Ground state in $k$-space convert back to real space - spinless non-interacting fermionic system

Question

Say, you have the following spinless non-interacting fermionic (1D) Hamiltonian: $$\hat H = -t \displaystyle \sum_{\langle i,j\rangle} (\hat c_i^+ \hat c_j + \hat c_j^+ \hat c_i)$$

Diagonalizing it (by going to k-space with Fourier) gives you:

$\hat H = -t\displaystyle \sum_{k} \epsilon_k c_k^+ c_k$, with $k = \frac{2\pi m}{N}$, $m$ is an integer, and $N$ total number of particles

The ground state for a fixed number of Fermions (satisfying Pauli exclusion principle) is:

$$|\psi^k_{GS}\rangle = \left(\displaystyle\prod_{|k|\leq k_F}c_k^+\right)|0\rangle = c_{k_F}^+\dotsc_1^+c_0^+|0, 0, 0, \dots, 0\rangle $$

Now, if we want to convert it back to real space (just to see how it looks like), we use the following Fourier transform:

$$c_k^+ = \frac{1}{\sqrt{N}} \sum_j e^{ikj}c_j^+ \qquad\text{and}\qquad c_j^+ = \frac{1}{\sqrt{N}} \sum_k e^{-ikj}c_k^+$$

$$|\psi^k_{GS}\rangle = \left(\frac{1}{\sqrt{N}} \sum_j e^{ik_{N-1}j}c_j^+\right)\dots\left(\frac{1}{\sqrt{N}} \sum_l e^{ik_1l}c_l^+\right)\left(\frac{1}{\sqrt{N}} \sum_m e^{ik_0m}c_m^+\right)|0, 0, 0,\dots, 0\rangle $$

How can we simplify this state?

Answer 1

• Relabeling in a systematic way ($j, l, m\rightarrow j_1, j_2, j_3...$): $$ |\psi_{GS}^{k_F}\rangle = \left(\frac{1}{\sqrt{N}}\sum_{j_{N-1}}e^{ik_{N-1}j_{N-1}}c_{j_{N-1}}^\dagger\right)...\left(\frac{1}{\sqrt{N}}\sum_{j_{0}}e^{ik_{0}j_{0}}c_{j_{0}}^\dagger\right)|0,0,...,0\rangle $$
• Moving all the exponents into one
• Using the fact that one cannot create two electrons on the same site.

Remark however that there are already problems with notation - first of all, the ground state wave function label - why $k$? Secondly, the number of the factors in the product should be determined by by $k_F$, not by the total number of sites, $N$.