Interactions; applications to susceptibility and effective mass

We continue using the assumption that while the interparticle interactions in a Fermi liquid may be strong, we consider only small perturbations of the quasiparticle occupancies from the equilibrium distribution. This is somewhat justified because the "most probable" configuration, as shown last time, still is given by the Fermi-Dirac distribution of quasiparticles, even though the quasiparticles are no longer free. The assumption that we could keep only the leading term in the quasiparticle interaction gives
$$\begin{equation} \delta \epsilon_{\alpha \beta}({\bf p}) = \int f_{\alpha \gamma, \beta \delta}({\bf p},{\bf p}^\prime) \delta n({\bf p^\prime})_{\gamma \delta}\,d{\bf p}^\prime. \end{equation}$$
Recall that $\delta \epsilon_{\alpha \beta}({\bf p})$ was defined as the change in the effective energy of quasiparticles of momentum ${\bf p}$ induced by a given nonequilibrium distribution of the other quasiparticles:
$$\begin{equation} {\delta E \over V} = \int \epsilon_{\alpha \beta} (p) \delta n_{\beta \alpha} d\tau. \end{equation}$$

The total value of ${\hat \epsilon}$ is now
$$\begin{equation} {\hat \epsilon}({\bf p}) - \epsilon_F = v_F (p-p_F) + {\rm tr}\ ^\prime \int {\hat f}({\bf p},{\bf p}^\prime) \delta {\hat n}({\bf p}^\prime). \end{equation}$$
In practice we will only be interested in small deviations near the Fermi surface, so the interaction ${\hat f}$ will only need to be evaluated when both momenta lie on the Fermi surface. We will assume spherical symmetry of the Fermi surface, so ${\hat f}$ depends only on the angle $\theta$ between ${\bf p}$ and ${\bf p}^\prime$.

The final simplifying assumption we will make is that the spin part of the interaction is rotationally symmetric in spin space, so that the Pauli matrices can only appear as a scalar product. Then we can finally write
$$\begin{equation} {p_F m^* \over \pi^2 \hbar^3} {\hat f}({\bf p},{\bf p}^\prime) = F(\theta) + {\bf \sigma} \cdot {\bf \sigma^\prime} G(\theta). \end{equation}$$
Here $F$ and $G$ are two dimensionless scalar functions of the angle $\theta$ between ${\bf p}$ and ${\bf p}^\prime$. The units work since ${\hat f}$ multipled by a density gives an energy; the term on the left side is the density of quasi-particle states per energy on the Fermi surface
$$\begin{equation} \nu(\epsilon_F) = {2 \cdot 4 \pi {p_F}^2 \over (2 \pi \hbar)^3} {dp \over d\epsilon}|_{p_F} = {{p_F}^2 \over \pi^2 \hbar^3 v_F} = {p_F m^* \over \pi^2 \hbar^3}. \end{equation}$$

Now we will show how the effective mass $m^*$ can be written in terms of $F$, using Galilean invariance: this is the first of many examples of how experimental quantities can be related to $F$ and $G$. (We also need to make use of our assumption that the total number of quasiparticles is equal to the total number of original particles, so that the Fermi surface is unchanged.) The number flux of quasiparticles is
$$\begin{equation} {\rm tr}\ \int {\hat n} {\partial \epsilon \over \partial {\bf p}}\,d\tau. \end{equation}$$
Since the number of particles moving is the same as the number of quasiparticles, the physical mass flux should be the above, multiplied by the {\bf bare} mass $m$:
$$\begin{equation} {\rm tr}\ \int {\bf p} {\hat n}\,d\tau = {\rm tr}\ \int m {\partial {\hat \epsilon} \over \partial {\bf p}} {\hat n}\,d\tau. \end{equation}$$
Now we assume that all the tensors are diagonal and vary both sides of the above equation. There are two terms on the right, and we integrate the second by parts and switch variables:
$$\begin{eqnarray} \int {\bf p} \delta n\,d\tau &=& m \int {\partial \epsilon \over \partial {\bf p}} \delta n\,d\tau + m \int {\partial f({\bf p},{\bf p}^\prime) \over \partial {\bf p}} n\,\delta n^\prime\,d\tau\,d\tau^\prime \cr &&=m \int {\partial \epsilon \over \partial {\bf p}} \delta n\,d\tau - m \int f({\bf p},{\bf p}^\prime) {\partial n^\prime \over \partial {\bf p}^\prime} \delta n\,d\tau\,d\tau^\prime. \end{eqnarray}$$
This implies, since $\delta n$ is so far arbitrary, that
$$\begin{equation} {{\bf p} \over m} = {\partial \epsilon \over \partial {\bf p}} - \int f({\bf p},{\bf p}^\prime) {\partial n^\prime \over \partial {\bf p}^\prime}\,d\tau^\prime. \end{equation}$$
At zero temperature, the derivative of $n^\prime$ is proportional to a delta-function:
$$\begin{equation} {\partial n^\prime \over {\bf p}^\prime} = - {{\bf p}^\prime \over p^\prime} \delta(p^\prime - p_F). \end{equation}$$
Now we substitute in the earlier equation
$$\begin{equation} {\hat \epsilon}({\bf p}) - \epsilon_F = v_F (p-p_F) + {\rm tr}\ ^\prime \int {\hat f}({\bf p},{\bf p}^\prime) \delta {\hat n}({\bf p}^\prime). \end{equation}$$
and get, again assuming that the momentum is at the Fermi surface and using $\cos \theta = {\bf \hat p} \cdot {\bf \hat p}^\prime$,
$$\begin{equation} {p_F {\bf \hat p} \over m} = v_F {\bf \hat p} - {{p_F}^2 \over (2 \pi \hbar)^3} \int f(\theta) \cos(\theta) d\Omega. \end{equation}$$
Here the integral is over the Fermi surface and $\Omega$ is the element of solid angle.
Dividing through by ${\bf \hat p} p_F$ and using the definition of $m^*$, we obtain
$$\begin{equation} {1 \over m} = {1 \over m^*} + {p_F \over (2 \pi \hbar)^3} \int f(\theta) \cos(\theta) d\Omega \end{equation}$$
which becomes finally
$$\begin{equation} {m^* \over m} = 1 + \langle F(\theta) \cos(\theta) \rangle. \end{equation}$$

This suggests that it is useful to parametrize $F$ and $G$ in terms of Legendre polynomials:
$$\begin{equation} F(\theta) = \sum_l (2l+1) F_l P_l(\cos \theta),\quad G(\theta) = \sum_l (2l+1) G_l P_l(\cos \theta). \end{equation}$$
There is a stability requirement that follows from the assumption that stationary perturbations of the Fermi surface not lower the energy. This can be simply expressed as
$$\begin{equation} F_l + 1 > 0, \quad G_l + 1 > 0, \end{equation}$$
and we see that this is exactly sufficient to ensure that the effective mass
$$\begin{equation} m^* = m (1 + F_1) \end{equation}$$
is positive. The most important other experimental quantities are also expressible in terms of the "Fermi liquid parameters" $F_l$ and $G_l$. The compressibility is the derivative of pressure with respect to mass density,
$$\begin{equation} {\partial P \over \partial \rho} = {{p_F}^2 \over 3 m m^*} (1 + F_0) = {{p_F}^2 \over 3 m^2} {1 + F_0 \over 1 + F_1}. \end{equation}$$
Hence positive compressibility is required for stability, as makes sense.

The Zeeman susceptibility is found, by a similar procedure of equating two forms for the same physical quantity, to be
$$\begin{equation} \chi = {\mu^2 p_F m^* \over \pi^2 \hbar^3 (1 + G_0)}. \end{equation}$$
Hence the instability at $G_0 = -1$ is a ferromagnetic one (diverging susceptibility),
even though the particles in the Fermi liquid are mobile. This is thus an example of itinerant ferromagnetism.

Numbers: for He$^3$, $m^* \approx 3.1 m$ at normal pressure (note that these are now atomic masses rather than electron masses). For some heavy fermion compounds, $m^* \approx 10^3 m$. For He$^3$, $G_0 \approx -2/3$ so the system is fairly close to a magnetic instability.

There is an interesting type of wave propagation that can exist in a Fermi liquid. The simplest form of zero sound appears when $F_0$ is positive, and is an example of an additional experimental quantities can be derived from the same few Fermi liquid parameters. However, the actual derivation of this wave is quite complicated, so the reader is referred to Landau and Lifshitz volume 9. The dimensionless velocity of zero sound $v / v_F$ in the simplest case where $F_0$ dominates is given by
$$\begin{equation} {s \over 2} \log {s+1 \over s-1} - 1 = {1 \over F_0}, \end{equation}$$
which suggests that some interesting physics must be involved since for small $F_0$,
this predicts
$$\begin{equation} s - 1 \approx {2 \over e^2} e^{-2/F_0}. \end{equation}$$
The propagating wave disappears once $F_0$ becomes negative.