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Momentum Space Representation of the Tight Binding Hamiltonian

Physics Asked on December 4, 2021

I am trying to represent the tight-binding Hamiltonian
begin{equation}
hat{H}_{TB} = sum_{sigma} sum_{alpha,beta} sum_{mathbf{R}_1,mathbf{R}_2}
t^{alpha,beta}_{mathbf{R}_1,mathbf{R}_2}
hat{c}^{dagger}_{alpha,mathbf{R}_1,sigma}
hat{c}_{beta,mathbf{R}_2,sigma}
label{eq:Htb}tag{1}
end{equation}

in the momentum space, and it is not clear this relation
begin{equation}
sum_{mathbf{R}_1,mathbf{R}_2}
e^{-imathbf{k}_1 cdot mathbf{R}_1}
e^{imathbf{k}_2 cdot mathbf{R}_2}
t_{mathbf{R}_1,mathbf{R}_2}^{alpha,beta}
=
frac{1}{M}
sum_{mathbf{R}_0}
sum_{mathbf{R}_1,mathbf{R}_2}
e^{-imathbf{k}_1 cdot mathbf{R}_1}
e^{imathbf{k}_2 cdot mathbf{R}_2}
t_{mathbf{R}_1 – mathbf{R}_0,mathbf{R}_2 – mathbf{R}_1 – mathbf{R}_0}^{alpha,beta}
label{eq:pass2}tag{2}
end{equation}

where $M$ is the number of lattice sites and the exponentials come out of the Fourier transform of the operators in the real space to those in the momentum space
begin{equation}
hat{c}_{n,mathbf{R},sigma} = frac{1}{sqrt{M}}
sum_{mathbf{k}}
e^{imathbf{k} cdot mathbf{R}}
hat{c}_{n,mathbf{k},sigma}
label{eq:c_R}tag{3}
end{equation}

Moreover the translational invariance of the lattice imply
begin{equation}
t_{mathbf{R}_1,mathbf{R}_2}^{alpha,beta} =
t_{mathbf{R}_1 – mathbf{R}_0,mathbf{R}_2 – mathbf{R}_0}^{alpha,beta} quad forall mathbf{R}_0
label{eq:hopping_transl} tag{4}
end{equation}

One Answer

In (2) we can substitute $t^{alpha,beta}_{mathbf{R}_{1}-mathbf{R}_{0},mathbf{R}_{2}-mathbf{R}_{0}}$. Then since the left hand side of (2) does not depend on $mathbf{R}_{0}$, if we sum on it we have M times the same thing, so if we divide by M, we have a relation equivalent to the previous one

Answered by Giovanni on December 4, 2021

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