# Fourier Transform of Green function

I have a question regarding the (discrete) Fourier Transform of the retarded Green function (I neglect hats on operators): $$G(i,j;t)=-i\theta(t)\langle \{c_i(t),c^\dagger_j \} \rangle$$

specifically the conversion from position, $$i$$, to momentum, $$\vec{k}$$ (note this is a vector). These k-vectors will live in the first BZ (I work with a periodic lattice, see below). I give a lot of details so sorry if the statement is long...

For illustration purposes let's take as our model the tightbinding Hamiltonian for nearest neighbors (nn) only on a $$N\times N$$, 2D square lattice. Then in this case we have $$i\in [1,N^2]$$. Also take $$t=.25$$ to give a band width of 1 unit of energy.

I will set this up in a numerical way, but this model can still be easily solved this way. Labeling the sites for a computer is arbitrary but we can take the top left corner as $$i=1$$ then the right nn as $$i=2$$, and in general the (x,y) position is $$i=x+N(y-1)$$. We will also need vectors for position and we just measure this in respect to the center (so keep N even), $$\vec{x}_i$$.

Ok, here's the actual question: do I have to be careful with the sign of the dFT and which combinations of signs should I use? That is: $$G(\vec{k}_a,\vec{k}_b;t)=\frac{1}{N^2}\sum_{i,j}e^{\pm i\vec{x}_i\vec{k}_a} e^{\pm i\vec{x}_j\vec{k}_b}G(i,j;t)$$

The $$\pm$$ are thought to be independent.

You can do a FT on time and it does not seem to matter which sign you use (just be careful when adding your $$i\epsilon$$ to the exponent), but I get something weird when I freely do this for the position. I always thought a FT or dFT was somewhat arbitrary but it seems that it is not the case here. The next paragraph explains why.

We can cast the tb Hamintonian into the following form:

$$\hat{H}=\vec{c}^\dagger H\vec{c}=\vec{c}^\dagger V E V^{-1}\vec{c}$$

The $$[\vec{c}]_i=c_i$$ and the last bit is just diagonalizing our matrix. This matrix is known to of the form: $$[V]_{ij}=e^{i\vec{k}_i\vec{x}_j}$$ and the corresponding energy in $$E$$ is of the form $$-2t(\text{cos}(ik_{ix})+\text{cos}(ik_{iy}))$$. We can express our position creation and annihilation operators in terms of our quasi particles with this matrix. The issue is I know the diagonal part of the fully Fourier Transformed Green function looks like: $$G(\vec{k};\omega)=\frac{1}{\omega-E_k+i\epsilon}$$ We only get this if we are careful with the combination of signs in the exponents since we arrive at a representation of the Kronecker delta to kill the summations. Furthermore, there needs to be a relative sign between the two exponents. Is this due to the fact ones a creation operator and the other and annihilation? So:

$$G(\vec{k}_a,\vec{k}_b;t)=\frac{1}{N^2}\sum_{i,j}e^{\mp i\vec{x}_i\vec{k}_a} e^{\pm i\vec{x}_j\vec{k}_b}G(i,j;t)$$

The signs are not independent here! Am I thinking of something wrong here or is there a good reason for doing this? Please do not tell me another method of doing this problem since I am more than likely aware. I need to adapt this to a case where there is no analytic solution. Thanks for any clarifications in advanced!

I answered my own question at the last step. Basically the FT needs to look at the definition of $$\hat{c}_k$$ in terms of combinations of $$e^{ix_ik}\hat{c}_i$$ (a FT). Taking the adjoint of this will flip the sign of the exponential. This means that: $$G(k,k';t)=-i\theta(t)\langle \{c_k(t),c^\dagger_{k'}\}\rangle=\frac{1}{N^2}\sum_{ij}e^{ix_ik} e^{-ix_jk'}G(i,j;t)$$
Although this is often discussed as a Fourier transform, in the context of a descrete lattice with periodic boundary conditions we can treat it simply as a canonical/unitary transformation $$S$$, $$S^{-1}=S^\dagger$$, to a new basis: $$|\psi'\rangle = S|\psi\rangle, \langle\psi'| = |\psi'\rangle^\dagger = (S|\psi\rangle)^\dagger =\langle\psi | S^\dagger ,\\ H'=SHS^\dagger,\\ H'|\psi'\rangle=SHS^\dagger S|\psi\rangle=SH|\psi\rangle$$ All the operators transform as the Hamiltonian. Indeed, the matrix elements are given by $$\langle\psi|A|\phi\rangle = \langle\psi|S^\dagger SAS^\dagger S|\phi\rangle = \langle\psi'|A'|\phi'\rangle \Longrightarrow A'=SAS^\dagger$$ The transformation operator is a matrix of $$e^{ix_ak_j}$$ factors, and the Green's function can be simply written in terms of the new operators in k-space.
• Yes I did basically answered my question but this is just some additional insight, so thanks for that! This $S$ is exactly how I did my dFT so this is a nice way of looking at it. Mar 9, 2021 at 19:31