I've been recently reading about the derivation of the Bleaney-Bowers equation from Van Vleck paramagnetism and I am looking forward to derive it but using a statistical mechanics formalism instead.
The Bleaney-Bowers equation for the magnetic susceptibility is:
$$\chi = \frac{2ng_S\mu_B^2}{k_BT(3+e^{\Delta/k_BT})}$$
Now, in order to apply the statistical mechanics formalism, I've come up with the following system: N independent systems that do not interact with each other made by two spin-1/2 electrons that interact between themselves. As a result, there is a triplet state (S=1) and a single state (S=0), separated by a energy gap $\Delta$.
This is kind of a Heisenberg interaction whose hamiltonian per pair is:
$$\hat{h_1} = K ·\hat{\vec{S^a}}·\hat{\vec{S^b}}$$
Where $\hat{\vec{S^a}}$ and $\hat{\vec{S^b}}$ are the spin operators (4×4 spin matrices) for each particle and K is a positive constant. I've assumed that because it must have units of energy, $K = g_S\mu_B B$, where $g_S$ is the Landé factor (L=0), $\mu_0$ is the Bohr magneton and B is the applied magnetic field, but I don't know if such an equality for K is at all correct. In matrix representation (I've taken $\hbar=1$),
$$\hat{\vec{S^a}}·\hat{\vec{S^b}}= \frac{1}{4}\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 2 & 0 \\
0 & 2 & -1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$$
Then, the partition function for this hamiltonian is
$$Z_1 = Tr(e^{-\frac{\hat{h_1}}{k_BT}}) = 3e^{-\frac{K}{4k_BT}}+e^{\frac{3K}{4k_BT}}$$
Where I've made use of the eigenvalues of the matrix. From here on, my idea is to obtain the Helmholtz free energy $F = -Nk_BT\ln{Z_1}$ to later obtain the magnetic susceptibility: $\chi = -\frac{\partial^2 F}{\partial B^2}$, but by doing that I arrive to the final expression:
$$\chi = \frac{3Ng_S^2\mu_B^2 e^{g_S\mu_BB/k_BT}}{k_BT(3+e^{g_S\mu_BB/k_BT})^2}$$
Which is not the Bleaney-Bowers equation. Do you know what further assumptions I should take? Or what is wrong in my approach?
Thanks
Best Answer
You have made an incorrect assumption that $K = g_S\mu_B B$. In fact, $K$ is related to the energy gap $\Delta$, not the external magnetic field $B$. To simplify the notation, we consider a two-spin Hamiltonian of the form $$ \hat{H}_2 = J \hat{\vec{S}_1}\hat{\vec{S}_2} - h(\hat{S}_1^z + \hat{S}_2^z), $$ where $J$ and $h$ are measured in energy units and characterize the spin-spin interaction and the interaction with an external magnetic field. For two spins of $1/2$, the partition function is $$ Z_2(\theta) = e^{\frac{3J}{4\theta}} + e^{-\frac{J}{4\theta}}\left(e^{\frac{h}{\theta}} + 1 + e^{-\frac{h}{\theta}} \right), $$ where $\theta$ is the temperature in energy units. The mean $z$-projection of the total spin is $$ \langle S^z \rangle = \theta\frac{\partial}{\partial h}\ln Z_2(\theta) = \frac{2\sinh(h/\theta)}{e^{\Delta/\theta} + 1 + 2\cosh(h/\theta)}, $$ where $\Delta = J$. The zero-field susceptibility of the two-spin system is $$ \tilde{\chi} = \left. \frac{\partial \langle S^z \rangle}{\partial h}\right|_{h = 0} = \frac{2}{\theta(e^{\Delta/\theta} + 3)}. $$ From this, it is easy to obtain a formula for quantities in the required units of measurement.