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Susceptibility of a simple metal (Problem 31.6 in Ashcroft's

  1. May 14, 2017 #1

    MathematicalPhysicist

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    1. The problem statement, all variables and given/known data
    The susceptibility of a simple metal has a contribution ##\chi_{c.c}## from the conduction electrons and a contribution ##\chi_{ion}## from the diamagnetic response of the closed-shell core electrons.
    Taking the conduction electron susceptibility to be given by the free electron values of the Pauli paramagnetic and Landau diamagnetic susceptibilities, show that:

    $$ \frac{\chi_{ion}}{\chi_{c.c}} = -\frac{1}{3} \frac{Z_c}{Z_v}\langle (k_F r)^2 \rangle $$

    where ##Z_v## is the valence, ##Z_c## is the number of core electrons, and ##<r^2>## is the mean square ionic radius defined in (31.26).


    2. Relevant equations
    $$(31.26) <r^2> = \frac{1}{Z_i} \sum <0|r_i^2 |0>$$

    $$\chi^{molar} = -Z_i (e^2/(\hbar c))^2 \frac{N_A a_0^3}{6}\langle (r/a_0)^2 \rangle$$

    $$\chi_{pauli} = \bigg(\frac{\alpha}{2\pi}\bigg)^2 (a_0k_F)$$

    $$\chi_{Landau} = -1/3 \chi_{Pauli}$$

    3. The attempt at a solution
    I thought that ##\chi_{molar}=\chi_{ion}## and that ##\chi_{c.c} = \chi_{Landau}+\chi_{Pauli} = 2/3 \chi_{Pauli}##.

    But when I divide between the two susceptibilities I don't get the right factors, has someone already done this exercise from Ashcroft and Mermin?

    I tried searching google for a solution but to a veil.
     
  2. jcsd
  3. May 14, 2017 #2

    Charles Link

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    Homework Helper

    I was able to find something in a google search. For the closed shell electrons, this form of diamagnetism is described by the Langevin theory, as opposed to free electron diamagnetism, which is the Landau theory.
     
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