Hydrogen Atom

Consider the hydrogen atom, or hydrogen-like ions (e.g., He$^+$, Li$^{2+}$, ... etc.), with nuclear charge $+ze$, and mass $M$, and the electron with charge $-e$, and mass $m$. The potential energy of the system is a central potential (e.g., the Coulombic potential), $$V=-\frac{ze^2 k}{r},$$ where $r$ is the electron-nucleus distance and $k= \begin{cases}1 \hspace{.2cm} \text{in a.u.} \\ 1/4 \pi \epsilon_0 \hspace{.2cm} \text{in SI units}\end{cases}$

The total Hamiltonian is, 

$$\hat{H}=-\frac{\hbar^2}{2(m+M)} \nabla_R^2-\frac{\hbar^2}{2\mu} \nabla_r^2 +V(r),$$ where $\mu=\frac{m_e m_n}{m_n+m_e}$. Note that $\boxed{\mu \approx m_e}$, since $m_e<<m_n$. The Hamiltonian that includes only the second and third terms of $\hat{H}$ is represented by the symbol $\hat{H}^{el}$ and is called the electronic Hamiltonian because it depends only on the electronic coordinate $r$. In order to find the electronic eigenvalues, we must solve the equation,

$$\hat{H}^{el} \psi_{el} = E_{el} \psi_{el}.$$ (43)

Eq. (43) is the eigenvalue problem of a one particle central-potential. We consider the factorizable solution,

$$\psi_{el} =R(r) Y_l^m(\theta, \phi),$$   with,$$\qquad l= 0, 1, 2, ... \qquad \vert m\vert<l,$$ where $R(r)$ satisfies the equation,

$$-\frac{\hbar^2}{2\mu} \left[ \frac{\partial^2 R}{\partial r^2} +\......ial r} - \frac{\hbar^2}{\hbar^2 r^2} l(l+1) R \right ] -\frac{Ze^2 R}{r} = E R.$$ (44)

This equation could be solved by first transforming it into the associated Laguerre equation, for which solutions are well-known. Here, however, we limit the presentation to note that Eq. (44) has solutions that are finite, single valued and square integrable only when

$$E=-\frac{Z^2\mu e^4}{2\hbar^2 n^2}, \hspace{.3cm} E=-\frac{Z^2e^2}{2 a n^2},$$ (45)

where $n=1, 2, 3, ...$, and $a=\frac{\hbar^2}{\mu e^2}$ is the Bohr radius.

These are the bound-state energy levels of hydrogen-like atoms responsible for the discrete nature of the absorption spectrum. In particular, the wavenumbers of the spectral lines are

$$\bar{\omega}=\frac{E_2-E_1}{hc}=-\frac{Z^2\mu e^4}{hc2\hbar^2} \left( \frac{1}{n_2^2} -\frac{1}{n_1^2}\right).$$ The eigenvalues can be represented by the following diagram:
 

Degeneracy: Since the energy E depends only on the principal quantum number $n$, and the wave function $\psi_{el}$ depends on $n$, $l$ and $m$, there are $n^2$ possible states with the same energy. States with the same energy are called degenerate states. The number of states with the same energy is the degeneracy of the energy level.

n=1, 2, 3, ...

l=0, 1, 2, ... n-1 } these are n states

m=-l, -l+1, ..., 0, 1, 2, ...l } these are 2 l + 1 states

Exercise 38: Prove that the degeneracy of the energy level $E_n$ is $n^2$.

The complete hydrogen-like bound-state wave functions with quantum numbers $n$, $l$ and $m$ are, 

$$\psi_{nlm}(r, \theta, \phi) = R_{nl}(r) P_l^m(\theta) \frac{1}{\sqrt{2\pi}} e^{im \phi},$$ where $P_l^m(\theta)$ are the associated Legendre polynomials (introduced in page 49), and $R_{nl}(r)$ are the Laguerre associated polynomials, $$R_{nl}(r) = r^l e^{-\frac{zr}{na}} \sum_{j=0}^{n-l-1} b_j r^j,$$   where$$\hspace{.20cm} \ a \equiv \frac{\hbar^2}{\mu e^2}=0.529177 \AA,$$ and,  $$b_{j+1}=\frac{2z}{na} \frac{j+l+1-n}{(j+1)(j+2l+2)} b_j.$$ Example 1: Consider the ground state wave function of the H atom with $n=1, l=0, m=0:$ $$\boxed{R_{10}(r)=e^{-\frac{z}{a}r}b_0},$$ where, $b_0^2= 1/\int_0^{\infty}dr r^2 e^{-\frac{2zr}{a}}$, and $\hspace{.20cm} b_0=2(\frac{z}{a})^{3/2}.$

Therefore,

$$\psi_{100}(r, \theta, \phi)= 2(\frac{z}{a})^{3/2} \frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{2}} e^{-\frac{z}{a} r}.$$ Note: An alternative notation for wave functions with orbital quantum number $l = 0, 1, 2, ...$ is
 

Example 2: The possible wave functions with $n=2$ are:

$2 s \qquad 2 p_0 \qquad 2 p_1 \qquad 2 p_{-1},$

$\psi_{200} \qquad \psi_{210}, \qquad \psi_{211} \qquad \psi_{21-1},$

Exercise 39: Show that,

$$\psi_{2s}= \frac{1}{\sqrt{\pi}}(\frac{z}{2a})^{3/2}(1-\frac{zr}{2......ace{.5cm} \psi_{2p_{-1}}=\frac{1}{8\sqrt{\pi}} (\frac{z}{a})^{5/2} r e^{-zr/2a}$$   sin$$\theta e^{-i \phi},$$ $$\psi_{2p_0}=\frac{1}{\sqrt{\pi}} (\frac{z}{2a})^{5/2} r e^{-zr/2a}$$   Cos$$\theta, \hspace{.5cm} \psi_{2p_1}=\frac{1}{8\sqrt{\pi}} (\frac{z}{a})^{5/2} r e^{-zr/2a}$$   sin$$\theta e^{i \phi}.$$ Exercise 40: Compute the ionization energy of He$^+$.

Exercise 41: Use perturbation theory to first order to compute the energies of states $\psi_{210}$, $\psi_{211}$, and $\psi_{21-1}$ when a hydrogen atom is perturbed by a magnetic field $\vec{B}=B \hat{z}$, according to $\omega= -\beta \vec{L}.\vec{B}$, where $\beta=\frac{e\hbar}{2mc}$. (The splitting of spectroscopic lines, due to the perturbation of a magnetic field, is called Zeeman effect).

Radial Distribution Functions

The probability of finding the electron in the region of space where $r$ is between $r$ to $r+dr$, $\theta$ between $\theta$ to $\theta +d \theta$ and $\phi$ between $\phi$ and $\phi +d\phi$ is, 

$$P=R^*(r) R(r) Y_l^m(\theta)^* Y_l^m(\theta) r^2$$   sin$$\theta dr d\theta d\phi.$$ Therefore, the total probability of finding the electron with $r$ between $r$ and $r+dr$ is,  $$P^{\tau}(r) = \left[ \int_0^{\pi}d\theta \int_0^{2\pi}d\phi Y_l^m(\theta)^* Y_l^m(\theta) \text{sin} \theta \right ] R^*(r)R(r) r^2 dr,$$ where $\int_0^{\pi}d\theta \int_0^{2\pi}d\phi Y_l^m(\theta)^* Y_l^m(\theta)$   sin$\theta = 1$. For example, the radial probability for m=0 and l=1, can be visualized as follows:

In order to visualize other radial distribution functions of an electron in a hydrogen atom click here.
Pictures of atomic orbitals with n<=10 are available here.

Real Hydrogen-like Functions

Any linear combination of degenerate eigenfunctions of energy $E$ is also an eigenfunction of the Hamiltonian with the same eigenvalue $E$. Certain linear combinations of hydrogen-like wavefunctions generate real eigenfunctions. For example, when $l =1$,

$$\frac{1}{\sqrt{2}} \left( \psi_{n11} + \psi_{n1-1} \right) =R_{n1}(r)$$   sin$$\theta$$   Cos$$\phi \equiv \psi_{P_{2x}},$$ $$\frac{1}{\sqrt{2}i} \left( \psi_{n11} - \psi_{n1-1} \right) =R_{n1}(r)$$   sin$$\theta$$   Sin$$\phi \equiv \psi_{P_{2y}},$$ $$\psi_{210} \equiv \psi_{2P_{z}},$$ are real and mutually orthogonal eigenfunctions.

Function $\psi_{2P_{z}}$ is zero in the $xy$ plane, positive above such plane, and negative below it. Functions $\psi_{2P_{x}}$ and $\psi_{2P_{y}}$ are zero at the $zy$ and $xz$ planes, respectively. $\psi_{2P_{-1}}$ and $\psi_{2P_{1}}$ are eigenfunctions of $\hat{L}^2$ with eigenvalue $2 \hbar^2$. However, since $\psi_{2P_{-1}}$ and $\psi_{2P_{1}}$ are eigenfunctions of $\hat{L}_z$ with different eigenvalues (e.g., with eigenvalues $\hbar$ and $-\hbar$, respectively), linear combinations $\psi_{2P_{x}}$, and $\psi_{2P_{y}}$, are eigenfunctions of $\hat{L}^2$ but not eigenfunctions of $L_z$.

Exercise 42: (A) What is the most probable value of r, for the ground state of a hydrogen atom? Such value is represented by $r_M$.

(B) What is the total probability of finding the electron at a distance $r \leq r_M$?

(C) Verify the orthogonality of functions $2P_x, \ 2P_y, \$   and$\ 2P_z$.

(D) Verify that the ground state of the hydrogen atom is an eigenstate of $\hat{H}$, but that such state is not an eigenstate of $\hat{T}$, or $\hat{V}$.