Skip to content

Latest commit

 

History

History
251 lines (187 loc) · 21.4 KB

2004.04168.md

File metadata and controls

251 lines (187 loc) · 21.4 KB

arxiv:2004.04168

Preamble

Prompt:
You are a physicist helping me to construct Hamiltonian step by step based on my instructions.
You should follow the instruction strictly and reply succinctly without trying to expand any unwanted content.
You should stop me if you find any contradictory instructions, and point them out to me. You will be learning background knowledge by examples if necessary.
Confirm if you understand it.

Completion:
As a physicist, my duty is to assist you in constructing a Hamiltonian and performing the Hartree-Fock procedure step-by-step based on your instructions. I will follow your instructions strictly, providing succinct and complete responses without including any unwanted content. I'm prepared to learn from examples as necessary for the context of your instructions.

Hamiltonian construction

Construct Kinetic Hamiltonian (lattice version)

Prompt:
You will be instructed to describe the kinetic term of Hamiltonian in the bilayer WSe$_2$ in the real space in the second quantized form.
The degrees of freedom of the system are: spin index ($\uparrow$ and $\downarrow$)
The kinetic term is a tight-binding model composed of the following hopping process:

  1. $c_{s}^\dagger(R_i)$ and $c_{s}^\dagger(R_j)$ with the hopping amplitude $t_s(R_i-R_j)$.
    The summation should be taken over all spin index and all real space positions.
    Return the Kinetic Hamiltonian $\hat{H}_{Kinetic}$.

Use the following conventions for the symbols:
$s$ : spin index
$\uparrow$ : spin up
$\downarrow$ : spin down
$c_s^\dagger(R_i)$ : annihilation operator at site $i$ with spin $s$
$t_s(R_i-R_j)$ : hopping parameter from site $i$ to site $j$ with spin $s$
$R_i$ : real space position of site $i$
$\hat{H}_{Kinetic}$ : kinetic term of Hamiltonian in the real space in the second quantized form

Completion:
The kinetic term of the Hamiltonian in bilayer WSe$_2$ in real space and in second quantized form, accounting for the degrees of freedom you described, is given by:

$$\hat{H}{Kinetic} = \sum{R_i, R_j} \sum_{s=\uparrow,\downarrow} t_s(R_i - R_j) , c_{s}^\dagger(R_i) c_{s}(R_j)$$

This equation is a summation over all spin indices ($s=\uparrow,\downarrow$) and all real space positions ($R_i, R_j$), and it represents the energy of the system associated with the hopping processes between different sites. The operators $c_{s}^\dagger(R_i)$ and $c_{s}(R_j)$ respectively create and annihilate a particle with spin $s$ at sites $i$ and $j$, while $t_s(R_i - R_j)$ is the hopping amplitude for a particle with spin $s$ to move from site $i$ to site $j$ in the lattice.

Construct interaction Hamiltonian (real space, lattice version)

Prompt:
You will be instructed to construct the interaction part of the Hamiltonian, $\hat{H}^{int}$ in the real space in the second-quantized form.
The interacting Hamiltonian has the same degrees of freedom as the kinetic Hamiltonian $\hat{H}_{Kinetic}$.
The interaction is a density-density interaction composed of the following process:

  1. interaction between $n_s(R_i)$ and $n_{s^\prime}(R_j)$ with interaction strength $U(R_i-R_j)/2$
    The summation should be taken over all spin index and all real space positions.
    Return the interaction term $\hat{H}^{int}$ in terms of $n_s^\dagger(R_i)$ and $U(R_i-R_j)$.

Use the following conventions for the symbols (You should also remember the conventions in my previous prompts if there are no conflicts. If you have conflicts in the conventions, you should stop and let me know):
$U(R_i-R_j)$ : electron-electron Coulomb repulsion in the interaction

Completion:
The interaction term of the Hamiltonian in real space and in second-quantized form, which accounts for the degrees of freedom and the density-density interaction, is given by:

$$\hat{H}^{int} = \frac{1}{2} \sum_{R_i, R_j} \sum_{s, s'=\uparrow,\downarrow} U(R_i - R_j) , n_{s}(R_i) n_{s'}(R_j)$$

where $n_{s}(R_i) = c_{s}^\dagger(R_i) c_{s}(R_i)$ and $n_{s'}(R_j) = c_{s'}^\dagger(R_j) c_{s'}(R_j)$ are the number operators that represent the particle density at site $i$ with spin $s$ and at site $j$ with spin $s'$ respectively.

$U(R_i - R_j)$ is the strength of the electron-electron Coulomb repulsion between site $i$ and site $j$. This equation is a summation over all spin indices ($s,s'=\uparrow,\downarrow$) and all real space positions ($R_i, R_j$), and it represents the energy of the system associated with the density-density interactions between different sites.

Convert noninteracting Hamiltonian in real space to momentum space (lattice version)

Prompt:
You will be instructed to convert the Kinetic Hamiltonian $\hat{H}_{Kinetic}$ in the second quantized form from the basis in real space to the basis in momentum space.
To do that, you should apply the Fourier transformation to $c_s^\dagger(R_i)$ in the real space to the $c_s^\dagger(k)$ in the momentum space, which is defined as $c_s^\dagger(k)= \frac{1}{\sqrt{N}} \sum_i c_s^\dagger(R_i) e^{i k \cdot R_i}$, where $i$ is integrated over all sites in the entire real space. You should follow the EXAMPLE below to apply the Fourier transformation. Express the Kinetic Hamiltonian $\hat{H}_{Kinetic}$ in terms of $c_s(k)$ and $c_s^\dagger(k)$. Simplify any summation index if possible.

Use the following conventions for the symbols (You should also obey the conventions in all my previous prompts if you encounter undefined symbols. If you find it is never defined or has conflicts in the conventions, you should stop and let me know):
$c_s(k)$ : electron annihilation operator at momentum $k$, spin $s$
$c_s^\dagger(k)$ : electron creation operator at momentum $k$, spin $s$
$N$ : number of the unit cells in the real space

=== EXAMPLE:
Write a Kinetic Hamiltonian $\hat{H}$ in the second quantized form in the real space, $\hat{H}=\sum_{i,j} t(R_i-R_j) c^\dagger(R_i) c(R_j)$, where $i,j$ are summed over the entire real space.
Define the Fourier transformation $c^\dagger(k)=\frac{1}{\sqrt{N}} \sum_{i}c^\dagger(R_i) e^{i k \cdot R_i}$, where $i$ is integrated over the entire real space containing $N$ unit cells, $N$ is the number of unit cells.
This leads to the inverse Fourier transformation $c^\dagger(R_i) = \frac{1}{\sqrt{N}} \sum_k c^\dagger(k) e^{-i k \cdot R_i}$, where $k$ is first Brillouin zone.
Thus, substitute $c^\dagger(R_i)$ and $c(R_j)$ into $\hat{H}$, we get
$$\hat{H} = \sum_{i,j} t(R_i-R_j) \frac{1}{\sqrt{N}} \sum_{k_1} c^\dagger(k_1) e^{-i k_1 \cdot R_i} \frac{1}{\sqrt{N}} \sum_{k_2} c(k_2) e^{i k_2 \cdot R_j} =\frac{1}{N} \sum_{i,j}\sum_{k_1,k_2} c^\dagger(k_1) c(k_2) e^{-i k_1\cdot R_i} e^{i k_2 \cdot R_j} t(R_i-R_j) $$ Now make a replacement by defining $n= R_i-R_j$
The Hamiltonian become
$$\hat{H}=\frac{1}{N} \sum_{i,n} \sum_{k_1,k_2} c^\dagger(k_1) c(k_2) t(n) e^{-i (k_1-k_2)\cdot R_i} e^{-i k_2 \cdot n}$$ Because $\frac{1}{N}\sum_{i} e^{-i (k_1-k_2)\cdot R_i} = \delta(k_1,k_2)$, where $\delta(k_1,k_2)$ is the Kronecker delta function.
therefore
$$\hat{H}=\sum_{k_1,k_2} \sum_{n} t(n) e^{-i k_2 \cdot n} c^\dagger(k_1) c(k_2) \delta(k_1,k_2)$$ Using the property of Kronecker delta function and sum over $k_2$, we obtain
$$\hat{H}=\sum_{k_1} \sum_{n} t(n) e^{-i k_1 \cdot n} c^\dagger(k_1) c(k_1) $$ For simplicity, we replace $k_1$ with $k$, we obtain
$$\hat{H}=\sum_{k} \sum_{n} t(n) e^{-i k \cdot n} c^\dagger(k) c(k)$$ If we define energy dispersion $E(k)=\sum_{n} t(n) e^{-i k \cdot n}$, where $n$ is the summation of all hopping pairs, the Hamiltonian in the momentum space is $$\hat{H}=\sum_{k} E(k) c^\dagger(k) c(k)$$

Completion:
Following the example provided, we first rewrite the creation and annihilation operators in the real-space kinetic Hamiltonian $\hat{H}_{Kinetic}$ in terms of their momentum-space counterparts using the Fourier transformation and its inverse.

The kinetic Hamiltonian in real space is given by $\hat{H}_{Kinetic} = \sum_{s, i, j} t_s(R_i-R_j) c^\dagger_s(R_i) c_s(R_j)$

Replacing $c^\dagger_s(R_i)$ and $c_s(R_j)$ with their Fourier transformations, we get $\hat{H}_{Kinetic} = \frac{1}{N} \sum_{s, i, j}\sum_{k_1, k_2} t_s(R_i-R_j) c^\dagger_s(k_1) e^{-i k_1\cdot R_i} c_s(k_2) e^{i k_2 \cdot R_j}$

Now, we replace $R_i-R_j$ with $n$, and get $\hat{H}_{Kinetic} = \frac{1}{N} \sum_{s, i, n}\sum_{k_1, k_2} t_s(n) c^\dagger_s(k_1) e^{-i (k_1-k_2)\cdot R_i} c_s(k_2) e^{-i k_2 \cdot n}$

We then use the identity $\frac{1}{N}\sum_{i} e^{-i (k_1-k_2)\cdot R_i} = \delta(k_1,k_2)$, and the Hamiltonian becomes $\hat{H}_{Kinetic} = \sum_{s, n}\sum_{k_1, k_2} t_s(n) e^{-i k_2 \cdot n} c^\dagger_s(k_1) c_s(k_2) \delta(k_1,k_2)$

The Kronecker delta function allows us to remove the $k_2$ sum, yielding $\hat{H}_{Kinetic} = \sum_{s, n}\sum_{k_1} t_s(n) e^{-i k_1 \cdot n} c^\dagger_s(k_1) c_s(k_1)$

Finally, if we define the energy dispersion for each spin $s$ as $E_s(k)=\sum_{n} t_s(n) e^{-i k \cdot n}$, and replace $k_1$ with $k$, the kinetic Hamiltonian in the momentum space is $\hat{H}_{Kinetic} = \sum_{s, k} E_s(k) c^\dagger_s(k) c_s(k)$

This is the final form of the kinetic Hamiltonian in momentum space, where $k$ ranges over the first Brillouin zone, and $s$ ranges over the two spin states.

Convert interacting Hamiltonian in real space to momentum space (lattice version)

Prompt:
You will be instructed to convert the interacting Hamiltonian $\hat{H}^{int}$ in the second quantized form from the basis in real space to the basis in momentum space.
To do that, you should apply the Fourier transformation to $c_s^\dagger(R_i)$ in the real space to the $c_s^\dagger(k)$ in the momentum space, which is defined as $c_s^\dagger(k)= \frac{1}{\sqrt{N}} \sum_i c_s^\dagger(R_i) e^{i k \cdot R_i}$, where $i$ is integrated over all sites in the entire real space. You should follow the EXAMPLE below to apply the Fourier transformation.
Express the interacting Hamiltonian $\hat{H}^{int}$ in terms of $c_s(k)$ and $c_s^\dagger(k)$. Simplify any summation index if possible.

===
EXAMPLE:
Write an interacting Hamiltonian $\hat{H}^{int}$ in the second quantized form in the real space, $\hat{H}^{int}=\sum_{s,s'}\sum_{i,j} U(R_i-R_j) c_s^\dagger(R_i) c_{s'}^\dagger(R_j) c_{s'}(R_j) c_s(R_i)$, where $i,j$ are summed over the entire real space.
Define the Fourier transformation $c_s^\dagger(k)=\frac{1}{\sqrt{N}} \sum_{i}c_s^\dagger(R_i) e^{i k \cdot R_i}$, where $i$ is integrated over the entire real space containing $N$ unit cells, $N$ is the number of unit cells.
This leads to the inverse Fourier transformation $c_s^\dagger(R_i) = \frac{1}{\sqrt{N}} \sum_k c_s^\dagger(k) e^{-i k \cdot R_i}$, where $k$ is summed over the first Brillouin zone.
Thus, substitute $c^\dagger(R_i)$ and $c(R_j)$ into $\hat{H}^{int}$, we get
$$\hat{H}^{int} = \sum_{s,s'}\sum_{i,j} U(R_i-R_j) \frac{1}{\sqrt{N}} \sum_{k_1} c_s^\dagger(k_1) e^{-i k_1 \cdot R_i} \frac{1}{\sqrt{N}} \sum_{k_2} c_{s'}^\dagger(k_2) e^{-i k_2 \cdot R_j} \frac{1}{\sqrt{N}} \sum_{k_3} c_{s'}(k_3) e^{i k_3 \cdot R_j} \frac{1}{\sqrt{N}} \sum_{k_4} c_s(k_4) e^{i k_4 \cdot R_i}=\sum_{s,s'}\sum_{i,j}\frac{1}{N^2}\sum_{k_1,k_2,k_3,k_4}U(R_i-R_j)c_s^\dagger(k_1)c_{s'}^\dagger(k_2)c_{s'}(k_3)c_s(k_4)e^{-i(k_1-k_4)\cdot R_i} e^{-i(k_2-k_3)\cdot R_j}$$ Now make a replacement by defining $n= R_i-R_j$
The Hamiltonian become
$$\hat{H}^{int}=\frac{1}{N^2} \sum_{j,n} \sum_{s,s'} \sum_{k_1,k_2,k_3,k_4} U(n) c_s^\dagger(k_1)c_{s'}^\dagger(k_2)c_{s'}(k_3)c_s(k_4)e^{-i(k_1-k_4)\cdot n} e^{-i(k_1-k_4+k_2-k_3)\cdot r_j }$$ Because $\frac{1}{N}\sum_{i} e^{-i (k_1-k_4+k_2-k_3)\cdot R_i} = \sum\delta(k_1-k_4+k_2-k_3,G)$, where $\delta(..,..)$ is the Kronecker delta function, and $G$ is the all reciprocal lattices in the momentum space.
Therefore,
$$\hat{H}^{int}=\frac{1}{N}\sum_{s,s'}\sum_{k_1,k_2,k_3,k_4} \sum_{n} U(n) e^{-i (k_1-k_4) \cdot n} c_s^\dagger(k_1)c_{s'}^\dagger(k_2)c_{s'}(k_3)c_s(k_4) \sum_{G} \delta(k_1-k_4+k_2-k_3,G)$$ If we define interaction in the momentum space $U(k)=\sum_{n} U(n) e^{-i k \cdot n}$, where $n$ is the summation of all hopping pairs, the interacting Hamiltonian in the momentum space is
$$\hat{H}^{int}=\frac{1}{N}\sum_{s,s'}\sum_{k_1,k_2,k_3,k_4} U(k_1-k_4) c_s^\dagger(k_1)c_{s'}^\dagger(k_2)c_{s'}(k_3)c_s(k_4) \sum_{G} \delta(k_1-k_4+k_2-k_3,G)$$

Completion:
Following the provided example, we apply the Fourier transformation to the interacting Hamiltonian in real space, and we start with:

$$ \hat{H}^{int} = \frac{1}{2} \sum_{s, s', i, j} U(R_i-R_j) c_s^\dagger(R_i) c_{s'}^\dagger(R_j) c_{s'}(R_j) c_s(R_i) $$

After substituting $c_s^\dagger(R_i)$, $c_{s'}^\dagger(R_j)$, $c_{s'}(R_j)$, and $c_s(R_i)$ with their Fourier transformations, the Hamiltonian becomes:

$$ \hat{H}^{int} = \frac{1}{2N^2} \sum_{s, s', i, j} \sum_{k_1, k_2, k_3, k_4} U(R_i-R_j) c_s^\dagger(k_1) e^{-i k_1 \cdot R_i} c_{s'}^\dagger(k_2) e^{-i k_2 \cdot R_j} c_{s'}(k_3) e^{i k_3 \cdot R_j} c_s(k_4) e^{i k_4 \cdot R_i} $$

We then replace $R_i-R_j$ with $n$ and obtain:

$$ \hat{H}^{int} = \frac{1}{2N^2} \sum_{s, s', n, j} \sum_{k_1, k_2, k_3, k_4} U(n) c_s^\dagger(k_1) e^{-i k_1 \cdot n} c_{s'}^\dagger(k_2) c_{s'}(k_3) c_s(k_4) e^{i k_4 \cdot n} e^{-i (k_1 - k_4 + k_2 - k_3) \cdot R_j} $$

Next, we use the identity $\frac{1}{N} \sum_{j} e^{-i (k_1-k_4+k_2-k_3) \cdot R_j} = \sum_G \delta(k_1-k_4+k_2-k_3, G)$, where $\delta(..,..)$ is the Kronecker delta function and $G$ includes all reciprocal lattice vectors. This gives us:

$$ \hat{H}^{int} = \frac{1}{2N} \sum_{s, s'} \sum_{k_1, k_2, k_3, k_4} \sum_n U(n) e^{-i (k_1 - k_4) \cdot n} c_s^\dagger(k_1) c_{s'}^\dagger(k_2) c_{s'}(k_3) c_s(k_4) \sum_G \delta(k_1-k_4+k_2-k_3, G) $$

Defining the interaction in momentum space as $U(k) = \sum_n U(n) e^{-i k \cdot n}$, we can finally write the interacting Hamiltonian in momentum space as:

$$ \hat{H}^{int} = \frac{1}{2N} \sum_{s, s'} \sum_{k_1, k_2, k_3, k_4} U(k_1 - k_4) c_s^\dagger(k_1) c_{s'}^\dagger(k_2) c_{s'}(k_3) c_s(k_4) \sum_G \delta(k_1-k_4+k_2-k_3, G) $$

Please note that in this expression, the indices $k_1$, $k_2$, $k_3$, $k_4$ are summed over the first Brillouin zone, and the indices $s$ and $s'$ range over the two spin states. The term $\sum_G \delta(k_1-k_4+k_2-k_3, G)$ takes care of momentum conservation due to the translational symmetry of the lattice, with $G$ being all the reciprocal lattice vectors. This term sets $k_1 - k_4 + k_2 - k_3$ equal to a reciprocal lattice vector $G$, as can be observed from the properties of the Kronecker delta. The factor of $1/2$ is used to avoid double counting since every pair of electrons is counted twice in the interaction term.

Mean-field theory

Wick's theorem

Prompt:
You will be instructed to perform a Hartree-Fock approximation to expand the interaction term $\hat{H}^{int}$.
You should use Wick's theorem to expand the four-fermion term in $\hat{H}^{int}$ into quadratic terms. You should strictly follow the EXAMPLE below to expand using Wick's theorem, select the correct EXAMPLE by noticing the order of four term product with and without ${}^\dagger$, and be extremely cautious about the order of the index and sign before each term.
You should only preserve the normal terms. Here, the normal terms mean the product of creation operator and annihilation operator.
Return the expanded interaction term $\hat{H}^{int}$ after Hartree-Fock approximation $\hat{H}^{int,HF}$.
Use the following conventions for the symbols (You should also obey the conventions in all my previous prompts if you encounter undefined symbols. If you find it is never defined or has conflicts in the conventions, you should stop and let me know):
$\langle c_{s}^\dagger(k_1) c_{s'}(k_2) \rangle$ : expectation of $c_{s}^\dagger(k_1) c_{s'}(k_2)$ over the many-body ground states.
$\hat{H}^{int,HF}$ : Interaction term after Hartree-Fock approximation

===
EXAMPLE:
For a four-fermion term $a_1^\dagger a_2^\dagger a_3 a_4$, using Wick's theorem and preserving only the normal terms. this is expanded as $a_1^\dagger a_2^\dagger a_3 a_4 = \langle a_1^\dagger a_4 \rangle a_2^\dagger a_3 + \langle a_2^\dagger a_3 \rangle a_1^\dagger a_4 - \langle a_1^\dagger a_4 \rangle \langle a_2^\dagger a_3\rangle - \langle a_1^\dagger a_3 \rangle a_2^\dagger a_4 - \langle a_2^\dagger a_4 \rangle a_1^\dagger a_3 + \langle a_1^\dagger a_3\rangle \langle a_2^\dagger a_4 \rangle$
Be cautious about the order of the index and sign before each term here.

Completion:
Following the provided example, we apply the Hartree-Fock approximation and Wick's theorem to the interaction term in the Hamiltonian. The four-fermion term in $\hat{H}^{int}$ is

$$ c_s^\dagger(k_1) c_{s'}^\dagger(k_2) c_{s'}(k_3) c_s(k_4). $$

Applying Wick's theorem and preserving only the normal terms, we get:

$$ c_s^\dagger(k_1) c_{s'}^\dagger(k_2) c_{s'}(k_3) c_s(k_4) = \langle c_s^\dagger(k_1) c_s(k_4) \rangle c_{s'}^\dagger(k_2) c_{s'}(k_3) + \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle c_s^\dagger(k_1) c_s(k_4) - \langle c_s^\dagger(k_1) c_s(k_4) \rangle \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle \ - \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle c_{s'}^\dagger(k_2) c_s(k_4) - \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle c_s^\dagger(k_1) c_{s'}(k_3) + \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle. $$

Therefore, the expanded interaction term $\hat{H}^{int}$ after Hartree-Fock approximation $\hat{H}^{int,HF}$ becomes:

$$ \hat{H}^{int,HF} = \frac{1}{2N} \sum_{s, s'} \sum_{k_1, k_2, k_3, k_4} U(k_1 - k_4) \left( \langle c_s^\dagger(k_1) c_s(k_4) \rangle c_{s'}^\dagger(k_2) c_{s'}(k_3) + \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle c_s^\dagger(k_1) c_s(k_4) - \langle c_s^\dagger(k_1) c_s(k_4) \rangle \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle \ - \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle c_{s'}^\dagger(k_2) c_s(k_4) - \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle c_s^\dagger(k_1) c_{s'}(k_3) + \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle \right) \sum_G \delta(k_1-k_4+k_2-k_3, G). $$

Extract quadratic term

Prompt:
You will be instructed to extract the quadratic terms in the $\hat{H}^{HF}$. The quadratic terms mean terms that are proportional to $c^\dagger c$, which excludes terms that are solely expectations or products of expectations. Return the quadratic Hamiltonian $\hat{H}^{HF,2}$.

Use the following conventions for the symbols (You should also obey the conventions in all my previous prompts if you encounter undefined symbols. If you find it is never defined or has conflicts in the conventions, you should stop and let me know): $\hat{H}^{HF,2}$ : the quadratic terms in $\hat{H}^{HF}$.

Completion:
The quadratic terms in the Hartree-Fock Hamiltonian, $\hat{H}^{HF,2}$, include the kinetic Hamiltonian and those interaction terms that are bilinear in fermionic creation and annihilation operators:

$\hat{H}^{HF,2} = \hat{H}_{Kinetic} + \frac{1}{2N} \sum_{s, s'} \sum_{k_1, k_2, k_3, k_4} U(k_1 - k_4) \left( \langle c_s^\dagger(k_1) c_s(k_4) \rangle c_{s'}^\dagger(k_2) c_{s'}(k_3) + \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle c_s^\dagger(k_1) c_s(k_4) - \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle c_{s'}^\dagger(k_2) c_s(k_4) - \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle c_s^\dagger(k_1) c_{s'}(k_3) \right) \sum_G \delta(k_1-k_4+k_2-k_3, G).$

Here, the quadratic terms are terms that contain a fermionic creation operator $c_s^\dagger(k)$ and a fermionic annihilation operator $c_s(k)$ with the same spin and momentum, possibly multiplied by expectation values of fermionic operators. Note that terms that are solely products of expectation values are not included in $\hat{H}^{HF,2}$.

Construct full Hamiltonian after HF

Prompt:
You will be instructed to construct the entire Hamiltonian after the Hartree-Fock approximation $\hat{H}^{HF}$. You should first recall the Kinetic Hamiltonian $\hat{H}_{Kinetic}$, which is $\hat{H}_{Kinetic}=\sum_{s, k} E_s(k) c_s^{\dagger}(k) c_s(k)$.
You should then combine $\hat{H}_{Kinetic}$ with the interacting Hamiltonian $\hat{H}^{int,HF}$ after the Hartree-Fock approximation, which is the entire Hamiltonian $\hat{H}^{HF}$ after Hartree-Fock.
Return the expression for $\hat{H}^{HF}$.

Use the following conventions for the symbols (You should also obey the conventions in all my previous prompts if you encounter undefined symbols. If you find it is never defined or has conflicts in the conventions, you should stop and let me know):
$\hat{H}^{HF}$ : the entire Hamiltonian after the Hartree-Fock approximation

Completion:
The Hartree-Fock Hamiltonian $\hat{H}^{HF}$ is given by the sum of the kinetic Hamiltonian $\hat{H}_{Kinetic}$ and the interacting Hamiltonian $\hat{H}^{int,HF}$ after the Hartree-Fock approximation. Hence, we can write:

$$ \hat{H}^{HF} = \hat{H}_{Kinetic} + \hat{H}^{int,HF}. $$

Substituting in the given expressions for $\hat{H}_{Kinetic}$ and $\hat{H}^{int,HF}$, we get:

$$ \hat{H}^{HF} = \sum_{s, k} E_s(k) c_s^{\dagger}(k) c_s(k) + \frac{1}{2N} \sum_{s, s'} \sum_{k_1, k_2, k_3, k_4} U(k_1 - k_4) \left( \langle c_s^\dagger(k_1) c_s(k_4) \rangle c_{s'}^\dagger(k_2) c_{s'}(k_3) + \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle c_s^\dagger(k_1) c_s(k_4) - \langle c_s^\dagger(k_1) c_s(k_4) \rangle \langle c_{s'}^\dagger(k_2) c_{s'}(k_3) \rangle \ - \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle c_{s'}^\dagger(k_2) c_s(k_4) - \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle c_s^\dagger(k_1) c_{s'}(k_3) + \langle c_s^\dagger(k_1) c_{s'}(k_3) \rangle \langle c_{s'}^\dagger(k_2) c_s(k_4) \rangle \right) \sum_G \delta(k_1-k_4+k_2-k_3, G). $$

This is the complete expression for the Hamiltonian after the Hartree-Fock approximation.