I hope this is not a "homework" question, but I'm having a bad time trying to figure this out. Available literature proposes 3 equations to calculate the coupling constant during a Broken-Symmetry approach:

\begin{align*} J(1) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}^{2}} \\ J(2) = \frac{-\left ( E_{HS}-E_{BS} \right )}{S_{max}\left ( S_{max}+1 \right )} \\ J(3) = \frac{-\left ( E_{HS}-E_{BS} \right )}{\left \langle S_{HS}^{2} \right \rangle-\left \langle S_{BS}^{2} \right \rangle} \end{align*}

I'm intrigued by the fact that, from the equations, the more the system have unpaired electrons, the minor will be $J_{ab}$. Why does this happen? Doesn't more unpaired electrons increase magnetic momenta (and an increase in magnetic coupling))?

  • $\begingroup$ What do you mean by minor and $J_{ab}$ (to which $J$ does this refer)? $\endgroup$ – Deathbreath Jun 29 '16 at 15:56

I hope I have understood your question correctly.

For illustrative purposes, I will restrict myself to the Hartree-Fock picture. In a broken spin system, e.g., UHF, the spin state is not a good quantum number. That is to say that the UHF wavefunction need not be an eigenfunction of $S^2$ or $S_z$. For example two complexes with a single unpaired electron each will combine three singlet states (two ionic states and one open-shell) and three triplet states ($m_s\in \{-1,0,1\}$).

The open-shell singlet ($|\alpha\phi_1 \beta\phi_2|-|\alpha\phi_2 \beta\phi_1|$) and the $m_s=0$ triplet ($|\alpha \phi_1 \beta \phi_2|+|\alpha\phi_2 \beta\phi_1|$) states are not single Slater determinants. The singlet states need not be of the same energy, whereas the triplet states have to be degenerate for a Hamiltonian without explicitly spin-dependent components such as spin-orbit coupling. Hence, you have to accommodate for the multiple counting.

For higher spins, the degeneracy grows with $S_{max}$ and must be corrected accordingly. This does not imply that coupling necessarily decreases with higher spins.

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  • $\begingroup$ Thank you for your answer, @Deathbreath. I'm sorry if I haven't formulated my question in a clear way before (but you answered correctly). The doubt appeared when I was studying an isostructural coordination polymer of the kind [M(L)] (M is Co(II) or Co(II)). The comparison in energies from E(high spin) - E(singlet) is remarkable: Cobalt is 8,065 cm**-1 and copper is 0,804 cm**-1, but the coupling constants are very close: -0.9 for Co and -0.8 for Cu (with cobalt having more unpaired electrons). $\endgroup$ – Henrique Junior Jun 30 '16 at 21:22
  • $\begingroup$ So, from your answer, this difference in energy exists because if an atom has a higher multiplicity you still have to remember that the Hamiltonian is representing a "spin ladder" and not a single spin state, Correct? $\endgroup$ – Henrique Junior Jun 30 '16 at 21:22
  • $\begingroup$ I'm not sure what you mean by soon ladder, but the Hamiltonian commutes with $S^2$ and $S_z$, whereas the wave function is not a pure spin state. $\endgroup$ – Deathbreath Jul 2 '16 at 14:30

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