- #1
gaiussheh
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- TL;DR Summary
- In solid state physics or statistical physics, the many-body spin Hamiltonian is written as ##-E_{i,j}\sum_{i,j} \hat{S}_{i}\cdot \hat{S}_j##. How is this generalised from the two-electron case?
In solid state physics or statistical physics, the many-body spin Hamiltonian is written as ##\sum_{i,j} \hat{S}_{i}\cdot \hat{S}_j##. I referred to many textbooks, and most of them just generalised this from the two-electron case ##\hat{S}_{1}\cdot \hat{S}_2##. While this seems natural, it is completely unlogical as this is not the magnetic dipole itself but the exchange interaction that arose from the symmetry of the spatial wave function.
Take a step back. Two electrons can form either a singlet or a triplet. For a singlet, ##\hat{S}_{1}\cdot \hat{S}_2=-\frac{3}{4}## and the spin part is symmetric, hence the spatial part is antisymmetric. For a triplet, ##\hat{S}_{1}\cdot \hat{S}_2=\frac{1}{4}## and the spin part is antisymmetric, hence the spatial part is asymmetric. This arise different energies ##E_{\rm S}=\langle\psi_{\rm S}|\hat{\mathcal{H}}|\psi_{\rm S}\rangle## and ##E_{\rm T}=\langle\psi_{\rm T}|\hat{\mathcal{H}}|\psi_{\rm T}\rangle##. However, both can be written as
##E_{\rm T} = \frac{1}{4}E_{\rm S}+\frac{3}{4}E_{T}-\frac{1}{4}(E_{\rm S}-E_{\rm T})##
##E_{\rm S} = \frac{1}{4}E_{\rm S}+\frac{3}{4}E_{T}+\frac{3}{4}(E_{\rm S}-E_{\rm T})##
Hence in any case ##E=\frac{1}{4}E_{\rm S}+\frac{3}{4}E_{T}-\hat{S}_{1}\cdot \hat{S}_2(E_{\rm S}-E_{\rm T})##
The energy due to spin is ##-E_{1,2}\hat{S}_{1}\cdot \hat{S}_{2}##, where ##E_{1,2} = E_{\rm S}-E_{\rm T}##.
I find it not that easy to generalise to a many-body system. In general, you will need a Slater determinant to describe the wave function and consider the effect of exchanging any two electrons. I don't even know if it is possible to write this into the form of ##\psi_{\rm spatial}[{\vec r}_1, {\vec r}_2, \cdots, {\vec r}_n]\cdot[{\rm spin~of~i~and~j}] \cdot[{\rm spin~of~other~electrons}]## etc. (Actually, I think you can't let the slater determinant be an eigenstate of ##\hat{S}_{i}\cdot \hat{S}_j## for all pair of ##i,j##).
How is this generalised at all?
Take a step back. Two electrons can form either a singlet or a triplet. For a singlet, ##\hat{S}_{1}\cdot \hat{S}_2=-\frac{3}{4}## and the spin part is symmetric, hence the spatial part is antisymmetric. For a triplet, ##\hat{S}_{1}\cdot \hat{S}_2=\frac{1}{4}## and the spin part is antisymmetric, hence the spatial part is asymmetric. This arise different energies ##E_{\rm S}=\langle\psi_{\rm S}|\hat{\mathcal{H}}|\psi_{\rm S}\rangle## and ##E_{\rm T}=\langle\psi_{\rm T}|\hat{\mathcal{H}}|\psi_{\rm T}\rangle##. However, both can be written as
##E_{\rm T} = \frac{1}{4}E_{\rm S}+\frac{3}{4}E_{T}-\frac{1}{4}(E_{\rm S}-E_{\rm T})##
##E_{\rm S} = \frac{1}{4}E_{\rm S}+\frac{3}{4}E_{T}+\frac{3}{4}(E_{\rm S}-E_{\rm T})##
Hence in any case ##E=\frac{1}{4}E_{\rm S}+\frac{3}{4}E_{T}-\hat{S}_{1}\cdot \hat{S}_2(E_{\rm S}-E_{\rm T})##
The energy due to spin is ##-E_{1,2}\hat{S}_{1}\cdot \hat{S}_{2}##, where ##E_{1,2} = E_{\rm S}-E_{\rm T}##.
I find it not that easy to generalise to a many-body system. In general, you will need a Slater determinant to describe the wave function and consider the effect of exchanging any two electrons. I don't even know if it is possible to write this into the form of ##\psi_{\rm spatial}[{\vec r}_1, {\vec r}_2, \cdots, {\vec r}_n]\cdot[{\rm spin~of~i~and~j}] \cdot[{\rm spin~of~other~electrons}]## etc. (Actually, I think you can't let the slater determinant be an eigenstate of ##\hat{S}_{i}\cdot \hat{S}_j## for all pair of ##i,j##).
How is this generalised at all?
