Applications of Second Quantization
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.
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 \[ \hat{h}(1) = \frac{\hat{\vb*{p}}_1^2}{2m} + \hat{V}_{\mathrm{ion}}(\vb*{r}_1). \]
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 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*}
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 \[ \langle a^\dagger_\nu a^\dagger_\lambda a_\beta a_\alpha \rangle = (\delta_{\nu\alpha}\delta_{\lambda\beta} - \delta_{\nu\beta}\delta_{\lambda\alpha}) \langle \hat{n}_\alpha \rangle \langle \hat{n}_\beta \rangle. \]
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 \(\nu\).
We evaluated the energy of uniform electron gas under the following prescription:
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.
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.
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:
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.