Apparently it is to be shown that for electrons under an external magnetic field, in the limit as $B\to 0 $ $$ \chi = \frac{dM}{dB} \approx \frac{n\,\mu^{*^2}}{k\,T}\,\frac{f_{1/2}(z)}{f_{3/2}(z)} $$

So this is the Pauli paramagnetism problem and I understand the effect and the physics, but cannot seem to get the requested ratio of $f_{\nu}(z)$ functions.

Here is how I have been doing it:

With the grand canonical ensemble formalism I have $$ \langle N_{\pm} \rangle = \int_0^{\infty} d\epsilon\, g(\epsilon)\,\frac{1}{\exp^{\beta(\epsilon-\mu\pm \mu_B\,B)}} $$ where $$ g(\epsilon) = \frac{2}{\sqrt{\pi}}\,\left(\frac{2\,\pi\,m}{h^2}\right)^{3/2}\;V\,\sqrt{\epsilon} $$ and is the density of states from translational motion. I then say that $$ \langle M \rangle = \mu_B\,(\langle N_- \rangle - \langle N_+ \rangle ) = \cdots = Const*V*[f_{1/2}(\beta\,\mu+\mu_B\,B)-f_{1/2}(\beta\,\mu-\mu_B\,B)] $$

This is fine I think except that when I calculate $\chi$ I will get fermi-dirac functions with -1/2 and not the ratio I need.

$$ \chi = \lim_{B\to 0} \frac{\partial\,M}{\partial\,B} = \cdots ? $$

  • $\begingroup$ Anybody? Or is this too....? $\endgroup$ – nate Apr 9 '13 at 4:33

I believe that you need to shift not Fermi distribution, but density of states. When electron with parallel spin feels magnetic field it's density of states shifts downward by energy $\mu_B B$.You also need to take integral with other limits. So the real quantities will be: $$ N_+ = \frac{1}{2}\int\limits^{\epsilon_F}_{-\mu_B B}d\epsilon f(\epsilon)D(\epsilon+\mu_B B)\\ N_- = \frac{1}{2}\int\limits^{\epsilon_F}_{\mu_B B}d\epsilon f(\epsilon)D(\epsilon-\mu_B B) $$ Then use equation: $M = \mu_{B}(N_+-N_-)$ and you'll get what you needed.

Information from C.Kittel "Introduction to Solid State Physics".

P.S. You need to evaluate integrals approximately, because you can't analytically evaluate them.


I would start with the derivation in the following way:
Your Hamiltonian has additionally to its kinetic part an extension by a Zeeman term: $$ H_Z = -\frac{g \mu _B }{\hbar} \sum_{i=1}^N s_i^z H$$ with $g=2$ the gyromagnetic ratio and $\mu _B$ the Bohr magneton. One can absorb the energy difference of the parallel and antiparallel spins into a spin-dependent fugacity: $$Z= \prod_{\vec{p}} \left( \sum_{n_p} \left( z e^{- \beta e_p +\beta \mu _B H} \right)^{n_p} \right) \left( \sum_{n_p} \left( z e^{- \beta e_p -\beta \mu _B H} \right)^{n_p} \right) = \prod _ {\vec{p}} \prod_{\sigma = +,-} \sum_{n_p} \left( z_{\sigma} e^{-\beta e_p}\right)$$ where i denoted $n_p$ as the occupation number of the state (compare the derivation of ideal fermi gas) and $e_p$ as the energy of the state. Therefore you find $$ z_{\pm} = e^{\beta (\mu \pm \beta \mu _B H)}$$ such that the density of the Fermions is given by: $$n = N/V = \frac{1}{\lambda^3}(f_{3/2}(z_+) + f_{3/2}(z_-)) = n_+ + n_-$$ which is derived using: This leads to the relation $$\Omega=-k_BTln(Z)= -\frac{k_B T}{\lambda ^3}(f_{5/2}(z_+) + f_{5/2}(z_-))$$ With this you should be able to calculate m: $$ m= V \mu_B (n_+ + n_-)$$ and $$\chi = \frac{\partial M}{\partial H} = \frac{2 \mu_B^2}{\lambda^3 k_B T} f_{1/2}(z)$$ I compared and found this result in some books and found it to be right. You can now calculate the high and low temperature limit and will receive the Pauli spin susceptibility. The function $$f_{n/2}= - \sum_{l=1}^{\infty} (-1)^l \frac{z^l}{l^{n/2}} $$. I am therefore not sure if the result above is correct. I hope this helps


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