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Very often one can view the solid as lattice model and then use the the language of second quantization,namely taking the occupation number representation, to express the Hamiltonian of the system. In this representation the creation and annihilation operators play the fundamental role. But in recently I just found some discrepancies for the definition of these operators for lattice model:

Vijay Shenoy Version (starting at 13:10):

For $3D$ lattice model the Fourier transformation of the creation operator between K space and real space will be:

$$c_{i\sigma}^{\dagger}=\dfrac{1}{\sqrt{N}}\sum_{k \in BZ}e^{-i\vec{k}\cdot\vec{r}}c_{\sigma}^{\dagger}(\vec{k}) \tag{1}$$ here the subindices $\sigma$ and $i$ are the spin index and site index, respectively and N stands for all K points in first Brillouin zone. And Vijay didn't show the inverse transformation.

KaiSun's Version (at page 78):

He stated that we have a continuous k-space and it is periodic (the Brillouin zone), but the real space is discrete for the lattice model.The Fourier transoformation relations are (for 1D system): $$c_k=\dfrac{\sqrt{a}}{\sqrt{2\pi}}\sum_i c_i e^{-ikx} \tag{2}$$ $$c_i=\dfrac{\sqrt{a}}{\sqrt{2\pi}}\int_{BZ}dkc_ke^{ikx} \tag{3}$$

I think the argument of professor KaiSun is very convincible. But what's the difference between $(1)$ and $(3)$ and are there something wrong for the coefficients in Sun's definition? I mean $$\dfrac{\sqrt{a}}{\sqrt{2\pi}} \rightarrow \dfrac{1}{\sqrt{2\pi a}}.$$

Thanks in advance.

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  • $\begingroup$ What are you summing over in equation (2)? Do you mean to have $e^{-ikx_i}$? $\endgroup$ – Aaron Mar 9 '17 at 15:00
  • $\begingroup$ (1) is essentially the discrete version of (3), it applies to systems of finite size. The coefficients simply ensure that applying the Fourier transform and then the inverse Fourier transform get you back where you started. Different choices are possible. $\endgroup$ – Tony Mar 9 '17 at 16:04
  • $\begingroup$ Are these actually different expressions for the same operators (as opposed to just different but related operators)? What commutation relations do each of the $c_i$ and $c_k$ obey? If it's just a difference in choice of the normalization of a Fourier transform, the most likely thing is that they're equally valid choices because the differences also show up in the corresponding commutation relations. $\endgroup$ – Emilio Pisanty Mar 9 '17 at 19:15
  • $\begingroup$ @ Aaron:Yes,because for the lattice model,your real space is discrete but the K space is continuous. $\endgroup$ – Jack Mar 10 '17 at 0:28
  • $\begingroup$ @Emilio Pisanty:They are fermion operators and hence satisfy the anticommutation relation.The (1) and (3) are talking the same thing just for different dimensions, but Vijay's version needs summation over all the K-space. For a lattice model, the K-space is continuous. $\endgroup$ – Jack Mar 10 '17 at 0:38

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