# Magnetic susceptibility and Curie's law

If an atom is exposed to an external magnetic field, its energy levels will be splitted due to (at least) the Zeeman effect. Thus the hamiltonian of the system is:

${\hat {H}}={\frac {e}{m}}B{\hat {S}}_{z}$ We want to evaluate $C_{v}={\frac {d\left\langle E\right\rangle }{dT}}$ and $\left\langle \mu _{z}\right\rangle$ . Being $\Delta$ the energy levels split, we have:

$Z=e^{\beta \Delta }+e^{-\beta \Delta }$ Using (Canonical ensemble) we get:

$\left\langle E\right\rangle =-\Delta \tanh(\beta \Delta )$ Hence:

$C_{v}=-{\frac {1}{kT^{2}}}{\frac {\partial \left\langle E\right\rangle }{\partial \beta }}$ Being ${\overrightarrow {\mu }}=-{\frac {e}{2m}}g{\overrightarrow {S}}$ , where $g$ is called Lande factor and it takes on the value $g\simeq 2$ for the spin operator, we get:

$\left\langle \mu _{z}\right\rangle =\mu _{B}{\frac {e^{-\beta \Delta }}{Z}}-\mu _{B}{\frac {e^{\beta \Delta }}{Z}}=-\mu _{B}\tanh(\beta \Delta )\simeq -\mu _{B}\beta \Delta$ $\Rightarrow \left\langle \mu _{z}\right\rangle ={\frac {\mu _{B}^{2}B}{ZkT}}$ In a more general case, that is if we do consider both the spin and the angular momentum, we have:

${\hat {H}}=g_{LSJ}{\frac {e}{2m}}{\underline {B}}{\underline {J}}$ Since this new hamiltonian is of the form ${\hat {H}}=-{\underline {\mu }}_{eff}{\underline {B}}$ , we have that ${\underline {\mu }}_{eff}=-g_{LSJ}{\frac {e}{2m}}{\underline {J}}$ . Hence $\mu _{eff}^{2}=g_{LSJ}^{2}\mu _{B}^{2}j(j+1)$ and thus $\mu ={\frac {\mu _{eff}^{2}B}{3kT}}$ . In conclusion we get the Curie's law:

$\chi ={\frac {\mu _{eff}^{2}}{3kT}}={\frac {C}{T}}$ Where the quantity $\chi$ is called magnetic susceptibility, while $C={\frac {\mu _{eff}^{2}}{3k}}$ is the Curie constant. Actually all these quantities have to be defined per number of particles per unit of volume, that is:

$\chi ={\frac {N}{V}}{\frac {\mu _{eff}^{2}}{3kT}}={\frac {C}{T}}$ This is a quantum result, but it can be derived classicaly. In this case we'd have:

$\left\langle \mu _{z}\right\rangle =\mu \left(\coth(\gamma )-{\frac {1}{\gamma }}\right)$ Being $\gamma =\beta \mu B$ , this function is called Langevin function.

1. We do have: ${\hat {H}}=-{\overrightarrow {\mu }}{\overrightarrow {B}}$ . Being ${\overrightarrow {\mu }}=-{\frac {e}{2m}}g{\overrightarrow {B}}{\overrightarrow {S}}$ (if we do consider only spin), and $g$ is the Lande (or gyromagnetic) factor which takes on the value $g=2$ for ${\overrightarrow {S}}$ . Concluding that: ${\hat {H}}={\frac {eB}{m}}{\hat {S}}_{z}$ .