Studying statistical physics, in particular studying paramagnets and ferromagnets, I found some quantum mechanical formulas that I don't understand how to interpret. I don't understand if they are equations between operators, eigenvalues or mean values (or something else that I don't know). Can anyone tell me how to interpret these quantum equations (i write examples of them below)? To be clear, I am not asking for an explanation of physical phenomena but an explanation of the notation, which to me looks like a mix of eigenvalue operators and expectation values together.
Example about paramagnets
magnetic moment of each single ion
$$\vec \mu=g \mu_B \vec J$$
where g is the Landé g-factor
$$g=1+\frac {j(j+1)+s(s+1)-l(l+1)}{2j(j+1)}$$
It follow that, if there is a magnetic field, the interaction energy is
$$H=-\vec \mu \cdot \vec B=-g \mu_B B m_j$$
where the possible values of $m_j$ are $(-j,-j+1,…0,…,j-1,j)$
Example about ferromagnets
Hamiltonian of the system in the presence of $\vec B$
$$H=-g \mu_B \vec B \cdot \sum \vec S_i – J \sum \vec S_i \cdot \vec S_j $$
Best Answer
Note: I equate expectation values below to emphasize which constants can be moved outside bra-kets. To get the operator equations from such these expectation equations follow, just mentally delete the bra-kets. Nothing here contradicts any other users' insights, given which operators are mutually diagonalizable with the Hamiltonian.
In this case it would be more helpful to write${}^\dagger$ $$\left\langle\hat{H}\right\rangle=\left\langle-\hat{\vec{\mu}}\cdot\vec{B}\right\rangle=-\mu_BB\langle gm_j\rangle.$$The first set of angle brackets were likely dropped because an eigenstate of the Hamiltonian is assumed. Why someone woud drop the second set will hopefully become clear from the footnote below.
${}^\dagger$ We probably put $\vec{B}$ in "by hand", making it classical rather than quantized, which is why I didn't give it an operator hat. The second expression can be rewritten can then be rewritten as e.g. $-\left\langle\hat{\vec{\mu}}\right\rangle\cdot\vec{B}$. The resulting empirical law, by which the eigenenergy is a dot product, allows us to identify $\vec{\mu}$ with its mean.
Edit to address new ferromagnet example: this should be clarified as$$\left\langle\hat{H}\right\rangle=-\mu_B\vec{B}\cdot\sum_i\left\langle g\hat{\vec{S}}_i\right\rangle-\sum_{ij}\left\langle\hat{J}\hat{\vec{S}}_i\cdot\hat{\vec{S}}_j\right\rangle.$$