Lecture 15

amo/lecture15.md

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+---
+author:
+  - Fred Jendrzejewski
+  - Selim Jochim
+  - Matthias Weidemüller
+order: 15
+title: Lecture 15 - Diatomic molecules
+---
+
+In this lecture we will start to put atoms together to build simple
+molecules. We will first use the Born-Oppenheimer approximation, to
+eliminate slow processes from the study of the fast electron dynamics.
+Then, we will study simple mechanisms of binding atoms.
+
+# Introduction
+
+Molecules add a new layer of complexity to the system. In atoms, we had
+different combinations of nuclei and electrons, leading to different
+kinds of atoms. In this lecture, we will use atoms as basic building
+block of more complex structures, the molecules. While this complexity
+makes it necessary to introduce new approximations, it also allows us to
+study new processes in nature.
+
+So we will start out with the simplest of all molecules, barely a
+molecule, the $H_2^+$ ion. We start out with a discussion of the
+Born-Oppenheimer approximation. Detailled discussions can be found in
+[Chapter 8 of Atkins](https://books.google.de/books?id=9KEPAQAAMAAJ), [Chapter 9 of Demtröder](http://dx.doi.org/10.1007/978-3-642-10298-1) and
+[Chapter 10 of Bransden](https://books.google.de/books?id=i5IPWXDQlcIC).
+
+# Molecular hydrogen ion
+
+In molecular hydrogen we have only three ingredients. A single electron,
+which is bound to two nuclei as shown in Fig.
+[1](#fig-hydrogen-ion).
+
+<figure id="fig-hydrogen-ion">
+<img src="./lecture15_pic1.png" width="95%" />
+<figcaption>The molecular hydrogen ion as discussed in the main text.</figcaption>
+</figure>
+
+The full Hamiltonian of the system at study would read:
+
+$$\hat{H} = - \frac{1}{2}\nabla_\mathbf{r}^2 - \frac{1}{2M}\left(\nabla_{\mathbf{R}_\mathrm{A}}^2 +\nabla_{\mathbf{R}_\mathrm{B}}^2\right) + V(\mathbf{r}, \mathbf{R}_\mathrm{A}, \mathbf{R}_\mathrm{B})$$
+We will further introduce the short-hand notations:
+
+$$
+\hat{T}_e = - \frac{1}{2}\nabla_\mathbf{r}^2 \\
+\hat{T}_n = - \frac{1}{2M}\left(\nabla_{\mathbf{R}_\mathrm{A}}^2 +\nabla_{\mathbf{R}_\mathrm{B}}^2\right)
+$$
+
+In a stark difference to atoms, we now have two charged nuclei. The
+relative distance between them and between the electron will be of major
+importance. Most importantly, we should answer the question, why this
+configuration should be stable at all given that the two protons repel
+each other. To handle the problem, we will once again separate out
+energy scales.
+
+## The Born-Oppenheimer approximation
+
+The idea of the **Born-Oppenheimer approximation** is to separate the
+fast electronic motion from the slow motion of the heavy nuclueus
+($M=1836$). So we will
+
+1.  Solve the electronic motion with the nuclear coordinates fixed.
+
+2.  Solve the nuclear motion, assuming that the electron wavefunction
+    adapts instantaneously.
+
+So the ansatz is:
+$$\Psi(\mathbf{R}_\mathrm{A}, \mathbf{R}_\mathrm{B}, \mathbf{r}) = \psi_e(\mathbf{R}_\mathrm{A}, \mathbf{R}_\mathrm{B}, \mathbf{r})\cdot \psi_n(\mathbf{R}_\mathrm{A}, \mathbf{R}_\mathrm{B})$$
+
+We will plug this into the Schrödinger equation to obtain:
+$$\psi_n\hat{T}_e\psi_e +\psi_e\hat{T}_n\psi_n + V(\mathbf{r}, \mathbf{R}_\mathrm{A}, \mathbf{R}_\mathrm{B})\psi_e \psi_n + W = E \psi_e\psi_n$$
+
+This transformation introduced the _non-adiabatic_ effects:
+$$W = -\frac{1}{2M}\sum_{i=A,B} \left[(\nabla_{\mathbf{R}_i}\psi_e)\cdot(\nabla_{\mathbf{R}_i}\psi_n)+\psi_n \nabla_{\mathbf{R}_i}^2 \psi_e\right]$$
+
+In the following we will neglect these effects. And obtain:
+$$\psi_e \hat{T}_n\psi_n + \left(\hat{T}_e\psi_e+V\psi_e\right)\psi_n = E \psi_e\psi_n$$
+So we will first solve the _electronic motion_:
+$$\left(\hat{T}_e+\hat{V}\right)\psi_e = E_e(\mathbf{R}_\mathrm{A}, \mathbf{R}_\mathrm{B}) \psi_e$$
+To be explicit we obtain for the ionic hydrogen:
+
+$$
+
+H_e = -\frac{1}{2} \nabla_\mathbf{r}^2-\frac{1}{r_A}-\frac{1}{r_B}+\frac{1}{R}
+$$
+
+At this stage we can just focus on the electronic part to understand the
+structure of simple diatomic molecules, while assuming that $R$ is an
+independent parameter. Most importantly, we will focus at usual on
+symmetries, which will tell us more about the allowed states in the
+system.
+
+In the second step we will solve the nuclear motion:
+$$\hat{T}_n\psi_n + E_e \psi_n = E \psi_n$$
+
+This nuclear motion will be at the origin of rotational and vibrational
+levels, which will be discussed in on of the next lectures.
+
+# Symmetries of the electronic wavefunction
+
+This discussion follows along similiar lines as for the hydrogen atom
+and the helium atom. We basically can categorize the different states by
+their properties. This will help us later enormously to understand
+allowed transition etc.
+
+## Angular momentum
+
+For any (diatomic) molecule we break the spherical symmetry that we
+relied on for the atomic systems. This means that angular momentum is
+not a conserved quantity anymore.
+
+However, the full Hamiltonian is invariant under the rotation around the
+axis of the diatomic molecule. One can verify that this implies that:
+
+$$
+[H_e, L_z] = 0\\
+\Rightarrow L_z \psi_e = \pm \Lambda \psi_e (a.u.)
+$$
+
+The reason is that
+$\hat{L}_z =\frac{1}{i}\partial_\varphi$ depends solely on the angle
+$\varphi$ and not on $R$. Here the quantum number can have the integer
+values $\Lambda= 0, 1, 2 , \cdots$. We also note them
+$\Sigma, \Pi, \Delta, \Phi$ or $\sigma, \pi, \delta, \phi$ for single
+electrons.
+
+## Parity
+
+We further have symmetry under parity operation for _homo-nuclear,
+diatomic_ molecules $A_2$. This means that we have once more:
+$$\hat{P}\psi_e(\mathbf{r}) = \pm\psi_e(\mathbf{r})$$ In the same way as
+in the lecture on the Helium atom we distinguish the states by _gerade_
+and _ungerade_. So we then end up with something like
+$\Lambda_{u,g}^\pm$.
+
+## Spin
+
+If the system does not have explicit spin-orbit coupling, the total spin
+$S$ of the system will be conserved. So the full notation for electronic
+states is typically: $$^{2S+1}\Lambda^{\pm}_{g,u}$$ Most of the time the
+ground state of the system is $^{1}\Sigma^{+}_{g}$.
+
+# Stability of the ground state molecule
+
+We have now studied the symmetries that the system should have, but
+until now we did not discuss the most important question: Is this
+molecule stable ? Within the Born-Oppenheimer approximation, we can
+actually solve the ionic hydrogen molecule analytically (see [Chapter 9 of
+Demtröder](http://dx.doi.org/10.1007/978-3-642-10298-1)). The resulting **molecular potential curves** are
+shown in Fig. [2](#fig-potentials).
+
+<figure id="fig-potentials">
+<img src="./lecture15_pic2.png" width="60%" />
+<figcaption>Molecular potential curves for the molecular hydrogen ion. Figure is
+taken from <a href="https://global.oup.com/academic/product/molecular-quantum-mechanics-9780199541423">here</a>.</figcaption>
+</figure>
+
+## Linear combination of atomic orbitals
+
+The analytical solutions are rather bulky and not particularly
+instructive. One powerful idea, and very good approximation, is to
+decompose the molecule wavefunction over the atomic orbitals of its
+components. Going back to Fig. [1](#fig-hydrogen-ion) we could make the simple Ansatz:
+$$\psi_e(\mathbf{r})= c_1 \psi_{1s}(\mathbf{r}_\mathrm{A})+c_2 \psi_{1s}(\mathbf{r}_\mathrm{B})$$
+
+Note, that we made a very simple Ansatz at this stage and we could
+decompose the system over a much larger set of excited states. But for
+pedagogical reason we will stick to the simple model at this stage.
+Going through the symmetry requirements, we find that we can write the
+full wavefunction as:
+
+$$
+\psi_{g,u}(\mathbf{r})= \frac{1}{\sqrt{2\pm2S}}\left(\psi_{1s}(\mathbf{r}_\mathrm{A})\pm \psi_{1s}(\mathbf{r}_\mathrm{B})\right)
+$$
+
+The contribution $S$ describes the overlap of the two atomic orbitals
+$$S =  \int d\mathbf{r}\psi_{1s}^*(\mathbf{r}_\mathrm{A})\psi_{1s}(\mathbf{r}_\mathrm{B})$$
+We can then evaluate the energy of the two states through the
+variational principle:
+
+$$
+E_{g,u} = \left\langle\psi_{g,u}\right|\hat{H}_e\left|\psi_{g,u}\right\rangle\\
+ = \frac{1}{2\pm2S}\left(\left\langle\psi_A\right|\pm \left\langle\psi_B\right|\right)\hat{H}_e\left(\left|\psi_A\right\rangle\pm \left|\psi_B\right\rangle\right)\\
+ = \frac{E_{AA}\pm E_{AB}}{1\pm S}\\
+$$
+
+The resulting energy surfaces are shown in Fig. [3](#fig-lcao). In the most
+simplistic interpretation the gerade state does not have a node in the
+middle and it is therefore of smaller kinetic energy.
+
+<figure id="fig-lcao">
+<img src="./lecture15_pic3.svg" width="70%" />
+<figcaption>The energy surface of the LCAO for the hydrogen molecule ion.</figcaption>
+</figure>
+
+# The neutral hydrogen molecule
+
+In the previous section we have seen how we can treat the coupling of
+the nuclei through the exchange of a single shared electron. However, we
+should now move on to the case of two neutral particles binding
+together. What is here the relevant mechanism ?
+
+<figure id="fig-hydrogen">
+<img src="./lecture15_pic4.svg" width="70%" />
+<figcaption>The hydrogen molecule.</figcaption>
+</figure>
+
+In the following we will only consider the electronic part, which adds
+up too:
+
+$$\hat{H} = -\frac{1}{2}\left(\nabla_{\mathbf{r}_1}^2+\nabla_{\mathbf{r}_2}^2\right)-\frac{1}{r_{A1}}-\frac{1}{r_{A2}}-\frac{1}{r_{B1}}-\frac{1}{r_{B2}}+\frac{1}{r_{12}}+\frac{1}{R}$$
+
+We can now rewrite this Hamiltonian in the more instructive form
+
+$$
+
+\hat{H} = H_{0,1}+H_{0,2}+\frac{1}{r_{12}}+\frac{1}{R}\\
+H_{0,i} = -\frac{1}{2}\nabla_{\mathbf{r}_i}^2-\frac{1}{r_{A,i}}-\frac{1}{r_{B,i}}
+
+
+$$
+
+We have can now use the results of the hydrogen ion to understand this
+system.
+
+- We have for each electron the solution of hydrogen molecule ion.
+- In the next step, we have to put the two electrons properly within
+  this orbit with $S=0$ and _ignoring_ the $e^-$ - $e^-$ interaction.
+
+So we can make the Ansatz:
+
+$$
+\psi(\mathbf{r}_1, \mathbf{r}_2) = \psi_{g}(\mathbf{r}_1)\cdot\psi_{g}(\mathbf{r}_2)\\
+= \frac{1}{2 + 2S}\left(\psi_{1s}(\mathbf{r}_{A1})+\psi_{1s}(\mathbf{r}_{B1})\right)\left(\psi_{1s}(\mathbf{r}_{A2})+\psi_{1s}(\mathbf{r}_{B2})\right)\\
+= \frac{1}{2 + 2S}\left(\psi_{1s}(\mathbf{r}_{A1})\psi_{1s}(\mathbf{r}_{B2})+\psi_{1s}(\mathbf{r}_{B1})\psi_{1s}(\mathbf{r}_{A2}) +\psi_{1s}(\mathbf{r}_{A1})\psi_{1s}(\mathbf{r}_{A2})+\psi_{1s}(\mathbf{r}_{B1})\psi_{1s}(\mathbf{r}_{B2}) \right)
+$$
+
+The first two terms describe **kovalent binding**. They
+describe situations where each electron is associated with one core. The
+last two terms describe **ionic binding** as one associated both
+electrons with a single atom and then looks one the attraction of
+another ionic core. This is quite similiar to the interaction in the
+$H_2^+$ molecule.
+
+Within this approach, one actually finds a binding energy of
+$E_b = {-2.64}\text{eV}$ at an equilibrium distance of $R_e = 1.4 a_0$.
+The experimentally measured values differs quite substantially as we
+have $E_b = {-4.7}\text{eV}$. A substantial approximation was here that we
+neglected the interaction between the electrons, which should repel.
+
+## The Heitler-London method
+
+As the two electrons should repel each other, we can assume that the
+ionic binding is strongly suppressed. So the wavefunction is now assumed
+to be:
+
+$$
+
+\psi_{HL} = \frac{1}{\sqrt{2 + 2S^2}}\left(\psi_{1s}(\mathbf{r}_{A1})\psi_{1s}(\mathbf{r}_{B2})+\psi_{1s}(\mathbf{r}_{B1})\psi_{1s}(\mathbf{r}_{A2})\right)
+$$
+
+Again, the wavefunction cannot be factorized and the two
+electrons are entangled because of the interactions. Recognize the
+common theme with the Helium atom. Calculation of the binding energy
+within this approximation leads to $E_b = {-3.14}\text{eV}$ and
+$R_e=1.6a_0$.
+
+In the next lecture we will discuss how we can move on
+from these extremely simple diatomic molecules to the assembly of richer
+systems.