# Origin of 2π normalization factor in chemical $J$-coupling compared to the Heisenberg model

In chemistry, particularly the field of NMR spectroscopy, the interaction between two (nucleic) spins (or so I guess?) is governed by the Hamiltonian:

\begin{align} \mathcal{H}=2\pi\cdot J_{ij}{\vec {S_{i}}}\cdot {\vec {S_{j}}} \label{eq1} \end{align} https://en.wikipedia.org/wiki/J-coupling

This interaction is called $$J$$-coupling and it is usually heuristically explained by interaction between two electronic spins which is governed by the Pauli principle and the electron-nucleus interaction which in the end favors antiparallel alignment of the nucleic spins over parallel and thereby leads to a splitting of energy levels. Note that mainly chemists cover this topic and their approach is usually very hands-on in the sense that the values of $$J_{ij}$$ are usually measured on a spectrometer.

Now the above equation looks very similar to the Heisenberg model, governing the general relationship between two spins:

\begin{align} H_{\text{Heis}}=-J\sum _{\langle i,j\rangle }{\vec {S_{i}}}\cdot {\vec {S_{j}}}\qquad {\text{with }}i,j\,{\text{next neighbors}} \label{eq2} \end{align}

https://en.wikipedia.org/wiki/Heisenberg_model_(quantum)

This surely looks similar, here $$J$$ is the value of the exchange integral, which I would assume is also what defines the scalar coupling tensor $$J_{ij}$$ in the first equation. However, I am by no means sure those two $$J$$s are the same and if they are, where does the normalization factor of $$2\pi$$ coming from? Is it just common among chemists to divide their exchange integrals by $$2\pi$$? I would be very grateful if someone could shed a little light on that.

• this is a guess, but pulling out 2$\pi$ would make it more convenient to express the coupling in units of Hz instead of inverse second. – wcc Apr 26 '19 at 14:04