The Unitary Gas and its Symmetry Properties

Abstract

The physics of atomic quantum gases is currently taking advantage of a powerful tool, the possibility to fully adjust the interaction strength between atoms using a magnetically controlled Feshbach resonance. For fermions with two internal states, formally two opposite spin states $\uparrow$ and $\downarrow$, this allows to prepare long lived strongly interacting three-dimensional gases and to study the BEC–BCS crossover. Of particular interest along the BEC–BCS crossover is the so-called unitary gas, where the atomic interaction potential between the opposite spin states has virtually an infinite scattering length and a zero range. This unitary gas is the main subject of the present chapter: it has fascinating symmetry properties, from a simple scaling invariance, to a more subtle dynamical symmetry in an isotropic harmonic trap, which is linked to a separability of the N-body problem in hyperspherical coordinates. Other analytical results, valid over the whole BEC–BCS crossover, are presented, establishing a connection between three recently measured quantities, the tail of the momentum distribution, the short range part of the pair distribution function and the mean number of closed channel molecules. The chapter is organized as follows. In Sect. 5.1, we introduce useful concepts, and we present a simple definition and basic properties of the unitary gas, related to its scale invariance. In Sect. 5.2, we describe various models that may be used to describe the BEC–BCS crossover, and in particular the unitary gas, each model having its own advantage and shedding some particular light on the unitary gas properties: scale invariance and a virial theorem hold within the zero-range model, relations between the derivative of the energy with respect to the inverse scattering length and the short range pair correlations or the tail of the momentum distribution are easily derived using the lattice model, and the same derivative is immediately related to the number of molecules in the closed channel (recently measured at Rice) using the two-channel model. In Sect. 5.3, we describe the dynamical symmetry properties of the unitary gas in a harmonic trap, and we extract their physical consequences for many-body and few-body problems.

Appendices

Appendix 1: Effective Range in a Lattice Model

To calculate the effective range $r_e$ [defined by Eq. 5.5] for the lattice model of Sect. 5.2.1, it is convenient to perform in the expression (5.46) of the scattering amplitude an analytic continuation to purely imaginary incoming wavevectors $k_0$, setting $k_0=i q_0$ with $q_0$ real and positive. Eliminating $1/g_0$ thanks to Eq. 5.47 we obtain the useful expression:

$$ -\frac{1}{f_{k_0}} = \frac{1}{a} +4\pi \int_{{\fancyscript{D}}} \frac{d^3k}{(2\pi)^3} \left[\frac{1}{q_0^2+2m\varepsilon_{\mathbf{k}}/\hbar^2}-\frac{1}{2m\varepsilon_{\mathbf{k}}/\hbar^2}\right]. $$

(5.180)

We first treat the case of the parabolic dispersion relation Eq. 5.48. A direct expansion of Eq. 5.180 in powers of $q_0$ leads to an infrared divergence. The trick is to use the fact that the integral over $\fancyscript{D}$ in Eq. 5.180 can be written as the integral of the same integrand over the whole space minus the integral over the supplementary space $\mathbb{R}^3\setminus \fancyscript{D}$. The integral over the whole space may be performed exactly using

$$ \int_{{\mathbb{R}}^3} \frac{d^3k}{(2\pi)^3} \left[\frac{1}{q_0^2+k^2}-\frac{1}{k^2}\right] = -\frac{q_0}{4\pi}. $$

(5.181)

This leads to the transparent expression, where the term corresponding to ik in Eq. 5.3, and which is non-analytic in the energy E, is now singled out:

$$ -\frac{1}{f_{k_0}^{\rm parab}} = \frac{1}{a} - q_0 -4\pi \int_{{\mathbb{R}}^3\setminus{\fancyscript{D}}} \frac{d^3k}{(2\pi)^3} \left[\frac{1}{q_0^2+k^2}-\frac{1}{k^2}\right]. $$

(5.182)

This is now expandable in powers of $q_0^{2},$ leading to the effective range for the parabolic dispersion relation:

$$ r_e^{\rm parab} = \frac{1}{\pi^2} \int_{{\mathbb{R}}^3\setminus{\fancyscript{D}}} \frac{d^3k}{k^4}. $$

(5.183)

We now turn back to the general case. The trick is to consider the difference between the inverse scattering amplitudes of the general case and the parabolic case with a common value of the scattering length:

$$ \frac{1}{f_{k_0}^{\rm parab}} - \frac{1}{f_{k_0}} = 4\pi \int_{{\fancyscript{D}}} \frac{d^3k}{(2\pi)^3} \left[\frac{1}{q_0^2+2m\varepsilon_{\mathbf{k}}/\hbar^2}-\frac{1}{q_0^2+k^2} -\frac{1}{2m\varepsilon_{\mathbf{k}}/\hbar^2} + \frac{1}{k^2}\right]. $$

(5.184)

This is directly expandable to second order in $q_0,$ leading to:

$$ r_e - r_e^{\rm parab} = 8\pi \int_{{\fancyscript{D}}} \frac{d^3k}{(2\pi)^3} \left[ \frac{1}{k^4} - \left(\frac{\hbar^2}{2m\varepsilon_{\mathbf{k}}}\right)^{2} \right]. $$

(5.185)

The numerical evaluation of this integral for the Hubbard dispersion relation Eq. 5.59 leads to the Hubbard model effective range Eq. 5.60.

Finally, we specialize the general formula to the parabolic plus quartic form Eq. 5.61. Setting $\mathbf{k}=(\pi/b)\mathbf{q}$ and using Eq. 5.183, we obtain

$$ \frac{\pi^3 r_e^{\rm mix}}{b} = \int_{{\mathbb{R}}^3\setminus[-1,1]^3} \frac{d^3q}{q^4} + \int_{[-1,1]^3} \frac{d^3q}{q^4} \left[1-\frac{1}{(1-Cq^2)^2}\right]. $$

(5.186)

The trick is to split the cube $[-1,1]^3$ as the union of $B(0,1),$ the sphere of center 0 and unit radius, and of the set $X=[-1,1]^3\setminus B(0,1).$ One has also $\left(\mathbb{R}^3\setminus[-1,1]^3\right) \cup X=\mathbb{R}^3\setminus B(0,1)$ so that

$$ \frac{\pi^3 r_e^{\rm mix}}{b} = \int_{{\mathbb{R}}^3\setminus B(0,1)} \frac{d^3q}{q^4} +\int_{B(0,1)} \frac{d^3q}{q^4} \left[1-\frac{1}{(1-Cq^2)^2}\right] - \int_X \frac{d^3q}{q^4} \frac{1}{(1-Cq^2)^2}. $$

(5.187)

One then moves to spherical coordinates of axis z. The first two terms in the right hand side may be calculated exactly. In particular, one introduces a primitive of $q^{-2}(1-C q^2)^{-2},$ given by $C^{1/2} \Upphi(C^{1/2}q)$ with

$$ \Upphi(x) = \frac{x}{2(1-x^2)} + \frac{3}{2}\hbox{arctanh}\,x -\frac{1}{x}. $$

(5.188)

In the last term of Eq. 5.187 one integrates over the modulus q of q for a fixed direction $(\theta,\phi)$ where $\theta$ is the polar angle and $\phi$ the azimuthal angle. One then finds that q ranges from 1 to some maximal value $Q(\theta,\phi),$ and the integral over q provides the difference $\Upphi(C^{1/2}Q)-\Upphi(C^{1/2}).$ Remarkably, the term $-\Upphi(C^{1/2})$ cancels the contribution of the first two integrals in the right hand side of Eq. 5.187, so that

$$ \frac{r_e^{\rm mix}}{b} = -\frac{C^{1/2}}{\pi^3} \int_0^{2\pi} d\phi \int_{-1}^{1} du\,\Upphi[C^{1/2}Q(\theta,\phi)] $$

(5.189)

where as usual we have set $u=\cos\theta$. Using the symmetry under parity along each Cartesian axis, which adds a factor 8, and restricting to the face $q_x=1$ of the cube, which adds a factor 3, the expression of $Q(\theta,\phi)$ is readily obtained, leading to

$$ \frac{r_e^{\rm mix}}{b} = -\frac{24 C^{1/2}}{\pi^3} \int_0^{\pi/4} d\phi \int_0^{\displaystyle\frac{\hbox{cos}\phi}{\sqrt{1+\hbox{cos}^2\phi}}} du \,\Upphi\left(\frac{C^{1/2}}{\hbox{cos}\phi \sqrt{1-u^2}}\right). $$

(5.190)

In the limit $C\to 0$, $r_e^{\rm mix}\to r_e^{\rm parab}$, and Eq. 5.190 may be calculated analytically with $\Upphi(x)\,{\sim}\, -1/x$ and with an exchange of the order of integration: This leads to Eq. 5.58. For a general value of $C\in [0,1/3]$ we have calculated Eq. 5.190 numerically, and we have identified the magic value of C leading to a zero effective range, see Eq. 5.62. With the same technique, we can calculate the value of K appearing in Eq. 5.49 from the expression

$$ K = \frac{12}{\pi C^{1/2}} \int_0^{\pi/4} d\phi \int_0^{\displaystyle\frac{\hbox{cos}\phi}{\sqrt{1+\hbox{cos}^2\phi}}} {\it du}\,\,\hbox{arctanh}\,\frac{C^{1/2}}{\hbox{cos}\phi \sqrt{1-u^2}}. $$

(5.191)

Appendix 2: What is the Domain of a Hamiltonian?

Let us consider a Hamiltonian H represented by a differential operator also called H. A naive and practical definition of the domain D(H) of H is that it is the set of wavefunctions over which the action of the Hamiltonian is indeed represented by the considered differential operator. In other words, if a wavefunction $\psi_{\rm bad}$ does not belong to D(H), one should not calculate the action of H on $\psi_{\rm bad}$ directly using the differential operator H. If H if self-adjoint, one should rather expand $\psi_{\rm bad}$ on the Hilbert basis of eigenstates of H and calculate the action of H in this basis.

For example, for a single particle in one dimension in a box with infinite walls in $x=0$ and $x=1$, so that $0\,{\leq}\, x \,{\leq}\, 1$, one has the Hamiltonian

$$ H = -\frac{1}{2} \frac{d^2}{dx^2}, $$

(5.192)

with the boundary conditions on the wavefunction

$$ \psi(0)=\psi(1)=0 $$

(5.193)

representing the effect of the box. To be in the domain, a wavefunction $\psi(x)$ should be twice differentiable for $0<x<1$ and should obey the boundary conditions (5.193). An example of a wavefunction which is not in the domain is the constant wavefunction $\psi(x)=1.$ An example of wavefunction in the domain is

$$ \psi(x) = 30^{1/2} x (1-x). $$

(5.194)

If one is not careful, one may obtain wrong results. Let us calculate the mean energy and the second moment of the energy for $\psi$ given by (5.194). By repeated action of H onto $\psi$, and calculation of elementary integrals, one obtains

$$ \langle H\rangle_\psi = 5 $$

(5.195)

$$ \langle H^2\rangle_\psi = 0 ?! $$

(5.196)

Eq. 5.195 is correct, but Eq. 5.196 is wrong (it would lead to a negative variance of the energy) because $H\psi$ is not in D(H) and the subsequent illicit action of H as the differential operator (5.192) gives zero.

How to calculate the right value of $\langle H^2\rangle_\psi$? One introduces the orthonormal Hilbert basis of eigenstates of H,

$$ \psi_n(x) = 2^{1/2} \hbox{sin} [\pi(n+1)x], \quad n\in {\mathbb{N}}, $$

(5.197)

with the eigenenergy $\varepsilon_n=\frac{\pi^2}{2} (n+1)^2$. Then $\psi$ of Eq. 5.194 may be expanded as $\sum_n c_n \psi_n(x)$, and the $k{\rm th}$ moment of the energy may be defined as

$$ \langle H^k\rangle_\psi = \sum_{n\in {\mathbb{N}}} (\varepsilon_n)^k |c_n|^2. $$

(5.198)

Since $c_n=4\sqrt{15}[1+(-1)^n]/[\pi(n+1)]^3,$ one recovers $\langle H\rangle_\psi =5$ and one obtains the correct value $\langle H^2\rangle_\psi=30,$ that leads to a positive energy variance as it should be. Also $\langle H^k\rangle_\psi=+\infty$ for $k\,{\geq}\, 3.$

The trick of expanding $\psi$ in the eigenbasis of H is thus quite powerful, it allows to define the action of H on any wavefunction $\psi$ in the Hilbert space (not belonging to the domain). It may be applied of course only if H is self-adjoint, as it is the case in our simple example.

Appendix 3: Separability and Jacobi Coordinates for Arbitrary Masses

We here consider $N\,{\geq}\, 2$ harmonically trapped particles interacting in the unitary limit, with possibly different masses $m_i$ but with the same isotropic angular oscillation frequency $\omega.$ The Hamiltonian reads

$$ H = \sum_{i=1}^{N} \left[ -\frac{\hbar^2}{2m_i} \Updelta_{{\mathbf{r}}_i} + \frac{1}{2} m_i \omega^2 r_i^2 \right] $$

(5.199)

and the unitary interaction is described by the Wigner-Bethe-Peierls contact conditions on the N-body wavefunction: For all pairs of particles (i, j), in the limit $r_{ij}=|\mathbf{r}_i-\mathbf{r}_j|\to 0$ with a fixed value of the centroid of the particles i and j, $\mathbf{R}_{ij}\equiv (m_i\mathbf{r}_i+m_j \mathbf{r}_j)/(m_i+m_j),$ that differs from the positions $\mathbf{r}_k$ of the other particles, $k\neq i,j,$ there exists a function $A_{ij}$ such that

$$ \psi({\mathbf{r}}_1,\ldots,{\mathbf{r}}_N) = \frac{A_{ij}({\mathbf{R}}_{ij};({\mathbf{r}}_k)_{k\neq i,j})}{r_{ij}} + O(r_{ij}). $$

(5.200)

As is well known and as we will explain below, the internal Hamiltonian $H_{\rm internal}= H - H_{\rm CM}$, where $H_{\rm CM}= -\frac{\hbar^2}{2M}\Updelta_\mathbf{C} + \frac{1}{2} M \omega^2 C^2,$ takes the form

$$ H_{\rm internal} = \sum_{i=1}^{N-1} \left[ -\frac{\hbar^2}{2\bar{m}} \Updelta_{{\mathbf{u}}_i} + \frac{1}{2} \bar{m} \omega^2 u_i^2 \right] $$

(5.201)

in suitably defined Jacobi coordinates [see Eqs. 5.205, 5.214]. Here $\mathbf{C}=\sum_{i=1}^{N} m_i\mathbf{r}_i/M$ is the center of mass position, $M= \sum_{i=1}^{N} m_i$ is the total mass, and $\bar{m}$ is some arbitray mass reference, for example the mean mass $M/N$. Then it is straightforward to express Eq. 5.201 in hyperspherical coordinates, the vector $(\mathbf{u}_1,\ldots,\mathbf{u}_{N-1})$ with $3N-3$ coordinates being expressed in terms of its modulus R and a set of $3N-4$ hyperangles $\Upomega,$ so that

$$ H_{\rm internal} = -\frac{\hbar^2}{2\bar{m}} \left[\partial_R^{2} + \frac{3N-4}{R} \partial_R +\frac{1}{R^2} \Updelta_{\Upomega} \right] +\frac{1}{2} \bar{m} \omega^{2} R^{2} $$

(5.202)

where $\Updelta_{\Upomega}$ is the Laplacian over the unit sphere of dimension $3N-4.$ As we shall see, the expression for the hyperradius is simply

$$ R^{2} \equiv \sum_{i=1}^{N-1} u_i^{2} = \frac{1}{\bar{m}} \sum_{i=1}^{N} m_{i} ({\mathbf{r}}_i - {\mathbf{C}})^{2}. $$

(5.203)

This form of the Hamiltonian is then useful to show the separability of Schrödinger’s equation for the unitary gas in hyperspherical coordinates [55, 89] for $N\,{\geq}\, 3$ and arbitrary masses. The separability Eq. 5.161 that was described for simplicity in the case of equal mass particles in Sect. 5.3.3 indeed still holds in the case of different masses, if the Wigner-Bethe-Peierls model defines a self-adjoint Hamiltonian. ²¹ We recall here the various arguments. First, for zero energy free space eigenstates, the form Eq. 5.159 is expected from scale invariance, if the Hamiltonian is self-adjoint [89]. Second, the form Eq. 5.161 for the general case, including non-zero energy and an isotropic harmonic trap, is expected because (i) the Hamiltonian (5.199), after separation of the center of mass, has the separable form (5.202) in hyperspherical coordinates, and (ii) Eq. 5.161 obeys the Wigner-Bethe-Peierls contact conditions if Eq. 5.159 does. This point (ii) results from the fact that the Wigner-Bethe-Peierls conditions are imposed, for each pair of particles (i,j), for $r_{ij}\to 0$ with a fixed value of $\mathbf{R}_{ij}$ that differs from the positions $\mathbf{r}_k$ of the other particles, $k\neq i,j.$ Using $\mathbf{r}_i=\mathbf{R}_{ij}+[m_j/(m_i+m_j)] \mathbf{r}_{ij}$ and $\mathbf{r}_j=\mathbf{R}_{ij}-[m_i/(m_i+m_j)] \mathbf{r}_{ij}$, with $\mathbf{r}_{ij}\equiv \mathbf{r}_i-\mathbf{r}_{j}$, we indeed find that

$$ \bar{m} R^2 = \frac{m_i m_j}{m_i+m_j} r_{ij}^2 + (m_i+m_j) ({\mathbf{R}}_{ij}-{\mathbf{C}})^2 + \sum_{k\neq i,j} m_k ({\mathbf{r}}_k-{\mathbf{C}})^2. $$

(5.204)

For $N\,{\geq}\, 3$, we see that $\lim_{r_{ij}\to 0} R^2 > 0,$ so that R varies only to second order in $r_{ij}$ in that limit. Provided that the function F(R) in Eq. 5.161 has no singularity at non-zero R, the Wigner-Bethe-Peierls contact conditions are preserved [similarly to the argument Eq. 5.115]. Third, bosonic or fermionic exchange symmetries imposed on the N-body wavefunction cannot break the separability in hyperspherical coordinates: Exchanging the positions of particles of same mass does not change the value of the hyperradius R, it only affects the hyperangles and thus the eigenvalues $[(3N-5)/2]^2-s^{2}$ of the Laplacian on the unit sphere.

To derive the form Eq. 5.201 of the internal Hamiltonian, we introduce the usual Jacobi coordinates given for example in [102]:

$$ {\mathbf{y}}_i \equiv {\mathbf{r}}_i -\frac{\sum_{j=i+1}^{N} m_j {\mathbf{r}}_j}{\sum_{j=i+1}^{N} m_j} \quad \hbox{for } 1\,{\leq}\, i \,{\leq}\, N-1. $$

(5.205)

We note that $\mathbf{y}_i$ simply gives the relative coordinates of particle i with respect to the center of mass of the particles from $i+1$ to N. To simplify notations, we also set $\mathbf{y}_N \equiv \mathbf{C}$. Here we derive Eq. 5.201 in a pedestrian way. Note that a more elegant derivation by recursion is given in page 63 of [55]. In compact form, the Jacobi change of variables corresponds to setting $\mathbf{y}_i = \sum_{j=1}^{N} M_{ij} \mathbf{r}_j$ for $1\,{\leq}\, i \,{\leq}\, N$, where the non-symmetric matrix M is such that:

• In the case $1\,{\leq}\, i<N$, one has: $M_{ij}=0$ for $1\,{\leq}\, j < i$, $M_{ij}=1$ for $j=i$, and $M_{ij}=-m_j/\left(\sum_{k=i+1}^{N} m_k\right)$ for $i< j \,{\leq}\, N$.

• $M_{Nj} = m_j/\left(\sum_{k=1}^{N} m_k\right)$ for $1 \,{\leq}\, j \,{\leq}\, N$.

From the formula giving the derivative of a composite function, the kinetic energy operator writes

$$ H_{\rm kin} \equiv \sum_{i=1}^{N} -\frac{\hbar^2}{2m_i} \Updelta_{{\mathbf{r}}_i} = -\frac{\hbar^2}{2} \sum_{j=1}^{N} \sum_{k=1}^{N} S_{jk} {\hbox{grad}}_{{\mathbf{y}}_j} \cdot {\hbox{grad}}_{{\mathbf{y}}_{k}}, $$

(5.206)

where the symmetric matrix S is defined as $S_{jk} = \sum_{i=1}^{N} M_{ji} M_{ki}/m_i.$ The explicit calculation of the matrix elements $S_{jk}$ is quite simple. Taking advantage of the fact that S is symmetric, one has to distinguish three cases, (i) $1\,{\leq}\, j,k\,{\leq}\, N-1$, with $j=k$ and $j<k$ as subcases, (ii) $j=k=N,$ and (iii) $j<N, k=N.$ One then finds that $S$ is purely diagonal, with $S_{ii}=1/\mu_i$ for $1\,{\leq}\, i\,{\leq}\, N-1$ and $S_{NN}=1/M$. Here $\mu_i$ is the reduced mass for the particle i and for a fictitious particle of mass equal to the sum of the masses of the particles from $i+1$ to N:

$$ \frac{1}{\mu_i} = \frac{1}{m_i} + \frac{1}{\sum_{j=i+1}^{N} m_j} \quad \hbox{for} \ \ 1\,{\leq}\, i\,{\leq}\, N-1. $$

(5.207)

This results in the following form

$$ H_{\rm kin} = -\frac{\hbar^2}{2M} \Updelta_{{\mathbf{C}}} - \sum_{i=1}^{N-1} \frac{\hbar^2}{2\mu_i} \Updelta_{{\mathbf{y}}_i}. $$

(5.208)

The next step is to consider the trapping potential energy term. Inspired by Eq. 5.208 one may consider the guess

$$ H_{\rm trap} \equiv \sum_{i=1}^{N} \frac{1}{2} m_i \omega^2 r_i^2 {\mathop = \limits^{?}} \frac{1}{2} M\omega^2 C^2 + \sum_{i=1}^{N-1} \frac{1}{2} \mu_i \omega^2 y_i^2. $$

(5.209)

Replacing each $\mathbf{y}_i$ by their expression in the guess gives

$$ M C^{2} + \sum_{i=1}^{N-1} \mu_i y_i^2 = \sum_{j=1}^{N} \sum_{k=1}^{N} Q_{jk} {\mathbf{r}}_j \cdot {\mathbf{r}}_k $$

(5.210)

where Q is uniquely defined once it is imposed to be a symmetric matrix. Setting $M_i=\sum_{j=i+1}^{N} m_j$ for $0\,{\leq}\, i \,{\leq}\, N-1,$ and $M_N=0,$ we find for the off-diagonal matrix elements

$$ Q_{jk} = -\frac{\mu_{{\rm min}(j,k)}m_{{\rm max}(j,k)}}{M_{{\rm min}(j,k)}} + \frac{m_j m_k}{M} +m_j m_k \sum_{i=1}^{{\rm min}(j,k)-1} \frac{\mu_i}{M_i^2} $$

(5.211)

where $1\,{\leq}\, j,k\,{\leq}\, N, \hbox{min}(j,k)$ and $\hbox{max}(j,k)$ respectively stand for the smallest and for the largest of the two indices j and k. The key relation is then that

$$ \frac{\mu_i}{M_i^2} = \frac{1}{M_i} - \frac{1}{m_i+M_i} =\frac{1}{M_i} - \frac{1}{M_{i-1}} $$

(5.212)

since $M_i + m_i = M_{i-1}$ for $1\,{\leq}\, i\,{\leq}\, N$. This allows to calculate the sum over $i$ of $\mu_i/M_i^2$, as all except the border terms compensate by pairs. E.g. for $j < k$:

$$ \sum_{i=1}^{j-1} \frac{\mu_i}{M_i^2} = \frac{1}{M_{j-1}}- \frac{1}{M} $$

(5.213)

since $M_0=M$. One then finds that the off-diagonal elements of the matrix Q vanish. The diagonal elements of Q may be calculated using the same tricks (5.212, 5.213), one finds $Q_{ii}= m_i$ for $1\,{\leq}\, i\,{\leq}\, N.$ As a consequence, the guess was correct and the question mark can be removed from Eq. 5.209.

The last step to obtain Eq. 5.201 is to appropriately rescale the usual Jacobi coordinates, setting

$$ {\mathbf{u}}_i \equiv (\mu_i/\bar{m})^{1/2} {\mathbf{y}}_i $$

(5.214)

where $\bar{m}$ is an arbitrarily chosen mass. A useful identity is the expression for the square of the hyperradius, Eq. 5.203. Starting from the definition [first identity in Eq. 5.203] we see that $R^2=\sum_{i=1}^{N-1} \frac{\mu_i}{\bar{m}} y_i ^2.$ Then the second identity in Eq. 5.203 results from the fact that the guess in Eq. 5.209 is correct.

Appendix 4: Hydrodynamic Equations

The hydrodynamic equations for a normal compressible viscous fluid are (see [98], ²² or Sect. 15 and Sect. 49 in [101]):

• the continuity equation

  $$ \frac{\partial \rho}{\partial t}+\varvec{\nabla}\cdot(\rho{\mathbf{v}}) =0, $$

  (5.215)

• the equation of motion

  $$ \begin{aligned} m \rho \left(\frac{\partial {\it v}_{i} }{\partial t}+{\mathbf{v}}\cdot{\varvec\nabla} {\it v}_{i}\right)&=-\frac{\partial p}{\partial x_i}-\rho\frac{\partial U}{\partial x_i}+\sum_k\frac{\partial}{\partial x_k}\left[\eta\left(\frac{\partial{\it v}_{i}}{\partial x_k}+\frac{\partial{\it v}_{k}}{\partial x_i}-\frac{2}{3}\delta_{ik}\varvec{\nabla}\cdot{\mathbf{v}}\right)\right] \\ &\enskip\quad+\frac{\partial}{\partial x_i}\left(\zeta\,\varvec{\nabla}\cdot{\mathbf{v}}\right) \end{aligned} $$

  (5.216)

  where m is the atomic mass, $\eta$ is the shear viscosity, $\zeta$ is the bulk viscosity, and the pressure $p(\mathbf{r},t)$ [as well as the temperature $T(\mathbf{r},t)$ appearing in the next equation] is as always expressible in terms of $\rho(\mathbf{r},t)$ and $s(\mathbf{r},t)$ via the equation of state, ²³

• the entropy-production equation

  $$ \rho T \left(\frac{\partial s}{\partial t}+{\mathbf{v}}\cdot\varvec{\nabla}s\right) = \varvec{\nabla}\cdot(\kappa\varvec{\nabla}T)+\frac{\eta}{2}\sum_{i,k}\left( \frac{\partial {\it v}_{i}}{\partial x_k}+\frac{\partial{\it v}_{k}}{\partial x_i}-\frac{2}{3}\delta_{ik}\varvec{\nabla}\cdot{\mathbf{v}} \right)^{2} +\zeta\|\varvec{\nabla}\cdot{\mathbf{v}}\|^2 $$

  (5.217)

  where $\kappa$ is the thermal conductivity.

Appendix 5: Alternative Derivation of the Vanishing Bulk Viscosity

Consider the particular case of a unitary gas initially prepared at thermal equilibrium in an isotropic harmonic trap at a temperature T above the critical temperature. When the harmonic trap becomes time dependent, $U(\mathbf{r},t)=\frac{1}{2} m \omega^2(t) r^2$, each many-body eigenstate of the statistical mixture evolves under the combination Eq. 5.109 of a time dependent gauge transform and a time dependent scaling transform of scaling factor $\lambda(t)$. The effect of the gauge transform is to shift the momentum operator $\mathbf{p}_i$ of each particle i by the spatially slowly varying operator $m\mathbf{r}_i \dot{\lambda}/\lambda$. In the hydrodynamic framework, this is fully included by the velocity field Eq. 5.178. ²⁴ Using the macroscopic consequences of a spatial scaling Eqs. 5.19–5.22, one a priori obtains a time dependent solution of the hydrodynamic equations:

$$ T({\mathbf{r}},t) = T(t=0)/\lambda^2(t) $$

(5.218)

$$ \rho({\mathbf{r}},t) = \rho({\mathbf{r}}/\lambda,0)/\lambda^3(t) $$

(5.219)

$$ s({\mathbf{r}},t) = s({\mathbf{r}}/\lambda,0) $$

(5.220)

$$ p({\mathbf{r}},t) = p({\mathbf{r}}/\lambda,0)/\lambda^5(t) $$

(5.221)

$$ {\it v}_{i}({\mathbf{r}},t) = x_i \dot{\lambda}(t)/\lambda(t). $$

(5.222)

One then may a posteriori check that Eq. 5.215 is inconditionally satisfied, and that Eq. 5.217 is satisfied if $\zeta\equiv 0.$ Setting $\zeta\equiv 0$ in Eq. 5.216, and using the hydrostatic condition ${\nabla}p=-\rho{\nabla} U$ at time $t=0,$ one finds that Eq. 5.216 holds provided that $\lambda(t)$ solves Eq. 5.111 as it should be.

Appendix 6: ${\boldsymbol{n}}$-Body Resonances

Usually in quantum mechanics one takes the boundary condition that the wavefunction is bounded when two particles approach each other; in contrast, the Wigner-Bethe-Peierls boundary condition (5.75) expresses the existence of a two-body resonance. If the interaction potential is fine-tuned not only to be close to a two-body resonance (i.e. to have $|a|\gg b$) but also to be close to a n-body resonance (meaning that a real or virtual n-body bound state consisting of $n_\uparrow$ particles of spin $\uparrow$ and $n_\downarrow$ particles of spin $\downarrow$ is close to threshold), then one similarly expects that, in the zero-range limit, the interaction potential can be replaced by the Wigner-Bethe-Peierls boundary condition, together with an additional boundary condition in the limit where any subset of $n_\uparrow$ particles of spin $\uparrow$ and $n_\downarrow$ particles of spin $\downarrow$ particles approach each other. Using the notations of Sect. 5.3.6, this additional boundary condition reads [55, 70, 89, 103]:

$$ \psi({\mathbf{r}}_1,\ldots,{\mathbf{r}}_N) = \left(R_J^{-s}-\frac{\varepsilon}{l^{2s}} R_J^s\right)\,R_J^{-\frac{3 n-5}{2}}\,\phi(\Upomega_J) \,A_J({\mathbf{C}}_j,{{\fancyscript{R}}}_J) +o(R_J^{\it v}) $$

(5.223)

where $s=s_{\rm min}(n_\uparrow,n_\downarrow),$ while $l>0$ and $\varepsilon=\pm1$ are parameters of the model playing a role analogous to the absolute value and the sign of the two-body scattering length. This approach is only possible if the wavefunction remains square integrable, i.e. if $0\le s<1$, which we assume in what follows. This condition is satisfied e.g. for $n_\uparrow=2, n_\downarrow=1$ for a mass ratio $m_\uparrow/m_\downarrow \in \, [8.62\ldots;13.6\ldots]$[5]. Moreover we are assuming for simplicity that $s\neq0.$

Let us now consider the particular case where the two-body scattering length is infinite, and the external potential is either harmonic isotropic, or absent. Then the separability in internal hyperspherical coordinates of Sect. 5.3.3 still holds for $n=N$. Indeed, Eq. 5.223 then translates into the boundary condition on the hyperradial wavefunction

$$ \exists A\in{\mathbb{R}}/\ F(R){\mathop = \limits_{{R \to 0}}} A \cdot \left( R^{-s} - \frac{\varepsilon}{l^{2s}} R^s \right) + O\left(R^{s+2}\right) $$

(5.224)

and does not affect the hyperangular problem. Consequently [55],

• For the n-body bound state, which exists if $\varepsilon=+1$:

  $$ E=-\frac{2\hbar^2}{m\,l^2} \left[\frac{\Upgamma(1+s)}{\Upgamma(1-s)}\right]^{\frac{1}{s}}, $$

  (5.225)
  $$ F(R)=K_s\left(R \sqrt{-2 E \frac{m}{\hbar^2}}\right). $$

  (5.226)

• For the eigenstates in a trap:

  $$ E { \hbox{solves:} } -\varepsilon\cdot\left(\frac{\hbar}{m\omega\,l^2}\right)^s = \frac{\Upgamma\left( \frac{1+s-E/(\hbar\omega)}{2} \right) \Upgamma(-s)} {\Upgamma\left( \frac{1-s-E/(\hbar\omega)}{2} \right) \Upgamma(s)}, $$

  (5.227)
  $$ F(R)=\frac{1}{R}\,W_{\frac{E}{2\hbar\omega},\frac{s}{2}}\left(R^2\frac{m\omega}{\hbar}\right). $$

  (5.228)

In particular, for $l=\infty$, we are exactly at the n-body resonance, since the energy of the n-body bound state vanishes. The spectrum in a trap then is $E=(-s+1+2 q)\hbar\omega$ with $q\in\mathbb{N}.$

Note that, most often, $s\,{\geq}\,1,$ in which case one would have to use an approach similar to the one developped by Pricoupenko for the case of two-body resonances in non-zero angular momentum channels, and to introduce a modified scalar product [22, 104].