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PHY.K02UF Molecular and Solid State Physics

## Review of atomic physics

Hydrogen
The simplest atom is hydrogen; it consists of one proton and one electron. The Hamiltonian for a hydrogen atom is,

$$$\label{eq:hydrogen} H_{\text{total}}^{H}= - \frac{\hbar^2}{2m}\nabla^2 - \frac{e^2}{4\pi\epsilon_0 r}.$$$

Here $m$ is the mass of an electron, $e$ is the elementary charge, $\epsilon_0$ is the permittivity constant, and $\hbar$ is the reduced Plank's constant. Hydrogen is the only atom for which the eigenstates can be found analytically. The first few eigenstates are,

$$$\begin{matrix} \psi_{1s} = \frac{1}{\sqrt{\pi a_0^3}}\exp\left(-\frac{r}{a_0}\right), \\ \psi_{2s} = \frac{1}{4\sqrt{2\pi a_0^3}}\left(2-\frac{r}{a_0}\right)\exp\left(-\frac{r}{2a_0}\right), \\ \psi_{2px} = \frac{1}{4\sqrt{2\pi a_0^3}}\frac{r}{a_0}\exp\left(-\frac{r}{2a_0}\right) \sin\theta \cos\varphi, \\ \psi_{2py} = \frac{1}{4\sqrt{2\pi a_0^3}}\frac{r}{a_0}\exp\left(-\frac{r}{2a_0}\right) \sin\theta \sin\varphi, \\ \psi_{2pz} = \frac{1}{4\sqrt{2\pi a_0^3}}\frac{r}{a_0}\exp\left(-\frac{r}{2a_0}\right) \cos\theta. \\ \end{matrix}$$$

Here $a_0$ is the Bohr radius. The energies of the eigenstates can be determined by calculating the expectation value,

$$$E= \frac{\langle \psi |H|\psi\rangle}{\langle \psi|\psi\rangle}=-\frac{me^4}{32\pi^2\epsilon_0^2\hbar^2n^2}=-\frac{13.6}{n^2}\text{ eV},$$$

where $n$ is the principle quantum number.

Helium
Helium has two electrons and a positively charged nucleus with a charge of $2e$. The Hamiltonian for helium is,

$$$\label{eq:helium} H_{\text{total}}^{\text{He}}= - \frac{\hbar^2}{2m}\nabla_1^2 - \frac{\hbar^2}{2m}\nabla_2^2 - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 |\vec{r}_1-\vec{r}_2| }.$$$

The first term is the kinetic energy of electron 1, the second term is the kinetic energy of electron 2, the third term is the attractive Coulomb interaction between electron 1 and the nucleus, the fourth term is the attractive Coulomb interaction between electron 2 and the nucleus, and the last term is the repulsive electron-electron interaction. The eigenstates of this Hamiltonian are two-electron wavefunctions that depend on the positions of both electrons $\Psi(\vec{r}_1,\vec{r}_2)$. It is not possible to find a simple analytic expression for this two-electron wavefunction.

To simplify this problem, the electron-electron interactions are neglected resulting in the reduced Hamiltonian for helium,

$$$\label{eq:red} H_{\text{red}}^{\text{He}}= - \frac{\hbar^2}{2m}\nabla_1^2 - \frac{\hbar^2}{2m}\nabla_2^2 - \frac{2e^2}{4\pi\epsilon_0 r_1} - \frac{2e^2}{4\pi\epsilon_0 r_2} .$$$

The Schrödinger equation for this reduced Hamiltonian is,

$$$- \frac{\hbar^2}{2m}\nabla_1^2\Psi - \frac{\hbar^2}{2m}\nabla_2^2\Psi - \frac{2e^2}{4\pi\epsilon_0 r_1}\Psi - \frac{2e^2}{4\pi\epsilon_0 r_2}\Psi=E\Psi .$$$

This can be solved by the separation of variables. Assume that $\Psi$ can be written as a product of two functions, $\Psi(\vec{r}_1,\vec{r}_2)=\phi_m(\vec{r}_1)\phi_n(\vec{r}_2)$. Substitute this form into the Schrödinger equation and divide by $\Psi$. The resulting terms can be rearranged so that all terms that depend on $\vec{r}_1$ can be put on the left side of the $=$ sign and all terms that depend on $\vec{r}_2$ can be put on the right side of the $=$ sign. Since a function of $\vec{r}_1$ cannot be equal to a function of $\vec{r}_2$ for all $\vec{r}_1$ and $\vec{r}_2$, both sides of the equation must be equal to a constant. Let's call the constant $E^{\prime}$. The separated equations are,

$$$\begin{matrix} - \frac{\hbar^2}{2m}\nabla_1^2\phi_m(\vec{r}_1) - \frac{2e^2}{4\pi\epsilon_0 |\vec{r}_1|}\phi_m(\vec{r}_1) = (E-E^{\prime})\phi_m(\vec{r}_1), \\ - \frac{\hbar^2}{2m}\nabla_2^2\phi_n(\vec{r}_2) - \frac{2e^2}{4\pi\epsilon_0 |\vec{r}_2|}\phi_n(\vec{r}_2) = E^{\prime}\phi_n(\vec{r}_2). \end{matrix}$$$

Both of these equations are nearly the same as the Schrödinger equation for hydrogen. A difference is that the Coulomb term is a factor of 2 greater than for hydrogen because of the $+2e$ charge of the helium nucleus. Nevertheless, slightly modified hydrogen wavefunctions called the atomic orbitals solve these equations.

Atomic orbitals
The first few atomic orbitals are,

$$$\begin{matrix} \phi_{1s}^Z = \sqrt{\frac{Z^3}{\pi a_0^3}}\exp\left(-\frac{Zr}{a_0}\right), \\ \phi_{2s}^Z = \frac{1}{4}\sqrt{\frac{Z^3}{2\pi a_0^3}}\left(2-\frac{Zr}{a_0}\right)\exp\left(-\frac{Zr}{2a_0}\right), \\ \phi_{2px}^Z = \frac{1}{4}\sqrt{\frac{Z^3}{2\pi a_0^3}}\frac{Zr}{a_0}\exp\left(-\frac{Zr}{2a_0}\right) \sin\theta \cos\varphi, \\ \phi_{2py}^Z = \frac{1}{4}\sqrt{\frac{Z^3}{2\pi a_0^3}}\frac{Zr}{a_0}\exp\left(-\frac{Zr}{2a_0}\right) \sin\theta \sin\varphi, \\ \phi_{2pz}^Z = \frac{1}{4}\sqrt{\frac{Z^3}{2\pi a_0^3}}\frac{Zr}{a_0}\exp\left(-\frac{Zr}{2a_0}\right) \cos\theta, \\ \end{matrix}$$$
$$$\label{eq:atomic_E} E= -\frac{Z^2me^4}{32\pi^2\epsilon_0^2\hbar^2n^2}=-\frac{13.6Z^2}{n^2}\text{ eV}.$$$

Many-electron wavefunctions
At this point it is necessary to discuss the properties of many-electron wavefunctions. This is because many-electron wavefunctions must satisfy an additional condition beyond being solutions to the Schrödinger equation: they must also be antisymmetric. A many-electron wavefunction must change sign when any two electrons are exchanged.

$$$\label{eq:antisymm} \Psi(\vec{r}_1,\cdots,\vec{r}_j,\cdots,\vec{r}_k,\cdots,\vec{r}_N)= -\Psi(\vec{r}_1,\cdots,\vec{r}_k,\cdots,\vec{r}_j,\cdots,\vec{r}_N).$$$

Many-electron wavefuntions are often written as products of atomic orbitals where it is necessary to include the spin in these products. The complete wavefunction of an electron is a product of a spatial part and a spin part. This is called a spin orbital. There are two spin orbitals for each of the atomic orbitals. The first few spin orbitals are,

$$$\phi_{1s}\uparrow, \phi_{1s}\downarrow, \phi_{2s}\uparrow , \phi_{2s}\downarrow,\cdots$$$

Here the up and down arrows denote the spin of the spin orbitals. A many-electron wavefunction can be written as an antisymmetrized product of spin orbitals.

$$$\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_N)= \mathcal{A}\phi_{1s}\uparrow\phi_{1s}\downarrow\cdots\phi_N\uparrow.$$$

Here $\mathcal{A}$ is the antisymmetrizing operator.

Slater determinants
One way to ensure the antisymmetry of the wavefunction is to construct a Slater determinant.

$$$\label{slater} \Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_N)= \frac{1}{\sqrt{N!}}\left|\begin{array}{slate} \phi_{1s}\uparrow (\vec{r}_1) & \phi_{1s}\downarrow (\vec{r}_1) & \cdots & \phi_N\uparrow (\vec{r}_1) \\ \phi_{1s}\uparrow (\vec{r}_2) & \phi_{1s}\downarrow (\vec{r}_2) & \cdots & \phi_N\uparrow (\vec{r}_2) \\ \vdots & \vdots & & \vdots \\ \phi_{1s}\uparrow (\vec{r}_N) & \phi_{1s}\downarrow (\vec{r}_N) & \cdots & \phi_N\uparrow (\vec{r}_N) \end{array}\right|.$$$

The rows in a Slater determinant are labeled by the electrons and the columns are labeled by the spin orbitals. If any two rows of a matrix are exchanged, the determinant changes sign. Therefore a Slater determiant always satisfies the antisymmetry condition. If any two columns of the matrix are identical, the determinant is zero. This is an expression of the Pauli exclusion principle. No two electrons can occupy the same spin orbital. For notational convenience, a Slater determinant is often expressed in Dirac notation,

$$$\Psi(\vec{r}_1,\vec{r}_2,\cdots,\vec{r}_N)= |\phi_{1s}\uparrow, \phi_{1s}\downarrow, \cdots, \phi_N\uparrow\rangle.$$$

The first spin orbital becomes the first column of the Slater determinant. The second spin orbital becomes the second column of the Slater determinant, and so forth. The order of the spin orbitals is important since exchanging the columns of the Slater determinant changes the sign of the wave function.

Returning to helium, the ground state of helium has two electrons in 1s orbitals. The two-electron ground state wavefunction is,

$$$\Psi_0^{\text{He}}(\vec{r}_1,\vec{r}_2)=\frac{1}{\sqrt{2}}\left|\begin{matrix} \phi_{1s}^{\text{He}}\uparrow(\vec{r}_1) & \phi_{1s}^{\text{He}}\downarrow(\vec{r}_1) \\ \phi_{1s}^{\text{He}}\uparrow(\vec{r}_2) & \phi_{1s}^{\text{He}}\downarrow(\vec{r}_2) \end{matrix}\right| = \frac{1}{\sqrt{2}}\phi_{\text{1s}}^{\text{He}}(\vec{r}_1)\phi_{\text{1s}}^{\text{He}}(\vec{r}_2)(\uparrow(\vec{r}_1)\downarrow(\vec{r}_2) - \uparrow(\vec{r}_2)\downarrow(\vec{r}_1)).$$$

This is an exact solution to $H_{\text{red}}^{\text{He}}$ with an energy that can be determined from Eq. \eqref{eq:atomic_E} to be -108.8 eV. A much better estimation for the energy can be obtained by numerically evaluating the energy using $H_{\text{total}}^{\text{He}}$ which includes the electron-electron interactions.

$$$E_0^{\text{He}}\approx \frac{\langle \Psi_0^{\text{He}} |H_{\text{total}}^{\text{He}}|\Psi_0^{\text{He}}\rangle}{\langle \Psi_0^{\text{He}}|\Psi_0^{\text{He}}\rangle}.$$$

This evaluates to -77.49 eV which is close to the actual ground state energy of -78.99 eV.

The first excited state of helium has one electron in a 1s orbital and one electron in a 2s orbital. There are four possible spin configurations for this state: $\uparrow\uparrow$, $\downarrow\downarrow$, $\downarrow\uparrow$, and $\uparrow\downarrow$. When the energy of these four states are evaluated using the reduced Hamiltonian \eqref{eq:red}, the energies of all four states are the same. Using Eq. \eqref{eq:atomic_E}, the energy of the first excited state is -68 eV. However, when the energy of the four states is evaluated using the total Hamiltonian \eqref{eq:helium} which includes the electron-electron interactions, three of the states have the same energy but the fourth one has a different energy. The three states with the same energy are called the triplet state. The fourth state with a different energy is called a singlet.

A description of the hydrogen atom and many-electron atoms can be found in The Physics of Atoms and Quanta by H. Haken and H. C. Wolf.