Fourier Transforming a HgTe 2D Hamiltonian

In summary, the conversation discusses deriving the Fourier transform of a 2D HgTe Hamiltonian with specific boundary conditions. The Hamiltonian is split into two parts, Hx and Hy, and the Fourier transform is taken in the y direction to real space while leaving the x direction in momentum space. The final result is a Hamiltonian with 2-D Nambu spinors and a combination of c and c^{\dagger} operators.
  • #1
DeathbyGreen
84
16
Hi! I am currently trying to derive the Fourier transform of a 2D HgTe Hamiltonian, with [itex] k_x [/itex] PBC and vanishing boundary conditions in the y direction at 0 and L. Here is the Hamiltonian:

[itex]
H = \sum_{k}\tilde{c_k}^{\dagger}[A\sin{k_x}\sigma_x + A\sin{k_y}\sigma_y + (M-4B+2B[\cos{k_x} + \cos{k_y}])\sigma_z]\tilde{c_k}
[/itex]

Where the [itex]\sigma_i [/itex] are the Pauli matrices, and A,M, and B are material dependent parameters, and the [itex]\tilde{c_k}^{\dagger}[/itex] and [itex]\tilde{c_k}[/itex] are the Nambu spinors ([itex]\tilde{c_k} = (c_k,c^{\dagger}_k))[/itex] and the c's are the fermion creation/annihilation operators. So what I want to do is Fourier transform in the [itex] y[/itex] direction to real space, and leave the [itex] x[/itex] direction in momentum space. My attempt was to split the [itex] H[/itex] into two parts,

[itex]
H_x = \sum_{k_x}\tilde{c_k}^{\dagger}[A\sin{k_x}\sigma_x + (M-4B+2B[\cos{k_x})\sigma_z]\tilde{c_k}
[/itex]

[itex]
H_y = \sum_{k_y}\tilde{c_k}^{\dagger}[ A\sin{k_y}\sigma_y + (M-4B+2B[\cos{k_y}])\sigma_z]\tilde{c_k}
[/itex]

I took the Fourier transform into real space for [itex]H_y[/itex] and got this:

[itex]
H_y = \sum_i^L\tilde{c_i}^{\dagger}D\tilde{c_i} + \sum_i^{L-1}(\tilde{c_{i+1}}^{\dagger}T\tilde{c_i} + \tilde{c_i}^{\dagger}T^{\dagger}\tilde{c_{i+1}})
[/itex]

With the matrices being equal to
[itex]
D = (M-4B)\sigma_z, \quad\quad T = \left(\frac{B}{a^2}\sigma_z - \frac{A}{2ia}\sigma_y\right)
[/itex]

But now it doesn't make sense to add the two parts together...I was thinking that maybe I Fourier transform the whole thing into real space, and then write it in Nambu notation and include boundary conditions on the matrix like for y direction j and x direction i [itex]c_i = c_L+1 [/itex] and [itex] c_{j-1} = 0, c_{L+1} = 0[/itex]. Could someone help point me in the right direction? I've worked on 1D matrices in PBC before but I'm not sure how it should look in 2D. Would the Nambu spinors be 4N length? And if I were to FT into frequency space in the y direction, would the same solution method apply?
 
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  • #2
I figured it out in case anyone is curious. The trick was to Fourier transform all of the [itex] c_k[/itex] operators into position space, giving 2-D Nambu spinors (length 4N) of [itex] c_{i,j}[/itex] and Fourier transforming the [itex] k_y [/itex] into position space (the Fourier transforms lead to Delta functions, which can be rewritten as c operators). Because the [itex] k_x[/itex] was PBC, I left that in momentum space, so the final product came out to

[itex]
H = \sum_{i,j}^{L}\sum_{k_x}[c_{i,j}^{\dagger}Dc_{i,j} + c_{i,j}^{\dagger}(A\sin{k_x}\sigma_x + 2B\cos{k_x}\sigma_z)c_{i,j}] + \sum_i^{L-1}(c_{i,j}^{\dagger}T_ic_{i,j} + c_{i,j}T^{\dagger}c_{i,j}^{\dagger})
[/itex]
 

1. What is a Fourier Transform?

A Fourier Transform is a mathematical operation that decomposes a function into its constituent frequencies. It is commonly used in signal processing to analyze and manipulate signals. In the context of a 2D Hamiltonian, it is used to transform the Hamiltonian from its original space to Fourier space.

2. Why is the HgTe 2D Hamiltonian being Fourier Transformed?

The HgTe 2D Hamiltonian is being Fourier Transformed because it allows for a more convenient representation of the Hamiltonian in terms of momentum space. This is because the HgTe 2D Hamiltonian is a periodic function, and Fourier Transforming it can reveal its underlying periodicity and symmetries.

3. What are the benefits of Fourier Transforming a HgTe 2D Hamiltonian?

Fourier Transforming a HgTe 2D Hamiltonian can provide valuable insights into the electronic properties of the material. It can help identify the band structure, energy bands, band gaps, and other important characteristics that are crucial for understanding the behavior of electrons in the material.

4. How is a Fourier Transform performed on a HgTe 2D Hamiltonian?

A Fourier Transform is performed by taking the integral of the Hamiltonian over all momentum values. This integral is then transformed into an integral over frequency space, resulting in a new representation of the Hamiltonian in terms of frequency or energy. This process is repeated for each dimension of the Hamiltonian to obtain a complete Fourier Transform.

5. Are there any limitations to Fourier Transforming a HgTe 2D Hamiltonian?

Yes, there are some limitations to Fourier Transforming a HgTe 2D Hamiltonian. One limitation is that it assumes the material is homogeneous and has a well-defined periodicity. In reality, many materials may have imperfections or variations in their periodicity, which can affect the accuracy of the Fourier Transform. Additionally, Fourier Transforming a Hamiltonian can be computationally expensive and may require specialized software or algorithms.

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