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### AuthorTopic: Understanding the equation for Spin Energy in Zero Field  (Read 804 times) !function(d,s,id){var js,fjs=d.getElementsByTagName(s)[0];if(!d.getElementById(id)){js=d.createElement(s);js.id=id;js.src="https://platform.twitter.com/widgets.js";fjs.parentNode.insertBefore(js,fjs);}}(document,"script","twitter-wjs"); (function() {var po = document.createElement("script"); po.type = "text/javascript"; po.async = true;po.src = "https://apis.google.com/js/plusone.js";var s = document.getElementsByTagName("script")[0]; s.parentNode.insertBefore(po, s);})();

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#### HenriqueCSJ

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##### Understanding the equation for Spin Energy in Zero Field
« on: October 16, 2018, 05:06:16 AM »

I’m working with a system that experiences spin frustration and for me this is the first time dealing with such a phenomenon. To grasp the concept, I’ve decided to go to the basics and read the classic book on Molecular Magnetism by Professor Olivier Kahn. On chapter 10.1 Kahn works to get the formula for the relative energies on Zero Field:

See a print of the page here: https://i.stack.imgur.com/Yyx4K.png

S' varies by an integer from 0 to $2S_{a}$ and for every S' value S varies by an integer from $|S'-S_{b}|$ to $S'+S_{b}$ My main interest is the chapter 10.1.1 Copper(II) trinuclear species where Kahn states that we have Sa = Sb = ½ and that the relative energies of the states can be deduced from equation 10.1.6 (the book says 10.6 but it was probably a typo).

E(1/2,1) = 0
E(3/2,1) = -3J/2
E(1/2,0) = -J + J’

Substituting the values for S and S’ I’m failing to get the same results of 0, -3J/2 and -J + J’ and I’m getting, instead, 0.625J - 1.0J′, -0.875J - 1.0J′, −0.375J. You can see how I’m working to get these strange results here in this Jupyter Notebook: https://nbviewer.jupyter.org/github/HenriqueCSJ/spin-frustration/blob/master/Spin_frustration.ipynb

Maybe someone with more experience could guide me on how do I get the same results as the book? Thanks in advance
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