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

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1. May 14, 2017

### MathematicalPhysicist

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. May 14, 2017