Average magnetic moment of atom in magnetic field ##B##

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SUMMARY

The discussion centers on deriving the average magnetic moment of an atom in a magnetic field, represented by the equation ##\langle \mu \rangle = \beta^{-1} (\partial \log Z / \partial B)##, where ##Z## is the canonical partition function defined as ##Z = \sum_{m=-j}^{j} \mathrm{exp}(\mu_0 \beta B m)##. The average energy is expressed as ##\langle E \rangle = - \frac{\partial}{\partial \beta} \log Z##, which relates to the magnetic moment through the equation ##\langle E \rangle = -\langle \mu \rangle B##. The discussion also references the Feynman-Hellman theorem to establish the connection between these results.

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from the partition function - am trying to show that ##\langle \mu \rangle = \beta^{-1} (\partial \log Z / \partial B)## where ##Z## is the canonical partition function for one atom, i.e. ##Z = \sum_{m=-j}^{j} \mathrm{exp}(\mu_0 \beta B m)##, and ##\mu = \mu_0 m##. The average energy:\begin{align*}
\langle E \rangle = - \frac{\partial}{\partial \beta} \log Z
\end{align*}and ##\langle E \rangle = -\langle \mu \rangle B ##. How do I get the derivative ##\partial B/ \partial \beta## to link the two results? Or is there another way.
 
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I guess you mean ##H_{\text{mag}}=-\vec{\mu}_{\text{mag}} \cdot \vec{B}## in the Hamiltonian; unfortunately ##\mu## is already reserved for the chemical potential of some charge or particle-number like conserved quantity (or quantities).

Then your formula follows from the Feynman-Hellman theorem as usual.
 

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