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

[MathematicalPhysicist]

Homework Statement

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).

Homework Equations

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

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

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

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

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.

 

[Charles Link]

Homework Statement

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).

Homework Equations

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

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

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

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

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.

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.