Ze Chen

## Advanced Solid State Physics (I)

Applications of Second Quantization

#### Solid State Physics

##### Heat Capacity

The heat capacity may be given by $C_V = T\qty(\pdv{S}{T})_V.$ The contribution from electrons may be given by $\delta S = \frac{V}{T} \int \dd{\epsilon_{\vb*{p}}} \delta f_{\vb*{p}} N(\epsilon_{\vb*{p}})(\epsilon_{\vb*{p}} - \mu),$ where the DoS of electrons is given by $N(\epsilon_{\vb*{p}}) = \frac{p^2}{\pi^2 \hbar^3} \dv{p}{\epsilon_{\vb*{p}}}.$ The integral of $$\delta S$$ may be evaluated using the Sommerfeld expansion.

#### Second Quantization

• The basis functions of our expansion doesn't have to cooperate with the exact eigenfunctions of the Hamiltonian.
• Field operator: \begin{align*} \hat{\psi}(\vb*{x}) &= \sum_{\vb*{k}} \psi_{\vb*{k}}(\vb*{x}) c_{\vb*{k}}, \\ \hat{\psi}{\vb*{x}} &= \sum_{\vb*{k}} \psi_{\vb*{k}}(\vb*{x})^\dagger c^\dagger_{\vb*{k}}. \end{align*}
• Second-quantized operator:
• Before second-quantization: $J = \sum_{i=1}^N J(\vb*{x}_i).$
• After second-quantization: \begin{align*} \color{orange} \hat{J} &= \sum_{rs} \bra{r}J\ket{s} c_{r}^\dagger c_s \\ &\color{orange}= \int \dd{^3 \vb*{x}} \hat{\psi}^\dagger(\vb*{x}) J(\vb*{x}) \hat{\psi}(\vb*{x}). \end{align*}
• Hamiltonian:
• $\color{orange} \hat H = \sum_{rs} a_{r}^\dagger \bra{r}T\ket{s} a_s + \frac{1}{2} \sum_{rstu} a_r^\dagger a_s^\dagger \bra{rs}V\ket{tu} a_u a_t.$
• $\color{orange} \hat{H} = \int \dd{^3 \vb*{x}} \hat{\psi}^\dagger(\vb*{x}) T \hat{\psi}(\vb*{x}) + \frac{1}{2} \iint \dd{^3 \vb*{x}} \dd{^3 \vb*{x}'} \hat{\psi}^\dagger(\vb*{x}) \hat{\psi}^\dagger(\vb*{x}') V(\vb*{x},\vb*{x}') \hat{\psi}(\vb*{x}')\hat{\psi}(\vb*{x}).$
• Number-density operator: $\color{orange} n(\vb*{x}) = \hat{\psi}^\dagger(\vb*{x}) \hat{\psi}(\vb*{x}).$
• Total-number operator: $\color{orange}\hat{N} = \int \dd{^3 \vb*{x}} \hat{\psi}^\dagger(\vb*{x}) \hat{\psi}(\vb*{x}).$
• The problem in the abstract Hilbert space separates into a sequence of problems in the subspaces corresponding to a fixed total number of particles. Nevertheless, the abstract Hilbert space contains states with any number of particles.

#### Hatree-Fock Approximation

The Hamiltonian is given by $\hat{H}_e = \sum_{\nu\lambda} \bra{\nu} \hat{h}(1) \ket{\lambda} a_\nu^\dagger a_\lambda + \frac{1}{2} \sum_{\nu\lambda\alpha\beta} \bra{\nu\lambda} \frac{e^2}{\abs{\vb*{r}_1 - \vb*{r}_2}}\ket{\alpha\beta} a_\nu^\dagger a_\lambda^\dagger a_\beta a_\alpha,$ where $\hat{h}(1) = \frac{\hat{\vb*{p}}_1^2}{2m} + \hat{V}_{\mathrm{ion}}(\vb*{r}_1).$

##### Non-Interacting Limit

The ket of non-interacting electron gas is given by $\ket{\psi_0} = \ket{\vb*{p}_0 \uparrow, \vb*{p}_0\downarrow, \cdots, \vb*{p}_{\mathrm{F}} \uparrow, \vb*{p}_{\mathrm{F}} \downarrow} = a^\dagger_{\vb*{p}_0\uparrow} a^\dagger_{\vb*{p}_0\downarrow} \cdots a^\dagger_{\vb*{p}_{\mathrm{F}} \uparrow} a^\dagger_{\vb*{p}_{\mathrm{F}} \downarrow} \ket{0}.$

Now we evaluate the expectation value of $$h_1$$.

• The expectation value of kinetic energy is given by $\langle \hat{T} \rangle = 2 \sum_{p<p_{\mathrm{F}}} \frac{p^2}{2m} = \frac{3}{5} \frac{p_{\mathrm{F}}^2 }{2m} N.$
• The expectation value of ion potential energy is given by $\langle \hat V_{\mathrm{ion}} \rangle = 2\sum_{p<p_{\mathrm{F}}}V_{\mathrm{ion}}(0).$ where $V_{\mathrm{ion}}(0) = \frac{1}{V} \int \dd{\vb*{r}} V_{\mathrm{ion}}(\vb*{r}).$

The Hamiltonian may be written in second-quantized form as \begin{align*} \hat{H}_1 &= \sum_\sigma \int \dd{\vb*{r}} \qty[-\frac{\hbar^2}{2m} \psi^\dagger_\sigma (\vb*{r}) \laplacian \psi_\sigma(\vb*{r}) + \psi^\dagger_\sigma(\vb*{r}) V_{\mathrm{ion}}(\vb*{r}) \psi_\sigma(\vb*{r})] \\ &= \sum_\sigma \int \dd{\vb*{r}} \qty[ \frac{\hbar^2}{2m} \abs{\grad \psi_\sigma}^2 + \hat{n}_\sigma(\vb*{r}) \psi_{\mathrm{ion}}(\vb*{r}) ]. \end{align*} This could also be written in the orbital basis as \begin{align*} \langle \hat{H}_1 \rangle &= \sum_\nu \bra{\nu}\hat{h}(1)\ket{\nu} n_\nu \\ &= \sum_{\nu \mathrm{\ (occ.)}} \int \qty[\frac{\hbar^2}{2m}\abs{\grad \phi_\nu}^2 + \phi_\nu^*(\vb*{r}) V_{\mathrm{ion}}(\vb*{r})\phi_\nu(\vb*{r})] \dd{\vb*{r}}. \end{align*}

##### Coulomb Interaction

The two-body part of $$H$$ is given by \begin{align*} \langle \hat{H}_2 \rangle &= \bra{\psi_0} \hat{V}_{\mathrm{ee}}\ket{\psi_0} \\ &= \frac{1}{2} \sum_{\alpha,\beta,\nu,\lambda} \bra{\nu}\bra{\lambda} \frac{e^2}{\abs{\vb*{r}_1 - \vb*{r}_2}} \ket{\alpha}\ket{\beta} \langle a^\dagger_\nu a^\dagger_\lambda a_\beta a_\alpha \rangle \\ &= \frac{}{2} \sum_{\nu,\lambda} \bra{\nu}\bra{\lambda} \frac{e^2}{\abs{\vb*{r}_1 - \vb*{r}_2}} \ket{\nu}\ket{\lambda} n_\lambda n_\nu - \frac{1}{2} \sum_{\nu,\lambda} \bra{\nu}\bra{\lambda} \frac{e^2}{\abs{\vb*{r}_1 - \vb*{r}_2}} \ket{\lambda}\ket{\nu} n_\lambda n_\nu \\ &= \frac{1}{2} \sum_{\nu,\lambda \mathrm{\ (occ.)}} (U_{\nu\lambda} - \delta_{\sigma_\nu \sigma_\lambda}J_{\nu\lambda}). \end{align*} Where we have used Wick's rules and found $\langle a^\dagger_\nu a^\dagger_\lambda a_\beta a_\alpha \rangle = (\delta_{\nu\alpha}\delta_{\lambda\beta} - \delta_{\nu\beta}\delta_{\lambda\alpha}) \langle \hat{n}_\alpha \rangle \langle \hat{n}_\beta \rangle.$

• $$U_{\nu\lambda}$$ is the direct Coulomb interaction, $U_{\nu\lambda} = \int \dd{\vb*{r}_1} \dd{\vb*{r}_2} \abs{\phi_\nu(\vb*{r}_1)}^2 \frac{e^2}{\abs{\vb*{r}_1 - \vb*{r}_2}} \abs{\phi_\lambda(\vb*{r}_2)}^2.$
• $$J_{\nu\lambda}$$ is the exchange interaction, $J_{\nu\lambda} = \int \dd{\vb*{r}_1} \dd{\vb*{r}_2} \phi^*_\nu(\vb*{r}_1) \phi^*_\lambda(\vb*{r}_2) \frac{e^2}{\abs{\vb*{r}_1 - \vb*{r}_2}} \phi_\nu(\vb*{r}_2) \phi_\lambda(\vb*{r}_1).$

The total Hatree-Fock energy is given by $E_{\mathrm{HF}} = \langle \hat{H}_1 \rangle + \frac{1}{2} \sum_{\lambda,\nu \mathrm{\ (occ.)}} [U_{\nu\lambda} - U_{\nu\lambda}].$ To obtain the orbitals, we minimize the ground state energy with respect to the $$\phi$$'s, under the condition that $$\phi$$'s are normalized. We introduce a Lagrange multiplier $$\epsilon_\nu$$ and write down the Euler-Lagrange equation $\frac{\delta E_{\mathrm{HF}}}{\delta \phi_\nu^*(\vb*{r})} = \epsilon_\nu \frac{\delta}{\delta \phi^*_\nu(\vb*{r})} \int \dd{\vb*{r}'} \abs{\phi_\nu(\vb*{r})}^2,$ which yields the Hatree-Fock equations \begin{align*} \qty[-\frac{\hbar^2}{2m}\grad^2 + \hat{V}_{\mathrm{ion}}(\vb*{r}) + \sum_\lambda \int \dd{\vb*{r}'} n_\lambda(\vb*{r}')\frac{e^2}{\abs{\vb*{r} - \vb*{r}'}}]\phi_\nu(\vb*{r}) & \\ -\sum_\lambda \int \dd{\vb*{r}'} \phi_\lambda^*(\vb*{r}') \phi_\nu(\vb*{r}') \frac{e^2}{\abs{\vb*{r} - \vb*{r}'}} \phi_\lambda(\vb*{r}) &= \epsilon_\nu \phi_\nu(\vb*{r}). \end{align*} Therefore we find $\epsilon_\nu = \bra{\nu} h_1 \ket{\nu} + \sum_{\lambda \mathrm{\ occ.}} (U_{\nu\lambda} - J_{\nu\lambda}),$ and $E_{\mathrm{HF}} = \sum_{\nu \mathrm{\ (occ.)}} \epsilon_\nu - \frac{1}{2} \sum_{\lambda,\nu \mathrm{\ (occ.)}} (U_{\nu\lambda} - J_{\nu\lambda}).$

The Hatree-Fock energy $$\epsilon_\nu$$ is the energy required to add a particle in the previously unoccupied orbital $$\nu$$.

#### Interacting Electron Gas

##### Uniform Electron Gas

We evaluated the energy of uniform electron gas under the following prescription:

• The basis functions are given by $\phi_{\vb*{p}}(\vb*{r}) = \frac{e^{i\vb*{p}}\cdot \vb*{r}}{\sqrt{V}}.$
• The charge density is uniform, i.e. $\sum_\lambda n_{\lambda}(\vb*{r}') = n_{\mathrm{e}}(\vb*{r}') = n_{\mathrm{e}}.$
• The ion potential is given by $V_{\mathrm{ion}}(\vb*{r}) = -\sum_i \frac{Ze^2}{\abs{\vb*{r} - \vb*{R}_i}} \rightarrow -e^2 n_{\mathrm{e}}\int \frac{\dd{\vb*{R}}}{\abs{\vb*{r} - \vb*{R}}},$ which cancels the direct Coulomb interaction.
• The exchange interaction is given by \begin{align*} -e^2 \sum_\lambda \int \dd{\vb*{r}'} \phi_\nu(\vb*{r}') \phi^*_\lambda(\vb*{r}') \phi_\lambda(\vb*{r})\frac{1}{\abs{\vb*{r} - \vb*{r}'}} &\rightarrow -e^2 \int \frac{\dd{\vb*{p}'}\dd{\vb*{x}}}{(2\pi\hbar)^3} \frac{e^{i(\vb*{p}-\vb*{p}')\cdot \vb*{x}/\hbar}}{\abs{\vb*{x}}} \phi_{\vb*{p}}(\vb*{r}). \end{align*}

Therefore, the energy of a particle in a momentum state $$\vb*{p}$$ is given by $\epsilon(\vb*{p}) = \frac{p^2}{2m} + \epsilon_{\mathrm{exch}}(\vb*{p})$ where \begin{align*} \epsilon_{\mathrm{exch}}(\vb*{p}) &= - e^2 \int_0^{p_{\mathrm{F}}} \frac{\dd{\vb*{p}'}}{(2\pi\hbar)^3} \int \dd{\vb*{x}} \frac{e^{i(\vb*{p} - \vb*{p}')\cdot \vb*{x}/\hbar}}{\abs{\vb*{x}}} \\ &= -\frac{e^2p_{\mathrm{F}}}{\pi\hbar} \qty(1 + \frac{(p_{\mathrm{F}}^2 - p^2)}{2pp_F}\ln \abs{\frac{p+p_{\mathrm{F}}}{p - p_{\mathrm{F}}}}). \end{align*} The total Hatree-Fock energy is therefore given by \begin{align*} \color{orange}E_{\mathrm{HF}} &= 2V\int_0^{p_{\mathrm{F}}} \frac{\dd{\vb*{p}'}}{(2\pi\hbar)^3} \qty(\frac{p^2}{2m} + \frac{1}{2}\epsilon_{\mathrm{exch}}\qty(\vb*{p})) \\ &= \frac{Ne^2}{2a_0}\qty[\frac{3}{5}\qty(\frac{p_{\mathrm{F}}a_0}{\hbar})^2 - \frac{3}{2\pi}\qty(\frac{p_{\mathrm{F}}a_0}{a_0})] \\ &\color{orange}= N\qty(\frac{2.21}{r_{\mathrm{s}}^2} - \frac{0.916}{r_{\mathrm{s}}}) \operatorname{Ry}, \end{align*} where $r_{\mathrm{s}} = \qty(\frac{9\pi}{4})^{1/3} \frac{e^2}{\hbar v_{\mathrm{F}}}.$

Introducing $F(x) = \frac{1}{2} + \frac{1-x^2}{4x} \ln \abs{\frac{1+x}{1-x}}$ we find $\frac{\epsilon(\vb*{p})}{\epsilon_{\mathrm{F}}^0} = x^2 - 0.663 r_{\mathrm{s}} F(x),$ where $x = \frac{p}{p_{\mathrm{F}}}.$

The exchange energy is negative because electrons of like spin avoid one another.

##### Hatree-Fock Excitation Spectrum

The DoS evaluates to $N(\epsilon) = \frac{p_{\mathrm{F}}^2}{\pi^2 \hbar^3} \pdv{p}{\epsilon} \approx \frac{mp_{\mathrm{F}}/\pi^2 \hbar^2}{1 - (me^2/\pi\hbar p_{\mathrm{F}})\ln\abs{(p-p_{\mathrm{F}})/2p_{\mathrm{F}}}}.$ With $$p^2/2m - p_{\mathrm{F}}^2/2m \approx T$$ we find $C_V \sim \frac{T}{\ln T},$ which deviates from the (correct) linear dependence, because the screening of Coulomb potential was not taken into account.

Hatree-Fock approximation overestimate the energy bandwidth to $\Delta = \epsilon(x=1) - \epsilon(x=0) = \epsilon_{\mathrm{F}}^0(1+0.331r_{\mathrm{s}}),$ while in the free-electron model $$\Delta = \epsilon_{\mathrm{F}}^0$$ fits well with experimental data.

##### Cohesive Energy of Metals

The cohesive energy is defined by $\epsilon_{\mathrm{coh}} = (\epsilon/\mathrm{atom})_{\mathrm{matal}} - (\epsilon/\mathrm{atom})_{\mathrm{free}}.$ We write the cohesive energy as $\epsilon_{\mathrm{coh}} = \epsilon_0 - \epsilon_{\mathrm{atom}} + \epsilon_{\mathrm{kin}} + \epsilon_{\mathrm{coul}}.$ Using the Wigner-Seitz method for alkali metals, we find the following:

• $$\epsilon_0$$ is given by $\qty{\frac{\hat{p}^2}{2m} + V_{\mathrm{ion}}} \varphi_0(\vb*{r}) = \epsilon_0 \varphi_0(\vb*{r})$ subjected to the boundary condition $\varphi_0'(\vb*{r}) = 0,$ which may be estimated to be $\epsilon_0 \approx -n_e \int_{a_0}^{r_{\mathrm{e}}} \dd{\vb*{r}} \frac{e^2}{r} = -\frac{40.82\, \mathrm{eV}}{r_{\mathrm{s}}}\qty(1-\frac{1}{r_{\mathrm{s}}^2}).$
• The kinetic energy may be given by $\epsilon_{\mathrm{kin}}^0 = \frac{3}{5} \frac{p_{\mathrm{F}}^2}{2m} = \frac{2.21}{r_{\mathrm{s}}^2} \operatorname{Ry}.$
• $$\epsilon_{\mathrm{coul}} = \epsilon_{\mathrm{direct}} + \epsilon_{\mathrm{exch}}$$ where $\epsilon_{\mathrm{direct}} = \int_0^{r_{\mathrm{e}}} \dd{\vb*{r}} \frac{e^2}{r} n_{\mathrm{e}} n(r) = \frac{6}{5r_{\mathrm{s}}}\operatorname{Ry}$ where $$n(r) = 4\pi r^3 n_{\mathrm{e}}/3$$ and $\epsilon_{\mathrm{exch}} = -\frac{0.916}{r_{\mathrm{s}}}\operatorname{Ry}.$

Using perturbation theory we add the correlation energy \begin{align*} \epsilon_{\mathrm{corr}} &\approx \begin{cases} \displaystyle -\frac{0.884}{r_{\mathrm{s}}} \operatorname{Ry},& r_{\mathrm{s}} \rightarrow \infty, \\ \displaystyle 0.062\ln r_{\mathrm{s}} - 0.096, & r_{\mathrm{s}} \rightarrow 0, \end{cases} \end{align*} which yields more accurate results.

##### Wigner Solid
• The Hatree-Fock approximation applies at $$r_{\mathrm{s}}\rightarrow 0$$.
• In the low-density limit $$r_{\mathrm{s}}\rightarrow \infty$$, we should use the Wigner solid model by allowing electrons to crystallize in a Wigner solid.
• Energy $\color{orange} \frac{E}{N} = \frac{e^2}{2a_0}\qty{-\frac{1.79}{r_{\mathrm{s}}} + \frac{2.66}{r_{\mathrm{s}}^{3/2}} + \cdots}.$

2021/3/16 10:56:01

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