Bose Gas

  • Noninteracting bosons form a Bose–Einstein condensate (BEC)
  • What do interactions do?
  • BEC closely related to superfluidity

Gross–Pitaevskii Approximation

  • Variational appraoch to Bose gas (c.f. Hartree–Fock)

  • Put all particles in same single particle state!

$$ \Psi(\br_1,\ldots \br_N) = \prod_{j=1}^N \varphi_0(\br_i)= \frac{1}{\sqrt{N!}}\left(\adop(\varphi_0)\right)^N\ket{\text{VAC}}. $$

  • State with macroscopic number of particles in a single particle state is a Bose condensate
  • For noninteracting Hamiltonian

$$ H = \sum\left[-\frac{\nabla_i^2}{2m} + V(\br_i)\right], $$

  • Ground state is exactly $$ \Psi(\br_1,\ldots \br_N) = \prod_{j=1}^N \varphi_0(\br_i)= \frac{1}{\sqrt{N!}}\left(\adop(\varphi_0)\right)^N\ket{\text{VAC}}. $$ with $\varphi_0(\br)$ the ground state of single particle Hamiltonian
  • Add interaction

$$ \begin{align*} H_\text{int.} &= \sum_{j<k} U(\br_j-\br_k) \\ &= \frac{1}{2}\int d\br_1 d\br_2\, U(\br_1-\br_2)\pdop(\br_1)\pdop(\br_2)\pop(\br_2)\pop(\br_1) \end{align*} $$

  • Ground state is more complicated, but can use BEC form with $\varphi_0(\br)$ as a variational function

  • Optimal $\varphi_0(\br)$ obeys Gross–Pitaevskii equation

Gross–Pitaevskii Equation

  • Take short-ranged interactions for simplicity

$$ U(\br-\br’) = U_0\delta(\br-\br’) $$

  • For variational calculation we need

$$ \langle E \rangle = \frac{\braket{\Psi|H|\Psi}}{\braket{\Psi|\Psi}} $$

  • Minimize $\braket{\Psi|H|\Psi}$ and fix norm. using Lagrange multiplier

$$ \begin{align*} \braket{\Psi|H|\Psi}=N \int d\br \left[\frac{1}{2m}|\nabla\varphi_0|^2+V(\br)|\varphi_0(\br)|^2 \right]\\ +\frac{1}{2}N(N-1)U_0\int d\br |\varphi_0(\br)|^4 \end{align*} $$

  • Neglect difference between $N$ and $N+1$ $$ \begin{align*} \braket{\Psi|H|\Psi}=N \int d\br \left[\frac{1}{2m}|\nabla\varphi_0|^2+V(\br)|\varphi_0(\br)|^2 \right]\\ +\frac{1}{2}N^2 U_0\int d\br |\varphi_0(\br)|^4 \end{align*} $$

  • Extremize the functional

$$ \braket{\Psi|H|\Psi} - \mu N \int d\br |\varphi_{0}(\br)|^{2}. $$

  • Calculus of variations yields

$$ \left[-\frac{1}{2m}\nabla^2-\mu+V(\br)+NU_0|\varphi_0(\br)|^2\right]\varphi_0(\br)=0 $$

  • Define $\varphi(\br)\equiv\sqrt{N}\varphi_{0}(\br)$

  • $\varphi(\br)$ is condensate wavefunction or order parameter. Obeys Gross–Pitaevskii equation

$$ \left[-\frac{1}{2m}\nabla^2-\mu+V(\br)+U_0|\varphi(\br)|^2\right]\varphi(\br)=0. $$

  • Fix Lagrange multiplier $\mu$ by $\int d\br\abs{\varphi(\br)}^{2}=N$

  • $\braket{\Psi|H|\Psi}- \mu \int d\br \abs{\varphi(\br)}^{2}=\braket{\Psi|H-\mu \mathsf{N}|\Psi}$ was extremized under general variations, including change in $N$ $$ \mu=\frac{\partial\braket{\Psi|H|\Psi}}{\partial N}, $$ $\mu$ is identified with the chemical potential.

Healing length

  • 1D, no potential

$$ \left[-\frac{1}{2m}\partial_x^2-\mu+U_0|\varphi|^2\right]\varphi(x)=0. $$

  • Dimensionless form $\varphi(\br)=\sqrt{n}\phi(\br/\xi)$ for $\mu=U_0n$

$$ -\frac{1}{2m \xi^2}\phi''+\mu (|\phi|^2-1)\phi(x)=0. $$

  • If we set $\xi\equiv \frac{1}{\sqrt{2m \mu}}=\frac{1}{\sqrt{2m n U_0}}$ then

$$ -\phi''+(|\phi|^2-1)\phi(x)=0. $$

  • Healing length $\xi$ is scale on which $\varphi(\br)$ disturbed by localized perturbation of scale $\ll \xi$

Near a wall

$$ \varphi(x)=\varphi_{\infty}\tanh \frac{x}{\sqrt{2}\xi} $$

where $x$ is distance from wall, and $\varphi_{\infty}=\sqrt{n}$ is fixed by the density far away.

Condensate in a Can

drawing

Some Observables

  • Density and current density

$$ \begin{align*} \rho(\br)&=|\varphi(\br)|^2,\\ \mathbf{j}(\br)&=-\frac{i}{2m}\left[\varphi^{*}(\br)\left(\nabla\varphi(\br)\right)-\left(\nabla\varphi^{*}(\br)\right)\varphi(\br)\right] \end{align*} $$

  • Useful decomposition into magnitude and phase

$$ \varphi(\br)=\sqrt{\rho(\br)}e^{i\chi(\br)}. $$

  • Using $\mathbf{j}=\rho \mathbf{v}$ we get superfluid velocity

$$ \mathbf{v}_{s}\equiv\frac{1}{m}\nabla\chi. $$

Example: Vortex

  • Expect $\mathbf{v}_{s}=\frac{1}{m}\nabla\chi$ to be irrotational $$ \nabla\times \mathbf{v}_s = 0, $$

  • Equivalently to have vanishing circulation around any closed loop

$$ \oint \mathbf{v}_s\cdot d\mathbf{l}=0 $$

  • Phase can increase by a multiple of $2\pi$ around a closed loop, so $$ \oint \mathbf{v}_s\cdot d\mathbf{l}=\frac{2\pi \ell}{m},\quad \ell\in\mathbb{Z}, $$ Onsager–Feynmann quantization condition.
  • Localized configuration with finite circulation is a vortex in fluid dynamics

  • In normal fluid there is no reason for the vorticity to be quantized $$ \oint \mathbf{v}_s\cdot d\mathbf{l}=\frac{h\ell}{m},\quad \ell\in\mathbb{Z}, $$ shows that this is a truly quantum phenomenon.

  • A non-zero winding of phase requires that $\rho(\br)$ vanishes at a point (in 2D) or on a line (in 3D)

  • Look for 2D solutions where phase winds $\ell$ times around origin

$$ \varphi(r,\theta)\xrightarrow{r\to\infty} \sqrt{n} e^{i\ell\theta} $$

  • Paremeterize $\varphi(r,\theta) = \sqrt{n} f(r/\xi)e^{i\ell\theta}$ to give an equation in $s\equiv r/\xi$ (set $\mu=U_0 n$ as before).

$$ -f'' -\frac{f'}{s} + \frac{\ell^2 f}{s^2} - f +f^3 =0. $$

Without finding the solution explicitly, show that $f(s)\sim s^\ell$ for small $s$, and $f(s\to\infty) \to 1$.

  • Region of suppressed density size $\xi$, is vortex core

  • In three dimensions, vortex core is a line

  • Find energy of vortex by substituting solution back into energy

$$ \begin{align*} \braket{\Psi|H|\Psi}=\int d\br \left[\frac{1}{2m}|\nabla\varphi|^2+V(\br)|\varphi(\br)|^2 \right]+\frac{1}{2}U_0\int d\br |\varphi(\br)|^4. \end{align*} $$

  • Excess energy (relative to uniform state of density $n$)

$$ \Delta E = \int d\br \left[\frac{n^2}{2m\xi^2}(f')^2+\frac{U}{2}n^2 \left(f^2-1\right)^2\right] + \frac{n}{2m}\int d\br\, f^2(\nabla\chi)^2 $$

$$ \Delta E = \int d\br \left[\frac{n^2}{2m\xi^2}(f')^2+\frac{U}{2}n^2 \left(f^2-1\right)^2\right] + \frac{n}{2m}\int d\br\, f^2(\nabla\chi)^2 $$

  • First integral is finite: due to density $\neq$ bulk value

  • Second is KE contribution from winding of vortex

$$ \nabla \chi = \frac{\ell}{r}\hat{\mathbf{e}}_\theta, $$

  • This contribution to the energy is logarithmically divergent.

$$ \Delta E = \text{const.} + \frac{\pi n \ell^2}{m}\log\left(\frac{L}{\xi}\right). $$

  • Far-reaching analogy between superfluid velocity fields of vortices and magnetostatics of current-carrying wires,
VorticesMagnetostatics
Vortex coresWires
Superfluid velocity $\mathbf{v}_s$Magnetic field, $\mathbf{B}$
Kinetic EnergyMagnetostatic Energy
  • Vortices with $\abs{\ell}>1$ generally unstable: break into multiple vortices of winding $\ell=\pm 1$
  • Like vortices repel each other, and can form spectacular vortex lattices, akin to crystals.

drawing

  • Vortices are one manifestation of superfluidity

Bogoliubov Theory

  • How can we improve Gross–Pitaevskii approximation?

  • What about excited states?

  • From now on… uniform condensate with no $V(\br)=0$

$$ H =\sum_\bk \epsilon(\bk)\adop_\bk\aop_\bk + \overbrace{\frac{U_0}{2V}\sum_{\bk_1+\bk_2=\bk_3+\bk_4} \adop_{\bk_1}\adop_{\bk_2}\aop_{\bk_3}\aop_{\bk_4}}^{\equiv H_\text{int}}, $$

  • $\epsilon(\bk)=\bk^2/2m$, and $V$ the volume

  • GP approximation to ground state is $\ket{\Psi_\text{GP}} = \frac{1}{\sqrt{N!}}\left(\adop_0\right)^N\ket{\text{VAC}}$

  • Act with $H_\text{int}$: only terms that contribute have $\bk_3=\bk_4=0$

$$ H_\text{int}\ket{\Psi_\text{GP}} = \frac{U_0}{2V}\sum_{\bk} \adop_{\bk}\adop_{-\bk}\aop_{0}\aop_{0}\ket{\Psi_\text{GP}}. $$

  • For a better wavefunction, need to add some $(\bk, -\bk)$ pairs!

Bogoliubov Hamiltonian

  • When interactions weak, expect true ground state close to $\ket{\Psi_\text{GP}}$

  • Most particles in zero momentum state, with few $(\bk, -\bk)$ pairs

  • Remember that

$$ \aop\ket{N} = \sqrt{N}\ket{N-1},\quad \adop\ket{N} = \sqrt{N+1}\ket{N+1}, $$

  • A term in $H_\text{int}$ with $\aop_0$ or $\adop_0$ is more important than one without. Arrange Hamiltonian by occurrences of $\aop_0$, $\adop_0$

$$ \begin{align*} H_\text{int} = \frac{U_0}{2V}\adop_0\adop_0\aop_0\aop_0 + \frac{U_0}{2V}\sum_{\bk\neq0}\left[\adop_{\bk}\adop_{-\bk}\aop_{0}\aop_{0} + \adop_{0}\adop_{0}\aop_{\bk}\aop_{-\bk}+4\adop_\bk\adop_0\aop_0\aop_\bk\right]\\\nonumber +\frac{U_0}{V}\sum_{\substack{\bk_1=\bk_2+\bk_3\\ \bk_{1,2,3}\neq 0}}\left[\adop_{\bk_3}\adop_{\bk_2}\aop_{\bk_1}\aop_0 +\adop_0\adop_{\bk_1}\aop_{\bk_2}\aop_{\bk_3}\right]+\frac{U_0}{2V}\sum_{\substack{\bk_1+\bk_2=\bk_3+\bk_4\\ \bk_{1,2,3,4}\neq 0}} \adop_{\bk_1}\adop_{\bk_2}\aop_{\bk_3}\aop_{\bk_4} \end{align*} $$

  • Gross–Pitaevskii approximation corresponds to the first term

  • We now keep second term, and neglect third and fourth

$$ \begin{align*} H_\text{pair} &= \sum_\bk \epsilon(\bk)\adop_\bk\aop_\bk +\frac{U_0}{2V}\adop_0\adop_0\aop_0\aop_0 \nonumber\\ &\quad+\frac{U_0}{2V}\sum_{\bk\neq0}\left[\adop_{\bk}\adop_{-\bk}\aop_{0}\aop_{0} + \adop_{0}\adop_{0}\aop_{\bk}\aop_{-\bk}+4\adop_\bk\adop_0\aop_0\aop_\bk\right] \end{align*} $$

$$ \begin{align*} H_\text{pair} &= \sum_\bk \epsilon(\bk)\adop_\bk\aop_\bk +\frac{U_0}{2V}\adop_0\adop_0\aop_0\aop_0 \nonumber\\ &\quad+\frac{U_0}{2V}\sum_{\bk\neq0}\left[\adop_{\bk}\adop_{-\bk}\aop_{0}\aop_{0} + \adop_{0}\adop_{0}\aop_{\bk}\aop_{-\bk}+4\adop_\bk\adop_0\aop_0\aop_\bk\right] \end{align*} $$

  • Rewrite second term using $\adop_0\aop_0 = N - N'$, with $N'\equiv \sum_{\bk\neq 0} N_\bk $
  • Then $\adop_0\adop_0\aop_0\aop_0 = N(N-1) - 2N'N_0+O(N_0^0 )$

$$ \begin{align*} H_\text{pair} = &N\epsilon(0)+\frac{U_0}{2V}N(N-1) +\sum_{\bk\neq 0}\left[\epsilon(\bk)-\epsilon(0)\right]\adop_\bk\aop_\bk\\ &+\frac{U_0}{2V}\sum_{\bk\neq 0}\left[\adop_{\bk}\adop_{-\bk}\aop_{0}\aop_{0} + \adop_{0}\adop_{0}\aop_{\bk}\aop_{-\bk}+2\adop_\bk\adop_0\aop_0\aop_\bk\right] \end{align*} $$

$$ \begin{align*} H_\text{pair} = &N\epsilon(0)+\frac{U_0}{2V}N(N-1) +\sum_{\bk\neq 0}\left[\epsilon(\bk)-\epsilon(0)\right]\adop_\bk\aop_\bk\\ &+\frac{U_0}{2V}\sum_{\bk\neq 0}\left[\adop_{\bk}\adop_{-\bk}\aop_{0}\aop_{0} + \adop_{0}\adop_{0}\aop_{\bk}\aop_{-\bk}+2\adop_\bk\adop_0\aop_0\aop_\bk\right] \end{align*} $$

Weird trick alert! Replace $\adop_0$, $\aop_0$ with $\sqrt{N}$, giving quadratic Hamiltonian (that we can solve)

  • Resulting Hamiltonian no longer conserves particle number
  • Let’s see why this is a good approximation.

  • Consider action of Hamiltonian on state of form $\ket{\Psi'}\otimes\ket{N_0}_0$

$$ \adop_\bk\aop_0\ket{\Psi'}\otimes\ket{N_0}_0 = \left(\adop_\bk \ket{\Psi'}\right)\otimes \aop_0\ket{N_0}_0 = \left(\adop_\bk \ket{\Psi'}\right)\otimes \sqrt{N_0}\ket{N_0-1}_0 $$ $$ \aop_\bk\adop_0\ket{\Psi'}\otimes\ket{N_0}_0 = \left(\aop_\bk \ket{\Psi'}\right)\otimes \adop_0\ket{N_0}_0 = \left(\aop_\bk \ket{\Psi'}\right)\otimes \sqrt{N_0+1}\ket{N_0+1}_0 $$

  • Ignore the difference between $N_0$ and $N_0+1$

  • If $N_0$ doesn’t fluctuate much matrix elements of $H_\text{pair}$ are approximately unchanged when replace operator with numbers

  • Result is Bogoliubov Hamiltonian $$ \begin{align*} H_\text{pair} &= \sum_\bk \epsilon(\bk)\adop_\bk\aop_\bk +\frac{U_0}{2V}N(N-1) \nonumber\\ &\quad+\frac{U_0n_0}{2}\sum_{\bk\neq0}\left[\adop_{\bk}\adop_{-\bk} + \aop_{\bk}\aop_{-\bk}+2\adop_\bk\aop_\bk\right] \end{align*} $$ $n_0 = N_0/V$ is density of particles in zero momentum state

  • Hamiltonian diagonalized by Bogoliubov transformation

Bogoliubov transformation

  • Suppose we have an operator

$$ h = \epsilon\left[\adop_1\aop_1+\adop_2\aop_2\right] + \delta\left[\adop_1\adop_2+\aop_1\aop_2\right] $$

  • Want to express $h$ in terms of some new bosons $\bop_{1,2}$ as $$ h = \Omega\left[\bdop_1\bop_1+\bdop_2\bop_2\right] +\text{const.} $$ (because then the spectrum is obvious).

$$ h = \begin{pmatrix} \adop_1 & \aop_2 \end{pmatrix} \begin{pmatrix} \epsilon & \delta \\ \delta & \epsilon \end{pmatrix} \begin{pmatrix} \aop_1 \\ \adop_2 \end{pmatrix}-\epsilon $$

  • Try to express new bosons linearly in terms of old $$ \begin{pmatrix} \bop_1 \\ \bdop_2 \end{pmatrix}= \Lambda \begin{pmatrix} \aop_1 \\ \adop_2 \end{pmatrix} $$ $\Lambda$ is some $2\times 2$ matrix. What conditions should it satisfy?

If $\bop_{1,2}$ satisfy usual commutation relations then $$ \Lambda^\dagger \sigma_3 \Lambda = \sigma_3 $$

  • Following parameterization is sufficient for our $h$

$$ \Lambda= \begin{pmatrix} \cosh\kappa & \sinh\kappa \\ \sinh\kappa & \cosh\kappa \end{pmatrix} $$

  • Notice differences from rotation matrix:

    1. Hyperbolic instead of trignometric functions
    2. No sign difference between the off digaonal elements.

$$ h = \begin{pmatrix} \adop_1 & \aop_2 \end{pmatrix} \begin{pmatrix} \epsilon & \delta \\ \delta & \epsilon \end{pmatrix} \begin{pmatrix} \aop_1 \\ \adop_2 \end{pmatrix}-\epsilon $$

$$ \begin{pmatrix} \bop_1 \\ \bdop_2 \end{pmatrix}= \begin{pmatrix} \cosh\kappa & \sinh\kappa \\ \sinh\kappa & \cosh\kappa \end{pmatrix} \begin{pmatrix} \aop_1 \\ \adop_2 \end{pmatrix} $$

$$ \begin{align*} \tanh 2\kappa = \frac{\delta}{\epsilon},\qquad \Omega = \sqrt{\epsilon^2-\delta^2\nonumber}\\ h = \Omega\left[\bdop_1\bop_1+\bdop_2\bop_2\right] + \Omega - \epsilon. \end{align*} $$ What happens if $\delta>\epsilon$?

What changes if $\begin{pmatrix} \epsilon & \delta \\ \delta & \epsilon \end{pmatrix}\longrightarrow \begin{pmatrix} \epsilon_1 & \delta \\ \delta & \epsilon_2 \end{pmatrix}$?

  • Apply to Bogoliubov Hamiltonian

$$ \begin{align*} H_\text{pair} &= \sum_\bk \epsilon(\bk)\adop_\bk\aop_\bk +\frac{U_0}{2V}N(N-1) \nonumber\\ &\quad+\frac{U_0n_0}{2}\sum_{\bk\neq0}\left[\adop_{\bk}\adop_{-\bk} + \aop_{\bk}\aop_{-\bk}+2\adop_\bk\aop_\bk\right] \end{align*} $$

$$ \begin{align*} \bop_\bp=\aop_\bp\cosh\kappa_\bp+\adop_{-\bp}\sinh\kappa_\bp\nonumber\\ \tanh2\kappa_\bp=\frac{n_0 U_0}{\epsilon(\bp)+n_0 U_0} \end{align*} $$

$$ H=E_0+\sum_{\bp\neq 0}\omega(\bp)\bdop_\bp \bop_\bp. $$

  • $\omega(\bp)$ is the Bogoliubov dispersion relation

$$ \omega(\bp) = \sqrt{\epsilon(\bp)\left(\epsilon(\bp)+2U_0n_0\right)} $$

  • Ground state energy is

$$ E_0=\frac{1}{2}nU_0 N+\sum_{\bp\neq 0}\frac{1}{2}\left[\omega(\bp)-\epsilon(\bp)-n_0U_0\right] $$

$$ E_0=\frac{1}{2}nU_0 N+\sum_{\bp\neq 0}\frac{1}{2}\left[\omega(\bp)-\epsilon(\bp)-n_0U_0\right] $$

  • Annoyingly, this integral is divergent at high momenta because

$$ \omega(\bp)\underset{|\bp| \to \infty}{\longrightarrow} \epsilon(\bp) + n_0 U_0 -\frac{(n_0 U_0)^2}{2\epsilon(\bp)} $$

  • We can cure the problem by writing

$$ \begin{align*} E_0&=\overbrace{\frac{1}{2}nU_0 N\left[1-\frac{1}{V}\sum_\bp \frac{U_0}{2\epsilon(\bp)}\right]}^{\text{1st and 2nd order PT}}\\ &+\overbrace{\sum_{\bp\neq 0}\frac{1} {2}\left[\omega(\bp)-\epsilon(\bp)-n_0U_0+ \frac{(n_0U_0)^2}{2\epsilon(\bp)}\right]}^{\text{finite}} \end{align*} $$

  • Term added and subtracted is contribution to the ground state energy in second order perturbation theory

  • See Appendix and Problem Set 2 for further discussion

The Ground State

  • So far, ground state defined by condition that it is a vacuum of Bogoliubov excitations

$$ \bop_\bk\ket{0}=\left(\cosh\kappa_\bk \aop_\bk+\sinh\kappa_\bk \adop_{-\bk}\right)\ket{0}=0. $$

  • A state (unnormalized) that satisfies this condition is

$$ \ket{0}=\prod_{\bk\neq 0} \exp\left(-\frac{1}{2}\tanh\kappa_\bk\adop_{\bk}\adop_{-\bk}\right)\ket{\Psi_\text{GP}} $$

Show this. If you’ve seen coherent states before, remember that the state $e^{\alpha \adop}\ket{\text{VAC}}$ is an eigenstate of $\aop$ with eigenvalue $\alpha$.

  • Recall that Fourier components of density operator are

$$ \rho_\bq = \sum_\bk \adop_{\bk-\bq}\aop_\bk $$

  • Most important terms: $\adop_{\bk-\bq}$ or $\aop_\bk$ acts on the condensate

$$ \rho_\bq \sim \sqrt{N}\left(\adop_{-\bq} + \aop_{\bq}\right) = \sqrt{N}e^{-\kappa_\bq} \left(\bdop_{-\bq} + \bop_{\bq}\right) $$

$$ e^{-\kappa_\bq} = \sqrt{\frac{\epsilon(\bq)}{\omega(\bq)}} $$

  • Density fluctuations in ground state $$ \braket{0|\rho_{-\bq}\rho_{\bq}|0} = N\frac{\epsilon(\bq)}{\omega(\bq)}\xrightarrow{\bq\to 0} \frac{N\abs{\bq}}{2mc}. $$ (Used low momentum form of Bogoliubov dispersion) $$ \omega(\bq)\xrightarrow{\bq\to 0} c\abs{\bq} $$ where $c = \sqrt{\frac{n_0U_0}{m}}$ is the speed of sound)

  • c.f Gross–Pitaevskii ground state (Poissonian fluctuations) $$ \braket{0|\rho_{-\bq}\rho_{\bq}|0} = N $$

Quantum Depletion

  • Finite fraction of particles have $\bp\neq 0$. Let’s find $\langle N_\bp \rangle$

  • $N_\bp=\adop_{\bp}\aop_{\bp}$ in terms of Bogoliubov quasiparticles

$$ \adop_{\bp}\aop_{\bp}=(\bdop_\bp\cosh\kappa_{\bp}-\bop_{-\bp}\sinh\kappa_{\bp})(\bop_\bp\cosh\kappa_{\bp}-\bdop_{-\bp}\sinh\kappa_{\bp}) $$

  • Then evaluate expectation

$$ \langle N_\bp\rangle=\langle \adop_{\bp}\aop_{\bp}\rangle = \sinh^2\kappa_{p}\xrightarrow{ \abs{\bp}\ll \xi^{-1}}\frac{mc_s}{2\abs{\bp}} $$

  • Radial density distribution $4\pi p^2 N_\bp$ peaked around $\xi^{-1}$
  • Fraction of atoms not in the condensate $$ \frac{1}{N}\sum_{\bp\neq 0} \langle N_\bp\rangle=\frac{8}{3\sqrt{\pi}}\sqrt{n a^3}, $$ Born approximation for scattering length $a=\frac{mU_0}{4\pi}$

  • Typical experimental conditions in experiments on ultracold atoms: depletion does not much exceed $0.01$, which justifies GP approximation

  • Liquid He$^{4}$ is an interacting Bose condensate, but depletion is much larger (condensate fraction $\sim 10\%$). Bogoliubov not accurate here

  • Applying a lattice can lead to total depletion and a quantum phase transition out of superfluid state