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 $V(r)$. 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 $r_{0}\sim \left(\mu C_{6}/\hbar ^{2}\right)\sim 50\,a_{0}$ 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 ${\vec {r}}_{1}-{\vec {r}}_{2}$ 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 $m/2$, distance vector ${\vec {r}}$, and initial relative wave vector ${\vec {k}}$) with

$(\nabla ^{2}+k^{2})\Psi _{\vec {k}}({\vec {r}})=v(r)\Psi _{\vec {k}}({\vec {r}})\quad {\mbox{with }}k^{2}={\frac {mE}{\hbar ^{2}}}\quad {\mbox{and }}v(r)={\frac {mV(r)}{\hbar ^{2}}}\,.$

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

$\Psi _{\vec {k}}({\vec {r}})\sim e^{i{\vec {k}}\cdot {\vec {r}}}+f({\vec {k}}',{\vec {k}}){\frac {e^{ikr}}{r}}\,.$

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

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 $(r,\theta ,\phi )$ instead of Cartesian basis $(x,y,z)$. 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 ${\vec {k}}$, and it is another natural reason to perform the usual expansion into partial waves with angular momentum $l$.

In the free space, the eigenfunctions satisfy:

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

We can therefore write down $\nabla ^{2}$ in the spherical coordinates solve the schrodinger equation for $r,\theta ,\phi$. As an example, the plane wave with $k$ along the $z$ 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 $V(r,\theta ,\phi )$ 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 $V(r)$ depends on $|r|$ 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 $\psi ({\vec {r}})$ can be separated into radial part and the angular part as $\psi ({\vec {r}})=\sum _{l,m}R_{l}(r)Y_{l}^{m}(cos(\theta ))$. 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:

$[\partial _{r}^{2}+{\frac {2}{r}}\partial _{r}-{\frac {l(l+1)}{r^{2}}}-V(r)+k^{2}]R_{l}(r)=0$

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 $-{\frac {l(l+1)}{r^{2}}}$ 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 $l\neq 0$, s-wave scattering $l=0$ is dominant over all other partial waves (if allowed by the Pauli principle). So we can consider only the s-wave. The corresponding one-dimensional Schrodinger equation takes the form

$[\partial _{r}^{2}+{\frac {2}{r}}\partial _{r}-V(r)+k^{2}]R_{l}(r)=0$

It is possible to directly solve it given boundary conditions and the general solution can be found on standard textbook. Comparing it with the scattering amplitude we defined, we have

$f\approx f_{s}={\frac {1}{2ik}}(e^{2i\delta _{s}}-1)={\frac {1}{k\cot \delta _{s}-ik}}\,.$

where $f_{s}$ and $\delta _{s}$ are the $s$-wave scattering amplitude and phase shift. The effect of a short range scattering potentail is nothing but a extra phase shift acting on the corresponding partial wave.

We take further look into it. 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 $k^{2}$ and define the scattering length:

$a=-\lim _{k\ll 1/r_{0}}{\frac {\tan \delta _{s}}{k}},\,$

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 $r_{0}$.

We now have

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

In the limit $k|a|\ll 1$ and $|r_{\rm {eff}}|\lesssim 1/k$, $f$ becomes independent on momentum and equals $-a$. For $k|a|>>1$ 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.

Pseudo-potentials

The scattering potential is usually hard to handle mathematically. We can think of a 'pseudopotential' which gives the correct "long range" physics (the correct $s$-wave scattering) but easier to handle mathematically, when the de Broglie wavelength of the colliding particles is much larger than the fine details of the interatomic potential. A candidate for such a "pseudo-potential" is a delta-potential $\delta ({\vec {r}})$.

With proper regulation, we have the form $V({\vec {r}})\psi ({\vec {r}})=V_{0}\delta ({\vec {r}}){\frac {\partial }{\partial r}}(r\psi ({\vec {r}}))$. With

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

It leads exactly to a scattering amplitude $f(k)=-a/(1+ika)$.

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

${\frac {1}{V_{0}}}={\frac {m}{4\pi \hbar ^{2}a}}-{\frac {m}{\hbar ^{2}}}\int {\frac {d^{3}q}{(2\pi )^{3}}}{\frac {1}{q^{2}}}$

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 $E_{R}=\hbar ^{2}/mR^{2}$ (momentum cut-off $\hbar /R$), taken to be much larger than typical scattering energies. Then we have

${\frac {1}{f(k)}}\approx -{\frac {4\pi }{v_{0}}}-{\frac {2}{\pi }}{\frac {1}{R}}+{\frac {2R}{\pi }}k^{2}-ik\,.$

This is now exactly of the form $f$ with the scattering length

$a={\frac {\pi }{2}}{\frac {R}{1+{\frac {2\pi ^{2}R}{v_{0}}}}}\,.$

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}).

Padagotical Example

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

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

with $K^{2}=mV/\hbar ^{2}$. which one can write 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 \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

${\frac {a}{R}}={\frac {K^{2}R^{2}}{{\frac {2}{\pi }}K^{2}R^{2}-3}}$

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.