Interacting Fermi Gases
Elastic collisions
Due to their diluteness, most properties of systems of ultracold atoms are related to two-body collisions. If we neglect the weak magnetic dipole interaction between the spins, the interatomic interaction is described by a central potential $V(r)$. At large distances from each other, atoms interact with the van der Waals-potential $-C_6/r^6$ as they experience each other's fluctuating electric dipole. At short distances on the order of a few Bohr radii $a_0$, the two electron clouds strongly repel each other, leading to ``hard-core repulsion. If the spins of the two valence electrons (we are considering alkali atoms) are in a triplet configuration, there is an additional repulsion due to Pauli's exclusion principle. Hence, the triplet potential $V_T(r)$ is shallower than the singlet one $V_S(r)$. The exact inclusion of the interatomic potential in the description of the gas would be extremely complicated. However, the gases we are dealing with are ultracold and ultradilute, which implies that both the de Broglie wavelength $\lambda_{dB}$ and the interparticle distance $n^{-1/3} \sim 5\,000-10\,000\, a_0$ are much larger than the range of the interatomic potential $r_0$ (on the order of the van der Waals length $r_0 \sim \left(\mu C_6 / \hbar^2\right) \sim 50\, a_0$ for \li). As a result, scattering processes never explore the fine details of the short-range scattering potential. The entire collision process can thus be described by a single quantity, the {\it scattering length}.
we quickly summarize some important results of scattering theory.
Reduced one-particle problem
The Schr\"odinger equation for the reduced one-particle problem in the center-of-mass frame of the colliding atoms (with reduced mass $m/2$, distance vector $r$, and initial relative wave vector $\vect{k}$) is
- Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\nabla^2 + k^2)\Psi_{\vect{k}}(\vect{r}) = v(r)\Psi_{\vect{k}}(\vect{r}) \quad\mbox{with } k^2 = \frac{m E}{\hbar^2} \quad \mbox{and } v(r) = \frac{m V(r)}{\hbar^2} \,. <math> Far away from the scattering potential, the wave function <math>\Psi_{\vect{k}}(\vect{r})} is given by the sum of the incident plane wave
Failed to parse (unknown function "\vect"): {\displaystyle e^{i \vect{k} \cdot \vect{r}}} and an outgoing scattered wave:
- Failed to parse (unknown function "\vect"): {\displaystyle \Psi_{\vect{k}}(\vect{r}) \sim e^{i \vect{k} \cdot \vect{r}} + f(\vect{k}',\vect{k}) \frac{e^{i k r}}{r} \,. }
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(\vect{k}',\vect{k})} is the scattering amplitude for scattering an incident plane wave with wave vector $\vect{k}$ into the direction $\vect{k}' = k\, \vect{r}/r$ (energy conservation implies $k' = k$).
Since we assume a central potential, the scattered wave must be axially symmetric with respect to the incident wave vector $\vect{k}$, and we can perform the usual expansion into partial waves with angular momentum $l$~\cite{land77qm}. For ultracold collisions, we are interested in describing the scattering process at {\it low momenta} $k \ll 1/r_0$, where $r_0$ is the range of the interatomic potential. In the absence of resonance phenomena for $l \ne 0$, {\it $s$-wave scattering} $\,l=0$ is dominant over all other partial waves (if allowed by the Pauli principle): \begin{equation} f \approx f_s = \frac{1}{2ik}(e^{2i\delta_s}-1) = \frac{1}{k \cot \delta_s - i k} \label{e:scattamp} \end{equation} where $f_s$ and $\delta_s$ are the $s$-wave scattering amplitude and phase shift, resp.~\cite{land77qm}. Time-reversal symmetry implies that $k\cot\delta_s$ is an even function of $k$. For low momenta $k \ll 1/r_0$, we may expand it to order $k^2$: \begin{equation} k \cot \delta_s \approx -\frac{1}{a} + r_{\rm eff} \frac{k^2}{2} \end{equation} which defines the {\it scattering length} \begin{equation} a = -\lim_{k \ll 1/r_0} \frac{\tan \delta_s}{k}, \end{equation} and the effective range $r_{\rm eff}$ of the scattering potential. For example, for a spherical well potential of depth $V \equiv \hbar^2 K^2/m$ and radius $R$, $r_{\rm eff} = R - \frac{1}{K^2 a} - \frac{1}{3} \frac{R^3}{a^2}$, which deviates from the potential range $R$ only for $|a| \lesssim R$ or very shallow wells. For van der Waals potentials, $r_{\rm eff}$ is of order $r_0$~\cite{flam99scatt}. With the help of $a$ and $r_{\rm eff}$, $f$ is written as~\cite{land77qm} \begin{equation} f(k) = \frac{1}{-\frac{1}{a} + r_{\rm eff} \frac{k^2}{2} - ik} \label{e:scattamplitude} \end{equation} In the limit $k|a| \ll 1$ and $|r_{\rm eff}| \lesssim 1/k$, $f$ becomes independent on momentum and equals $-a$. For $k|a| \gg 1$ and $r_{\rm eff} \ll 1/k$, the scattering amplitude is $f = \frac{i}{k}$ and the cross section for atom-atom collisions is $\sigma = \frac{4\pi}{k^2}$. This is the so-called unitarity limit. Such a divergence of $a$ occurs whenever a new bound state is supported by the potential (see section~\ref{s:squarewell}).
\subsection{Pseudo-potentials} \label{s:renormalization} If the de Broglie wavelength $\frac{2\pi}{k}$ of the colliding particles is much larger than the fine details of the interatomic potential, $1/k \gg r_0$, we can create a simpler description by modifying the potential in such a way that it is much easier to manipulate in the calculations, but still reproduces the correct $s$-wave scattering. An obvious candidate for such a ``pseudo-potential is a delta-potential $\delta(\vect{r})$.
However, there is a subtlety involved which we will address in the following. The goal is to find an expression for the scattering amplitude $f$ in terms of the potential $V(r) = \frac{\hbar^2 v(r)}{m}$, so that we can try out different pseudo-potentials, always ensuring that $f \rightarrow -a$ in the $s$-wave limit. For this, let us go back to the Schr\"odinger equation Eq.~\ref{e:schrodinger}. If we knew the solution to the following equation: \begin{equation}
(\nabla^2 + k^2)G_k(\vect{r}) = \delta(\vect{r})
\label{e:defGreen}
\end{equation} we could write an integral equation for the wave function $\Psi_{\vect{k}}(\vect{r})$ as follows: \begin{equation}
\Psi_{\vect{k}}(\vect{r}) = e^{i \vect{k}\cdot\vect{r}} + \int d^3 r' G_k(\vect{r}-\vect{r}')v(\vect{r}')\Psi_{\vect{k}}(\vect{r'})
\label{e:integralequation} \end{equation} This can be simply checked by inserting this implicit solution for $\Psi_{\vect{k}}$ into Eq.~\ref{e:schrodinger}. $G_k(\vect{r})$ can be easily obtained from the Fourier transform of Eq.~\ref{e:defGreen}, defining $G_k(\vect{p}) = \int d^3 r e^{-i \vect{p} \cdot \vect{r}} G_k(\vect{r})$: \begin{equation} (-p^2 + k^2)G_k(\vect{p}) = 1 \end{equation} The solution for $G_k(\vect{r})$ is \begin{equation} \label{e:Green} G_{k,+}(\vect{r}) = \int \frac{d^3 p}{(2\pi)^3} \frac{e^{i \vect{p} \cdot \vect{r}}}{k^2 - p^2 + i \eta} = -\frac{1}{4\pi}\frac{e^{ikr}}{r} \end{equation} where we have chosen (by adding the infinitesimal constant $i \eta$, with $\eta>0$ in the denominator) the solution that corresponds to an outgoing spherical wave. $G_{k,+}(\vect{r})$ is the {\it Green's function} of the scattering problem. Far away from the origin, $|\vect{r}-\vect{r'}| \sim r - \vect{r'}\cdot \vect{u}$, with the unit vector $\vect{u} = \vect{r}/r$, and \begin{equation} \Psi_{\vect{k}}(\vect{r}) \sim e^{i \vect{k} \cdot \vect{r}} - \frac{e^{i k r}}{4\pi r} \int d^3 r' e^{-i \vect{k}'\cdot \vect{r}'} v(\vect{r}')\Psi_{\vect{k}}(\vect{r'}) \end{equation} where $\vect{k}' = k \vect{u}$. With Eq.~\ref{e:psiasymptotic}, this invites the definition of the scattering amplitude via \begin{equation} f(\vect{k}',\vect{k}) = -\frac{1}{4\pi} \int d^3 r\, e^{-i \vect{k}'\cdot \vect{r}} v(\vect{r})\Psi_{\vect{k}}(\vect{r}) \end{equation} Inserting the exact formula for $\Psi_{\vect{k}}(\vect{r})$, Eq.~\ref{e:integralequation}, combined with Eq.~\ref{e:Green}, leads to an integral equation for the scattering amplitude \begin{eqnarray} f(\vect{k}',\vect{k}) = -\frac{v(\vect{k}'-\vect{k})}{4\pi} +\int \frac{d^3 q}{(2\pi)^3} \, \frac{v(\vect{k}'-\vect{q})f(\vect{q},\vect{k})}{k^2-q^2+i\eta} \label{e:lippmannschwinger} \end{eqnarray} where $v(\vect{k})$ is the Fourier transform of the potential $v(\vect{r})$ (which we suppose to exist). This is the Lippmann-Schwinger equation, an exact integral equation for $f$ in terms of the potential $v$, useful to perform a perturbation expansion. Note that it requires knowledge of $f(\vect{q},\vect{k})$ for $q^2 \ne k^2$ (``off the energy shell). However, the dominant contributions to the integral do come from wave vectors $\vect{q}$ such that $q^2 = k^2$. For low-energy $s$-wave scattering, $f(\vect{q},\vect{k}) \rightarrow f(k)$ then only depends on the magnitude of the wave vector $\vect{k}$. With this approximation, we can take $f(k)$ outside the integral. Taking the limit $k \ll 1/r_0$, dividing by $f(k)$ and by $v_0 \equiv v(\vect{0})$, we arrive at \begin{eqnarray} \frac{1}{f(k)} \approx -\frac{4\pi}{v_0} + \frac{4\pi}{v_0} \int \frac{d^3 q}{(2\pi)^3}\, \frac{v(-\vect{q})}{k^2-q^2+i\eta} \label{e:scattampintegral} \end{eqnarray} If we only keep the first order in $v$, we obtain the scattering length in {\it Born approximation}, $a = \frac{v_0}{4\pi}$. For a delta-potential $V(\vect{r}) = V_0\, \delta(\vect{r})$, we obtain to first order in $V_0$ \begin{equation} V_0 = \frac{4\pi \hbar^2 a}{m} \end{equation} However, already the second order term in the expansion of Eq.~\ref{e:scattampintegral} would not converge, as it involves the divergent integral $\int \frac{d^3 q}{(2\pi)^3} \frac{1}{q^2}$. The reason is that the Fourier transform of the $\delta$-potential does not fall off at large momenta. Any physical potential {\it does} fall off at some large momentum, so this is not a ``real problem. For example, the van-der-Waals potential varies on a characteristic length scale $r_0$ and will thus have a natural momentum cut-off $\hbar/r_0$. A proper regularization of contact interactions employs the pseudo-potential~\cite{huan87} $V(\vect{r})\psi(\vect{r}) = V_0 \delta(\vect{r})\frac{\partial}{\partial r} (r \psi(\vect{r}))$. It leads exactly to a scattering amplitude $f(k) = -a/(1+ i k a)$ if $V_0 = \frac{4\pi\hbar^2 a}{m}$.
Here we will work with a Fourier transform that is equal to a constant $V_0$ at all relevant momenta in the problem, but that falls off at very large momenta, to make the second order term converge. The exact form is not important. If we are to calculate physical quantities, we will replace $V_0$ in favor of the observable quantity $a$ using the formal prescription \begin{equation} \frac{1}{V_0} = \frac{m}{4\pi\hbar^2 a} - \frac{m}{\hbar^2}\int \frac{d^3 q}{(2\pi)^3} \frac{1}{q^2} \label{e:renormalize} \end{equation} We will always find that the diverging term is exactly balanced by another diverging integral in the final expressions, so this is a well-defined procedure~\cite{melo93,haus99}.
Alternatively, one can introduce a ``brute force energy cut-off $E_R = \hbar^2/m R^2$ (momentum cut-off $\hbar/R$), taken to be much larger than typical scattering energies. Eq.~\ref{e:scattampintegral} then gives \begin{eqnarray} \frac{1}{f(k)} \approx -\frac{4\pi}{v_0} - \frac{2}{\pi} \frac{1}{R} + \frac{2 R}{\pi} k^2 - i k \label{e:scattampcutoff} \end{eqnarray} This is now exactly of the form Eq.~\ref{e:scattamplitude} with the scattering length \begin{eqnarray} a = \frac{\pi}{2}\frac{R}{1+\frac{2\pi^2 R}{v_0}} \label{e:acutoff} \end{eqnarray} For any physical, given scattering length $a$ we can thus find the correct strength $v_0$ that reproduces the same $a$ (provided that we choose $R \ll a$ for positive $a$). This approach implies an effective range $r_{\rm eff} = \frac{4}{\pi}R$ that should be chosen much smaller than all relevant distances. Note that as a function of $v_0$, only one pole of $a$ and therefore only one bound state is obtained, at $v_0 = -2\pi^2 R$.
This prompts us to discuss the relation between Eq.~\ref{e:renormalize} and Eq.~\ref{e:lippmannschwinger}: The Lippmann-Schwinger equation is an exact reformulation of Schr\"odinger's equation for the scattering problem. One can, for example, exactly solve for the scattering amplitude in the case of a spherical well potential~\cite{bray71}. In particular, all bound states supported by the potential are recovered. However, to arrive at Eq.~\ref{e:renormalize}, one ignores the oscillatory behavior of both $v(\vect{q})$ and $f(\vect{q},\vect{k})$ and replaces them by $\vect{q}$-independent constants. As a result, Eq.~\ref{e:renormalize}, with a cut-off for the diverging integral at a wave vector $1/R$, only allows for {\it one} bound state to appear as the potential strength is increased (see Eq.~\ref{e:acutoff}).
We will analyze this approximation for a spherical well of depth $V$ and radius $R$. The true scattering length for a spherical well is given by~\cite{land77qm}
with $K^2 = m V/\hbar^2$. which one can write as \begin{eqnarray}
\frac{a}{R} &=& 1 - \frac{\prod_{n=1}^\infty (1 - \frac{K^2 R^2}{n^2 \pi^2})}{\prod_{n=1}^\infty(1 - \frac{4K^2 R^2}{(2n-1)^2\pi^2})} \quad \left.%
\begin{array}{ll}
\leftarrow \mbox{Zeros of }$a-R$ &\\
\leftarrow \mbox{Resonances of }a &\\
\end{array}% \right. \end{eqnarray} In contrast, Eq.~\ref{e:renormalize} with $V_0 = - \frac{4\pi}{3} V R^3$ and the ``brute force cut-off at $1/R$ gives \begin{equation}
\frac{a}{R} = \frac{K^2 R^2}{\frac{2}{\pi}K^2 R^2 - 3}
\end{equation} The sudden cut-off strips the scattering length of all but one zero (at $V = 0$) and of all but one resonance. For a shallow well that does not support a bound state, the scattering length still behaves correctly as $a = -\frac{1}{3} \frac{V}{E_R} R$. However, the sudden cut-off $v(\vect{q}) \approx {\rm const.}$ for $q \le \frac{1}{R}$ and 0 beyond results in a shifted critical well depth to accommodate the first bound state, $V = \frac{3\pi}{2} E_R$, differing from the exact result $V = \frac{\pi^2}{4} E_R$. This could be cured by adjusting the cut-off. But for increasing well depth, no new bound state is found and $a$ saturates at $\sim R$, contrary to the exact result.
At first, such an approximation might be unsettling, as the van-der-Waals potentials of the atoms we deal with contain many bound states. However, the gas is in the ultracold regime, where the de Broglie-wavelength is much larger than the range $r_0$ of the potential. The short-range physics, and whether the wave function has one or many nodes within $r_0$ (i.e. whether the potential supports one or many bound states), is not important. All that matters is the phase shift $\delta_s$ {\it modulo $2\pi$} that the atomic wave packets receive during a collision. We have seen that with a Fourier transform of the potential that is constant up to a momentum cut-off $\hbar/R$, we can reproduce any low-energy scattering behavior, which is described by the scattering length $a$. We can even realize a wide range of combinations of $a$ and the effective range $r_{\rm eff}$ to capture scattering at finite values of $k$. An exception is a situation where $0 < a \lesssim r_{\rm eff}$ or potentials that have a negative effective range. This can be cured by more sophisticated models.
