As we know, no long range order in a one dimensional electron system is expected due to quantum fluctuation. A typical 'phase diagram' for a system with short-range interactions is shown on page 69 of Giamarchi's Quantum Physics in One Dimension.

My question is about a basic claim (page 11-12) in his book that a repulsive interaction in a general system can drive an instability in the charge sector by using the RPA calculation (summing over only bubbles). His argument is based on Eq (1.22): $$\langle \rho{\uparrow}(q)\rangle = \frac{\chi^0(q) V^{\uparrow}(q)}{1+U(q)\chi^0(q)}.$$ Note that this expression is for spins but he seems to use the same expression for charges. This is in contradiction with my understanding of RPA calculation which, for charges, has a denominator in the form of $$ 1-V(q)\chi^0(q).$$ Please refer to Mahan's Many Particle Physics, section 6.3, even though Mahan deals with the extended Hubbard-like model. Do you have a clear explanation?

Edit: I feel obligated to attach the frequency $\omega$ dependence of $\chi^0(q,\omega)$ because we know that it can change sign at a finite $\omega$ which is associated with plasmons. Please explain the sign convention.

  • $\begingroup$ kindly explain RPA $\endgroup$ – hsinghal Jun 22 '16 at 14:13
  • $\begingroup$ @hsinghal Random Phase Approximation. $\endgroup$ – DarKnightS Jun 22 '16 at 14:16
  • $\begingroup$ Thanks, always define abbreviations in scientific writing. RPA also stan for radiation pressure acceleration. $\endgroup$ – hsinghal Jun 22 '16 at 15:13
  • $\begingroup$ Have you carefully checked signs ? And what is $U(q)$ ? $\endgroup$ – Dimitri Jun 27 '16 at 9:51
  • $\begingroup$ @Dimitri Yes. $U(q)$ is the Hubbard interaction. If we use the convention making $\chi^0(q)$ opposite of the Lindhard function, then it is negative, and repulsive $U(q)$ is positive. $\endgroup$ – DarKnightS Jun 27 '16 at 14:15

Your Answer

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

Browse other questions tagged or ask your own question.