Fourier Transform of Green function

Question

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!

Answer 1

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: $$G(k,k';t)=-i\theta(t)\langle \{c_k(t),c^\dagger_{k'}\}\rangle=\frac{1}{N^2}\sum_{ij}e^{ix_ik} e^{-ix_jk'}G(i,j;t) $$

Answer 2

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.

Sure, this requires working through the math, but it is a consistent way to go about it (probably you already see where the sign switch in question comes from).