§1.2 One- and Two- Particle Operators, Common Representations and Pictures
In the previous post, we have obtained the expression for the field operators,
\[\hat{\psi}_{\alpha}(\vb{x}) := \sum_{k} \psi_{k}(\vb{x}) \hat{a}_{k\alpha},\quad \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) := \sum_{k} \psi_{k}^\dagger(\vb{x}) \hat{a}_{k\alpha}^\dagger,\]where $k$ denotes different quantum numbers. This can help us calculate observables in second quantization form. In this post, we will learn how “first-quantized” operators (operators expressed in terms of particle coordinates $\vb{x}_i$, momentum $\hat{\vb{p}}_i$ and spin $\hat{s}_i$) can be translated into their second-quantized versions.
There are two common types of operators, single-particle operators and two-particle operators. Note that the expressions for operators in second-quantized form can be derived from the requirement that they yield the same result as the first quantized-form in a state with a given particle number.
1. Single particle operators
The operators which involve sum over only single particles are known as single-particle operators
\[\begin{equation} \hat{F}^{(1)}_{} = \sum_{i=1}^{N}\hat{f}(\vb{x}_i,\hat{\vb{p}}_i,\hat{s}_i). \end{equation}\]This is its expression in first quantization. In second quantization (or occupation number formalism), it can be expressed as
Proof
This can be proven using the method we discussed earlier. Let us choose $\qty{\ket{\tilde{\alpha}}}$ to be the single-particle basis of eigenstates of $\hat{f}$, i.e. $\hat{f}\ket{\tilde{\alpha}} = f_{\tilde{\alpha}} \ket{\tilde{\alpha}}$, where $f_{\tilde{\alpha}}$ is the eigenvalue of $\hat{f}$ associated with the eigenstate $\ket{\tilde{\alpha}}$. Then we claim that the second-quantized representation of $\hat{F}^{(1)}$ can be written as
\[\hat{F}^{(1)} = \sum_{\tilde{\alpha}}f_{\tilde{\alpha}} \hat{n}_{\tilde{\alpha}} = \sum_{\tilde{\alpha}} f_{\tilde{\alpha}} \hat{a}_{\tilde{\alpha}}^{\dagger} \hat{a}_{\tilde{\alpha}}.\]We can calculate the matrix elements of $\hat{F}^{(1)}$ in both first and second quantization to prove this. Consider $\mel{\Phi’}{\hat{F}^{(1)}}{\Phi}$ where $\ket{\Phi}$ and $\ket{\Phi’}$ are two arbitrary basis states. In second quantization, this is
\[\begin{align*} \mel{\Phi'}{\hat{F}^{(1)}}{\Phi} &= \mel{n_1'n_2'\cdots n_\infty'}{\sum_{\tilde{\alpha}}f_{\tilde{\alpha}}\hat{n}_{\tilde{\alpha}}}{n_1n_2\cdots n_{\infty}}\\[.2cm] &=\sum_{\tilde{\alpha}}f_{\tilde{\alpha}}\braket{n_{1}'n_2'\cdots n_{\infty}'}{n_1n_2\cdots n_{\infty}}=\delta_{\Phi \Phi'} \sum_{\tilde{\alpha}}f_{\tilde{\alpha}}n_{\tilde{\alpha}}. \end{align*}\]In first quantization, this is
\[\begin{align*} \mel{\Phi'}{\hat{F}^{(1)}}{\Phi} &= \int \dd[3]{x_1}\cdots \dd[3]{x_N} \Phi'^{*}(\vb{x}_1\cdots\vb{x}_N)\qty(\sum_{i=1}^{N} \hat{f}) \Phi(\vb{x}_1\cdots \vb{x}_N)\\[.2cm] &=\sum_{\tilde{\alpha}}f_{\tilde{\alpha}} n_{\tilde{\alpha}}\int \int \dd[3]{x_1}\cdots \dd[3]{x_N} \Phi'^{*}(\vb{x}_1\cdots\vb{x}_N)\Phi(\vb{x}_1\cdots \vb{x}_N)=\delta_{\Phi\Phi'}\sum_{\tilde{\alpha}} f_{\tilde{\alpha}}n_{\tilde{\alpha}}, \end{align*}\]which is consistent with the result in second quantization. For other basis $\qty{\ket{\alpha}}$, we can find that
\[\hat{a}_{\alpha}^{\dagger}\ket{0} = \ket{\alpha} = \sum_{\tilde{\alpha}}\ket{\tilde{\alpha}}\braket{\tilde{\alpha}}{\alpha} = \sum_{\tilde{\alpha}}\braket{\tilde{\alpha}}{\alpha}\hat{a}_{\tilde{\alpha}}^\dagger\ket{0}.\]So
\[\begin{align*} \hat{a}_{\alpha}^{\dagger} = \sum_{\tilde{\alpha}} \braket{\tilde{\alpha}}{\alpha} a_{\tilde{\alpha}}^{\dagger},\quad \hat{a}_{\alpha} = \sum_{\tilde{\alpha}}\braket{\alpha}{\tilde{\alpha}}a_{\tilde{\alpha}}. \end{align*}\]Then the single-particle operator $\hat{F}^{(1)}$ in other basis $\qty{\ket{\alpha}}$ can be written as
\[\begin{align*} \hat{F}^{(1)} &= \sum_{\tilde{\alpha}} f_{\tilde{\alpha}}\hat{a}_{\tilde{\alpha}}^{\dagger}\hat{a}_{\tilde{\alpha}} = \sum_{\tilde{\alpha}} \qty(\sum_{\alpha}\braket{\alpha}{\tilde{\alpha}} \hat{a}_{\alpha})\qty(\sum_{\beta}\braket{\tilde{\alpha}}{\beta}\hat{a}_{\beta})\\[.2cm] &=\sum_{\alpha,\beta} \bra{\alpha} \qty(\sum_{\tilde{\alpha}}\ket{\tilde{\alpha}}f_{\alpha}\bra{\tilde{\alpha}}) \ket{\beta} \hat{a}_{\alpha}^\dagger\hat{a}_{\beta} = \sum_{\alpha,\beta}\mel{\alpha}{\hat{f}}{\beta} \hat{a}_{\alpha}^\dagger\hat{a}_{\beta}. \end{align*}\]So, in other bases, it can also be written in a diagonal form. The matrix element $\mel{\alpha}{\hat{f}}{\beta}$ can be written as
\[\mel{\alpha}{\hat{f}}{\beta} = \int\dd[3]{x}\dd[3]{x'} \braket{\alpha}{\vb{x}} \mel{\vb{x}}{\hat{f}}{\vb{x}'}\braket{\vb{x}'}{\beta} = \int\dd[3]{x} \braket{\alpha}{\vb{x}}\hat{f} \braket{\vb{x}}{\beta} = \int \dd[3]{x} \psi_{\alpha}^*(\vb{x})\hat{f} \psi_{\beta}(\vb{x}).\]By substituting the expression for the field operators, we can finally prove $\eqref{equ:single-particle operator}$.
Next, we will discuss the second quantization form of some single-particle operators.
1.1. Total kinetic energy operator and total momentum operator
1.1.1. Total kinetic energy operator
The total kinetic energy operator in first quantization is $\displaystyle{\hat{T} = \sum_{i=1}^{N}-\frac{\hbar^2 \nabla_{i}^2}{2m}}$, and in second quantization, it is
\[\begin{align} \hat{T} &= \sum_{\alpha}\sum_{\beta} \int \dd[3]{x}\hat{\psi}^\dagger_{\alpha}(\vb{x}) \qty(-\frac{\hbar^2\nabla^2}{2m}) \braket{\alpha}{\beta} \hat{\psi}_{\beta}(\vb{x})\notag\\[.2cm] &= \sum_\alpha \int \dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) \qty(-\frac{\hbar^2 \nabla^2}{2m}) \hat{\psi}_{\alpha}(\vb{x}). \end{align}\]1.1.2. Total momentum operator
Similarly, we can obtain the representation of the total momentum operator
\[\begin{equation} \hat{\mathbf{P}}= \sum_{\alpha}\int\dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) (-i\hbar \nabla) \hat{\psi}_{\alpha}(\vb{x}). \end{equation}\]1.2. Density operator
1.2.1. Local density operator
Local density operator is defined as $\displaystyle{\hat{n}(\vb{y}) = \sum_{i=1}^{N}\delta(\vb{y}-\vb{x}_i)}$. In second quantization, it is
\[\begin{align} \hat{n}(\vb{y}) &= \sum_{\alpha}\sum_{\beta}\int\dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) \qty[\delta(\vb{y}-\vb{x})\delta_{\alpha\beta}]\hat{\psi}_{\beta}(\vb{x})\notag\\[.2cm] &=\sum_{\alpha} \int \dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) \delta(\vb{y}-\vb{x})\hat{\psi}_{\alpha}(\vb{x}) = \sum_{\alpha} \hat{\psi}_{\alpha}^{\dagger}(\vb{y})\hat{\psi}_{\alpha}(\vb{y}). \end{align}\]And total particle number operator is
\[\begin{equation} \hat{N} = \int \dd[3]{y} \hat{n}(\vb{y}) = \sum_{\alpha} \int \dd[3]{y} \hat{\psi}_{\alpha}^{\dagger}(\vb{y})\hat{\psi}_{\alpha}(\vb{y}). \end{equation}\]1.2.2. Spin density operator
Spin density operator is defined as $\displaystyle{\hat{S}(\vb{y})} = \sum_{i=1}^{N} \hat{s}_i \delta(\vb{y}-\vb{x}_i)$. In second quantization, it is
\[\begin{align} \hat{S}(y)&= \sum_{\alpha}\sum_{\beta} \int\dd[3]{x} \hat{\psi}_{\alpha}^{\dagger}(\vb{x})\mel{\alpha}{\delta(\vb{y}-\vb{x})\hat{s}}{\beta} \hat{\psi}_{\beta}(\vb{x})\notag\\[.2cm] &=\sum_{\alpha}\sum_{\beta}\int \dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) \delta(\vb{y}-\vb{x})\mel{\alpha}{\hat{s}}{\beta}\hat{\psi}_{\beta}(\vb{x})\notag \\[.2cm] &=\sum_{\alpha}\sum_{\beta} \hat{\psi}_{\alpha}^{\dagger} \mel{\alpha}{\hat{s}}{\beta}\hat{\psi}_{\beta}(\vb{y}). \end{align}\]Similarly, the total spin operator is
\[\begin{equation} \hat{S} = \int\dd[3]{y} \hat{S}(\vb{y}) = \sum_{\alpha}\sum_{\beta}\int\dd[3]{y} \hat{\psi}^{\dagger}_{\alpha}(\vb{y}) \mel{\alpha}{\hat{s}}{\beta}\hat{\psi}_{\beta}(\vb{y}). \end{equation}\]1.2.3. Electrical current density operator
Suppose that all identical particles have the same charge $e$. In classical case, we know that electrical current is \(\displaystyle{\vb{J} (\vb{y}) = e \sum_{i=1}^{N} \vb{v}_i\delta(\vb{y}-\vb{x}_i)}\). If we switch to quantum case, we will find \([\vb{v}_i,\delta(\vb{y}-\vb{x}_i)] \neq 0\), therefore \(e \sum_{i} \vb{v}_i \delta(\vb{y} - \vb{x}_i)\) is not a Hermitian operator, so we defined electrical current density operator as
\[\begin{align} \hat{\vb{J}}(\vb{y}) &= \frac{e}{2} \sum_{i=1}^{N} [\vb{v}_i \delta(\vb{y}-\vb{x}_i) + \delta(\vb{y}-\vb{x}_i)\vb{v}_i]\notag\\[.2cm] &= \frac{e}{2} \sum_{i=1}^{N} \qty[\qty(-\frac{i\hbar}{m}\nabla)\delta(\vb{y}-\vb{x}_i) + \delta(\vb{y}-\vb{x}_i)\qty(-\frac{i\hbar}{m}\nabla)]. \end{align}\]The corresponding second-quantized form is
\[\begin{align} \hat{\vb{J}}(\vb{y}) &= \frac{e}{2} \sum_{\alpha}\sum_{\beta} \int \dd[3]{x} \hat{\psi}^\dagger_{\alpha}(\vb{x}) \qty[\bra{\alpha}\qty{-\frac{i\hbar\nabla}{m}\delta(\vb{y}-\vb{x}) + \delta(\vb{y}-\vb{x}) \frac{-i\hbar\nabla}{m}}\ket{\beta}]\hat{\psi}_{\beta}(\vb{x})\notag\\[.2cm] &=\frac{e}{2} \sum_\alpha\sum_\beta \int\dd[3]{x} \hat{\psi}_\alpha^\dagger (\vb{x}) \braket{\alpha}{\beta} \qty[- \frac{i\hbar\nabla}{m}\delta(\vb{y}-\vb{x}) + \delta(\vb{y}-\vb{x}) \frac{-i\hbar \nabla}{m}] \hat{\psi}_{\beta}(\vb{x})\notag\\[.2cm] &=-\frac{i\hbar e}{2m}\sum_{\alpha}\Bigg\{ \underbrace{\int \dd[3]{x} \hat{\psi}^\dagger_\alpha(\vb{x})\nabla_{x}\qty[\delta(\vb{y}-\vb{x})\hat{\psi}_\alpha(\vb{x})]}_{\mathrm{IBP}} + \int\dd[3]{x} \hat{\psi}^\dagger_\alpha(\vb{x}) \delta(\vb{y}-\vb{x})\nabla_{x} \hat{\psi}_{\alpha}(\vb{x})\Bigg\}\notag\\[.2cm] &=-\frac{i\hbar e}{2m} \sum_{\alpha}\qty[\hat{\psi}_{\alpha}^{\dagger}(\vb{y})\nabla_y \hat{\psi}_{\alpha}(\vb{y}) - (\nabla_y \hat{\psi}_{\alpha}^\dagger(\vb{y})) \hat{\psi}_{\alpha}(\vb{y})]. \end{align}\]1.2.4. Free electrons in a magnetic field
Suppose we have $N$ electrons (each with a charge of $-e$) in a magnetic field. The system’s Hamiltonian (neglecting their Coulomb interactions) is
\[H = \sum_{i=1}^{N} \frac{1}{2m} \qty(\vb{p}_i + \frac{e}{c} \mathbf{A}(\vb{x}_i))^2 + \frac{g\mu_B}{\hbar} \sum_{i=1}^{N} \vb{S}_i\cdot \vb{B}(\vb{x}_i),\]where $g$ is the Landé g-factor, $\mu_B$ is the Bohr magneton and $\vb{S}$ is the spin of the electron. Then its second-quantized form is
\[\begin{equation} \begin{split} \hat{H} &= \sum_{\alpha}\int\dd[3]{x} \hat{\psi}^\dagger_{\alpha}(\vb{x}) \frac{1}{2m}\qty(-i\hbar \nabla + \frac{e}{c} \vb{A})^2\hat{\psi}_{\alpha}(\vb{x}) \\[.2cm] &\quad + \frac{g\mu_B}{\hbar} \sum_{\alpha}\sum_{\beta} \int \dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x})\mel{\alpha}{\vb{S}\cdot \vb{B}}{\beta} \hat{\psi}_{\beta}(\vb{x}). \end{split} \end{equation}\]We are using Gaussian units.
2. Two particle operators
The operators which involve sum over two particles are known as two-particle operators
\[\begin{equation} \hat{F}^{(2)} = \frac{1}{2} \sum_{\substack{i,j=1\\[.2cm] i\neq j}}^{N} \hat{f}(\vb{x}_i,\vb{x}_j;\hat{\vb{p}}_i,\hat{\vb{p}}_j;\hat{s}_i,\hat{s}_j). \end{equation}\]For simplicity, we will omit the summation over indices $i$ and $j$ in the future, only keep $i\neq j$.
In second quantization, it is
\[\begin{equation} \hat{F}^{(2)} = \frac{1}{2}\sum_{\alpha\alpha'\beta\beta'} \int\dd[3]{x} \int\dd[3]{x'} \hat{\psi}^{\dagger}_\alpha (\vb{x}) \hat{\psi}^{\dagger}_{\beta}(\vb{x}') \hat{f}_{\alpha\alpha',\beta\beta'}(\vb{x},\vb{x}')\hat{\psi}_{\beta'}(\vb{x}')\hat{\psi}_{\alpha'}(\vb{x}), \end{equation}\]where $\hat{f}_{\alpha\alpha’,\beta\beta’}(\vb{x},\vb{x}’) = \bra{\alpha}\otimes \bra{\beta}\hat{f}\ket{\alpha’}\otimes \ket{\beta’}$. Note that we have left out $\hat{\mathbf{p}}$ and $\hat{s}$ here.
If the two-particle operator is independent of spin, we can find
\[\begin{align*} \hat{f}_{\alpha\alpha',\beta\beta'}(\vb{x},\vb{x}') = \hat{f}(\vb{x},\vb{x}')\qty(\bra{\alpha}\otimes\bra{\beta})\qty(\ket{\alpha'}\otimes\ket{\beta'}) = \hat{f}(\vb{x},\vb{x}') \delta_{\alpha\alpha'}\delta_{\beta\beta'}, \end{align*}\]and the two-particle operator can be simplified as
\[\hat{F}^{(2)} = \frac{1}{2}\sum_{\alpha}\sum_{\beta} \int\dd[3]x \int\dd[3]{x'} \hat{\psi}_{\alpha}^\dagger(\vb{x})\hat{\psi}_{\beta}^\dagger(\vb{x}')\hat{f}(\vb{x},\vb{x}') \hat{\psi}_{\beta}(\vb{x}')\hat{\psi}_{\alpha}(\vb{x}).\]Hamiltonian of the jellium model
We will talk about the jellium model in detail later. The Hamiltonian is
\[\begin{align*} \hat{H} &= \sum_{i=1}^{N} \qty(-\frac{\nabla_i^2}{2m}) + \frac{1}{2}\sum_{i\neq j} \hat{V}(\vb{x}_i,\vb{x}_j) \\[.2cm] &=\sum_{i=1}^{N}\qty(-\frac{\nabla_i^2}{2m}) + \frac{1}{2V}\sum_{i\neq j} \sum_{\vb{q}\neq 0} \frac{4\pi e^2}{q^2} \mathrm{e}^{i\vb{q}\cdot(\vb{x}_i - \vb{x}_j)}. \end{align*}\]The second-quantized form of the interaction is
\[\hat{H}^{(2)} = \frac{1}{2}\sum_{\alpha}\sum_{\beta} \int\dd[3]{x}\int\dd[3]{x'} \hat{\psi}^\dagger_{\alpha}(\vb{x})\hat{\psi}_{\beta}^\dagger(\vb{x}') \qty[\frac{1}{V}\sum_{\vb{q}\neq 0} \frac{4\pi e^2}{q^2}\mathrm{e}^{i\vb{q}\cdot(\vb{x}-\vb{x}')}] \hat{\psi}_{\beta}(\vb{x}')\hat{\psi}_{\alpha}(\vb{x}).\]3. Common representations
So far, we have provided the second-quantized form for one- and two-particle operators without giving specific forms. Next, we will introduce three commonly used representations in many-body theory.
3.1. Plane wave representation
Consider box normalization and periodic boundary condition, the plane wave solution of Schrödinger equation is $\psi_{\vb{k}}(\vb{x}) = \frac{1}{\sqrt{V}}\mathrm{e}^{i\vb{k}\cdot\vb{x}}$, where $V$ is volume of the box and $\vb{k}$ is wave vector of particle. $\qty{\psi_{\vb{k}}(\vb{x})}$ is a set of orthonormal and complete functions, which means
\[\begin{equation} \begin{cases} \displaystyle{\int\dd[3]{x}\psi^*_{\vb{k}}(\vb{x})\psi_{\vb{k}'}(\vb{x}) = \delta_{\vb{k}\vb{k}'}, \quad {\text{Orthonormalization,}}}\\[.2cm] \displaystyle{\sum_{\vb{k}}\psi_{\vb{k}}^*(\vb{x}')\psi_\vb{k}(\vb{x}) = \delta(\vb{x}-\vb{x}'), \quad {\text{Completeness.}}} \end{cases} \end{equation}\]We can introduce creation and annihilation operators for electron in $\ket{\vb{k},\alpha}$ state, so
\[\begin{equation} [\hat{C}_{\vb{k}\alpha},\hat{C}^\dagger_{\vb{k}'\alpha'}]_+ = \delta_{\vb{k}\vb{k}'}\delta_{\alpha\alpha'}, \quad [\hat{C}_{\vb{k}\alpha},\hat{C}_{\vb{k}'\alpha'}]_+ = [\hat{C}_{\vb{k}\alpha}^\dagger,\hat{C}^{\dagger}_{\vb{k}'\alpha'}]_+=0. \end{equation}\]Then the field operators can be written as
\[\begin{equation} \hat{\psi}_{\alpha}(\vb{x}) = \frac{1}{\sqrt{V}}\sum_{\vb{k}} \hat{C}_{\vb{k}\alpha}\mathrm{e}^{i\vb{k}\cdot\vb{x}}, \quad \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) = \frac{1}{\sqrt{V}}\sum_{\vb{k}} \hat{C}_{\vb{k}\alpha}^\dagger \mathrm{e}^{-i\vb{k}\cdot\vb{x}}. \end{equation}\]Total kinetic energy operator:
\[\begin{equation} \hat{T} = \sum_{\alpha} \int\dd[3]{x} \hat{\psi}_{\alpha}^\dagger(\vb{x}) \qty(-\frac{\hbar^2\nabla^2}{2m})\hat{\psi}_{\alpha}(\vb{x}) = \sum_{\vb{k}}\sum_{\alpha} \frac{\hbar^2 k^2}{2m} \hat{C}_{\vb{k}\alpha}^\dagger \hat{C}_{\vb{k}\alpha}. \end{equation}\]Total momentum operator:
\[\begin{equation} \hat{\mathbf{P}} = \sum_{\vb{k}}\sum_{\alpha} \hbar \vb{k} \hat{C}_{k\alpha}^\dagger \hat{C}_{k\alpha}. \end{equation}\]For local density operator, we consider its Fourier transform and inverse Fourier transform,
\[\begin{gather} \hat{n}(\vb{y}) = \frac{1}{V}\sum_\vb{q} \hat{n}(\vb{q}) \mathrm{e}^{i\vb{q}\cdot\vb{y}},\\[.2cm] \hat{n}(\vb{q}) = \int\hat{n}(\vb{y})\mathrm{e}^{-i\vb{q}\cdot\vb{y}}\dd[3]{y} = \sum_{\vb{k}}\sum_{\alpha}\hat{C}_{\vb{k}\alpha}^{\dagger}\hat{C}_{\vb{k}+\vb{q},\alpha}. \end{gather}\]Similarly for spin density operator:
\[\begin{gather} \hat{S}(\vb{y}) = \frac{1}{V}\sum_\vb{q} \hat{S}(\vb{q})\mathrm{e}^{i\vb{q}\cdot\vb{y}},\\[.2cm] \hat{S}(\vb{q}) = \int \dd[3]{y} \hat{S}(\vb{y}) \mathrm{e}^{-i\vb{q}\cdot\vb{y}} = \sum_\vb{k}\sum_\alpha\sum_\beta \hat{C}_{\vb{k}\alpha}^\dagger \mel{\alpha}{\hat{s}}{\beta}\hat{C}_{\vb{k}+\vb{q},\beta}. \end{gather}\]And electrical current density operator:
\[\begin{gather} \hat{\mathbf{J}}(\vb{y}) = \frac{1}{V}\sum_\vb{q} \hat{\mathbf{J}}(\vb{q}) \mathrm{e}^{i\vb{q}\cdot\vb{y}},\\[.2cm] \hat{\mathbf{J}}(\vb{q}) = \int\dd[3]{y} \hat{\mathbf{J}}(\vb{y})\mathrm{e}^{-i\vb{q}\cdot\vb{y}} = -\frac{e}{m} \sum_{\vb{k}}\sum_{\alpha}\qty(\vb{k}+\frac{\vb{q}}{2}) \hat{C}_{\vb{k}\alpha}^\dagger \hat{C}_{\vb{k}+\vb{q},\alpha}. \end{gather}\]
Jellium model in plane wve presentation
As an example for two-particle operator, the interaction of the jellium model in plane wave representation is
\[\begin{align} \hat{H}^{(2)} &= \frac{1}{2}\sum_{\alpha}\sum_{\beta}\int \dd[3]{x}\dd[3]{x'} \hat{\psi}^\dagger_\alpha(\vb{x})\hat{\psi}_\beta^\dagger(\vb{x}')\qty[\sum_{\vb{q}\neq 0} \frac{4\pi e^2}{q^2}\mathrm{e}^{i\vb{q}\cdot(\vb{x}-\vb{x}')}]\hat{\psi}_{\beta}(\vb{x}')\hat{\psi}_\alpha(\vb{x})\notag\\[.2cm] &=\frac{1}{2V}\sum_{\vb{k}}\sum_{\vb{k}'}\sum_{\vb{q}\neq 0}\sum_{\alpha}\sum_{\beta}\qty(\frac{4\pi e^2}{q^2})\hat{C}_{\vb{k}+\vb{q},\alpha}^\dagger\hat{C}_{\vb{k}'-\vb{q},\beta}^\dagger\hat{C}_{\vb{k}'\beta}\hat{C}_{\vb{k}\alpha}. \end{align}\]3.2. Bloch representation
Bloch’s theorem tell us that solutions to the Schrödinger equation in a periodic potential can be expressed as plane waves modulated by periodic functions. The Bloch function is defined as $\psi_{\vb{k}}^{B}(\vb{x}) = \mathrm{e}^{i\vb{k}\cdot\vb{x}}u_{\vb{k}}(\vb{x})$, where \(u_{\vb{k}}(\vb{x}) = u_{\vb{k}}(\vb{x}+\vb{R}_{l})\), \(\vb{R}_{l}\) is lattice vector. \(\qty{\psi_{\vb{k}}^{B}(\vb{x})}\) is also a set of orthonormal and complete functions,
\[\begin{equation} \begin{cases} \displaystyle{\int\dd[3]{x} \psi_{\vb{k}'}^{B*}(\vb{x})\psi_{\vb{k}}^{B}(\vb{x}) = \delta_{\vb{k}\vb{k}'},\quad \text{Orthonormalization,}}\\[.2cm] \displaystyle{\sum_{\vb{k}}\psi_{\vb{k}}^{B*}(\vb{x}')\psi_{\vb{k}}^{B}(\vb{x}) = \delta(\vb{x}-\vb{x}'), \quad \text{Completeness.}} \end{cases} \end{equation}\]Introduce creation and annihilation operators for $\ket{\vb{k},\sigma}$ which satisfies
\[\begin{equation} [\hat{b}_{\vb{k}\sigma},\hat{b}_{\vb{k}'\sigma'}]_+ = \delta_{\vb{k}\vb{k}'}\delta_{\sigma\sigma'}, \quad [\hat{b}_{\vb{k}\sigma},\hat{b}_{\vb{k}'\sigma'}]_+ = [\hat{b}_{\vb{k}\sigma}^\dagger,\hat{b}^\dagger_{\vb{k}'\sigma'}]_+ =0. \end{equation}\]The field operators in this representation are
\[\begin{equation} \hat{\psi}_{\sigma}(\vb{x}) = \sum_\vb{k} \hat{b}_{\vb{k}\sigma}\psi_{k}^{B}(\vb{x}),\quad \hat{\psi}^{\dagger}_{\sigma}(\vb{x}) = \sum_{\vb{k}}\hat{b}_{\vb{k}\sigma}^\dagger \psi_{\vb{x}}^{B*}(\vb{x}). \end{equation}\]Total kinetic energy operator in Bloch representation is
\[\begin{equation} \hat{T} = \sum_{\sigma} \int\dd[3]{x} \hat{\psi}_{\sigma}(\vb{x}) \qty(-\frac{\hbar^2\nabla^2}{2m}) \hat{\psi}_{\sigma}(\vb{x}) = \sum_{\sigma}\sum_{\vb{k}}\sum_{\vb{k}'}\hat{b}_{\vb{k}\sigma}^\dagger\hat{b}_{\vb{k}'\sigma}V_{\vb{k}\vb{k}'}, \end{equation}\]where $\displaystyle{V_{\vb{k}\vb{k}’} = \int\dd[3]x \psi_{\vb{k}}^{B*}(\vb{x})\qty[-\frac{\hbar^2\nabla^2}{2m}]\psi_{\vb{k’}\sigma}^{B}(\vb{x})}$.
3.3. Wannier representation
We can expand Bloch function with lattice vector $\vb{R}_l$,
\[\begin{equation} \hat{\psi}_{\vb{k}}^{B}(\vb{x}) = \frac{1}{\sqrt{N}} \sum_{\vb{R}_l} \mathrm{e}^{i\vb{k}\cdot\vb{R}_l} a(\vb{x} - \vb{R}_{l}). \end{equation}\]This is also a Fourier transform. The coefficient $a(\vb{x}-\vb{R}_l)$ is Wannier function. $\qty{a(\vb{x}-\vb{R}_l)}$ is also a set of orthonormal and complete functions, so
\[\begin{equation} \begin{cases} \displaystyle{\int\dd[3]{x} a^{*}(\vb{x}-\vb{R}_l)a({\vb{k}-\vb{R}_{l'}}) = \delta_{ll'},\quad \text{Orthonormalization,}}\\[.2cm] \displaystyle{\sum_{\vb{R}_l}a^*(\vb{x}'-\vb{R}_l)a(\vb{x}-\vb{R}_l) = \delta(\vb{x}-\vb{x}'), \quad \text{Completeness.}} \end{cases} \end{equation}\]For field operator $\hat{\psi}_{\alpha}(\vb{x})$ in Bloch representation, we can derive that
\[\begin{align*} \hat{\psi}_{\sigma}(\vb{x}) &= \sum_{\vb{k}} \hat{b}_{\vb{k}\sigma}\psi^{B}(\vb{x}) = \sum_{k} \hat{b}_{\vb{k}\sigma}\qty[\frac{1}{\sqrt{N}} \sum_{\vb{R}_l} \mathrm{e}^{i\vb{k}\cdot\vb{R}_l} a(\vb{x} - \vb{R}_{l}) ]\\[.2cm] &=\sum_{\vb{R}_l} a(\vb{x}-\vb{R}_l) \Big[\underbrace{\frac{1}{\sqrt{N}}\sum_{\vb{k}}\mathrm{e}^{i\vb{k}\cdot\vb{R}_l}\hat{b}_{\vb{k}\sigma}}_{\hat{a}_{l\sigma}}\Big]. \end{align*}\]So we can define a new creation operator \(\hat{a}_{l\sigma}\) (the same as annihilation operator \(\hat{a}_{l\sigma}^{\dagger}\)). Thus the field operators in Wannier representation are:
\[\begin{equation} \hat{\psi}_{\sigma}(\vb{x}) = \sum_{\vb{R}_{l}} \hat{a}_{l\sigma} a(\vb{x} - \vb{R}_l) , \quad \hat{\psi}^{\dagger}_{\sigma}(\vb{x}) = \sum_{\vb{R}_l} \hat{a}_{l\sigma}^\dagger a^*(\vb{x}-\vb{R}_l). \end{equation}\]Total kinetic energy operator in Wannier representation is
\[\begin{align} \hat{T} &= \sum_\alpha \int \dd[3]{x} \hat{\psi}^{\dagger}_{\alpha}(\vb{x}) \qty(-\frac{\hbar^2\nabla^2}{2m}) \hat{\psi}_{\alpha}(\vb{x})\notag\\[.2cm] &= \sum_{\alpha}\sum_{\vb{R}_l}\sum_{\vb{R}_{l'}} \hat{a}_{l'\alpha}^{\dagger}\hat{a}_{l\alpha} \underbrace{\int\dd[3]{x} a^*(\vb{x}-\vb{R}_{l'})\qty(-\frac{\hbar^2\nabla^2}{2m}) a(\vb{x}-\vb{R}_l)}_{T_{l'l}} \\[.2cm] &= \sum_{\alpha}\sum_{l}\sum_{l'}T_{l'l}\hat{a}_{l'\alpha}^{\dagger}\hat{a}_{l\alpha}. \end{align}\]4. Pictures
All our previous discussions have been in the Schrödinger picture, but employing different pictures such as the Heisenberg picture and the interaction picture might yield remarkable results when dealing with other problems. Here, we briefly introduce the Heisenberg picture and the interaction picture.
4.1. Heisenberg picture
In the case where $\hat{H}$ is independent of $t$, the time evolution operator is $\hat{U}(t,t_0) = \mathrm{e}^{-i\hat{H}(t-t_0)/\hbar}$. And the field operators in the Heisenberg picture are
\[\begin{equation} \hat{\psi}_{H\alpha} (\vb{x},t) = \mathrm{e}^{i\hat{H}(t-t_0)/\hbar}\hat{\psi}_{\alpha}(\vb{x})\mathrm{e}^{-i\hat{H}(t-t_0)/\hbar}, \quad \hat{\psi}_{H\alpha}^\dagger (\vb{x},t) = \mathrm{e}^{i\hat{H}(t-t_0)/\hbar}\hat{\psi}_{\alpha}^{\dagger}(\vb{x})\mathrm{e}^{-i\hat{H}(t-t_0)/\hbar}. \end{equation}\]The commutation (or anticommutation) relations are still true,
\[\begin{gather} [\hat{\psi}_{H\alpha}(\vb{x},t),\hat{\psi}_{H\beta}^{\dagger}(\vb{x}',t)]_{\mp} = \delta_{\alpha\beta}\delta(\vb{x}-\vb{x}'),\\[.2cm] [\hat{\psi}_{H\alpha}(\vb{x},t),\hat{\psi}_{H\beta}(\vb{x}',t)]_{\mp} = [\hat{\psi}_{H\alpha}^\dagger(\vb{x},t),\hat{\psi}_{H\beta}^{\dagger}(\vb{x}',t)]_{\mp}=0. \end{gather}\]This can be demonstrated using the commutation (or anticommutation) relations in the Schrödinger picture. And a general operator in the Heisenberg picture is given by
\[\begin{equation} \hat{O}_{H}(t) = \e^{i\hat{H}(t-t_0)/\hbar} \hat{O}_{S}(t_0) \e^{-i\hat{H}(t-t_0)/\hbar} = \hat{U}^{\dagger}(t,t_0)\hat{O}_{S}(t_0)\hat{U}(t,t_0), \end{equation}\]where $\hat{O}_{S}$ is the same operator in the Schrödinger picture. It satisfies the Heisenberg equation
\[\begin{equation} i\hbar \pdv{t} \hat{O}_{H}(t) = [\hat{O}_{H}(t),\hat{H}]. \end{equation}\]Any state vectors in the Heisenberg picture do not depend on time,
\[\begin{equation} \ket{\psi_H(t)} = \hat{U}^{\dagger}(t,t_0)\ket{\psi_S(t)} = \hat{U}^\dagger(t,t_0)\hat{U}(t,t_0)\ket{\psi_S(t_0)} = \ket{\psi_S(t_0)} = \ket{\psi_H(t_0)}. \end{equation}\]In general, the Hamiltonian $\hat{H}$ depend on time $t$, so the time evolution operator should be generalized as
\[\begin{equation} \hat{U}(t,t_0) = \hat{\mathsf{T}}\exp\qty[- \frac{i}{\hbar} \int_{t_0}^{t}\dd{t'}\hat{H}(t')],\quad (t>t_0) \end{equation}\]where $\hat{\mathsf{T}}$ is the time-ordering operator.
4.2. Interaction picture
The Hamiltonian is $\hat{H} = \hat{H}_0 + \hat{V}$ where $\hat{V}$ is the interaction operator. We can solve Schrödinger equation exactly when $\hat{V} =0$. And $\hat{H}_0\ket{\phi^0} = E_g^0 \ket{\phi^0}$. Then the field operators in the interaction picture are
\[\begin{equation} \hat{\psi}_{I\alpha}(\vb{x},t) = \mathrm{e}^{i\hat{H}_0(t-t_0)/\hbar} \hat{\psi}_{\alpha}(\vb{x}) \mathrm{e}^{-i\hat{H}_0(t-t_0)/\hbar},\quad \hat{\psi}_{I\alpha}^\dagger(\vb{x},t) = \mathrm{e}^{i\hat{H}_0(t-t_0)/\hbar} \hat{\psi}_{\alpha}^\dagger(\vb{x}) \mathrm{e}^{-i\hat{H}_0(t-t_0)/\hbar}. \end{equation}\]The commutation (or anticommutation) relations are
\[\begin{gather} [\hat{\psi}_{I\alpha}(\vb{x},t),\hat{\psi}_{I\beta}^{\dagger}(\vb{x}',t)]_{\mp} = \delta_{\alpha\beta}\delta(\vb{x}-\vb{x}'), \\[.2cm] [\hat{\psi}_{I\alpha}(\vb{x},t),\hat{\psi}_{I\beta}(\vb{x}',t)]_{\mp} = [\hat{\psi}_{I\alpha}^\dagger(\vb{x},t),\hat{\psi}_{I\beta}^{\dagger}(\vb{x}',t)]_{\mp}=0. \end{gather}\]For a general operator $\hat{O}_{I}(t)$ in the interaction picture, it can be expressed as
\[\begin{equation} \hat{O}_I(t) = \e^{i\hat{H_0}(t-t_0)/\hbar} \hat{O}_{S}(t_0) \e^{-i\hat{H_0}(t-t_0)/\hbar}, \end{equation}\]Its time evolution equation is
\[\begin{equation} i\hbar \pdv{t} \hat{O}_I(t) = [\hat{O}_I(t),\hat{H}_0]. \end{equation}\]And state vectors in the interaction picture are defined as
\[\begin{equation} \ket{\psi_I(t)} = \e^{i\hat{H_0}(t-t_0)/\hbar}\ket{\psi_{S}(t)} := \hat{S}(t,t_0)\ket{\psi_I(t_0)}, \end{equation}\]where $\hat{S}(t,t_0)$ is the time evolution operator. And
\[\begin{equation} \hat{S}(t,t') = \e^{i\hat{H_0}(t-t_0)/\hbar} \hat{U}(t,t') \e^{-i\hat{H}_0(t-t_0/\hbar).} \end{equation}\]The time evolution operator in the interaction picture can be expressed as
\[\begin{equation} \hat{S}(t,t_0) = \hat{\mathsf{T}} \exp\qty[-\frac{i}{\hbar}\int_{t_0}^{t} \hat{V}_I(t')\dd{t'}], \quad (t>t_0) \end{equation}\]where $\hat{\mathsf{T}}$ is the time-ordering operator.
If we want to switch Heisenberg picture to the interaction picture, we have the following relations:
\[\begin{gather} \ket{\psi_H} = \hat{S}(t_0,t)\ket{\psi_I(t)},\\[.2cm] \hat{O}_{H}(t) = \hat{S}^\dagger(t,t_0)\hat{O}_I(t)\hat{S}(t,t_0) = \hat{S}(t_0,t)\hat{O}_I(t)\hat{S}(t,t_0). \end{gather}\]5. References
- Fetter, A. L., Walecka, J. D. Quantum Theory of Many-Particle Systems. Courier Corporation (2002).
- Mahan, G. D., Many-Particle Physics. Springer Science+Business Media, LLC. (2000).
- Altland, A., Simons, B. D., Condensed Matter Field Theory. Cambridge University Press. (2010)
- 金彪. 量子多体理论. 2024年春