Jayden Mckay

2022-05-14

Given the quantum Heisenberg model with Hamiltonian

$\hat{H}=-\frac{1}{2}\sum _{i,j}{J}_{ij}{\hat{\mathbf{\text{S}}}}_{i}\cdot {\hat{\mathbf{\text{S}}}}_{j}$,

the uniform mean-field approximation ${\hat{\mathbf{\text{S}}}}_{i}=\u27e8\hat{\mathbf{\text{S}}}\u27e9+({\hat{\mathbf{\text{S}}}}_{i}-\u27e8\hat{\mathbf{\text{S}}}\u27e9)$ allows to rewrite it as

${\hat{H}}_{MF}=\sum _{i}{\mathbf{\text{B}}}_{eff}\cdot {\hat{\mathbf{\text{S}}}}_{i}+\text{const.}$,

in order to perform a diagonalization by means of a Fourier transform. To do so, I am told to choose the z-axis so to align with the effective field ${\mathbf{\text{B}}}_{eff}$, which I find to be ${\mathbf{\text{B}}}_{eff}=-\u27e8\hat{\mathbf{\text{S}}}\u27e9\sum _{j}({J}_{ij}+{J}_{ji})$. That's where I remain stuck. First of all, how should the Fourier transform which allows me to diagonalize ${\hat{H}}_{MF}$ look like? And how is this related to the alignment of the effective field?

$\hat{H}=-\frac{1}{2}\sum _{i,j}{J}_{ij}{\hat{\mathbf{\text{S}}}}_{i}\cdot {\hat{\mathbf{\text{S}}}}_{j}$,

the uniform mean-field approximation ${\hat{\mathbf{\text{S}}}}_{i}=\u27e8\hat{\mathbf{\text{S}}}\u27e9+({\hat{\mathbf{\text{S}}}}_{i}-\u27e8\hat{\mathbf{\text{S}}}\u27e9)$ allows to rewrite it as

${\hat{H}}_{MF}=\sum _{i}{\mathbf{\text{B}}}_{eff}\cdot {\hat{\mathbf{\text{S}}}}_{i}+\text{const.}$,

in order to perform a diagonalization by means of a Fourier transform. To do so, I am told to choose the z-axis so to align with the effective field ${\mathbf{\text{B}}}_{eff}$, which I find to be ${\mathbf{\text{B}}}_{eff}=-\u27e8\hat{\mathbf{\text{S}}}\u27e9\sum _{j}({J}_{ij}+{J}_{ji})$. That's where I remain stuck. First of all, how should the Fourier transform which allows me to diagonalize ${\hat{H}}_{MF}$ look like? And how is this related to the alignment of the effective field?

Superina0xb4i

Beginner2022-05-15Added 17 answers

One could write:

$H=-\frac{1}{2}\sum _{i,j}{J}_{ij}[\u27e8{S}_{i}\u27e9+({S}_{i}-\u27e8{S}_{i}\u27e9)][\u27e8{S}_{j}\u27e9+({S}_{j}-\u27e8{S}_{j}\u27e9)]\approx -\frac{1}{2}\sum _{i,j}{J}_{ij}[\u27e8{S}_{i}\u27e9\u27e8{S}_{j}\u27e9+({S}_{i}-\u27e8{S}_{i}\u27e9)\u27e8{S}_{j}\u27e9+\u27e8{S}_{i}\u27e9({S}_{j}-\u27e8{S}_{j}\u27e9)]=\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\sum _{i,j}{J}_{ij}[{S}_{i}\u27e8{S}_{j}\u27e9+\u27e8{S}_{i}\u27e9{S}_{j}-\u27e8{S}_{i}\u27e9\u27e8{S}_{j}\u27e9]=\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\sum _{i,j}{J}_{ij}[{S}_{i}\u27e8{S}_{j}\u27e9+\u27e8{S}_{i}\u27e9{S}_{j}]+const=\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\sum _{i,j}[{J}_{ij}\u27e8{S}_{j}\u27e9+{J}_{ji}\u27e8{S}_{j}\u27e9]{S}_{i}+const=\phantom{\rule{0ex}{0ex}}-\sum _{i,j}{B}_{i}^{eff}{S}_{i}+const,$

where

${B}_{i}^{eff}=\frac{1}{2}\sum _{j}[{J}_{ij}\u27e8{S}_{j}\u27e9+{J}_{ji}\u27e8{S}_{j}\u27e9].$

Note that strictly speaking the effective field is inhomogeneous. It can be assumed homogeneous only under specific assumptions about ${J}_{ij}$, e.g., that it depends only on the distance between the sites, but not on their absolute position or their arrangement on the lattice.

Fourier transform on a discrete lattice becomes a discrete transform, which depends on the specific lattice vectors. E.g., in 1D one could write :

${b}_{k}^{eff}=\sum _{j}{e}^{ijk}{B}_{i}^{eff}.$

(note that i here is the imaginary unit) However, mean field expansion is known to give incorrect results in 1D, and the problem in question is probably done in multiple dimensions. Therefore one cannot say more without knowing more about the underlying lattice. Such expansions are extensively used in solid state physics, transforming from the crystal lattice to the reciprocal lattice.

$H=-\frac{1}{2}\sum _{i,j}{J}_{ij}[\u27e8{S}_{i}\u27e9+({S}_{i}-\u27e8{S}_{i}\u27e9)][\u27e8{S}_{j}\u27e9+({S}_{j}-\u27e8{S}_{j}\u27e9)]\approx -\frac{1}{2}\sum _{i,j}{J}_{ij}[\u27e8{S}_{i}\u27e9\u27e8{S}_{j}\u27e9+({S}_{i}-\u27e8{S}_{i}\u27e9)\u27e8{S}_{j}\u27e9+\u27e8{S}_{i}\u27e9({S}_{j}-\u27e8{S}_{j}\u27e9)]=\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\sum _{i,j}{J}_{ij}[{S}_{i}\u27e8{S}_{j}\u27e9+\u27e8{S}_{i}\u27e9{S}_{j}-\u27e8{S}_{i}\u27e9\u27e8{S}_{j}\u27e9]=\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\sum _{i,j}{J}_{ij}[{S}_{i}\u27e8{S}_{j}\u27e9+\u27e8{S}_{i}\u27e9{S}_{j}]+const=\phantom{\rule{0ex}{0ex}}-\frac{1}{2}\sum _{i,j}[{J}_{ij}\u27e8{S}_{j}\u27e9+{J}_{ji}\u27e8{S}_{j}\u27e9]{S}_{i}+const=\phantom{\rule{0ex}{0ex}}-\sum _{i,j}{B}_{i}^{eff}{S}_{i}+const,$

where

${B}_{i}^{eff}=\frac{1}{2}\sum _{j}[{J}_{ij}\u27e8{S}_{j}\u27e9+{J}_{ji}\u27e8{S}_{j}\u27e9].$

Note that strictly speaking the effective field is inhomogeneous. It can be assumed homogeneous only under specific assumptions about ${J}_{ij}$, e.g., that it depends only on the distance between the sites, but not on their absolute position or their arrangement on the lattice.

Fourier transform on a discrete lattice becomes a discrete transform, which depends on the specific lattice vectors. E.g., in 1D one could write :

${b}_{k}^{eff}=\sum _{j}{e}^{ijk}{B}_{i}^{eff}.$

(note that i here is the imaginary unit) However, mean field expansion is known to give incorrect results in 1D, and the problem in question is probably done in multiple dimensions. Therefore one cannot say more without knowing more about the underlying lattice. Such expansions are extensively used in solid state physics, transforming from the crystal lattice to the reciprocal lattice.

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