1.1 Probability within the Bohr radius
1.5 The many-electron Hamiltonian
1.6 Memory needed to store the many-electron wavefunction
1.7 Atomic Orbital Hamiltonian
1.9 Energies evaluated with the Reduced Hamiltonian
1.10 Consequences of electrons being fermions
1.12 First excited state of lithium
1.13 Electron configuration of nickel
1.14 An atom with $Z$ protons and $Z$ electrons
1.17 Expanding a wavefunction in a complete set of states
What is the probability of finding an electron in the 1s (2s) atomic orbital in the region $|r| < a_0/Z$? Here $a_0$ is the Bohr radius.
1s atomic orbital 2s atomic orbital
1s: 0.323323
2s: 0.034316
The Schrödinger equation for a hydrogen atom in spherical coordinates is,
\begin{equation} H\psi= \frac{-\hbar^2}{2m}\left[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \psi}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \psi}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right]-\frac{e^2}{4\pi \epsilon_0 r} \psi= E\psi. \end{equation}The ground state wave function has the form,
\begin{equation} \psi= \exp\left(\frac{-r}{a_0}\right). \end{equation}By substituting this wave function into the Schrödinger equation, determine the energy of this state and the value of the Bohr radius $a_0$.
Often in molecular or solid state physics we know the Hamiltonian but we can't solve the Schrödinger equation associated with this Hamiltonian. In these cases we often guess a solution and then calculate the corresponding energy.
Consider the Hamiltonian for a hydrogen atom. In spherical coordinates it is,
\begin{equation} H\psi= \frac{-\hbar^2}{2m}\left[\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial \psi}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial \psi}{\partial\theta}\right)+\frac{1}{r^2\sin^2\theta}\frac{\partial^2\psi}{\partial\varphi^2}\right]-\frac{e^2}{4\pi \epsilon_0 r} \psi= E\psi. \end{equation}Find the expectation value of the energy for the following wavefunction,
\begin{equation} \psi= \exp\left(\frac{-r^2}{a^2}\right), \end{equation}where $a$ is a parameter. Note that this wavefunction is not an eigenfunction of the Hamiltonian. Determine the value of $a$ that minimizes the energy. Compare $a$ to the Bohr radius a0 = 5.3 × 10-11 m.
(a) Draw the $\phi_{3d_{x^2-y^2}}$ atomic orbital looking down the $z-$axis.
(b) What do the different colors mean in the drawing of the atomic orbital?
(c) For which elements would you expect the $\phi_{3d_{x^2-y^2}}$ atomic orbital to be a valence orbital?
(d) What is the energy of the $\phi_{3d_{x^2-y^2}}$ orbital if the effective nuclear charge is $Z_{eff} =3$?
(d) $E=-13.6$ eV.
Write down the many-electron Hamiltonian for
The many-electron wave function for $N$ electrons is defined in $3N$ dimensions. It is a complex number for every value of the coordinates $x_1,y_1,z_1,x_2,y_2,z_2,\cdots,x_N,y_N,z_N$. Suppose the wavefunction is peaked around the origin and you decide to save the value of the wavefunction at a discrete set of points centered around the origin. Divide $x_1$ into 100 intervals, $y_1$ into 100 intervals, etc. For one electron you need store $100\times100\times100=10^6$ complex numbers to store the state of the wavefunction. How many electrons must the many-electron wavefunction contain before you could not store the state of the many-electron wavefunction on a computer with 100 TB of memory?
What is the atomic orbital Hamiltonian $H_{\text{ao}}$ for
Show that the product of atomic orbitals is an eigenstate of the corresponding reduced Hamiltonian and find the eigen energies.
(a) What is the energy of the following electron configurations evaluated in the reduced Hamiltonian? Use Slater's rules.
(b) The antisymmetrized solutions to the reduced Hamiltonian can be written as a Slater determinant. What is the Slater determinant for
(c) How can you determine which of the following electron configurations for a nickel atom has the lower energy?
1s2 2s2 2p6 3s2 3p63d10 or 1s2 2s2 2p6 3s2 3p63d94s1 or 1s2 2s2 2p6 3s2 3p63d84s2
(a) Without Slater's rules:
With Slater's rules:
Because electrons are fermions, there is an additional condition on a many-electron wave function besides being a solution of the Schrödinger equation. Explain this extra condition.
The antisymmetrized solutions to the reduced Hamiltonian can be written as a Slater determinant. What is the Slater determinant for
How many terms are there when each of these determinants are written out?
The solutions to the many-electron Hamiltonian are difficult to determine. There are techniques that can be used to numerically determine the solutions such as the Hartree-Fock method. A simpler approach to find the approximate energies of a multi-electron wave function is to take the antisymmetrized product of atomic orbitals that the reduced Hamiltonian and evaluate them in the many-electron Hamiltonian that includes the electron-electron interactions.
\begin{equation} E= \frac{\langle \Psi |H_{\text{me}}|\Psi\rangle}{\langle \Psi |\Psi\rangle}. \end{equation}The energies of these solutions evaluated in the reduced Hamiltonian do not depend on spin but the energies evaluated with the many-electron Hamiltonian do depend on spin. The difference in energy between two wave functions that only differ in their spin component is called the exchange energy.
Write down the integral that must be performed to evaluate the expectation value of the energy of the first excited state of lithium assuming that the wave function can be written as an antisymmetrized product of atomic orbitals and the energy is evaluated using the many-electron Hamiltonian.
How can you determine which of the following electron configurations for a nickel atom has the lower energy?
1s2 2s2 2p6 3s2 3p63d10 or 1s2 2s2 2p6 3s2 3p63d94s1
The starting point for the quantum description of an atom with $Z$ protons and $Z$ electrons is the many-electron Hamiltonian,
\begin{equation} H_{\text{me}}= - \sum \limits_{i=1}^Z \frac{\hbar^2}{2m_e}\nabla_i^2 - \sum \limits_{i=1}^Z \frac{Ze^2}{4\pi\epsilon_0 | \vec{r}_i |} + \sum \limits_{i\lt j}\frac{e^2}{4\pi\epsilon_0 |\vec{r}_i - \vec{r}_j | }. \end{equation}An approximate solution of the many-electron Hamiltonian for an atom can be found by neglecting the electron-electron interactions. The resulting Hamiltonian is called the reduced Hamiltonian. The reduced Hamiltonian is the sum of $Z$ identical atomic orbital Hamiltonians.
\begin{equation} H_{\text{red}}= \sum \limits_{i=1}^Z H_{\text{ao}}. \end{equation}(a) What is the atomic orbital Hamiltonian $H_{\text{ao}}$ for silicon $Z =$ 14?
(b) The antisymmetrized solutions to the reduced Hamiltonian can be written as a Slater determinant. What is the Slater determinant for He $1s^2$?
Lithium has three protons and three electrons.
(a) Write down the many-electron Hamiltonian for lithium.
(b) Write down the reduced Hamiltonian for lithium.
(c) Show that the Schrödinger equation using the reduced Hamiltonian can be solved by the separation of variables.
(d) Calculate the energies of the ground state and the first excited state of lithium using the reduced Hamiltonian.
(e) Construct a Slater determinant for the ground state.
Helium has two protons and two electrons.
(a) Write down the total Hamiltonian for helium.
(b) Write down the reduced Hamiltonian for helium.
(c) Calculate the energies of the ground state and the first excited state of helium using the reduced Hamiltonian neglecting Slater's rules.
(e) Construct a Slater determinant for the first excited state.
In quantum mechanics, when we have a potential where we don't know the eigenstates, we often seek an approximate form of the solution in terms of some complete set of states. This will be done several times in this course. One class of problems that can be solved like this are infinite well potentials where $V(x)=\infty$ for $x<-L/2$ and for $x>L/2$. The time-independent Schrödinger equation for this case is,
\[ \begin{equation} -\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi =E\psi . \end{equation} \]For these potentials we expand the wavefunction in terms of $N$ square-well eigenfunctions,
\begin{equation} \psi= c_1\phi_1+c_2\phi_2+c_3\phi_3+\cdots+c_N\phi_N, \end{equation}where $c_i$ are unknown coefficients and $\phi_i$ are the normalized square-well eigenfunctions,
\[ \begin{equation} \phi_n = \sqrt{\frac{2}{L}}\sin\left(n\pi\left(\frac{x}{L}+\frac{1}{2}\right)\right) \hspace{1cm}n=1,2,3,\cdots. \end{equation} \]The square-well eigenfunctions solve the Schrödinger equation for a potential $V(x)=0$. Substitute this wave function into the time-independent Schrödinger equation,
\begin{equation} H\psi= E\psi. \end{equation}Multiply the Schrödinger equation from the left by each of the square-well eigenfunctions and integrate from $-L/2$ to $L/2$. This results in a set of $N$ algebraic equations,
\[ \begin{equation} \begin{matrix} \langle \phi_1 | H |\psi\rangle =E\langle\phi_1|\psi \rangle \\ \langle \phi_2 | H |\psi\rangle =E\langle\phi_2|\psi \rangle\\ \vdots \\ \langle \phi_N | H |\psi\rangle =E\langle\phi_N|\psi \rangle \end{matrix} \end{equation} \].These equations can be written in matrix form,
\[ \begin{equation} \left[ \begin{matrix} H_{11} & H_{12} & \cdots & H_{1N} \\ H_{21} & H_{22} & \cdots & H_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ H_{N1} & H_{N2} & \cdots & H_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right] = E\left[ \begin{matrix} S_{11} & S_{12} & \cdots & S_{1N} \\ S_{21} & S_{22} & \cdots & S_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ S_{N1} & S_{N2} & \cdots & S_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right]. \end{equation} \]Here the elements of the Hamiltonian matrix and the overlap matrix are,
\[ \begin{equation} H_{mn}= \langle\phi_{m}|H|\phi_{n}\rangle \hspace{1cm}\text{and}\hspace{1cm} S_{mn}= \langle\phi_{m}|\phi_{n}\rangle. \end{equation} \]Since the square well eigenfunctions are orthonormal, the overlap matrix $\textbf{S}$ is the identity matrix. The matrix elements $H_{mn}$ are found by evaluating the following integral numerically.
\[ \begin{equation} H_{mn} = \int \limits_{-L/2}^{L/2}\frac{2}{L}\sin\left(m\pi\left(\frac{x}{L}+\frac{1}{2}\right)\right)\left(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right)\sin\left(n\pi\left(\frac{x}{L}+\frac{1}{2}\right)\right)dx . \end{equation} \]The equations that need to be solved reduce to an eigenvalue problem,
\[ \begin{equation} \left[ \begin{matrix} H_{11} & H_{12} & \cdots & H_{1N} \\ H_{21} & H_{22} & \cdots & H_{2N} \\ \vdots & \vdots & \ddots & \vdots \\ H_{N1} & H_{N2} & \cdots & H_{NN} \end{matrix} \right] \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right] = E \left[ \begin{matrix} c_1 \\ c_2 \\ \vdots \\ c_N \end{matrix} \right]. \end{equation} \]This $N\times N$ Hamiltonian matrix has $N$ eigenvalues and their corresponding $N$ eigenvectors. The eigenvalue with the lowest energy is the best approximate ground state energy that can be found for the potential using this set of $N$ square well eigenfunctions. The corresponding eigenvector contains the coefficients $c_i$ needed to construct the ground state wave function $\psi= c_1\phi_1+c_2\phi_2+\cdots+c_N\phi_N$. The other eigenvalues and their eigenvectors describe the excited states of the potential.
The form below will construct the Hamiltonian matrix for a square well potential. The units of the matrix elements are eV.
|
The Hamiltonian matrix is,
Consider an electron moving in potential where $V(x)=\infty$ for $x<-L/2$ and for $x>L/2$. In the range $-L/2 < x < L/2$, the potential is $V(x)=0.5\frac{x}{L} -10\left(\frac{x}{L}\right)^2 +50\left(\frac{x}{L}\right)^4$ eV.
(a) Calculate one element of the 3×3 Hamiltonian matrix.
(b) What are the first three eigenfunctions and their eigenenergies for this potential where $L=3$ nm?
(c) Recalculate the first three eigenenergies using a 4×4 Hamiltonian matrix.
You need to use a program such as Matlab, Mathematica, or Python. Python is free and it is probably worthwhile to install a copy. Python code to find the eigenvalues of a matrix is,
Open this python program in Colab. You will need to paste the correct Hamiltonian matrix into the code.
In Wolfram Alpha type,
Consider the transition metals in the periodic table. These are the elements in the d-block in columns 3-12. The electron configurations of these elements mostly follow the aufbau principle. What are the exceptions to the aufbau principle? Do you see a pattern in the exceptions?