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I think most of you are familiar with this model (sum runs over nearest neighbours):
H = -J ∑S_iz * S_jz
It demonstrates one of the succeses of meanfield theory as one can succesfully introduce:
S_iz = <S_iz> + S_iz - <S_iz> = <S_iz> + δS_iz
Such that:
S_iz*S_jz ≈ 2S_iz<S_jz> + const
Where I have neglected the second order term. Now my question:
In the presence of an external magnetic field the Hamiltonian gets introduced a second term which couples the spins to the external magnetic field:
H=H_ising + ∑ S_iz * B
In this case can I still use the mean-field approximation separately for the Ising term? My book certainly does it, but I am a bit confused because in my head the average <S_iz> is affected by the external magnetic field's effect on the spin, and in this case it is not for me obvious that just because the deviation from the average is small in the case of no external field, it should be too in the presence of one.
H = -J ∑S_iz * S_jz
It demonstrates one of the succeses of meanfield theory as one can succesfully introduce:
S_iz = <S_iz> + S_iz - <S_iz> = <S_iz> + δS_iz
Such that:
S_iz*S_jz ≈ 2S_iz<S_jz> + const
Where I have neglected the second order term. Now my question:
In the presence of an external magnetic field the Hamiltonian gets introduced a second term which couples the spins to the external magnetic field:
H=H_ising + ∑ S_iz * B
In this case can I still use the mean-field approximation separately for the Ising term? My book certainly does it, but I am a bit confused because in my head the average <S_iz> is affected by the external magnetic field's effect on the spin, and in this case it is not for me obvious that just because the deviation from the average is small in the case of no external field, it should be too in the presence of one.