Advanced Chemistry - Mono Mole

February 7, 2023April 22, 2025

Thermodynamic property, state, equilibrium and process

A thermodynamic property is a measurable or computable characteristic of a system, such as pressure, volume and temperature. Two systems with all thermodynamic properties being quantitatively equivalent are said to be in the same thermodynamic state. The state of a system is therefore defined by the values of the system’s thermodynamic properties.

A system is in a state of thermodynamic equilibrium when there is no net transfer of energy or matter with its surroundings, resulting in its thermodynamic properties being constant over time.

We are mainly concerned with systems in thermodynamic equilibrium in our discussion of chemical thermodynamics. This is because the properties of the system are only well defined and no longer time-dependent when the system is at equilibrium. For example, when we subject a system to a change (e.g. by increasing the temperature of the system), we are interested in analysing the initial state of the system before the change when the system is in a state of thermodynamic equilibrium at an initial temperature, and the final state of the system a certain time after the change when all regions of the system have the same final temperature (i.e. the system is again in a state of thermodynamic equilibrium).

Finally, the path taken by the system from its initial state to its final state when a change is initiated is known as a thermodynamic process. A process can also be viewed as an integrated sequence of thermodynamic states at equilibrium, starting with an initial state and ending with a final state.

Question

Consider two systems that are initially at different temperatures. They are brought in contact via an immovable thermally conducting boundary. Do the two systems necessarily have the same state when they are in thermal equilibrium?

Answer

No. Even though the two systems have the same temperature over time, they may have different volumes and pressures. Hence, two systems at thermal equilibrium may not be in the same thermodynamic state.

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February 7, 2023November 26, 2024

Systems, surroundings and boundaries

A system is the part of the universe being studied, while the surroundings are the remaining parts of the universe, with the two regions separated by a boundary. For example, if we are investigating the relationship between the volume and pressure of a gas using a J-shaped tube immersed in a constant-temperature bath, the system is the gas trapped in the left arm of the tube (see diagram below).

The surroundings include the entire length of mercury in the tube and the constant temperature bath. The boundary consists of the wall of the J-shaped tube and the mercury meniscus in the left arm of the tube.

We need not consider the air in the right arm of the tube, the wall of the container of the bath and the air in the atmosphere since the system is in thermal equilibrium with the mercury and the bath.

The example mentioned above describes a closed system, where energy, but not matter, can be transferred between the system and its surroundings. This implies that the boundary of a closed system is thermally conductive. An open system is one in which both energy and matter can be transferred between the system and its surroundings. The boundary in this case is either physically permeable or notional. If neither energy nor matter can be transferred between the system and its surroundings, the system is called an isolated system, where the boundary is both thermally non-conductive and impermeable—that is, an adiabatic boundary.

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February 7, 2023April 14, 2023

Chemical thermodynamics (overview)

Chemical therrmodynamics is the branch of Chemistry that studies the interaction of chemical species with energy on a macroscopic level, with the objective of understanding and predicting chemical reactions.

Concepts such as system, surrounding, equilibrium, heat, work, entropy, enthalpy, state functions and path functions are elaborated in this topic along with the four laws of thermodynamics.

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December 27, 2022September 26, 2024

Solution of the Hartree-Fock-Roothaan equations for helium

To solve the Hartree-Fock-Roothaan equations for He, we begin by noting that $n'=2$ in the spin orbitals $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ and $\phi_i(x)=\sum_{\beta=1}^{n'}c_{i\beta}\chi_{\beta}(x)$ . If the basis wavefunctions $\chi_{\alpha}(x)$ and $\chi_{\beta}(x)$ are real, $\phi_i^*(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}^*\chi_{\alpha}^*(x)$ becomes $\phi_i(x)=\sum_{\alpha=1}^{n'}c_{i\alpha}\chi_{\alpha}(x)$ . With reference to eq157, we have

$(H_{11}-\lambda_iS_{11})c_{i1}+(H_{12}-\lambda_iS_{12})c_{i2}=0\;\;\;\;when\;\alpha=1$

$(H_{21}-\lambda_iS_{21})c_{i1}+(H_{22}-\lambda_iS_{22})c_{i2}=0\;\;\;\;when\;\alpha=2$

where $H_{\alpha\beta}=\int\chi_{\alpha}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_{\beta}(x)dx+\sum_{j\neq i}\sum_{\gamma,\delta}c_{j\gamma}c_{j\delta}\int\int\chi_{\alpha}(x)\chi_{\beta}(x)\frac{\chi_{\gamma}(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$ (assuming no contribution from the exchange integral) and $S_{\alpha\beta}=\int\chi_{\alpha}(x)\chi_{\beta}(x)dx$ .

The pair of simultaneous equations can be represented by the following matrix equation:

$\begin{pmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{21}-\lambda_iS_{21} & H_{22}-\lambda_iS_{22} \end{pmatrix}\begin{pmatrix} c_{i1}\\c_{i2}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}\;\;\;\;\;\;\;\;(158a)$

Let the basis wavefunctions be $\chi_1=\sqrt{\frac{\zeta_1^3}{\pi}}e^{-\zeta_1r}$ and $\chi_x=\sqrt{\frac{\zeta_x^3}{\pi}}e^{-\zeta_xr}$ . Since these Slater-type orbitals are real, $S_{12}=S_{21}$ . Eq158 becomes

$\begin{pmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{21}-\lambda_iS_{12} & H_{22}-\lambda_iS_{22} \end{pmatrix}\begin{pmatrix} c_{i1}\\c_{i2}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}\;\;\;\;\;\;\;\;(159)$

Question

Show that $H_{12}=H_{21}$ .

Answer

Since $\chi_1$ and $\chi_2$ are real,

$H_{12}=\int\chi_1(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_2(x)dx+$ $\sum_{j\neq i}\sum_{\gamma,\delta}c_{j\gamma}c_{j\delta}\int\int\chi_1(x)\chi_2(x)\frac{\chi_{\gamma}(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$
Using the Hermitian property of the KE operator,

$H_{12}=\int\chi_2(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\chi_1(x)dx+$ $\sum_{j\neq i}\sum_{\gamma,\delta}c_{j\gamma}c_{j\delta}\int\int\chi_2(x)\chi_1(x)\frac{\chi_{\gamma}(x')\chi_{\delta}(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'=H_{21}$

Therefore, eq159 becomes

$\begin{pmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{12}-\lambda_iS_{12} & H_{22}-\lambda_iS_{22} \end{pmatrix}\begin{pmatrix} c_{i1}\\c_{i2}\end{pmatrix}=\begin{pmatrix} 0\\0\end{pmatrix}\;\;\;\;\;\;\;\;(160)$

Eq160 is a linear homogeneous equation, which has non-trivial solutions if

$\begin{vmatrix}H_{11}-\lambda_iS_{11}& H_{12}-\lambda_iS_{12} \\ H_{12}-\lambda_iS_{12} & H_{22}-\lambda_iS_{22} \end{vmatrix}=0$

Expanding the determinant and noting that $S_{ii}=1$ , we get the characteristic equation:

$(1-S_{12}^2)\lambda_i^2+(2H_{12}S_{11}-H_{11}-H_{22})\lambda_i+H_{11}H_{22}-H_{12}^2 =0\;\;\;\;\;\;\;\;(161)$

The computation process is as follows:

Evaluate the integrals $H_{11}$ , $H_{12}$ , $H_{22}$ and $S_{12}$ in eq161 either analytically or numerically by letting $\zeta_1=1.6$ and $\zeta_2=2.0$ , and using the initial guess values of $c_{21}=c_{22}=0.50$ .
Substitute the evaluated integrals in eq161, solve for $\lambda_1$ and retain the lower root. Substitute $\lambda_1$ back in eq159 to obtain an expression of $c_{11}$ in terms of $c_{12}$ .
Use the expression derived in part 2, together with $\int\phi_i^*(x)\phi_i(x)d\tau=c_{11}^2S_{11}+2c_{11}c_{12}S_{12}+c_{22}^2S_{22}=1$ , to solve for $c_{11}$ and $c_{12}$ , which are then used as improved values of $c_{21}$ and $c_{22}$ for the next iteration.
Repeat steps 1 through 4 until $c_{11}$ , $c_{12}$ and $\lambda_1$ are invariant up to 6 decimal places.

The results are as follows:

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December 7, 2022September 26, 2024

Properties of the antisymmetriser

The properties of the antisymmetriser that are useful in deriving the Hartree-Fock equations are:

$\hat{A}^2=(n!)^{\frac{1}{2}}\hat A\;\;\;\;\;\;\;\;(63)$

$\hat{A}=\hat A^{\dagger}\;\;\;\;\;\;\;\;(64)$

$\hat{A}\hat H=\hat H\hat A\;\;\;\;\;\;\;\;(65)$

To proof eq63, we substitute eq62 in $\hat{A}^2$ to give $\hat{A}^2=\frac{1}{n!}\sum_{r=1}^{n!}(-1)^r\hat{P}_r\sum_{q=1}^{n!}(-1)^q\hat{P}_q$ .

Question

Show that

$\sum_{r=1}^{n!}(-1)^r\hat{P}_r\sum_{q=1}^{n!}(-1)^q\hat{P}_q\Psi_s(x_1,x_2,\cdots,x_n)=n!\sum_{t=1}^{n!}(-1)^t\hat{P}_t\Psi_s(x_1,x_2,\cdots,x_n)$

Answer

When $\sum_{q=1}^{n!}(-1)^q\hat{P}_q$ acts on $\Psi_s$ , we have $n!$ permutated terms. $\sum_{r=1}^{n!}(-1)^r\hat{P}_r$ then acts on each of these terms, resulting in $n!$ sets of $n!$ permutated terms, with the $n!$ terms in one set being identical to the $n!$ terms in another set. An illustration with $n=3$ is given below:

where the top row is the result of $\sum_{r=1}^{3!}(-1)^r\hat{P}_r\Psi_s(x_1,x_2,\cdots,x_n)$ , and each column is the outcome of $\sum_{q=1}^{3!}(-1)^q\hat{P}_q$ acting on each term in the top row.

In other words,

$\sum_{r=1}^{n!}(-1)^r\hat{P}_r\sum_{q=1}^{n!}(-1)^q\hat{P}_q\Psi_s(x_1,x_2,\cdots,x_n)=n!\sum_{t=1}^{n!}(-1)^t\hat{P}_t\Psi_s(x_1,x_2,\cdots,x_n)$

Therefore, $\hat{A}^2=\sum_{t=1}^{n!}(-1)^t\hat{P}_t\$ , which when substituted with eq62 gives eq63.

Eq64 states that $\hat{A}$ is Hermitian. To prove that $\int\Psi_s^*\hat A\Psi_s d\tau=\left[\int\Psi_s^*\hat A\Psi_s d\tau\right]^*$ , we note that

$\int\Psi_s^*\hat A\Psi_s d\tau=\frac{1}{\sqrt{n!}}\int\cdots\int\phi_1^*(x_1)\phi_2^*(x_2)\cdots\phi_n^*(x_n)$
$\times\left[\phi_1(x_1)\phi_2(x_2)\cdots\phi_n(x_n)-\phi_2(x_1)\phi_1(x_2)\cdots\phi_n(x_n)+\cdots\right]dx_1\cdots dx_n$
$=\small{\frac{1}{\sqrt{n!}}=\left(\frac{1}{\sqrt{n!}}\right)^*}$

Finally, eq65 states that $\hat{A}$ commutes with $\hat H$ . This is because

$[\hat{A}\hat H]\Psi_s=\hat A\hat H\Psi_s-\hat H\hat A\Psi_s=E\hat A\Psi_s-\hat H\Psi_a=E\Psi_a-E\Psi_a=0$

$\Psi_s$ and $\Psi_a$ correspond to the same state with energy $E$ because the way we label identical particles cannot affect the state of the system.

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December 7, 2022November 27, 2024

Slater determinant

A Slater determinant represents a multi-electron wavefunction that satisfies the Pauli exclusion principle. It was named after the American physicist John Slater for his contribution to quantum mechanics.

The normalised anti-symmetric wavefunction for an n-electron system with spin-orbitals $\phi_i(x_j)$ can be expressed as $\hat{A}\Psi_s(x_1,x_2,\cdots,x_n)$ , where $\Psi_s(x_1,x_2,\cdots,x_n)=\phi_1(x_1)\phi_2(x_2)\cdots\phi_n(x_n)$ and $\hat{A}$ is the antisymmetriser. From eq59 and 62,

$\hat{A}\Psi_s=\frac{1}{\sqrt{n!}}\sum_{r=1}^{n!}(-1)^r\hat{P}_r\Psi_s=\Psi_a$

Comparing the above equation with the definition of a determinant, i.e. $\vert G\vert=\sum_{i_1\cdots i_n}^n\varepsilon_{i_1\cdots i_n}a_{1,i_1}\cdots a_{n,i_n}$ , we have $\Psi_a=\frac{1}{\sqrt{n!}}\vert G\vert$ or

$\Psi_a=\frac{1}{\sqrt{n!}}\begin{vmatrix} \phi_1(x_1)&\phi_2(x_1)&\cdots &\phi_n(x_1)\\ \phi_1(x_2)&\phi_2(x_2)& \cdots &\phi_n(x_2)\\ \vdots&\vdots &\ddots &\vdots \\ \phi_1(x_n)&\phi_2(x_n)&\cdots &\phi_n(x_n) \end{vmatrix}\;\;\;\;\;\;\;\;(66)$

which is called an n-electron Slater determinant.

Since $\vert G\vert=\vert G^T\vert$ (see this article for proof), the Slater determinant can also be written as

$\Psi_a=\frac{1}{\sqrt{n!}}\begin{vmatrix} \phi_1(x_1)&\phi_1(x_2)&\cdots &\phi_1(x_n)\\ \phi_2(x_1)&\phi_2(x_2)& \cdots &\phi_2(x_n)\\ \vdots&\vdots &\ddots &\vdots \\ \phi_n(x_1)&\phi_2(n_2)&\cdots &\phi_n(x_n) \end{vmatrix}\;\;\;\;\;\;\;\;(67)$

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December 7, 2022September 26, 2024

Unitary transformation

A unitary transformation of a set of vectors to another set of vectors preserves the lengths of the vectors and the angles between the vectors.

In other words, a unitary transformation is a rotation of axes in the Hilbert space. This implies that if the transformation involves matrices of eigenvectors, the eigenvalues of the eigenvectors are preserved. Consider 2 complete sets of orthonormal bases:

$basis\;1:\;\;\;\left\langle\phi_i\vert\phi_j\right\rangle=\delta_{ij}\;\;\;\sum_{i=1}^n\vert\phi_i\rangle\langle\phi_i\vert=I$

$basis\;2:\;\;\;\left\langle\phi_q^{'}\vert\phi_s^{'}\right\rangle=\delta_{qs}\;\;\;\sum_{q=1}^n\vert\phi_q^{'}\rangle\langle\phi_q^{'}\vert=I$

where I is the identity matrix.

The relation between the two basis sets are

$\vert\phi_q^{'}\rangle=I\vert\phi_q^{'}\rangle=\sum_{i=1}^n\vert\phi_i\rangle\langle\phi_i\vert\phi_q^{'}\rangle=\sum_{i=1}^n\vert\phi_i\rangle u_{iq}=\sum_{i=1}^n\vert\phi_i\rangle (U)_{iq}\;\;\;\;\;\;\;\;(68)$

where $u_{iq}=\langle\phi_i\vert\phi_q^{'}\rangle$ are elements of the transformation matrix U.

We say that $\vert\phi_i\rangle$ is transformed to $\vert\phi_q^{'}\rangle$ by $u_{iq}$ . Similarly, we have $\vert\phi_s^{'}\rangle=\sum_{j=1}^n\vert\phi_j\rangle u_{js}$ .

The reverse transformation of eq68 is:

$\vert\phi_i\rangle=I\vert\phi_i\rangle=\sum_{q=1}^n\vert\phi_q^{'}\rangle\langle\phi_q^{'}\vert\phi_i\rangle=\sum_{q=1}^n\vert\phi_q^{'}\rangle u_{iq}^*=\sum_{q=1}^n\vert\phi_q^{'}\rangle (U^\dagger)_{qi}\;\;\;\;\;\;\;\;(69)$

Question

Show that $U^{-1}=U^{\dagger}$ .

Answer

$I=\delta_{ij}=\langle\phi_i\vert\phi_j\rangle=\sum_{q=1}^n\langle\phi_i\vert\phi_q^{'}\rangle\langle\phi_q^{'}\vert\phi_j\rangle$ $=\sum_{q=1}^nu_{iq}u_{jq}^*=\sum_{q=1}^n(U)_{iq}(U^\dagger)_{qj}=(UU^\dagger)_{ij}$

$I=UU^\dagger$

$U^{-1}I=U^{-1}UU^\dagger$

$U^{-1}=U^\dagger\;\;\;\;\;\;\;\;(70)$

Similarly,

$I=\delta_{qs}=\langle\phi_q^{'}\vert\phi_s^{'}\rangle=\sum_{i=1}^n\langle\phi_q^{'}\vert\phi_i\rangle\langle\phi_i\vert\phi_s^{'}\rangle$ $=\sum_{i=1}^nu_{iq}^*u_{is}=\sum_{i=1}^n(U^\dagger)_{qi}(U)_{is}=(U^\dagger U)_{qs}\;\;\;\;\;\;\;\;(71)$

$I=U^{\dagger}U$

$IU^{-1}=U^\dagger UU^{-1}$

$U^{-1}=U^\dagger$

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December 7, 2022October 27, 2024

Slater determinant of unitarily transformed spin orbitals

The Slater determinant of unitarily transformed spin orbitals $\Psi_a^{'}$ is related to the Slater determinant of the original spin orbitals $\Psi_a$ by the product of $\Psi_a$ and the determinant of the unitary matrix $U$ .

A unitary matrix is defined by eq70, where $U^{\dagger}=U^{-1}$ . Let $G'=GU$ , where

$G=\begin{pmatrix} \phi_1(x_1)&\phi_2(x_1)&\cdots &\phi_n(x_1)\\ \phi_1(x_2)&\phi_2(x_2)& \cdots &\phi_n(x_2)\\ \vdots&\vdots &\ddots &\vdots \\ \phi_1(x_n)&\phi_2(x_n)&\cdots &\phi_n(x_n) \end{pmatrix}$

Substituting $G$ and $G'$ in eq66 yields

$\Psi_a=\frac{1}{\sqrt{n!}}\vert G\vert\;\;\;\;\;\;\;\;(71a)$
and
$\Psi_a^{'}=\frac{1}{\sqrt{n!}}\vert G'\vert\;\;\;\;\;\;\;\;(71b)$
respectively.

Using the identity of $\vert AB\vert=\vert A\vert\vert B\vert$ for square matrices (see this article for proof) results in

$\vert G'\vert=\vert GU\vert=\vert G\vert\vert U\vert\;\;\;\;\;\;\;\;(72)$

Substituting eq71a and eq71b in eq72 gives

$\Psi_a^{'}=\Psi_a\vert U\vert$

From eq70, $\vert I\vert=\vert U^{\dagger}U\vert=\vert U^{\dagger}\vert\vert U\vert=\vert U^T\vert^*\vert U\vert=\vert U\vert^*\vert U\vert$ . Since $\vert I\vert=1$ (see this article for proof), we have $\vert U\vert^*\vert U\vert=1$ . Therefore, $\vert U\vert=e^{i\phi}$ (or $\vert U\vert=e^{-i\phi}$ ) and

$\Psi_a^{'}=e^{i\phi}\Psi_a$

The average value $O$ of a physical property of an electron, which is described by $\Psi_a^{'}$ , is $\langle e^{i\phi}\Psi_a\vert\hat O\vert e^{i\phi}\Psi_a\rangle=O\langle e^{i\phi}\Psi_a\vert e^{i\phi}\Psi_a\rangle=O$ , which is the same when the electron is described by $\Psi_a$ . In other words, any expectation value of a physical property of an electron is invariant to a unitary transformation of the electron’s wavefunction that is expressed as a Slater determinant. Such a consequence is used to derive the Hartree-Fock equations.

Lastly, if $\phi_q^{'}(x_j)$ , $\phi_i(x_j)$ and $u_{iq}$ are the matrix elements of $G'$ , $G$ and $U$ , we have

$\phi_q^{'}(x)=\sum_{i=1}^n\phi_i(x)u_{iq}$

$\phi_r^{'*}(x)=\sum_{i=1}^n\phi_i^*(x)u_{ir}^*$

$\phi_s^{'}(x)=\sum_{j=1}^n\phi_j(x)u_{js}$

$\phi_t^{'*}(x)=\sum_{j=1}^n\phi_j^*(x)u_{jt}^*$

where we have replaced the dummy variable within the parentheses with $x$ .

The reverse transformations are

$\phi_i(x)=\sum_{q=1}^n\phi_q^{'}(x)u_{iq}^*\;\;\;\;\;\;\;\;(74)$

$\phi_i^{*}(x)=\sum_{r=1}^n\phi_r^{'*}(x)u_{ir}\;\;\;\;\;\;\;\;(75)$

$\phi_j(x)=\sum_{s=1}^n\phi_s^{'}(x)u_{js}^*\;\;\;\;\;\;\;\;(76)$

$\phi_j^{*}(x)=\sum_{t=1}^n\phi_t^{'*}(x)u_{jt}\;\;\;\;\;\;\;\;(77)$

With reference to eq71,

$\sum_{i=1}^nu_{iq}^*u_{ir}=\sum_{i=1}^n(U^{\dagger})_{qi}(U)_{ir}=(U^{\dagger}U)_{qr}=\delta_{qr}\;\;\;\;\;\;\;\;(82)$

$\sum_{j=1}^nu_{js}^*u_{jt}=\sum_{j=1}^n(U^{\dagger})_{sj}(U)_{jt}=(U^{\dagger}U)_{st}=\delta_{st}\;\;\;\;\;\;\;\;(83)$

$\sum_{i=1}^n\sum_{j=1}^nu_{js}^*\lambda_{ji}^*u_{ir}=\sum_{i=1}^n\sum_{j=1}^n(U^{\dagger})_{sj}\Lambda_{ji}^*(U)_{ir}=(U^{\dagger}\Lambda U)_{sr}=\lambda_{sr}^{'}\;\;\;\;\;\;\;\;(84)$

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December 7, 2022May 16, 2026

Slater-Condon rule for a one-electron operator

The Slater-Condon rule for a one-electron operator is an expression of the expectation value of the one-electron operator involving Slater determinants.

Let’s consider the expectation value of the one-electron Hamiltonian operator: $\langle\Psi_a\vert\hat h\vert\Psi_a\rangle$ , where $\hat h=\sum_{i=1}^n\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)$ and $\Psi_a$ is given by eq66.

Question

Show that the product of $\hat h$ and $\hat A$ , where $\hat A$ is the antisymmetriser, is Hermitian.

Answer

$(\hat h\hat A)_{ij}^{\dagger}=[(\hat h\hat A)_{ij}^T]^*=(\hat h\hat A)_{ji}^*=\left(\sum_kh_{jk}a_{ki}\right)^*=\left[\sum_k(h^T)_{kj}(a^T)_{ik}\right]^*$ $=\left[\sum_k(a^T)_{ik}(h^T)_{kj}\right]^*=(\hat A^T\hat h^T)_{ij}^*=(\hat A^{\dagger}\hat h^{\dagger})_{ij}$

or $(\hat h\hat A)_{ij}^{\dagger}=(\hat A^{\dagger}\hat h^{\dagger})_{ij}$ .

With reference to eq64 and the fact that $\hat h=\hat h^{\dagger}$ , we have $(\hat h\hat A)^{\dagger}=\hat A^{\dagger}\hat h^{\dagger}=\hat A\hat h$ . Using eq65, we have $(\hat h\hat A)^{\dagger}=\hat h\hat A$ , i.e. the product of $\hat h$ and $\hat A$ is Hermitian. This implies that $\hat h\hat O$ , where $\hat O=\hat A^2$ is also Hermitian.

Similarly, the product of the dipole moment operator $\hat{\boldsymbol{\mathit{\mu}}}=\sum_iq_i\hat{\boldsymbol{\mathit{r}}}_i$ and $\hat A$ is Hermitian. This is useful in evaluating the electronic transition selection rules for atoms.

Substituting eq59 in $\langle\Psi_a\vert\hat h\vert\Psi_a\rangle$ and using

i) eq65
ii) the Hermitian property of $\hat h\hat A$
iii) the Hermitian property of $\hat h\hat A^2$ twice

yields

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\langle\hat A\Psi_s\vert\hat h\hat A\vert\Psi_s\rangle=\langle\hat h\hat A^2\Psi_s\vert\Psi_s\rangle=\langle\Psi_s\vert\hat h\hat A^2\vert\Psi_s\rangle$

Substitute eq62 and eq63 in the above equation,

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\langle\Psi_s\vert\hat h\sum_{t=1}^{n!}(-1)^t\hat P_t\Psi_s\rangle\;\;\;\;\;\;\;\;(85)$

Substituting $\Psi_s=\phi_1(x_1)\phi_2(x_2)\cdots\phi_x(x_n)$ and $\hat h=\sum_{i=1}^n\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)$ in eq85 and simplifying gives

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\sum_{i=1}^n\int\phi_1^*(x_1)\phi_1(x_1)dx_1\cdots\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i$ $\cdots\int\phi_n^*(x_n)\phi_n(x_n)dx_n$

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\sum_{i=1}^{n}H_i\;\;\;\;\;\;\;\;(86)$

where $H_i=\int\phi_i^*(x_i)\left(-\frac{1}{2}\nabla_i^2-\frac{Z}{r_i}\right)\phi_i(x_i)dx_i$ .

Given that any expectation value of a physical property of an electron is invariant to a unitary transformation of the electron’s wavefunction expressed as a Slater determinant, we substitute eq74 and eq75 in eq86 to give

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\int\sum_{r=1}^n\sum_{q=1}^n\sum_{i=1}^nu_{iq}^*u_{ir}\phi_r^{'*}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_q'(x)dx\;\;\;\;\;\;\;\;(87)$

where we have changed the dummy variable from $x_i$ to $x$ .

Substitute eq82 in eq87,

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\int\sum_{r=1}^n\sum_{q=1}^n\delta_{qr}\phi_r^{'*}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_q'(x)dx$

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\int\sum_{r=1}^n\phi_r^{'*}(x)\left(-\frac{1}{2}\nabla^2-\frac{Z}{r}\right)\phi_r'(x)dx\;\;\;\;\;\;\;\;(88)$

Eq88 is used in the derivation of the Hartree-Fock equations.

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Slater-Condon rule for a two-electron operator

The Slater-Condon rule for a two-electron operator is an expression of the expectation value of the two-electron operator involving Slater determinants.

Let’s consider the expectation value of the two-electron Hamiltonian operator: $\langle\Psi_a\vert\hat h\vert\Psi_a\rangle$ , where $\hat h=\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\frac{1}{r_{ij}}$ and $\Psi_a$ is given by eq66.

Substituting $\hat h=\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\frac{1}{r_{ij}}$ in eq85,

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\int\cdots\int\Psi_s^*\frac{1}{r_{ij}}\sum_{t=1}^{n!}(-1)^t\hat P_t\Psi_s dx_1\cdots dx_n$

Substituting $\Psi_s=\phi_1(x_1)\phi_2(x_2)\cdots\phi_x(x_n)$ in the above equation and simplifying, we have

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\biggr[\int\phi_1^*(x_1)\phi_1(x_1)dx_1\cdots\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{1}{r_{ij}}\phi_i(x_i)\phi_j(x_j)dx_idx_j$ $\cdots\int\phi_n^*(x_n)\phi_n(x_n)dx_n-\int\phi_1^*(x_1)\phi_1(x_1)dx_1$ $\cdots\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{1}{r_{ij}}\phi_j(x_i)\phi_i(x_j)dx_idx_j\cdots\int\phi_n^*(x_n)\phi_n(x_n)dx_n\biggr]$

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\left(J_{ij}-K_{ij}\right)=\frac{1}{2}\sum_{j=i}^n\sum_{i=1}^n\left(J_{ij}-K_{ij}\right)\;\;\;\;\;\;\;\;(89)$

where $J_{ij}=\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{1}{r_{ij}}\phi_i(x_i)\phi_j(x_j)dx_idx_j$ and $K_{ij}=\int\int\phi_i^*(x_i)\phi_j^*(x_j)\frac{1}{r_{ij}}\phi_j(x_i)\phi_i(x_j)dx_idx_j$ .

$J_{ij}$ and $K_{ij}$ are known as the Coulomb integral and the exchange integral, respectively.

Question

Show that $\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\left(J_{ij}-K_{ij}\right)=\frac{1}{2}\sum_{j=i}^n\sum_{i=1}^n\left(J_{ij}-K_{ij}\right)$ .

Answer

Since $J_{kk}-K_{kk}=0$ , we can add $n$ terms of $J_{kk}-K_{kk}$ , where $k=1,\cdots,n$ to $\frac{1}{2}\sum_{j \neq i}\sum_{i=1}^n\left(J_{ij}-K_{ij}\right)$ , resulting in $\frac{1}{2}\sum_{j=i}^n\sum_{i=1}^n\left(J_{ij}-K_{ij}\right)$ , i.e.

$\sum_{j=i}^n\sum_{i=1}^nJ_{ij}=\int\int\sum_{r,t,s,j,q,i=1}^nu_{iq}^*u_{ir}\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_q'(x)\phi_s'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}u_{js}^*u_{jt}dxdx'$

where we have changed the dummy variables from $x_i$ and $x_j$ to $x$ and $x'$ respectively.

Substituting eq82 in the above equation and simplifying

$\sum_{j=i}^n\sum_{i=1}^nJ_{ij}=\int\int\sum_{r,t,s,j,q=1}^n\delta_{qr}\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_q'(x)\phi_s'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}u_{js}^*u_{jt}dxdx'$ $=\int\int\sum_{r,t,s,j=1}^nu_{js}^*u_{jt}\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_r'(x)\phi_s'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$
Similarly, substituting eq83 in the above equation and simplifying

$\sum_{j=i}^n\sum_{i=1}^nJ_{ij}=\int\int\sum_{r,t}^n\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_r'(x)\phi_t'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'\;\;\;\;\;\;\;\;(90)$

Substituting eq74, eq75, eq76 and eq77 in $\sum_{j=i}^n\sum_{i=1}^nK_{ij}$ of eq89 and repeating the above logic, we have

$\sum_{j=i}^n\sum_{i=1}^nK_{ij}=\int\int\sum_{r,t}^n\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_t'(x)\phi_r'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'\;\;\;\;\;\;\;\;(91)$

Substitute eq90 and eq91 back in eq89

$\langle\Psi_a\vert\hat h\vert\Psi_a\rangle=\frac{1}{2}\sum_{r=i}^n\sum_{t=1}^n\int\int\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_r'(x)\phi_t'(x')-\phi_t'(x)\phi_r'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'$ $=\frac{1}{2}\sum_{t\neq r}\sum_{t=1}^n\int\int\phi_r^{'*}(x)\phi_t^{'*}(x')\frac{\phi_r'(x)\phi_t'(x')-\phi_t'(x)\phi_r'(x')}{\vert\boldsymbol{\mathit{r}}-\boldsymbol{\mathit{r}}'\vert}dxdx'\;\;\;\;\;\;\;\;(92)$

Eq92 is used in the derivation of the Hartree-Fock equations.

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