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I am studying the BCS-BEC crossover in atomic Fermi gases from this reference https://arxiv.org/pdf/1402.5171.pdf , but I am having hard times understanding some of the details. In particular I can't understand equation 1.2: $$ \frac{m}{4\pi\hbar^2 a_s} = \frac{1}{U} + \frac{1}{V}\sum_k^{\Lambda} \frac{1}{2\varepsilon_k}, $$ where $m$ is the particle mass, $a_s$ is the scattering length, $U$ is the "bare interaction", $\varepsilon_k = \hbar^2k^2/2m$ is the dispersion relation for the non interacting particles, and $\Lambda$ is a suitable cutoff, which is inversely proportional to the range of the particle-particle interaction potential (correct me if I'm wrong), $V$ is the volume of the system (I added the factor $1/V$ myself to make a correct dimensional analysis).

Now I have many questions about this equation:

  1. where does this come from? If a thorough demonstration is too long or difficult, at least I would like to have a qualitative explanation, because this comes out of the blue to me.

  2. how should I use this equation? To my understanding, I should use it to find out the value of $U$ given $a_s$ and $\Lambda$. As far as I know, $a_s$ can be tuned experimentally, e.g. via a Feshbach resonance, and $\Lambda$ should be fixed for a given atomic species, so the only "theoretical" parameter is $U$. Am I right?

  3. For a 3d system of particles with no external potential, I would simplify the last term as: $$ \frac{1}{V}\sum_k^{\Lambda}\frac{1}{2\varepsilon_k} = \frac{1}{(2\pi)^3} \int_0^{\Lambda} 4\pi k^2 dk \frac{m}{\hbar^2k^2} = \frac{m\Lambda}{2\pi^2 \hbar^2} $$ is this correct?

  4. In case point 2. and point 3. are correct, then the equation can be rewritten as $$ \frac{1}{U} = \frac{m}{4\pi\hbar^2}\left( \frac{1}{a_s} - \frac{1}{r_0} \right) $$ where $r_0 = \pi/2\Lambda$. Now since the BCS-BEC theory assumes that $U<0$, we have that this condition is encountered just if $a_s > r_0$, or if $a_s<0$, am I right? What is the physical meaning of the former condition?

Thank you in advance for any help! :)

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The scattering length equation you show is for a contact interaction with a $k$ cutoff at $\Lambda$.

One way to get this form is to i) Write down the Schrödinger equation ii) enforce the usual boundary conditions that the incoming part is from the free particle or plane wave solution to get the Lippmann-Schwinger equation. At low energy, the scattering is s-wave, and since the interaction is separable in momentum space, and you can solve for the s-wave scattering amplitude $f$. The scattering length and effective range are defined from the low energy s-wave scattering amplitude $f$ as $$ Re \frac{1}{f(k)} = -\frac{1}{a}+\frac{1}{2}r_e k^2+... $$ At $k=0$, the energy is $E=0$, and the energy denominator in the relative coordinate Lippmann Schwinger equation becomes the $2\epsilon_k$, leading to the expression for your first equation. If you keep the next term, you find that changing your cutoff $\Lambda$ changes the effective range, with larger cutoffs having smaller effective ranges. For atomic systems, the physical effective range is of order the size of the atoms, but since the spacing between the atoms is generally much larger, for numerical computations, you can choose a smaller $\Lambda$ and still get good predictions. Any standard quantum book that covers the Lippmann-Schwinger equation and separable potentials will have this calculation.

Yes for a spherical cutoff in the large volume limit you can perform the spherical integral this way. The equation you wrote can also be used for lattice calculations which gives a nonspherical cutoff in this limit.

In principle both $\Lambda$ and $U$ should be adjusted to match the the scattering parameters as you change the magnetic field.

A positive scattering length with a negative potential can occur when the potential is strong enough to have a bound state. The unitary limit where $a$ diverges is when the potential is just at the point of having bound state at zero energy. So as you make $U$ more negative starting from a small negative value, with $a^{-1}$ a large negative value, as you increase the magnitude of $U$, $a^{-1}$ approaches zero from the negative values. At the unitary limit $a^{-1}$ is zero. For larger negative $U$ values you get a bound state (which is then the BEC side) and $a^{-1}$ is positive.

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  • $\begingroup$ Ok! So in the Lippmann-Schwinger equation $U$ is the depth of the potential well, while in the second-quantized Hamiltonian presented in the reference, the same $U$ becomes the interaction energy scale, right? For a 1d lattice system $\varepsilon_k=-2t\cos{ka}$, where the quasi-momentum has boundaries $-\pi/a<k \leq\pi/a$. What is a reasonable choice for $\Lambda$ here? How do I deal with the divergent terms at $k=\pm\pi/(2a)$? $\endgroup$
    – Matteo
    Apr 1, 2023 at 11:34
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    $\begingroup$ The energy in the Lippmann-Schwinger equation for low energy scattering should be the energy at k=0. Usually you take $\epsilon(k)=0$ to be zero and then $E=0$. So usually for a 1-d lattice with nearest neighbor hopping you would get $2t(1-\cos ka)$ so that at low k the energy goes like $k^2$. With this lattice choice, you sum over all available quasi-momenta. The lattice provides the cutoff, i.e. $\Lambda = \pi/a$. $\endgroup$
    – user200143
    Apr 1, 2023 at 18:48
  • $\begingroup$ Thanks! One last thing: should I sum over all momenta or just those corresponding to occupied states? $\endgroup$
    – Matteo
    Apr 2, 2023 at 1:09
  • $\begingroup$ This is the scattering calculation, so it corresponds to 2 particles. The momentum sum is from introducing a completeness relation in the Lippman-Schwinger equation, so you sum over all momenta. The momentum sum is cut off either from the lattice for lattice problems or because the potential matrix elements are taken to be zero beyond the cutoff for continuum problems. $\endgroup$
    – user200143
    Apr 2, 2023 at 1:44

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