2
$\begingroup$

In Ashcroft/Mermin's solid state physics, in equation (17.64) they argued that:

We expect from lowest-order perturbation theory (Born approximation) that $\tau$ will depend on the electron-electron interaction through the square of the Fourier transform of the interaction potential.

$\tau$ is the e-e scattering time, which means the average time for a pair of particles to scatter. The inverse of it, $1/\tau$ is the scattering rate.

Then they use the Tomas-Fermi screening potential so that:

$$ \frac{1}{\tau} \sim (k_BT)^2(\frac{4\pi e^2}{k_0^2})^2 $$

After the dimensional analysis, they argued:

$$ \frac{1}{\tau}\sim A(k_BT)^2\frac{\hbar}{E_F} $$

Where $A$ is dimensionless quantities and $E_F$ is Fermi energy. Then they argued that A is of order of a power or two of ten which I can't understand, why should $A$ be this order?

Another question very related to this is that in graphene at half filling. $E_F$ is zero which cause the the scattering rate to be zero as it seems. However in graphene the independent electron approximation is rather good. So can anyone give a physics picture of this?

$\endgroup$
1
$\begingroup$

I'll comment on the second issue that $E_F$ should always be greater than zero, which leads to a finite scattering rate. However, that's beside the point because in an ideal Fermi liquid at $T\rightarrow 0, |E_F-E|\rightarrow0$, the scattering rate becomes zero, which leads to an infinite lifetime. If in graphene, as you say, the independent particle picture is good then the scattering should be low (or zero in the limit). Whenever the scattering rate is non-zero it means that the single-particle states are not the exact eigenstates of the system and thus the independent particle picture is not entirely valid. The particles are scattering in and out of the single-particle levels, which are only approximately stationary.

I agree that the value of the $A$ factor seems to be taken from thin air in the book. If I follow their insertion of $m/\hbar^7$ into Eq. 17.65, I get $A=\pi^4/2\approx49$. Perhaps the range $A=1\ldots100$ is given as a crude measure of uncertainty in their simplistic approach. They only need it in the next paragraph to make a rough estimation on the importance of e-e scattering.

$\endgroup$
  • $\begingroup$ At $0$ T for an ideal Fermi liquid, the scattering rate is zero means that they can be treated as exactly independent particles? $\endgroup$ – 喵喵是我的猫猫 Jul 14 '15 at 1:45
  • $\begingroup$ Your solution to the graphene problem is by claiming that in graphene half filling is impossible, right? Ok, if it can't be exactly half-filling, then at least $E_f$ should much smaller than that in ordinary metal, the scattering rate should also much larger than in normal metal. Then why treating it as single particle is good. Your answer seems not counting this problem. $\endgroup$ – 喵喵是我的猫猫 Jul 14 '15 at 1:51
  • $\begingroup$ Actually I am not claiming that half-filling is impossible. Quick googling shows that indeed people sometimes put $E_F=0$ for graphene. It's strange because I always assumed $E_F$ must be positive. Perhaps it's a matter of definition. In the book $E_F$ refers to the kinetic energy of the highest occupied state and is defined via the Fermi velocity, $E_F = mv_F^2/2$ (in graphene $v_F \sim 10^8$ ms$^{-1}$). $\endgroup$ – Raul Laasner Jul 14 '15 at 20:32
  • $\begingroup$ Note that the equations presented in the book are strictly valid only for the uniform electron gas. For real systems the behavior can sometimes be quite different. $\endgroup$ – Raul Laasner Jul 14 '15 at 20:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.