Yet another question on the Lindhard function

Here's another question concerning the Lindhard function as used in the physical description of metals.

First we define the general Lindhard function in the Random Phase approximation as $\chi(q,\omega)=\frac{\chi_{0}}{1-\frac{4\pi e^2}{q^2}\chi_{0}}$ where $\chi_{0}$ denotes the "bare" Lindhard function" excluding any feedback effects of electrons. A reference is Kittel's book "Quantum Theory of solids" Chapter 6, however he is using the dielectric function instead of the Lindhard function.

By writing $\chi_{0}(q,\omega)=\chi_{01}+i\chi_{02}$ as a complex number and using the relation $\frac{1}{z-i\eta} = P(1/z) + i\pi\delta(z)$ in the limit that $\eta$ goes to zero from above and P(..) denotes the principal value it is straight forward to see that the imaginary part of the lindhard function is zero except of the case when we have electron-hole excitations of the form $\hbar\omega=\epsilon_{k+q}-\epsilon_{k}$.

Now let's assume we study plasma excitations using the actual Lindhard function (not the bare one). These plasma excitations appear for small q outside of the electron hole continuum. Therefor we have a vanishing imaginary part of $\chi_{0}$ in this case and consequently since also a vanishing part of $\chi$ itself, since its a function of $\chi_{0}$ which is "real" except of the "$\chi_{0}$"-part.

However when deriving the plasma excitations using a small q expansion we obtain a none vanishing imaginary part of $\chi$ as given in equation (3.48) of the following document: http://www.itp.phys.ethz.ch/education/lectures_fs10/Solid/Notes.pdf

I don't see why this is consistent.

I'd be more than happy if you could help me.

Thanks in advance.

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The relation $z = \epsilon_{k+q} - \epsilon_{k} - \hbar\omega$ does not have to occur between an occupied and an unoccupied state, it can occur between two occupied states (or two unoccupied states). It is the density terms in the numerator that make occupied-occupied or unoccupied-unoccupied contributions zero - not sure if that helps... – Brendan Jan 18 '12 at 23:32
Your definition of the polarizability function was wrong, now it is fixed. – DaniH Apr 29 '12 at 16:36

1 Answer

There is no inconsistency. Indeed, when using the full polarizability in the random phase approximation

$\chi^{\text{RPA}}(\mathbf{q},\omega)=\dfrac{\chi^{\text{RPA}}_0(\mathbf{q},\omega)}{1-V(q)\chi^{\text{RPA}}_0(\mathbf{q},\omega)}$

you see that the imaginary part of the polarizability has been renormalized when taken into account the [renormalized] electron-electron interactions because $V(q)$ is no longer the bare Coulomb interaction. Note that, from the previous expression, if we set $\text{Im}\,\chi_0(\mathbf{q},\omega)=-i\delta$

$-\text{Im} \,\chi(\mathbf{q},\omega)=\pi \delta(1-V(q)\text{Re}\,\chi_0(\mathbf{q},\omega))$

The fact that the imaginary part of the polarizability function $\text{Im} \,\chi(\mathbf{q},\omega)$ is non-zero is related to the damping of the excitations and it is known as Landau damping.

It is possible to understand this linking $\text{Im} \,\chi(\mathbf{q},\omega)$ to the conductivity of the electron gas. Using the definition of the conductivity and the continuity equation we can show that

$e^2 \text{Im} \,\chi(\mathbf{q},\omega) = -\dfrac{1}{\omega} \mathbf{q}\cdot[\text{Re}\,\bar\sigma(\mathbf{q},\omega)]\cdot \mathbf{q}$

The real part of the conductivity is related to dissipation in the system [Joule heating] when a current $\mathbf{J}$ is flowing.

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