Quantum Scattering Theory

In this section, we review the basics of the quantum scattering theory.

Due to their diluteness, most properties of systems of ultracold atoms are related to two-body collisions. As an approximation, the interatomic interaction is described by a central potential 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 V(r)} . At large distances from each other, $V(r)=-C_{6}/r^{6}$ 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 gases we are dealing with are ultracold and ultradilute, which implies that both the de Broglie wavelength and the interparticle distance are much larger than the range of the interatomic potential $r_0$ (on the order of the van der Waals length 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 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 first observation is that the potential usually depends only on the relative position of the two particles (for example in a homogeneous system), therefore two-body problem can be decomposed into the motion of the relative coordinates 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 \vec{r}_1-\vec{r}_2} and the center of mass of the system.

In the following discussions, we ignore the motion of the center of mass. The Schrodinger equation for the relative coordinates can be further reduced to a one-particle problem in the center-of-mass frame of the colliding atoms (with reduced mass 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 m/2} , distance vector 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 \vec{r}} , and initial relative wave vector 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 \vec{k}} ) with

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_{\vec{k}}(\vec{r}) = v(r)\Psi_{\vec{k}}(\vec{r}) \quad\mbox{with } k^2 = \frac{m E}{\hbar^2} \quad \mbox{and } v(r) = \frac{m V(r)}{\hbar^2} \,. }

Far away from the scattering potential, the wave function $\Psi _{\vec {k}}({\vec {r}})$ is given by the sum of the incident plane wave 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 e^{i \vec{k} \cdot \vec{r}}} and an outgoing scattered wave:

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 \Psi_{\vec{k}}(\vec{r}) \sim e^{i \vec{k} \cdot \vec{r}} + f(\vec{k}',\vec{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(\vec{k}',\vec{k})} is the scattering amplitude for scattering an incident plane wave with wave vector 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 \vec{k}} into the direction ${\vec {k}}'=k\,{\vec {r}}/r$ (energy conservation implies 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 |\vec{k}|' = |\vec{k}|} ).

Partial Wave Decomposition

Partial Waves as the basis

There are two major reasons partial wave decomposition will be beneficial in this case:

• Our scatterer is at the origin of the coordinate. It is therefore natural to consider things on the Spherical basis 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 (r,\theta,\phi)} instead of Cartesian basis 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 (x,y,z)} . The partial waves are nothing but the angular part of the eigenstates of the free Schrodinger equation with the spherical coordinates.
• Since we assume a central potential, the scattered wave must be axially symmetric with respect to the incident wave vector 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 \vec{k}} , and it is another natural reason to perform the usual expansion into partial waves with angular momentum 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 l} .

In the free space, the eigenfunctions satisfy:

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 -\frac{\hbar^2}{2m}\nabla^2 \psi(r, \theta, \phi) = E\psi(r, \theta, \phi) }

We can therefore write down 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} in the spherical coordinates solve the schrodinger equation for $r,\theta ,\phi$. As an example, the plane wave with 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 k} along the 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 z} direction can be decomposed on this new basis as

$e^{ikz}=\sum _{\ell =0}^{\infty }(2\ell +1)i^{\ell }j_{\ell }(kr)P_{\ell }(\cos \theta )$.

Therefore the role of a scattering potential 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 V(r,\theta, \phi)} is clear :

• It changes the radial component of the wavefunction.
• It couples different partial waves and therefore different angular momentum components.

or both, depending on the form of the scattering potential.

Effective one-dimensional Schrodinger equation

In most cases, the potential we are dealing with is isotropic, that 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 V(r)} depends on 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 |r|} only. There are important several consequences originating from this assumption:

• The potential will not couple different partial waves. Angular momentum is then a good quantum number.
• Any states 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 \psi(\vec{r})} can be separated into radial part and the angular part as 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 \psi(\vec{r}) = \sum_{l,m}R_l(r)Y_l^m(cos(\theta))} . And the angular part is independent of the scattering potential.

The Schrodinger equation can, therefore, be decomposed into two decoupled parts where the radial components of the states satisfy:

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 [\partial_r^2+\frac{2}{r}\partial_r-\frac{l(l+1)}{r^2}-U(r)+k^2]R_l(r) = 0 }

This is the famous effective one-dimensional Schrodinger equation for the scattering problem. It governs the 'motion' of the radial part of the relative coordinates while the angular part is independent and will not change during the collisional process. The angular momentum of the collision acts as an effective potential 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 -\frac{l(l+1)}{r^2}} and therefore will modify the behavior of the scattering.

Be very careful about the fundamental assumption that V(r) is isotropic. It is not valid for many recent experimental developments including Dysprosium or dipolar molecules.

s-wave scattering

For ultracold collisions, we are interested in describing the scattering process at low momenta $k<<1/r0$. In the absence of resonance phenomena for 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 l\neq 0} , s-wave scattering 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 l=0} is dominant over all other partial waves (if allowed by the Pauli principle). So we can consider only the s-wave.

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 \approx f_s = \frac{1}{2ik}(e^{2i\delta_s}-1) = \frac{1}{k \cot \delta_s - i k} \,. }

where 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_s} and 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 \delta_s} are the 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 s} -wave scattering amplitude and phase shift.

Time-reversal symmetry implies that 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 k\cot\delta_s} is an even function of 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 k} . For low momenta $k\ll 1/r_{0}$, we may expand it to order 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 k^2} and define the scattering length:

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 a = -\lim_{k \ll 1/r_0} \frac{\tan \delta_s}{k}, \, }

and the effective range 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 r_{\rm eff}} of the scattering potential.

For example, for a spherical well potential of depth 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 V \equiv \hbar^2 K^2/m} and radius 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 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 r_{\rm eff} = R - \frac{1}{K^2 a} - \frac{1}{3} \frac{R^3}{a^2}} , which deviates from the potential range 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 R} only for 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 |a| \lesssim R} or very shallow wells. For van der Waals potentials, 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 r_{\rm eff}} is of order $r_{0}$.

We now have

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(k) = \frac{1}{-\frac{1}{a} + r_{\rm eff} \frac{k^2}{2} - ik} \,. }

In the limit 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 k|a| \ll 1} and 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 |r_{\rm eff}| \lesssim 1/k} , 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} becomes independent on momentum and equals 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 -a} . For 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 k|a| >> 1} and 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 r_{\rm eff} \ll 1/k} , the scattering amplitude 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 f = \frac{i}{k}} and the cross section for atom-atom collisions 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 \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.

Pseudo-potentials

When the de Broglie wavelength of the colliding particles is much larger than the fine details of the interatomic potential， we can create a pseudo potential which gives the correct 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 s} -wave scattering (the correct phase shift) but easier to handle mathematically. An candidate for such a "pseudo-potential" is a delta-potential 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 \delta(\vec{r})} .

Since the potential is short-range, it is intuitive to assume that the pseudopotential could be a delta function. With proper regulation, we have the form 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 V(\vec{r})\psi(\vec{r}) = V_0 \delta(\vec{r})\frac{\partial}{\partial r} (r \psi(\vec{r}))} . With

$V_{0}={\frac {4\pi \hbar ^{2}a}{m}}\,.$

It leads exactly to a scattering amplitude 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(k) = -a/(1+ i k a)} .

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

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 \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} }

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. Alternatively, one can introduce a "brute force energy cut-off 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 E_R = \hbar^2/m R^2} (momentum cut-off $\hbar /R$), taken to be much larger than typical scattering energies. Then we have

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 \frac{1}{f(k)} \approx -\frac{4\pi}{v_0} - \frac{2}{\pi} \frac{1}{R} + \frac{2 R}{\pi} k^2 - i k \,. }

This is now exactly of the form 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} with the scattering length

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 a = \frac{\pi}{2}\frac{R}{1+\frac{2\pi^2 R}{v_0}} \,. }

For any physical, given scattering length 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 a} we can thus find the correct strength 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 v_0} that reproduces the same 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 a} (provided that we choose 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 R \ll a} for positive 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 a} ). This approach implies an effective range 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 r_{\rm eff} = \frac{4}{\pi}R} that should be chosen much smaller than all relevant distances. Note that as a function of 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 v_0} , only one pole of $a$ and therefore only one bound state is obtained, at 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 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}

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 \frac{a}{R} = 1 - \frac{\tan(K R)}{K R} }

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.