The quantum hamiltonian of Heisenberg model is usually given in the form: $$\textit{H}_{heisenberg}=-\sum_{i,j}J_{ij}\vec{S}_{i}\cdot\vec{S}_{j}-g\mu_{B}\vec{H}\cdot\sum_{i}\vec{S}_{i}$$ such that first term corresponds to exchange interaction between spin, and second term corresponds to interaction of spin with external magnetic field.

I am confused about the second term, because most electrodynamics course say the interaction of magnetic dipole and field is given by $-\vec{m}\cdot\vec{B}$, not $-\vec{m}\cdot\vec{H}$. On the other hand, many magnetism textbook describe interaction as $-\vec{m}\cdot\vec{H}$, usually based on Coulombian view of magnetic monopole. Since this model is used in description of ferromagnetic material, the difference between $\vec{B}$ and $\vec{H}$ can't be just omitted, isn't it?

Another confusing point is that the term $-g\mu_{B}\vec{H}\cdot\vec{S}_i$ is describing interaction of individual atom with magnetic field, so it seems that the field must be local field $\vec{H}_{loc}=\vec{\mathfrak{H}}+\vec{H}_{d}$ (or $\vec{B}_{loc}=\vec{\mathfrak{B}}+\vec{B}_{d}$), rather than external field $\vec{\mathfrak{H}}$ (or $\vec{\mathfrak{B}}$). I failed to find clear manifestation about this distinction, and this makes me even hesitating. Since the difference between Coulombian view of monopole with field H and Amperian view of current loop with field B causes different internal structure of local field, it will directly effect local field we are considering.

In summary, my question is: what is the field H in interaction term $-g\mu_{B}\vec{H}\cdot\vec{S}_i$? Is it magnetic field $\vec{H}$, or magnetic induction $\vec{B}$? Is it external field given in experimental setup, or internal local field directly acting in individual atom?

P.S. I'd consulted the book "Electrodynamics of Materials: Forces, Stresses, and Energies in Solids and Fluids" from Scipione Bobbio, which argues both of Coulombian and Amperian view is identical in thermodynamic sense describing whole field and matter, and the difference in energy term just arises from difference between thermodynamic internal free energy. I'm not sure this view is widely accepted...

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    $\begingroup$ In microscopic electrodynamics there is no difference between $B$ and $H$, so one often uses the latter letter. The distinction between $B/H$ and $D/E$ arises when we work with macroscopic Maxwell equations (check in Wikipedia, also here). $\endgroup$
    – Roger V.
    Jun 6, 2022 at 13:46
  • $\begingroup$ @RogerVadim Thank you for replying! I see, there are no microscopic d and h, since they are introduced to deal with macroscopic average of bound charge and current. In that case, is it appropriate to assign same microscopic field h with every spin Si of atoms in the Heisenberg model? Each microscopic field must vary a lot from external field, since neighboring atom will produce field of order $\frac{\mu_B}{a^3}\approx0.1T$. Or may I just treat Heisenberg model as a crude model, considering statistical average of field-moment interaction as H*$S_i$? $\endgroup$ Jun 6, 2022 at 16:36
  • $\begingroup$ @RogerVadim "In microscopic electrodynamics there is no difference between 𝐵 and 𝐻". Well, I would not agree completely. It is enough to think about what happens if one has to write the equations using SI units. The annoying $\mu_0$ factor introduces a difference. $\endgroup$ Jun 6, 2022 at 20:32
  • $\begingroup$ @GiorgioP to be more precise, there are no fields D and H in the microscopic equations, but H was historically often used instead of B. The difference that you are talking about is just a difference in scale/units, used for the same quantity. This is why SI units are rarely used in science, when working on microscopic problems. $\endgroup$
    – Roger V.
    Jun 7, 2022 at 5:05

1 Answer 1


You are right; the SI units are just a pain in the $\vec{B}$ (!). And also you are right about the distinction between $\vec{H}$ and $\vec{B}$, for the former couples with ''monopoles'' but the latter couples with ''Amperian currents''. Although such distinction is of no relevance in a microscopic model, as pointed out in the comments, $H$ in these models is almost always ''the field filling the space before introducing the system to it'', the ''external field'' if I may. One can denote it best as $B_0$ or $B_{ext}$. And when it should couple to the magnetic moments $m_i$, the corresponding term is a Zeeman one, $-B_0m_i$, and then you can sum over all moments. Its spatial variation is so slow that it can appear as a mere parameter in the Hamiltonian, not as an operator or a degree of freedom. No ''local field'' concept has run into this definition. In fact, such term will be present in a coarse-grained model through the local and, perhaps, nonlocal interactions given by the $J_{ij}$ couplings.

Lastly, you can of course denote the external field equivalently with $\mu_0H_0 = B_0$.

  • $\begingroup$ As a side note about your statement about the local field, I should mention that while the Zeeman term appears as $-B_0m$, upon coarse-graining you can write the effective Hamiltonian as some local potential plus $-\frac{1}{2}\int H_d \cdot M dv$ plus the external coupling $-\int B_0 \cdot M dv$. Note the one-half factor. $\endgroup$
    – Bjaam
    Sep 4, 2022 at 7:59

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