Is is possible to extract an effective Hamiltonian from a Boltzmann equation (or any other kinetic theories)?

Question

I got kind of confused when reading Xiaogang Wen's famous textbook Quantum Field Theory of Many-body Systems. In Section 5.3.3 the book claims that

• From a kinetic theory of Fermi liquid (a Boltzmann equation with a self-energy correction which in turn relies on the distribution function, which people sometimes call the "Landau's equation") we can derive a bosonized Hamiltonian of the Fermi liquid, by considering the kinetic equation $$ \frac{\partial \tilde{\rho}}{\partial t}+v_{F}^{*} \hat{\boldsymbol{k}} \cdot \frac{\partial \tilde{\rho}}{\partial \boldsymbol{x}}+\frac{k_{F}}{(2 \pi)^{2}} \int \mathrm{d} \theta^{\prime} f\left(\theta, \theta^{\prime}\right) \hat{\boldsymbol{k}} \cdot \frac{\partial \tilde{\rho}\left(\theta^{\prime}, \boldsymbol{x}, t\right)}{\partial \boldsymbol{x}}=0 \tag{5.3.11} $$ as the equation of motion of the Hamiltonian $$ \begin{aligned} H &=E_{g}+\int \mathrm{d}^{2} \boldsymbol{x} \mathrm{d} \theta \tilde{\rho} \frac{h v_{F}^{*}}{2}+\frac{1}{2} \int \mathrm{d}^{2} \boldsymbol{x} \mathrm{d} \boldsymbol{k} \mathrm{d} \boldsymbol{k}^{\prime} f\left(\boldsymbol{k}, \boldsymbol{k}^{\prime}\right) \delta n_{\boldsymbol{k}} \delta n_{\boldsymbol{k}^{\prime}} \\ &=E_{g}+\int \mathrm{d}^{2} \boldsymbol{x} \mathrm{d} \theta \frac{2 \pi^{2} v_{F}^{*}}{k_{F}} \tilde{\rho}^{2}+\frac{1}{2} \int \mathrm{d}^{2} \boldsymbol{x} \mathrm{d} \theta \mathrm{d} \theta^{\prime} f\left(\theta, \theta^{\prime}\right) \tilde{\rho}(\theta) \tilde{\rho}\left(\theta^{\prime}\right), \end{aligned} \tag{5.3.12} $$ and
• The field in the bosonized Hamiltonian is labeled by $\theta$, which is the direction of the momentum in the distribution function appearing in the Boltzmann equation.

I understand that Boltzmann equation may be viewed as an equation about the Wigner function of the two-point correlation function, so it can be used to describe quantum systems. However it should then be noted that after all we are dealing with $\langle \phi(x) \phi(y) \rangle$ instead of $\phi(x) \phi(y)$. Treating the Boltzmann equation for Fermi liquid as a quantum EOM is therefore not well justified, as the real EOM may contain certain operators that vanishe when being averaged.

So my question is what he is doing in that section, and when the kinetic equations is nonlinear for collision, whether it is equivalent to adding a higher order term ("interaction channels between density modes") to $(5.3.12)$.

Answer 1

I think what we are dealing with here is more of an educated guess than a rigorous derivation - both in case of extracting the Hamiltonian from the Boltzmann equation and in case of quantizing hydrodynamic equations (or, more generally, any equations for a continuous media - hydrodynamic, elasticity, plasma, magnetization, etc.)

There are however a few more points to make:

• The OP mentions that Boltzmann equation can be derived as an equatioin for a Wigner function - the approach described in the classical Kadanoff&Baym's Quantum statistical mechanics, although there are more recent treatments. I suggest taking a look at Rammer&Smith's Quantim field-theoretical methods in transport theory of metals for rigorous derivation of the quantum corrections to the Boltzmann equation and, in particular, their diagrammatic representation.
• I don't know well the context in the Wen's book, but when the treatment is geared twoards treatment of critical phenomena, the underlying quantum nature is often irrelevant (There is a discussion in Goldenfeld's Lectures on phase transitions and the renormalization group. In fact, critical phenomena are often correctly captured by ad-hoc models such as Ising model, Heisenberg model, Hubbard model - which have little to do with the details of the underlying interactions and geometries (which typically vanish during renormalization).

Answer 2

It seems this is indeed possible, after all. Consider the following lines of reasoning:

1. A Boltzmann equation is just a quantum master equation after gradient expansion, and a quantum master equation is just a single-electron Green's function's EOM. The Boltzmann equation of Fermi liquid differs from ordinary Boltzmann equations in that it contains a self-energy correction ($f \delta n$).
2. The linear response of a single-electron Green's function's EOM is equivalent to the Bethe-Salpeter equation (BSE) for the four-point correlation function, where the kernel used in the latter is the functional derivative of the self-energy used in the former w.r.t. the single-electron Green's function. For example see https://www.pnas.org/doi/full/10.1073/pnas.1906938118, https://journals.aps.org/prb/abstract/10.1103/PhysRevB.84.245110, and https://journals.aps.org/prb/pdf/10.1103/PhysRevB.62.4927. This, by the way, is how excitons are treated in first-principles calculations: we know the GW approximation works incredibly well in weakly-correlated systems, so the Bethe-Salpeter equation derived from GW should also work incredibly well. In first-principles calculations people actually use the term BSE to refer to the BSE derived from GW.
3. Therefre the linear response of the single-electron Boltzmann equation is equivalent to the long-wavelength limit of the BSE corresponding to the self-energy correction used in the Boltzmann equation.
4. We know the singularities in the linear response of the Boltzmann equation, which corresponds to excitons, plasmons, etc., are given by oscillation modes of the linearized Boltzmann equation without any external driving. And now the $\rho$ in the Boltzmann equation, formally, has the same status with some sort of electron-hole pair annihilation operators. That's what Wen is doing in his textbook.
5. And if you inspect Eq. (5.3.12), now you can see the structure of a BSE: the first term, proportional to $v_{\text{F}}$, is nothing but $\epsilon_{k+q} - \epsilon_k$, the "bare propagator" part in BSE, and the second term, proportional to $f$, is just the kernel part which is the self-energy divided by single-electron Green's function.

Note that in order to form an exciton we need two bands, so if we want to have excitonic behaviors in a Boltzmann equation, we have to include the band index into $\rho$ and we need to indices (because it comes from the density matrix). This is often known as "including the band index coherently", because if we only include one band index to $\rho$, what we get is a rate equation that ignores non-diagonal terms in the single-electron density matrix. And now since we have "included the band index coherently", the Boltzmann equation should have an additional term $[\rho, H]$, and from this term we get things like $E_c - E_v$ (c means conduction band and v means valence band) that are essential for exciton formation. This subtlety - we need $\rho_{nn'}$ and not just $\rho_n$ - is mentioned in Eq. (5.5) in Statistical Physics II in Landau's famous Course of Theoretical Physics, where he's considering what's essentially an exciton between two energy levels split by an external magnetic field.

When the band index is not considered at all we don't have excitons, so what remains can only be some sort of plasmon; in Landau's original formulation of Fermi liquid theory there is no long-range interaction, so we don't have plasmons and only zero sounds. But if the only thing you can probe is bosonic modes made of electron-hole pairs and you're sure you don't have excitons, this is enough. Therefore Wen's "quantization" of Boltzmann equation leads to a kind of bosonization of Fermi liquid. This doesn't have much theoretical significance because this bosonization happens after all essential calculations are already done, unlike the case of bosonization in, say, 1D Hubbard model, at first we don't even know what the low-energy degrees of freedom are in the system.

Some further remarks: people doing first-principles calculations often ignore plasmon modes in BSE by removing the so-called long-range exchange term $K_{\text{x}}(\mathbf{q}=0, \mathbf{G}=0)$ (corresponding to the long-range term in Landau-Silin equation: this term has singularity and therefore in Landau-Silin eqation is written as a real space term) from BSE. This is because usually we're only interested in the absorption spectrum, which only contains transverse modes while the plasmon is longitudinal, and it can be shown that transverse excitons with zero overall momenta are not influenced by this term. In finite-momentum measurements, like EELS, plasmon modes are important, and then we add the long-range exchange term back to BSE. Since in all measurements the electromagnetic field's wave length is usually within the first Brillouin zone, when the long-range exchange term is removed, what is calculated by BSE is the irreducible polarizability, and when the long-range exchange term is not removed, what is calculated by BSE is the reducible polarizability, because the long-range exchange term is also considered in the theory about the probing fields so we have a double counting here. Consequently, using Fermi's golden rule, the first type of BSE gives us $\mathrm{Im} \epsilon$ while the second type of BSE gives us $\mathrm{Im} (\epsilon^{-1})$, where $\epsilon$ serves as the effective dielectric function of the probing field in the medium, often known as the macroscopic dielectric function (see https://journals.aps.org/prb/pdf/10.1103/PhysRevB.88.155113). But you can see, in the first type of BSE, the modes calculated using Wen's "hydrodynamics" are ignored, while in the second type of BSE, the modes are not ignored but since the overall momentum of an excitation can be large, his hydrodynamics breaks down at a certain point and has to be replaced by full BSE. This probably explains why this approach is only mentioned in textbooks.

Also, because of the many-body nature of the ground state, components like $\langle c_c c^\dagger_v \rangle$ are non-zero, and they turn out to be important in plasmonic calculations (this is known as "deexcitation", and ignroing them is known as Tamm-Dancoff approximation, which is not quite right for serious plasmon calculations). They are however not considered in most exciton calculations. In the Boltzmann equation approach, however, TDA is not made: it's possible for $n$ to slightly exceed 1 for a given state and therefore deexcitation is rightly considered here. This is a lesson this approach teaches us when we proceed to more quantitatively accurate calculations.