'push MOs'

ch08/note01.md

@@ -1,7 +1,18 @@
 ## BO approximation
 
-Because nuclei are much heavier than electrons, the Schrodinger equation can be approximately separated into the nuclear and the electron parts. Thus the electronic Schr\"odinger equation for a molecule can be solved separately at each fixed nuclear configuration. This is called the Born-Oppenheimer approximation.
+:::{admonition} What you need to know
+:class: note
 
+-  **Born-Oppenheimer (BO) approximation**, Because nuclei are much heavier than electrons, the Schrodinger equation can be approximately separated into the nuclear and the electron parts. Thus the electronic Schrodinger equation for a molecule can be solved separately at each fixed nuclear configuration. This is called the Born-Oppenheimer approximation.
+- BO approximation allows us to extend multi-electron treatment of atoms to molecules. 
+- Single electron wavefunctions of molecules are called **Molecular Orbitals (MO)**
+- Geometry of $H_2$ molecule defined by its bond length $R$ is fixed when determining molecular orbitals. 
+- Nuclear geometry is a varyable parameter in the problem: For different values of R one gets different energies and MOs. 
+:::
+
+
+
+### Simplest molecule
 
 In the following, we will consider the simplest molecule $H_2^+$, which contains only one electron. This simple system will demonstrate the basic concepts in chemical bonding. The Schr\"odinger equation for H$_2^+$ is:
 
@@ -14,7 +25,6 @@ $${\hat{H} = -\frac{\hbar^2}{2M}(\Delta_A + \Delta_B) - \frac{\hbar^2}{2m_e}\Del
 
 where $M$ is the proton mass, $m_e$ is the electron mass, $r_{1A}$ is the distance between the electron and nucleus A, $r_{1B}$ is the distance between the electron and nucleus B and $R$ is the A - B distance.
 
-
 Note that the Hamiltonian includes also the quantum mechanical kinetic energy for the protons. As such the wavefunction depends on $\vec{r}_1$, $\vec{R}_A$ and $\vec{R}_B$. Because the nuclear mass $M$ is much larger than the electron mass $m_e$, the wavefunction can be separated (Born-Oppenheimer approximation):
 
 $${\psi(\vec{r}_1,\vec{R}_A,\vec{R}_B) = \psi_e(\vec{r}_1, R)\psi_n(\vec{R}_A, \vec{R}_B)}$$
@@ -23,7 +33,7 @@ where $\psi_e$ is the electronic wavefunction that depends on the distance $R$ b
 
 $${\hat{H}_e\psi_e = E_e\psi_e}$$
 
-Note that Eq. (\ref{eq11.4}) depends parametrically on $R$ (``one equation for each value of $R$''). The \textit{electronic Hamiltonian} is:
+Note that equation depends parametrically on $R$ (``one equation for each value of $R$''). The \textit{electronic Hamiltonian} is:
 
 $${\hat{H}_e = -\frac{\hbar^2}{2m_e}\Delta_e + \frac{e^2}{4\pi\epsilon_0}
 \left(\frac{1}{R} - \frac{1}{|r_1 - R_A|} - \frac{1}{|r_1 - R_B|}\right)}$$

ch08/note02.md

@@ -1,20 +1,32 @@
 ## The hydrogen molecule ion
 
+:::{admonition} What you need to know
+:class: note
 
-The electronic Schr\"odinger equation for H$_2^+$ can be solved exactly because the equation contains only one particle. However, the  involved math is very complicated and here we take another simpler but  approximate approach. This method will reveal all the important features of chemical bond. An approximate (trial) wavefunction is written as (real functions):
+-  We construct trial wavefunction of the molecule from atomic orbitals.
+- Applying variational method we find the two MOs to have distinct spatial profiles with one increasing probability of electron between the two nuclei and the other depleating. 
+- These MOs are called bonding and anibonding orbitals. They lower and raise eneryg of molecule relative to two separated atoms. 
+:::
+
+### Constructing MOs from AOs
+
+- The electronic Schrodinger equation for H$_2^+$ can be solved exactly because the equation contains only one particle. However, the  involved math is very complicated and here we take another simpler but  approximate approach. This method will reveal all the important features of chemical bond. An approximate (trial) wavefunction is written as (real functions):
 
 $${\psi_\pm(\vec{r}_1) = c_11s_A(\vec{r}_1) \pm c_21s_B(\vec{r}_1)}$$
 
-where $1s_A$ and $1s_B$ are hydrogen atom wavefunctions centered at nucleus A 
+- where $1s_A$ and $1s_B$ are hydrogen atom wavefunctions centered at nucleus A 
 and B, respectively, and $c_1$ and $c_2$ are constants. This function is
 essentially a linear combination of the atomic orbitals (LCAO molecular
-orbitals). Because the two protons are identical, we must have $c_1 = c_2 \equiv c$ (also $c > 0$).
-The $\pm$ notation in Eq. (\ref{eq11.7}) indicates that two different wavefunctions can be 
-constructed, one with ``+'' sign and the other with ``$-$'' sign. Normalization of the wavefunction requires:
+orbitals). 
+- Because the two protons are identical, we must have $c_1 = c_2 \equiv c$ (also $c > 0$).
+
+- The $\pm$ notation indicates that two different wavefunctions can be 
+constructed, one with + sign and the other with $-$ sign. Normalization of the wavefunction requires:
 
 $${\int{\psi_\pm^*\psi_\pm d\tau} = 1}$$
 
 
+### Overlap integral
 
 In the following, we consider the wavefunction with a ``+'' sign and evaluate the normalization integral ($S$ = overlap integral, which depends on $R$):
 
@@ -32,17 +44,19 @@ and the complete ``+'' wavefunction is then:
 
 $${\psi_+ \equiv \psi_g = \frac{1}{\sqrt{2(1 + S)}}(1s_A + 1s_B)}$$
 
-In exactly the same way, we can get the ``$-$'' wavefunction:
+In exactly the same way, we can get the $-$ wavefunction:
 
 $${\psi_- \equiv \psi_u = \frac{1}{\sqrt{2(1 - S)}}(1s_A - 1s_B)}$$
 
 
 
 Note that the antibonding orbital has \underline{zero} electron density between the nuclei.
 
-Recall that the square of the wavefunction gives the electron density. In the left hand side figure (the ``+'' wavefunction), the electron density is amplified between the nuclei whereas in the ``$-$'' wavefunction the opposite happens. \textit{The main feature of a chemical bond is the increased electron  density between the nuclei.} This identifies the ``+'' wavefunction as
-a \underline{bonding orbital} and ``$-$'' as an \underline{antibonding orbital}.
+### Bonding vs antibonding orbitals
 
+- Recall that the square of the wavefunction gives the electron density. In the left hand side figure (the + wavefunction), the electron density is amplified between the nuclei whereas in the $-$ wavefunction the opposite happens. 
+
+- The main feature of a chemical bond is the increased electron  density between the nuclei. This identifies the + wavefunction as a  **bonding orbital** and $-$ as an **antibonding orbital**.
 
 - When a molecule has a center of symmetry (here at the half-way between the 
 nuclei), the wavefunction may or may not change sign when it is inverted
@@ -55,11 +69,10 @@ $g$ (even parity) and for $\psi(x, y, z) = -\psi(-x, -y, -z)$ we have $u$ label
 bonding orbital and the $u$ symmetry corresponds to the antibonding orbital.
 Later we will see that this is \underline{reversed} for $\pi$-orbitals!
 
-
 - The overlap integral $S(R)$ can be evaluated analytically (derivation not shown):
 
 $${S(R) = e^{-R}\left( 1 + R + \frac{R^3}{3}\right)}$$
 
-> Note that when $R = 0$ (i.e. the nuclei overlap), $S(0) = 1$ (just a check to see that the expression is reasonable).
+- Note that when $R = 0$ (i.e. the nuclei overlap), $S(0) = 1$ (just a check to see that the expression is reasonable).
 
 

ch08/note03.md

@@ -1,7 +1,7 @@
 ## Energy of the hydrogen molecule ion
 
 
-Using a linear combination of atomic orbitals, it is possible to calculate the best values, in terms of energy, for the coefficients $c_1$ and $c_2$. Remember that this linear combination can only provide an approximate solution to the $H_2^+$ Schr\"odinger equation. The variational principle (see Eq. (\ref{eq10.58})) provides a systematic way to calculate the energy when $R$ (the distance between the nuclei) is fixed:
+- Using a linear combination of atomic orbitals, it is possible to calculate the best values, in terms of energy, for the coefficients $c_1$ and $c_2$. Remember that this linear combination can only provide an approximate solution to the $H_2^+$ Schr\"odinger equation. The variational principle provides a systematic way to calculate the energy when $R$ (the distance between the nuclei) is fixed:
 
 $${E = \frac{\int\psi_g^*\hat{H}_e\psi_gd\tau}{\int\psi_g^*\psi_gd\tau} = 
 \frac{\int (c_11s_A + c_21s_B)\hat{H}_e(c_11s_A + c_21s_B)d\tau}
@@ -11,104 +11,96 @@ $${E = \frac{\int\psi_g^*\hat{H}_e\psi_gd\tau}{\int\psi_g^*\psi_gd\tau} =
 
 
 - where $H_{AA}$, $H_{AB}$, $H_{BB}$, $S_{AA}$, $S_{AB}$ and $S_{BB}$ have been 
-used to denote the integrals occurring in Eq. (\ref{eq11.14}). The integrals $H_{AA}$
-and $H_{BB}$ are called the \textit{Coulomb integrals} (sometimes generally termed as matrix elements). 
+used to denote the integrals occurring in variational treatment of the problem. The integrals $H_{AA}$ and $H_{BB}$ are called the Coulomb integrals (sometimes generally termed as matrix elements). 
 
-- This interaction is attractive and therefore its numerical value must be negative. Note that by symmetry 
-$H_{AA} = H_{BB}$. The integral $H_{AB}$ is called the \textit{resonance integral}
-and also by symmetry $H_{AB} = H_{BA}$.
+- This interaction is attractive and therefore its numerical value must be negative. Note that by symmetry $H_{AA} = H_{BB}$. The integral $H_{AB}$ is called the resonance integral and also by symmetry $H_{AB} = H_{BA}$.
 
+### Variational solution 
 
-To minimize the energy expectation value in Eq. (\ref{eq11.14}) with respect to $c_1$ and $c_2$, we have to calculate the partial derivatives of energy with respect to these parameters:
+- To minimize the energy expectation value in Eq. (\ref{eq11.14}) with respect to $c_1$ and $c_2$, we have to calculate the partial derivatives of energy with respect to these parameters:
 
 
 $${E\times (c_1^2 + 2c_1c_2S + c_2^2) = c_1^2H_{AA} + 2c_1c_2H_{AB} + c_2^2H_{BB}}$$
 
-Both sides can be differentiated with respect to $c_1$ to give:
+- Both sides can be differentiated with respect to $c_1$ to give:
 
 $${E\times (2c_1 + 2c_2S) + \frac{\partial E}{\partial c_1}\times (c_1^2 + 2c_1c_2S + c_2^2) = 2c_1H_{AA} + 2c_2H_{AB}}$$
 
-In similar way, differentiation with respect to $c_2$ gives:
+- In similar way, differentiation with respect to $c_2$ gives:
 
 $${E\times (2c_2 + 2c_1S) + \frac{\partial E}{\partial c_2}\times (c_1^2 + 2c_1c_2S + c_2^2) = 2c_2H_{BB} + 2c_1H_{AB}}$$
 
-At the minimum energy (for $c_1$ and $c_2$), the partial derivatives must be zero:
+- At the minimum energy (for $c_1$ and $c_2$), the partial derivatives must be zero:
 
 $${c_1(H_{AA} - E) + c_2(H_{AB} - SE) = 0}$$
 $${c_2(H_{BB} - E) + c_1(H_{AB} - SE) = 0}$$
 
-In matrix notation this is (a generalized matrix eigenvalue problem):
+- In matrix notation this is (a generalized matrix eigenvalue problem):
 
 $${\begin{pmatrix}H_{AA} - E & H_{AB} - SE\\
 H_{AB} - SE & H_{BB} - E\\
 \end{pmatrix}\begin{pmatrix} c_1\\ c_2\\\end{pmatrix} = 0}$$
 
-From linear algebra, we know that a non-trivial solution exists only if:
+- From linear algebra, we know that a non-trivial solution exists only if:
 
 $${\begin{vmatrix}H_{AA} - E & H_{AB} - SE\\
 H_{AB} - SE & H_{BB} - E\\
 \end{vmatrix} = 0}$$
 
 
+### Energies of bonding and antibonding
 
-It can be shown that $H_{AA} = H_{BB} = E_{1s} + J(R)$, where $E_{1s}$ is the energy
-of a single hydrogen atom and $J(R)$ is a function of internuclear distance $R$:
+- It can be shown that $H_{AA} = H_{BB} = E_{1s} + J(R)$, where $E_{1s}$ is the energy of a single hydrogen atom and $J(R)$ is a function of internuclear distance $R$:
 
 $${J(R) = e^{-2R}\left( 1 + \frac{1}{R}\right)}$$
 
-Furthermore, $H_{AB} = H_{BA} = E_{1s}S(R) + K(R)$, where $K(R)$ is also a function of $R$:
+- Furthermore, $H_{AB} = H_{BA} = E_{1s}S(R) + K(R)$, where $K(R)$ is also a function of $R$:
 
 $${K(R) = \frac{S(R)}{R} - e^{-R}\left( 1 + R\right)}$$
 
-If these expressions are substituted into the previous secular determinant, we get:
+- If these expressions are substituted into the previous secular determinant, we get:
 
 $${\begin{vmatrix}E_{1s} + J - E & E_{1s}S + K - SE\\
 E_{1s}S + K - SE & E_{1s} + J - E\\
 \end{vmatrix} = (E_{1s} + J - E)^2 - (E_{1s}S + K - SE)^2 = 0}$$
 
-This equation has two roots:
+- This equation has two roots:
 
 $${E_g(R) = E_{1s} + \frac{J(R) + K(R)}{1 + S(R)}}$$
 
 $${E_u(R) = E_{1s} + \frac{J(R) - K(R)}{1 - S(R)}}$$
 
 
-Since energy is a relative quantity, it can be expressed relative to separated
+- Since energy is a relative quantity, it can be expressed relative to separated
 nuclei:
 
 $${\Delta E_g(R) = E_g(R) - E_{1s} = \frac{J(R) + K(R)}{1 + S(R)}}$$
 
 $${\Delta E_u(R) = E_u(R) - E_{1s} = \frac{J(R) - K(R)}{1 - S(R)}}$$
 
-The energies of these states are plotted in the figure below.\vspace{0.45cm}
 
 
-These values can be compared with experimental results. The calculated 
-ground state equilibrium bond length is 132 pm whereas the experimental value is 106 pm. The
-binding energy is 170 kJ mol$^{-1}$ whereas the experimental value is 258 kJ 
-mol$^{-1}$. The excited state (labeled with $u$) leads to repulsive behavior at
-all bond lengths $R$ (i.e. antibonding). Because the $u$ state lies higher
-in energy than the $g$ state, the $u$ state is an excited state of H$_2^+$.
-This calculation can be made more accurate by adding more than two terms to
+
+### Comparison of MO energies with experiments 
+
+- These values can be compared with experimental results. The calculated 
+ground state equilibrium bond length is 132 pm whereas the experimental value is 106 pm. The binding energy is 170 kJ mol$^{-1}$ whereas the experimental value is 258 kJ  mol$^{-1}$. 
+- The excited state (labeled with $u$) leads to repulsive behavior at all bond lengths $R$ (i.e. antibonding). Because the $u$ state lies higher in energy than the $g$ state, the $u$ state is an excited state of H$_2^+$.
+- This calculation can be made more accurate by adding more than two terms to
 the linear combination. This procedure would also yield more excited state 
 solutions. These would correspond $u$/$g$ combinations of $2s$, $2p_x$,
 $2p_y$, $2p_z$ etc. orbitals.
 
+### MO diagrams
 
-It is a common practice to represent the molecular orbitals by molecular 
-orbital (MO) diagrams:
-
-
-The formation of bonding and antibonding orbitals can be visualized as follows:
-
+- It is a common practice to represent the molecular orbitals by molecular 
+orbital (MO) diagrams. The formation of bonding and antibonding orbitals can be visualized as follows:
 
 - $\sigma$ orbitals When two $s$ or $p_z$ orbitals interact,
 a $\sigma$ molecular orbital is formed. The notation $\sigma$ specifies
 the amount of angular momentum about the molecular axis (for $\sigma$, $\lambda = 0$ with $L_z = \pm\lambda\hbar$). In many-electron systems, both bonding and antibonding $\sigma$  orbitals can each hold a maximum of two electrons. Antibonding orbitals are often  denoted by *.
 
-
 - $\pi$ orbitals. When two $p_{x,y}$ orbitals interact, a $\pi$ molecular orbital forms. $\pi$-orbitals are doubly degenerate: $\pi_{+1}$ and $\pi_{-1}$ (or alternatively $\pi_x$ and $\pi_y$), where the $+1/-1$ refer to the eigenvalue of the $L_z$ operator ($\lambda = \pm1$). In many-electron systems a bonding $\pi$-orbital can therefore hold a maximum of 4 electrons (i.e. both $\pi_{+1}$ and $\pi_{-1}$ each can hold two electrons). The same holds for the antibonding $\pi$ orbitals. Note that only the atomic orbitals of the same symmetry mix to form molecular orbitals (for example, $p_z - p_z$, $p_x -  p_x$ and $p_y - p_y$). When atomic $d$ orbitals mix to form molecular orbitals, $\sigma (\lambda = 0)$, $\pi (\lambda = \pm 1)$ and $\delta (\lambda = \pm 2)$ MOs form. 
 
-
 - Excited state energies of $H_2^+$ resulting from a calculation employing an extended basis set (e.g. more terms in the LCAO) are shown on the left below. The MO energy diagram, which includes the higher energy molecular orbitals, is shown on the right hand side. Note that the energy order of the MOs depends on the molecule.
 

ch08/note04.md

@@ -1,4 +1,5 @@
-##Molecular orbital description of hydrogen molecule
+## Molecular orbital description of hydrogen molecule
+
 
 
 Using the Born-Oppenheimer approximation, the electronic Hamiltonian for H$_2$ molecule can be written as:
@@ -118,9 +119,13 @@ atoms):
 
 $${^{2S+1}\Lambda}$$
 
+### Examples of molecular term symbols 
+
 **Example** What is the term symbol for ground state $H_2$?
 
-**Solution** Both electrons are on a $\sigma$ orbital and hence $\lambda_1 = \lambda_2
+:::{dropdwon} **Solution** 
+
+Both electrons are on a $\sigma$ orbital and hence $\lambda_1 = \lambda_2
 = 0$. This gives $\Lambda = 0$, which corresponds to $\Sigma$. The electrons 
 occupy the same molecular orbital with opposite spins and hence $2S + 1 = 1$.
 This gives the term symbol as $^1\Sigma$.
@@ -136,11 +141,14 @@ label reflects the symmetry of the unpaired electron orbital. Closed shell
 molecules have always $g$. With more than one unpaired electron, the overall
 parity should be calculated using the following rules: $g \times g = g$, $g \times 
 u = u$, $u \times g = u$ and $u \times u = g$.
+:::
 
 **Example** What is the term symbol for ground state $O_2$?
 
-**Solution** Ground state $O_2$ has two electrons with parallel spins on the $\pi_{+1}$ and $\pi_{-1}$ orbitals. Thus this is a triplet state molecule with the orbital angular momentum from the two $\pi$-electrons being cancelled. This gives a $^3\Sigma$ term.  The two $\pi$'s are anti-bonding and as such they are desginated as $g$ and further $g\times g = g$ (remember that for $\pi$ orbigals the $g/u$ vs. bonding/anti-bonding is reversed from that of $\sigma$ orbitals). To see the $+/-$ symmetry, it is convenient to think about $\pi_x$ and $\pi_y$ Cartesian orbitals (draw a picture!) and see that one of them is $+$ and the other is $-$ (they are perpendicular to each other). Again $+ \times - = -$ and we have the complete term symbol as $^3\Sigma_g^-$.
+::{dropdwon} **Solution** 
 
+:Ground state $O_2$ has two electrons with parallel spins on the $\pi_{+1}$ and $\pi_{-1}$ orbitals. Thus this is a triplet state molecule with the orbital angular momentum from the two $\pi$-electrons being cancelled. This gives a $^3\Sigma$ term.  The two $\pi$'s are anti-bonding and as such they are desginated as $g$ and further $g\times g = g$ (remember that for $\pi$ orbigals the $g/u$ vs. bonding/anti-bonding is reversed from that of $\sigma$ orbitals). To see the $+/-$ symmetry, it is convenient to think about $\pi_x$ and $\pi_y$ Cartesian orbitals (draw a picture!) and see that one of them is $+$ and the other is $-$ (they are perpendicular to each other). Again $+ \times - = -$ and we have the complete term symbol as $^3\Sigma_g^-$.
 
 - When spin-orbit interaction is small, the above term symbols are  adequate (Hund's case (a)).
 - When spin-orbit interaction is large, $S$ and $\Lambda$ can no longer be  specified but their sum $J = |S + \Lambda|$ is a good quantum number.
+:::