Is there a physical meaning of the Fermi liquid parameters

Question

In Fermi liquid theory we define two parameters $F_l^s = VN(\epsilon_F)u_l^s$ and $F_l^a = VN(\epsilon_F)u_l^a$ where V is the fermi-volume, $N(\epsilon_F)$ the density of states at the Fermi energy and $u_l^s$ and $u_l^a$ are spin symmetric and anti-symmetric energy densities resulting of the interactions of the electrons or quasiparticles. With this parameter it is possible to write the effective mass in a compact way but what I'm wondering is if there is any physical meaning to the parameter, since it is a non-dimensional quantity?

Answer 1

The Landau parameters appear in the expansion of the free energy functional:

$$ \delta F=\frac{1}{V}\sum_{k\sigma}(\epsilon_{k\sigma}-\mu)\delta n_{k\sigma}+\frac{1}{2V}\sum_{k\sigma}f_{kk'\sigma\sigma'} \delta n_{k\sigma} \delta n_{k'\sigma'} $$ where $k$ is the wave vector, $\sigma$ is the spin, and $\delta n_{k\sigma}=n_{k\sigma}-n_{k\sigma}^0$, where the former is the interacting distribution function and the latter is the non-interacting distribution function. The Landau parameter can be interpreted as the second functional derivative of the total energy $E$ (or, equivalently, the first derivative of the quasiparticle energy $\epsilon$):

$$ f_{k\sigma}=\frac{\partial^2 E[n_{k\sigma}]}{\delta n_{k\sigma} \delta n_{k'\sigma'}}=\frac{\delta\epsilon_{k\sigma}}{\delta n_{k'\sigma'}} $$

The Landau parameter is a way of incorporating interactions into my description of the many-body state. Normally, I expand in terms of Legendre polynomials:

$$ f_{kk'\sigma \sigma'}=\sum_{\ell}P_{\ell}(\cos\chi)f_{\ell} $$

where $\chi$ is the angle between the directions $\theta$, $\phi$ of $k$ and $\theta'$, $\phi'$ of $k'$. Furthermore, we normally define

$$ N(0)f_\ell=F_\ell $$

where $N(0)$ is the density of states and $F_\ell$ is, therefore, dimensionless. We can then extract the symmetric and anti-symmetric form of the Landau parameter as such:

$$ f_{pp'\sigma\sigma'}=f_{pp'}^s+\sigma\cdot \sigma' f_{pp'}^a $$

The effective mass can be found via imposing Galilean invariance on the system. The standard procedure yields

$$ \frac{m^*}{m}=1+\frac{F_1^s}{3} $$

where $m^*$ is the effective mass and $m$ is the bare mass.

EDIT: I can give a more physical meaning to the Landau parameter by connecting it to a simple Hamiltonian model (this example is taken by Piers Coleman's excellent book on many-body physics, on page 137):

$$ H=\sum_{p\sigma}E_p n_{p\sigma}+\frac{\lambda}{2}\sum_{p\sigma',\,p'\sigma',\,q}V(q)c_{p-q,\sigma}^\dagger c_{p'+q,\,\sigma'}^\dagger c_{p'\sigma'}c_{p\sigma} $$

When we consider the energy, we can find that

$$ E=\sum_{p}E_p n_{p\sigma}+\frac{\lambda}{2}\sum_{p\sigma,\,p'\sigma'}\left\{ V(0)-V(p-p')\delta_{\sigma \sigma'} \right\}n_{p'\sigma'}n_{p\sigma} $$

This enables us to read off the Landau parameter:

$$ f_{p\sigma,\,p'\sigma'}\approx \lambda\left\{V(q=0)-V(p-p')\delta_{\sigma\sigma'})\right\} $$

So you see, Landau-Fermi liquid theory doesn't tell me what the Landau parameter is. Instead, I have to consider a specific model or experiment and then read off the Landau parameter from the data or the solution; i.e., the term that is quadratic in the fluctuations of the distribution function. This is because, by definition, the Landau parameters are just the quadratic terms in the expansion of the total energy $E$ of the fermionic system. I can then use Fermi liquid to find other physical observables from the Landau parameter; i.e., the effective mass, the specific heat, the magnetic susceptibility, the spin current, etc.