The purpose of this post is to document my progress for my summer internship in the Ultracold Quantum Matter Lab at Yale University working with the group of Professor Nir Navon to study bichromatic Rabi oscillations of a driven Fermi polaron.
Problem: Annotate the following Nature Physics paper from the group.
Solution:
Everything in the first half of the paper (steady state part) can be summed up like:
And the latter half discussing the resonant dynamics of the pre-steady state can be visualized qualitatively as:
Finally, a few points worth emphasizing:
- Clearly \(\uparrow\) is unstable, wants to decay back to \(\downarrow\) with transition rate \(\Gamma\), so the zero-detuning \(\delta_0<0\) must contain information about the \(\uparrow,B\) interactions.
- A Rabi experiment is different from say Ramsey in that it really is about a continuous drive so \(t\gg 1/\Omega\), as mentioned in the paper’s Figure \(1a\).
- To emphasize again, there is the implicit hyperparameters that \(a_{\uparrow B}=\infty\) is unitarily/strongly interacting while \(a_{\downarrow B}\approx 0\) is weakly/non-interacting; the paper gives exact values). So only the attractive polaron is present. At the end the paper mentions extending into the BEC regime, where both would coexist for a sufficiently broad Feshbach resonance.
And some random thoughts I have about the paper:
- Instead of the \(3D\) surface graph, maybe a heat map would be instructive too?
- At large \(\Omega\), how does the AC Stark shift associated with a dressed atom-photon affect the physics (since it seems it would affect the internal energies of the polaron)?
Problem: Prove the Sokhotski-Plemelj theorem on the real line \(\textbf R\):
\[\frac{1}{\omega-\omega’\pm i0^+}=\mathcal P\frac{1}{\omega-\omega’}\mp i\pi\delta(\omega-\omega’)\]
Solution: (in this solution, the function \(\chi(\omega)\) is assumed to be analytic on the integration interval in \(\textbf R\) to be able to apply the Cauchy integral formula):
Problem: Explain why any linear response function \(\chi\) should obey in the time domain \(\chi(t)=0\) for \(t<0\). Hence, what is the implication of this for \(\chi(\omega)\) in the frequency domain?
Solution: The fundamental definition of the linear response function \(\chi\) that gives it its name is that in the time domain \(\chi=\chi(t)\), the response \(x(t)\) should be proportional to the perturbation \(f(t)\) with \(\chi\) essentially acting as the proportionality constant according to the convolution:
\[x(t)=\int_{-\infty}^{\infty}dt’\chi(t-t’)f(t’)\]
(or equivalently the local behavior \(x(\omega)=\chi(\omega)f(\omega)\) in Fourier space). But if the “force” \(f(t’)\) is applied at time \(t’\), on causality grounds this can only affect the response \(x(t)\) at times \(t\geq t’\). In other words, it should be possible to change the limits on the integral from \(\int_{-\infty}^{\infty}dt’\) to \(\int_{-\infty}^tdt’\) without affecting the result. This therefore requires \(\chi(t-t’)=0\) for \(t<t’\), or more simply \(\chi(t)=0\) for \(t<0\). In light of this, \(\chi\) is also called a causal/retarded Green’s function.
Now then:
\[\chi(t)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}e^{i\omega t}\chi(\omega)\]
For \(t<0\), Jordan’s lemma asserts that one should close the contour in the lower half-plane \(\Im\omega<0\). But the fact that \(\chi(t)=0\) for \(t<0\) suggests that the sum of all residues of \(\chi(\omega)\) in the lower half-plane \(\Im\omega<0\) should be “traceless”. A sufficient condition for this is if \(\chi(\omega)\) is analytic in the lower half-plane \(\Im\omega<0\), and henceforth this will be assumed.
Problem: Qualitatively, what do the Kramers-Kronig relations assert? What about quantitatively?
Solution: Qualitatively, for a linear response function like \(\chi(\omega)\) which is analytic in the lower half-plane \(\Im\omega<0\), knowing its reactive part \(\Re\chi(\omega)\) is equivalent to knowing its absorptive/dissipative spectrum \(\Im\chi(\omega)\) which in turn is equivalent to knowing \(\chi(\omega)\) itself.
\[\Re\chi(\omega)\Leftrightarrow\Im\chi(\omega)\Leftrightarrow\chi(\omega)\]
Quantitatively, the bridges \(\Leftrightarrow\) are provided by the Kramers-Kronig relations:
Note: sometimes the discussion of Kramers-Kronig relations are phrased in terms of \(\chi(\omega)\) being analytic in the upper half-plane. This stems from an unconventional definition of the Fourier transform \(\chi(t)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}e^{-i\omega t}\chi(\omega)\) rather than the more conventional definition \(\chi(t)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}e^{i\omega t}\chi(\omega)\) used above. Consequently, the minus signs in the Kramers-Kronig relations may also appear flipped around.
Problem: If the response \(x(t)\) and the driving force \(f(t)\) are both real-valued, what are the implications of this for \(\Re\chi(\omega)\) and \(\Im\chi(\omega)\)?
Solution: Then \(\chi(t)\in\textbf R\) must also be real-valued, so \(\chi(\omega)\) is Hermitian:
\[\chi^{\dagger}(\omega)=\chi(-\omega)\]
Consequently, the reactive response \(\Re\chi(-\omega)=\Re\chi(\omega)\) is even while the absorptive/dissipative response \(\Im\chi(-\omega)=-\Im\chi(\omega)\) is odd.
Problem: State the thermodynamic sum rule.
Solution: The sum rule asserts that if one knows how much a system absorbs/dissipates at all frequencies \(\omega\in\textbf R\), then one can deduce the system’s DC linear response \(\chi(\omega=0)\), called its susceptibility (of course the Kramers-Kronig relations actually show that knowing \(\Im\chi(\omega)\) allows complete reconstruction of \(\chi(\omega)\) at all frequencies \(\omega\in\textbf R\), not just \(\omega=0\)).
Problem: Write an essay that summarizes the key points learned from the following papers/slides:
- Fermi polarons and beyond (Parish and Levinsen)
- The quantum impurity problem and beyond (Parish)
- Exact theory of the finite-temperature spectral function of Fermi polarons with multiple particle-hole excitations: Diagrammatic theory versus Chevy ansatz (Hu, Wang, Liu)
- A single impurity in an ideal atomic Fermi gas: current understanding and some open problems (Lan, Lobo)
- Many-particle physics with ultracold gases (Punk)
- Polarons, dressed molecules, and itinerant ferromagnetism in ultracold Fermi gases (Massignan, Zaccanti, Bruun)
- Ultrafast many-body interferometry of impurities coupled to a Fermi sea (Cetina et al.)
- https://static-content.springer.com/esm/art%3A10.1038%2Fs41567-025-02799-8/MediaObjects/41567_2025_2799_MOESM1_ESM.pdf
- file:///C:/Users/weidu/Downloads/Double_Mode_in_Driven_Fermi_Polaron.pdf
Solution: For a generic \(2\)-component Fermi gas whose \(2\) components may be called \(\uparrow\) and \(\downarrow\) (this could be \(2\) hyperfine states of the same atom, or \(2\) hyperfine states of different atoms) the Hamiltonian is \(H=H_0+V_{\downarrow\uparrow}\) where the kinetic energy is:
\[H_0=\sum_{\textbf k}\frac{\hbar^2|\textbf k|^2}{2m_{\uparrow}}c^{\dagger}_{\textbf k\uparrow}c_{\textbf k\uparrow}+\frac{\hbar^2|\textbf k|^2}{2m_{\downarrow}}c^{\dagger}_{\textbf k\downarrow}c_{\textbf k\downarrow}\]
and the short-range scattering pseudopotential \(V_{\downarrow\uparrow}\) of “bare strength” \(g_{\uparrow\downarrow}\) describes momentum-conserving collisions between the \(2\) components \(\uparrow,\downarrow\) of the Fermi gas in a volume \(V\) (note that interactions among the components themselves are neglected, i.e. they are separately ideal Fermi gases. That is, one assumes there is no \(\uparrow\uparrow\) or \(\downarrow\downarrow\) scattering. This is justified by the fact that identical spin parts of \(2\) identical fermions could only interact via an odd-\(\ell\) scattering channel, the lowest of which is \(\ell=1\) \(\p\)-wave scattering whose cross-section \(\sigma_{\ell}\sim k^{2\ell}\) is suppressed at low \(k\)):
\[V_{\downarrow\uparrow}=\frac{g_{\uparrow\downarrow}}{V}\sum_{\textbf k_1,\textbf k_2,\textbf k’_1,\textbf k’_2}\delta_{\textbf k’_1+\textbf k’_2,\textbf k_1+\textbf k_2}c^{\dagger}_{\textbf k’_2,\downarrow}c^{\dagger}_{\textbf k’_1,\uparrow}c_{\textbf k_2,\downarrow}c_{\textbf k_1,\uparrow}\]
The Fermi polaron is the limit \(N_{\downarrow}/N_{\uparrow}\to 0\) of the \(2\)-component Fermi gas, in fact typically one just takes \(N_{\downarrow}=1\). In light of this population imbalance between the \(2\) components \(\uparrow,\downarrow\) of the Fermi gas, the standard terminology is to call the majority \(\uparrow\) component as the bath and the minority \(\downarrow\) component as the impurity. Through the short-range interaction \(V_{\downarrow\uparrow}\), the \(\downarrow\) impurity polarizes the \(\uparrow\) bath in its \(\textbf x\)-space vicinity (hence the name polaron!), and it is common to say that the \(\downarrow\) impurity is dressed by the polarized \(\uparrow\) cloud that it “carries” along with it. This composite object of the \(\downarrow\) impurity together with the \(\uparrow\) cloud is a quasiparticle called the (Fermi) polaron (in particular, it is important to emphasize that polaron is not synonymous with \(\downarrow\) impurity; the interaction \(V_{\downarrow\uparrow}\) is essential and instead polaron is synonymous with \(\downarrow\) impurity + \(\uparrow\) polarized cloud).
This discussion has been an intuitive/qualitative picture in \(\textbf x\)-space (aka real space). In \(\textbf k\)-space (aka reciprocal space), one can get more quantitative. Here, rather than starting from a \(2\)-component \(\uparrow\downarrow\) Fermi gas, one can first visualize a single-component \(\uparrow\) ideal Fermi gas in its ground state where the Fermi sea is occupied up to the Fermi wavenumber \(k_F=(6\pi^2 n_{\uparrow})^{1/3}\):
\[|\text{FS}\rangle:=\prod_{|\textbf k|\leq k_F}c^{\dagger}_{\textbf k,\uparrow}|\space\space\rangle\]
The excited states are then particle-hole excitations of this ground state Fermi sea \(|\text{FS}\rangle\). When adding a single \(\downarrow\) impurity to the bath, one would expect that the impurity would scatter bath fermions from inside to outside the Fermi sea. The modified ground state \(|\text{FP}\rangle\) of the Fermi polaron system (i.e. the \(\downarrow\) impurity + \(\uparrow\) bath) would therefore be expected to be of the form (working in the ZMF classically, or quantum mechanically fixing an eigenstate of total momentum \(\textbf 0\)):
\[|\text{FP}\rangle=\alpha_0c^{\dagger}_{\textbf 0,\downarrow}|\text{FS}\rangle+\sum_{|\textbf k|\leq k_F,|\textbf k’|\geq k_F}\alpha_{\textbf k,\textbf k’}c^{\dagger}_{\textbf k-\textbf k’,\downarrow}c^{\dagger}_{\textbf k’,\uparrow}c_{\textbf k,\uparrow}|\text{FS}\rangle+…\]
where the \(…\) indicates \(N\) particle-hole excitations for \(N\geq 2\). Ignoring the \(…\) terms, this is called the Chevy ansatz and can be used as a trial ground state with \(\alpha_0,\alpha_{\textbf k,\textbf k’}\) the fitting parameters to be tuned such as to minimize the Rayleigh-Ritz energy quotient \(E=\frac{\langle\text{FP}|H|\text{FP}\rangle}{\langle \text{FP}|\text{FP}\rangle}\) in the variational method. The ground state energy eigenvalue \(E\) obtained in this manner is an estimate of the polaron energy. Explicitly:
As another corollary, the fitting parameter \(\alpha_0\), once fitted, gives the polaron residue:
\[Z:=|\alpha_0|^2\leq 1\]
The (unobservable) bare strength \(g_{\uparrow\downarrow}=g_{\uparrow\downarrow}(k^*)\) should be taken to run with the (unobservable) UV cutoff \(k^*\to\infty\) such as to keep \(a_s\) fixed! Specifically, through the renormalization condition:
\[\frac{1}{g_{\uparrow\downarrow}}=\frac{\mu}{2\pi\hbar^2 a_{\uparrow\downarrow}}-\frac{2\mu}{\hbar^2V}\sum_{|\textbf k|\leq k^*}\frac{1}{|\textbf k|^2}\]
Dippy renormalization derivation:
Since:
\[f(\textbf k,\textbf k’)=-\frac{\mu}{2\pi\hbar^2}\int d^3\textbf x’e^{-i\textbf k’\cdot\textbf x’}V(\textbf x’)\psi(\textbf x’)\]
For \(V(\textbf x’)=g\delta^3(\textbf x’)\), this becomes:
\[f(\textbf k,\textbf k’)=-\frac{\mu}{2\pi\hbar^2}g\psi(\textbf 0)\]
But \(\psi(\textbf x)=e^{ikz}+f(\textbf k,\textbf k’)\frac{e^{ikr}}{r}\), and it’s that divergent \(1/r\) piece in the scattered spherical wave that’s gonna cause trouble. This is because \(\psi(\textbf 0)\) seems to blow up due to it. But rather than let it blow up, allow it to be some large number call it \(k^*/\pi\) (clearly dimensionally okay). Then \(\psi(\textbf 0)=1+f(\textbf k,\textbf k’)k^*/\pi\). Substituting gives and isolating for \(f\):
\[\frac{1}{f}+\frac{k^*}{\pi}=-\frac{2\pi\hbar^2}{\mu g}\]
The point is now, suddenly, you introduce a new low-energy parameter into the game, the \(s\)-wave scattering length \(a_s\)! Notice it didn’t appear in any equations yet! But since we only care about low-energy/low-momentum \(\textbf k\to\textbf 0\), and we know we have the limit \(f(\textbf k,\textbf k’)\to -a_s\) as \(\textbf k\to \textbf 0\). It’s a sort of limit/correspondence principle-like knot at the end of a string that the theory has to approach. So making that substitution, one obtains the running of the bare coupling with the UV cutoff. This turns out the be the same as the above renormalization condition.
More precise derivation:
Write the Born series for the scattering amplitude:
\[f(\textbf k,\textbf k’)=-\frac{1}{4\pi}\frac{2\mu}{\hbar^2}\langle\textbf k’|V_{\uparrow\downarrow,s}|\psi_{\textbf k}\rangle\]
By defining the transition operator \(T_{\uparrow\downarrow,s}|\textbf k\rangle:=V_{\uparrow\downarrow,s}|\psi_{\textbf k}\rangle\) which can be easily checked to obey \(T=V+VG_0T\) with \(G_0=(E_{\textbf k}1-H_0)^{-1}\) the free particle resolvent, then because \(\langle k|V_{\uparrow\downarrow,s}|\textbf k’\rangle=g\) for \(V_{\uparrow\downarrow,s}=g\delta^3(\textbf X)\), then actually \(\textbf k’\) doesn’t even matter (i.e. \(s\)-wave scattering is isotropic!) so it can be used as a dummy index for the summation. Furthermore to the isotropy, \(f(\textbf k)=f(k)\):
\[f(k)=-\frac{\mu}{2\pi\hbar^2}\left(g+\frac{g^2}{V}\sum_{|\textbf k’|\leq k^*}\frac{1}{E_{\textbf k}-E_{\textbf k’}}+\text{geometric series}\right)\]
In the end, once you set \(\textbf k:=\textbf 0\) so that \(f(0)=-a_{\uparrow\downarrow,s}\), you get the same thing. Here, the subtleties are that the series part can be summed by letting \(V\to\infty\) so \(\frac{1}{V}\sum_{\textbf k’}\to\int\frac{d^3\textbf k’}{(2\pi)^3}\), and if you take \(\langle\textbf x|\textbf k\rangle=e^{i\textbf k\cdot\textbf x}\) then the correct identity resolution is \(\frac{1}{V}\sum_{\textbf k}|\textbf k\rangle\langle\textbf k|\) for quantization volume \(V\) and also remember \(G_0|\textbf k’\rangle=\frac{1}{E_{\textbf k}-E_{\textbf k’}}|\textbf k’\rangle\) is an eigenstate.
(there are both attractive and repulsive Fermi polarons so this polarization effect can go either way). In the attractive case, if the attraction is strong enough, the the polaron can dimerize with a bath fermion, forming a molecule; this polaron-molecule transition is interesting.
Surprisingly, the Chevy ansatz works remarkably well (i.e. agrees with state-of-the-art diagrammatic quantum Monte Carlo stuff)! Seems to include the dimer bound state in it?
———————-
There are \(2\) key assumptions about the typical regime of ultracold atomic gases, namely \(n^{-1/3},\lambda_T\gg r_{vdW}\sim 100a_0\).
In the vicinity of a broad Feshbach resonance, the scattering amplitude may be approximated by the Mobius transformation \(f_s(k)=-\frac{1}{ik+a^{-1}_s}\). However, in the vicinity of a narrow Feshbach resonance, need to also parameterize it with the effective range \(r_{\text{eff}}\) so that \(f_s(k)=-\frac{1}{ik+a^{-1}_s-\frac{1}{2}r_{\text{eff}}k^2}\). Although \(a_s\) and \(r_{\text{eff}}\) are determined by microscopic details of \(V_{\uparrow\downarrow}(r)\), different microscopic details in another potential \(\tilde V_{\uparrow\downarrow}(r)\) can lead to the same low-energy scattering amplitude \(f_s(k)\). The practical corollary of this observation is that one do just that, namely substitute \(V_{\uparrow\downarrow}(r)\) for a suitable pseudopotential.
Problem: Consider a toy model of the Fermi polaron in which the \(\downarrow\) impurity interacts with only the nearest \(\uparrow\) impurity in the Fermi sea, the rest of the \(\uparrow\) Fermi sea serving to exert a pressure that effectively confines the relative distance between the \(\downarrow\) and \(\uparrow\) impurities to a radius \(R\). By equating the ground state energy of the infinite spherical potential well with the Fermi energy \(E_F\), show that:
\[R=\sqrt{\frac{m_{\uparrow}}{\mu}}\]
Hence, show that for a positive-energy eigenstate \(E=\frac{\hbar^2k^2}{2m}\) the wavenumber \(k\) is determined through the \(s\)-wave scattering length \(a_s\) by:
\[k\cot kR=a^{-1}_s+R^*k^2\]
(where the Bethe-Peierls boundary condition is used). Show that for \(m_{\uparrow}=m_{\downarrow}\) and \(R^*=0\), this simplifies to:
\[-\frac{1}{k_Fa_s}=-\frac{k}{k_F}\cot\sqrt{2}\pi\frac{k}{k_F}\]
By considering the scaled energy from the Fermi energy \(\frac{E-E_F}{E_F}\) which in this case amounts to \(2(k/k_F)^2-1\), plot this as a function of \(-1/k_Fa_s\).
Problem: Explain how Ramsey interferometry works.
Solution: Applying two \(\pi/2\)-pulses separated by some time \(\Delta t\); then Ramsey fringes are seen as a function of this temporal separation \(\Delta t\); it is a bit like a time-domain analog of a Mach-Zender interferometer.