for spontaneous creation of vortex lines. See Chapter 5 of Stringari and Pitaevskii.

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

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

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

Superfluid Hydrodynamics

The notion of superfluid and Bose-Einstein condensation are intimately related. Superfluids can flow through narrow tubes without dissipating heat due to their zero viscosity. A well-known example of superfluid is low-temperature helium-4 (and helium-3, which is a fermionic superfluid). What characterizes a superfluid is that it can only dissipate heat via creation of elementary excitations, and if the flow velocity is slower than the characteristic speed of sound in the superfluid, those elementary excitations cannot be spontaneously created, thus ensuring a frictionless flow (Landau's criterion). Indeed, dilute BEC gases also have a linear dispersion with non-zero speed of sound at low momentum (Bogoliubov dispersion) and feature superfluidity. But it is important to note some differences between a superfluid and a BEC. An ideal BEC will have a quadratic dispersion (zero speed of sound), and there is no energy gap to protect the ground state from excitations, so an ideal BEC cannot be a superfluid. Also, even for the interacting BEC, the condensate density of BEC is not equivalent to superfluid density, for at zero temperature, by definition the superfluid fraction is 100% and normal fraction is 0%, but the condensate fraction is less than 100%, due to quantum depletion.

Now let's try to obtain an equation that describe the superfluid flow of BEC, which will require two variables: density and velocity. We again emphasize that the density in this case is not equivalent to superfluid density because of quantum depletion. In this article, we try a bit formal approach to arrive at the continuity equation.

The original Hamiltonian is invariant under the U(1) transformation $\psi \to \psi e^{i\theta }$. Indeed, the equation of motion obeys this symmetry, even though the superfluid itself has broken the symmetry by selecting a particular phase. This is characteristic of spontaneously broken symmetry. Noether's theorem (from classical mechanics / field theory) tells us that every continuous symmetry is associated with a conserved current $j^{\mu },(\partial _{\mu }j^{\mu }=0)$, and we will see that that the conserved charge is actually the average number of atoms, $N$. To make use of Noether's theorem, we need an action term that produces the GPE. We can obtain the GPE via the action principle if we suppose the action is given by the functional

$S[\psi ,\psi ^{*}]=\int d^{3}rdt\,\psi ^{*}{\Big (}-i\hbar {\frac {\partial }{\partial t}}{\Big )}\psi +{\frac {\hbar ^{2}}{2m}}\nabla \psi ^{*}\cdot \nabla \psi +V\psi ^{2}+{\frac {U_{0}}{2}}|\psi |^{4}=\int d^{3}rdt\,{\mathcal {L}}[\psi ,\psi ^{*}]$

If the field $\psi$ transforms infinitesimally under symmetry with parameter $\theta$ as $\delta \psi =\theta \Delta \psi$, then the Noether current $j^{\mu },\mu =t,x,y,z$ is given by ${\frac {\partial {\mathcal {L}}}{\partial (\partial _{\mu }\psi )}}\Delta \psi +({\text{similar for other independent field = }}\psi ^{*})$. The result is

Failed to parse (syntax error): {\displaystyle j^0 = n = |\psi|^2\\ j^{x,y,z} = \vec{j} = - \frac{i\hbar}{2m} (\psi^{*} \nabla \psi - \psi \nabla \psi^{*} )\\ \partial_{\mu} j^{\mu} =0 \Rightarrow \partial_t n + \nabla \cdot \vec{j} = 0 }

By using the fact that the equation of motion for the superfluid obeys U(1) symmetry, we have derived the continuity equation, which implies that the average number of atoms $N=\int d^{3}r\,n$ is conserved. Now consider writing the complex order parameter in terms of its amplitude and phase: $\psi ={\sqrt {n}}\exp(iS)$. You will find that the current density ${\vec {j}}$ can be expressed as proportional to gradient of the phase term:

Failed to parse (syntax error): {\displaystyle \vec{j} = n \frac{\hbar}{m} \nabla S = n \vec{v}_s\\ \vec{v}_s = \frac{\hbar}{m} \nabla S }

Since the curl of a gradient term is zero, the expression for the velocity field shows that the condensate flow is irrotational. If you plug in the phasor expression for the order parameter into the GPE, you obtain a complex equation whose real and imaginary part need to set to zero separately. The imaginary part gives back the continuity equation, and the real part gives

$\hbar {\frac {\partial S}{\partial t}}+{\frac {1}{2}}mv_{s}^{2}+V_{trap}+U_{0}n-{\frac {\hbar ^{2}}{2m{\sqrt {n}}}}\nabla ^{2}{\sqrt {n}}=0$

The last term is known as the quantum pressure term and emphasizes the importance of quantum effects in inhomogeneous condensates. Take the gradient of the above equation to obtain the equation of motion for the velocity field. Now we have a set of coupled equations for two hydrodynamic variables (density and velocity field):

${\begin{array}{rcl}m{\frac {\partial {\vec {v}}_{s}}{\partial t}}&=&-\nabla {\Big (}\mu +{\frac {1}{2}}mv_{s}^{2}{\Big )}\\\mu &=&-{\frac {\hbar ^{2}}{2m}}{\frac {\nabla ^{2}{\sqrt {n}}}{\sqrt {n}}}+V_{trap}+U_{0}n\\{\frac {\partial n}{\partial t}}&=&-\nabla \cdot (n{\vec {v}}_{s})\end{array}}$

Note these resemble the Euler equations from fluid dynamics. If the quantum pressure term were zero, the $\hbar$ dependence disappears (except implicitly in $U_{0}$) and the equations describe the potential flow of a non-viscous gas with pressure $U_{0}n^{2}/2$.

In the Thomas-Fermi approximation (ignoring the quantum pressure term), we can linearize the coupled hydrodynamic equations by expanding the variables around their stationary values

${\begin{array}{rcl}n(r,t)&=&n_{0}(r)+\delta n(r,t)\\{\vec {v}}_{s}(r,t)&=&{\vec {v}}_{s,0}(r)+{\vec {\delta v}}_{s}(r,t)\end{array}}$

If we assume ${\vec {v}}_{s,0}(r)=0$, then we obtain the following equations:

${\begin{array}{rcl}m{\frac {\partial {\vec {\delta v}}_{s}}{\partial t}}&=&-\nabla (U_{0}\delta n)\\{\frac {\partial \delta n}{\partial t}}&=&-\nabla (n_{0}{\vec {\delta v}}_{s})\end{array}}$

We see that these produce a wave equation for the density fluctuation:

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

For homogeneous $n_{0}$, the speed of sound $c$ is ${\sqrt {U_{0}n_{0}/m}}$. A droplet of condensate can have shape resonances, waves, and many other physical behaviors, captured by the solutions to the hydrodynamic equations. For an early discussion of collective excitations in trapped BEC, see PRL 77, 2360 (1996). For the description of the time-of-flight expansion of BEC out of a time-dependent harmonic trap, see PRL 77, 5315 (1996).

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Superfluid to Mott-Insulator transition

A condenstate in a shallow standing wave potential is a BEC, well described by a Bogoliubov 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 discussion of superfluid to Mott insulator (SF-MI) transition involves some basic knowledge of periodic potentials from solid state physics. We will review the concept briefly. To observe SF-MI transition, ultracold bosons are placed in an optical lattice, which is formed by a standing wave of light. A variety of optical lattice geometry can be generated by how the laser beams interfere with each other. In this article we only consider the simple (and most common) case of separable, sinusoidal potential, generated by a pair of retroreflected beams in each axis:

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

The Hamiltonian $-\hbar ^{2}\nabla ^{2}/2m+V$ is periodic in space (ignoring interactions for now) with period $a$, so we can use Bloch's theorem, which says that the wavefunctions of the eigenstates $|\psi \rangle$ of a periodic Hamiltonian can be written as

${\begin{array}{rcl}\psi (x)&=&e^{ikx}u_{n,k}(x)\,,{\Big (}-{\frac {\pi }{a}}\leq k\leq {\frac {\pi }{a}},n=1,2,\ldots {\Big )}\\u_{n,k}(x)&=&u_{n,k}(x+a)\end{array}}$

What this means is that if we translate the wavefunction by distance $a$, then the wavefunction can only change up to a complex phase. From a more formal point of view, because the continuous translation symmetry is lost and only discrete translation symmetry remains, momentum is only conserved modulo $\hbar K=\hbar {\frac {2\pi }{a}}$. For this reason the quantity $\hbar k$ is called quasimomentum. If the lattice potential were zero, the quasimomentum and the momentum are related by $p=\hbar k+n\hbar K$. The integer $n$ is called the band index, because if you draw the free particle dispersion $E=\hbar ^{2}k^{2}/2m$ and "folding" the parabola inside the region $-{\frac {\pi }{a}}\leq k\leq {\frac {\pi }{a}}$, you get a discrete number of energy bands. The region $-{\frac {\pi }{a}}\leq k\leq {\frac {\pi }{a}}$ in reciprocal space is called the first Brillouin zone, or simply the Brillouin zone. When the lattice potential is zero ($V_{0}\to 0$), there are points where some of the energy bands become degenerate (e.g. the first and second bands touch at the boundaries of the Brillouin zone). When the lattice potential is nonzero, the degeneracy is broken and an energy gap opens between each band.

The discussion above becomes clearer if we apply Bloch's theorem to express the Hamiltonian in reciprocal space. We can write $\psi (x)=\sum _{G}c_{k-G}e^{i(k-G)x},G=n(2\pi )/a$ using Bloch's theorem and similarly $V(x)=\sum _{G}V_{G}e^{-iGx}$. The net effect is that the Schrodinger equation becomes an algebraic equation in reciprocal space, with $V_{G}$ appearing as off-diagonal matrix element that couples the amplitude $c_{k+nG}$ to $c_{k+(n-1)G}$ (intuitively, the modulation due to lattice potential transfers momentum to the wavefunction). If we write the eigenvalue equation for 1-dimensional case, it looks like

${\begin{pmatrix}\ddots &\vdots &\vdots &\vdots &\vdots &\vdots \\\dots &\lambda _{k+G}-E&V_{-G}&V_{-2G}&V_{-3G}&\dots \\\dots &V_{G}&\lambda _{k}-E&V_{-G}&V_{-2G}&\dots \\\dots &V_{2G}&V_{G}&\lambda _{k-G}-E&V_{-G}&\dots \\\dots &\vdots &\vdots &\vdots &\vdots &\ddots \end{pmatrix}}{\begin{pmatrix}\vdots \\c_{k+G}\\c_{k}\\c_{k-G}\\\vdots \end{pmatrix}}=0$

The solutions to this equation, i.e. the eigenstates of the periodic Hamiltonian, are called the Bloch states. Similar to plane wave solutions for free particle Hamiltonian, the Bloch states are delocalized over the lattice sites. But for our application, where interaction term appears as a on-site term, it would be more convenient to work with a basis set with states localized at each lattice site. The wavefunction for such states, $\phi (r)$ are called Wannier functions, which is constructed as a complex linear combination of Bloch states (analogous to Fourier transform).

$\phi (r-R)=\sum _{k\in BZ}{\frac {1}{\sqrt {N}}}e^{-ikR}e^{i\phi _{n}(k)}\psi _{n,k}(r)\,$

where $R$ is any lattice vector. Wannier functions are not unique in general (because of the freedom to choose "gauge" $\phi _{n}(k)$), but one can define a localization function and find out that there exists a unique set of maximally localized Wannier functions (MLWF) - see the review paper by Marzari and Vanderbilt (RMP 84, 1419).

We are now ready to introduce the Bose-Hubbard Hamiltonian, which sets the framework of the SF-MI transition. Assume that all the atoms are in the lowest band, and the lattice is deep so that we work in the tight-binding limit - the atoms can tunnel only between nearest-neighboring sites.

$H_{BH}=-t\sum _{\langle i,j\rangle }{\Big (}{\hat {a}}_{i}^{\dagger }{\hat {a}}_{j}+{\hat {a}}_{j}^{\dagger }{\hat {a}}_{i}{\Big )}+U\sum _{i}{\hat {a}}_{i}^{\dagger }{\hat {a}}_{i}^{\dagger }{\hat {a}}_{i}{\hat {a}}_{i}-\mu \sum _{i}{\hat {a}}_{i}^{\dagger }{\hat {a}}_{i}$

This Hamiltonian can be derived from the Hamiltonian written in terms of field operator ${\hat {\psi }}(r)$ (see the notes on "Inhomogeneous Bose gases") and expanding the field operator in the Wannier function basis (${\hat {\psi }}=\sum _{i}\phi (r-R_{i}){\hat {a}}_{i}$). The three parameters that appear in the Bose-Hubbard Hamiltonian are tunneling constant $t$, on-site interaction energy $U$, and chemical potential $\mu$ which fixes the number of atoms in the system. In terms of Wannier functions,

${\begin{array}{rcl}t_{\langle i,j\rangle }&=&\int d^{3}r\,\phi (r-R_{i})^{*}{\Big (}-{\frac {\hbar ^{2}\nabla ^{2}}{2m}}+V_{lattice}{\Big )}\phi (r-R_{j})=t\\U&=&{\frac {4\pi \hbar ^{2}a_{s}}{m}}\int d^{3}r\,|\phi (r)|^{4}\end{array}}$

If the ratio of interaction energy to tunneling energy is low, the atoms stay as a superfluid (and occupy the $k=0$ point in the BZ). However, when the ratio becomes high, it becomes energetically costly to tunnel between sites due to the repulsive on-site interaction, such that the atoms cannot tunnel anymore and fail to maintain long-range coherence. This is the Mott insulating state, in which the tunneling is frozen and each lattice site has a fixed number of atoms. The Mott insulating state is characterized by the filling factor $n$: a $n=2$ Mott insulator has two atoms occupying every lattice site, no more, no less. In the superfluid state, the site occupation number fluctuates, but each site has a well-defined phase relationship with respect to other sites ($\langle {\hat {a}}_{i}\rangle \neq 0$).

Experimentally, the phase transition can be observed in the time-of-flight pattern of atoms released from the optical lattice. The time-of-flight pattern reveals the momentum distribution of the atoms, and below a critical lattice depth, we see well-defined peaks (at zero quasimomentum and elsewhere, translated by the reciprocal lattice vectors). One can think of this as a multi-slit diffraction pattern of a coherent matter wave. Above the critical lattice depth, the long-range coherence is lost and the sharp quasimomentum peaks are no longer visible.

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