1
$\begingroup$

How should I picture the Fourier transform of a fermionic creation (annihilation) operator acting on a site of a periodic, say one-dimensional, lattice? I mean, in a real-space picture, what are the operators $\hat{c}_{k}^{(\dagger)}$ occurring in the Fourier transform $$\hat{c}_{r}^{(\dagger)}=\frac{1}{\sqrt{N}}\sum_{k}e^{-(+)ikr}\hat{c}^{(\dagger)}_{k}$$ where the index $r\in\{1,2,..,N\}$ labels a specific lattice site on the chain, "creating" ("annihilating")?

$\endgroup$
3
  • $\begingroup$ They create a single particule whose wavefunction is a plane wave with the wavevector $k$. $\endgroup$
    – Christophe
    Apr 28, 2020 at 11:00
  • $\begingroup$ @Christophe which is then delocalized over the lattice, isn’t it? $\endgroup$
    – Milarepa
    Apr 28, 2020 at 11:52
  • 1
    $\begingroup$ Yes. If the wavefunction of a single particle created by $c_r^+$ is $\psi_r(x)=\langle x|c_r^+|0\rangle$ then the wavefunction of a single particle created by $c_k^+={1\over\sqrt N}\sum_r e^{ikr}c_r^+$ is $\psi_k(x)=\langle x|c_k^+|0\rangle={1\over\sqrt N}\sum_r e^{ikr}\langle x|c_k^+|0\rangle={1\over\sqrt N}\sum_r e^{ikr}\psi_r(x)$. $\endgroup$
    – Christophe
    Apr 29, 2020 at 12:34

1 Answer 1

1
$\begingroup$

In the example you posted, $c^{\dagger}_r$ means that you are creating a particle at one dimensional site $r$. Similarly, $c^{\dagger}_k$ means that you are creating a particle with a momentum $k$. If you want to visualise how Fourier transform actually work, we have to go back to Heisenberg Uncertainty Principle. By creating a particle at location $r$, you know the position of the particle with very high certainty, therefore the certainty of knowing the momentum is very low. Therefore, it is a liner combination of operator in momentum space which span from $-\infty$ to $\infty$ and vice-versa.

The Fourier Transformation of an operator or any quantity is useful when you have periodic constraint. If you have a periodic condition on real space i.e. $c^{\dagger}_{r+R}=c^{\dagger}_r$, then you can verify by taking Fourier transform both sides that $k=\frac{2m\pi}{R}$, where m is any integer. This quantizes the momentum space and we get our beautiful Energy momentum spectrum, Brillouin zone etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.