Hydrogen Atom

The hydrogen atom is one of the few real physical systems for which the allowed quantum states of a particle and corresponding energies can be solved for exactly (as opposed to approximately) in non-relativistic quantum mechanics. In the most basic quantum mechanical model of hydrogen, the proton is taken to be a fixed source of an electric potential and the Schrödinger equation for the wavefunction of the electron is solved.

Different possible wavefunctions of the electron in the hydrogen atom [1].

The quantum mechanics of hydrogen was hailed as one of the great early successes of the theory for its correct prediction of the Rydberg energy levels of the electron seen in photoelectron spectroscopy experiments. The quantum description of the electron in hydrogen fully resolved the classical problem in which an orbiting electron would radiate energy in EM waves and inspiral into the nucleus, resulting in instability. Although Bohr and Sommerfeld originally were able to correctly calculate the allowed states and energies of an electron in hydrogen without using the Schrödinger equation, only the full quantum-mechanical treatment allows generalization to arbitrary single-electron atoms and consideration of the electron spin.

To write down an equation that can successfully describe an electron in the hydrogen atom, it is necessary to convert the one-dimensional Schrödinger equation into an equation that works in three dimensions. The transition towards three dimensions mimics the corresponding generalization in classical mechanics, where equations are expressed in the $x,y,z$ coordinates in three dimensions. In classical mechanics, the Hamiltonian of a system is prescribed by

$$H=\frac{1}{2m}\left({{p}_{x}}^{2}+{{p}_{y}}^{2}+{{p}_{z}}^{2}\right) +V(x,y,z).$$

This is also true in QM, where now the time-dependent Schrödinger equation is

$$i\hbar \frac{\partial \Psi}{\partial t}= -\frac{{\hbar}^{2}}{2m}{\nabla}^{2}\Psi + V\Psi$$

or the time-independent case

$$E\Psi= -\frac{{\hbar}^{2}}{2m}{\nabla}^{2}\Psi + V\Psi$$

where ${\nabla}^{2}=\frac{{\partial}^{2}}{\partial {x}^{2}}+\frac{{\partial}^{2} }{\partial {y}^{2}}+\frac{{\partial}^{2} }{\partial {z}^{2}}$ is the Laplacian operator.

Since linearity still applies in 3D, all of the linear operators used in 1D that describe the momentum operators are still valid:

$${\hat{p}}_{x} = -i\hbar \frac{\partial}{\partial x}$$

$${\hat{p}}_{y} = -i\hbar \frac{\partial}{\partial y}$$

$${\hat{p}}_{z} = -i\hbar \frac{\partial}{\partial z}$$

It is customary in physics to break a higher-dimensional problem into its directional components, solve each component and reassemble the solutions. Since it is cumbersome to work in each $x,y,z$ component and the hydrogen atom is spherically symmetric, it is convenient to switch to a more efficient notation that will easily permit use of spherical coordinates. To this end, one can compile all spatial coordinates into a vector $\vec{r} =(x,y,z)$. A general wavefunction is now expressed as a superposition of energy eigenstates $\phi_n$ in three dimensions:

$$\Psi(\vec{r} , t) = \sum{ {c}_{n}{\psi}_{n}(\vec{r}) {e}^{-i{E}_{n}t/\hbar}}$$

where the wavefunctions evolve in some arbitrary potential $V(\vec{r})$.

As before, wavefunctions must be continuous and normalizable. In three dimensions, the single-dimensional integral for computing the normalization becomes a volume integral over all of space:

$$\int {\Psi}^{*}{\Psi}dV =1.$$

Since the hydrogen atom is spherically symmetric, at this point it is convenient to write the 3D Schrödinger equation in spherical coordinates $(r,\theta, \phi)$, which is the basis for solving the hydrogen atom and atomic orbitals. The geometric meaning of the spherical coordinates is displayed below:

Spherical coordinates $(r,\theta, \phi)$. The radius from the origin is given by $r$ while $\theta$ and $\phi$ give the angle with respect to the $z$ axis and angle of rotation around the $z$ axis, respectively.

Since the Laplacian in spherical coordinates is given by:

$${\nabla}^{2} = \frac{1}{{r}^{2}}\frac{\partial}{\partial r}\left({r}^{2} \frac{\partial}{\partial r}\right) + \frac{1}{{r}^{2}\sin{\theta}} \frac{\partial}{\partial \theta} \left(\sin{\theta} \frac{\partial}{\partial \theta}\right) + \frac{1}{{r}^{2}\sin^{2}{\theta}}\left(\frac{{\partial}^{2}}{\partial {\phi}^{2}}\right),$$

the time-independent Schrödinger equation can be written:

$$\frac{-{\hbar}^{2}}{2m}\left[\frac{1}{{r}^{2}}\frac{\partial}{\partial r}\left({r}^{2} \frac{\partial \Psi}{\partial r}\right) + \frac{1}{{r}^{2}\sin{\theta}} \frac{\partial }{\partial \theta} \left(\sin{\theta} \frac{\partial \Psi}{\partial \theta}\right) + \frac{1}{{r}^{2}\sin^{2}{\theta}}\left(\frac{{\partial}^{2} \Psi}{\partial {\phi}^{2}}\right)\right] + V\Psi = E\Psi.$$

Separation of variables is a common and powerful method when confronting higher-dimensional differential equations. In spherical coordinates, consider the ansatz of a solution that separates out the radial variable of a wavefunction from the angular variables:

$$\Psi(r, \theta, \phi) = R(r)Y(\theta, \phi).$$

Substituting into the time-independent Schrödinger equation,

$$\frac{-{\hbar}^{2}}{2m}\left[\frac{Y}{{r}^{2}}\frac{\partial}{\partial r}\left({r}^{2} \frac{\partial R}{\partial r}\right) + \frac{R}{{r}^{2}\sin{\theta}} \frac{\partial }{\partial \theta} \left(\sin{\theta} \frac{\partial Y}{\partial \theta}\right) + \frac{R}{{r}^{2}\sin^{2}{\theta}}\left(\frac{{\partial}^{2} Y}{\partial {\phi}^{2}}\right)\right] + VRY = ERY.$$

Rearranging this equation into its radial and angular parts:

$$\left[\frac{1}{R}\frac{\partial }{\partial r}\left({r}^{2}\frac{dR}{dr}\right)-\frac{2m{r}^{2}}{{\hbar}^{2}}[V-E] \right]+ \frac{1}{Y}\left[\frac{1}{\sin{\theta}}\frac{\partial}{\partial \theta}\left(\sin{\theta}\frac{\partial Y}{\partial \theta}\right) + \frac{1}{\sin^2{\theta}} \frac{{\partial}^{2}{Y}}{\partial{\phi}^{2}}\right] = 0$$

Separation of the above equation proceeds by assuming that the two bracketed terms are equal in magnitude to some constant and opposite in sign. Conventionally (for convenience later) the constant is taken to have the value $\ell (\ell + 1)$. Then each equation can be written separately:

Radial equation $$\frac{1}{R}\frac{d}{dr}\left({r}^{2}\frac{dR}{dr}\right)-\frac{2m{r}^{2}}{{\hbar}^{2}}[V-E] = \ell(\ell+1)$$

Angular equation $$\frac{1}{Y}\left[\frac{1}{\sin{\theta}}\frac{\partial}{\partial \theta}\left(\sin{\theta}\frac{\partial Y}{\partial \theta}\right) + \frac{1}{\sin^2{\theta}} \frac{{\partial}^{2}{Y}}{\partial{\phi}^{2}}\right] = -\ell(\ell+1).$$

Since the radial equation includes the energy $E$, one should expect the allowed energies of the Hydrogen atom to be determined by the radial solutions. Start with the radial equation:

$$\frac{1}{R}\frac{d}{dr}\left({r}^{2}\frac{dR}{dr}\right)-\frac{2m{r}^{2}}{{\hbar}^{2}}[V-E] = \ell(\ell+1)$$

The solutions of the radial equation gives the radial probability density. The radial equation is solved by defining the new function $u(r) = r R(r)$ and substituting:

$$-\frac{\hbar^2}{2m} \frac{d^2 u}{dr^2} + \left(V + \frac{\hbar^2}{2m} \frac{\ell (\ell + 1)}{r^2} \right) u = Eu.$$

This is the one-dimensional Schrödinger equation with the potential shifted by a centrifugal term.

In the hydrogen atom, the potential is the electric potential of the fixed proton:

$$V = -\frac{e^2}{4\pi \epsilon_0 r}.$$

Computation of the solutions via the standard method involves asymptotic analysis and series methods, a long and arduous procedure that can be found in [2]. Another, less well known method called the supersymmetric or factorization method, uses techniques of supersymmetric quantum mechanics. For brevity, the discussion here will skip to the the final solutions for $R(r)$, written in terms of two indicies $n$ and $\ell$, where $\ell$ is the separation constant used above [2]. Defining $\rho = \frac{rme^2}{4\pi \epsilon_0 \hbar^2 n}$, they are:

$$R_{n \ell} (r) = \frac{1}{r} \rho^{\ell + 1} e^{-\rho} v(\rho),$$

with $v(\rho)$ given by the series:

$$v(\rho) = \sum_{j=0}^{\infty} c_j \rho^j, \qquad c_{j+1} = \frac{2(j+ \ell + 1 - n)}{(j+1)(j + 2\ell + 2)} c_j.$$

Providing an overall normalization fixes $c_0$ and determines all of the coefficients that determine $v(\rho)$ from the above recursion relation. Importantly, all $c_j$ can be shown to be zero after $j_{\text{max}} = n-\ell - 1$ as proved in [2] in order for the solutions to be finite everywhere in space. That is, $n$ is defined to be the integer at which this is true. It is discussed in the solutions to the angular equation why $\ell$ is an integer.

The recurrence relation for the $c_j$ can be solved in closed form. The solutions involve the associated Laguerre polynomials:

$$L^p_{q-p} (x) = (-1)^p \left(\frac{d}{dx} \right)^p L_q (x),$$

with $L_q (x)$ the $q$th Laguerre polynomial:

$$L_q (x) = e^x \left( \frac{d}{dx} \right)^q (e^{-x} x^q).$$

The closed-form solutions for $v(\rho)$ are then:

$$v(\rho) = L^{2\ell + 1}_{n - \ell - 1} (2\rho),$$

up to normalization factor.

Compute $R_{21}(r)$, the radial wavefunction of an electron in hydrogen with $n=2$ and $\ell = 1$.

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Solution:

Plugging into the definition,

$$R_{21} (r) = \frac{1}{r} \rho^{2} e^{-\rho} v(\rho),$$

with:

$$\begin{aligned} v(\rho) &= L^{3}_{0} (2\rho) = L^3_{3-3} (2\rho) = (-1)^3 \left(\frac{d}{d(2\rho)} \right)^3 L_3 (2\rho) \\ &= - \frac{1}{8} \frac{d^3}{d\rho^3} \left(e^{2\rho} \left( \frac{d}{d(2\rho)} \right)^3 (e^{-2\rho} (2\rho)^3)\right) \end{aligned}$$ The result of the repeated differentiation is:

$$v(\rho) = 6 .$$

A constant value of $v(\rho)$ is consistent with the statement that $j_{\text{max}} = n - \ell - 1 = 2-1-1 = 0$. Substituting in above, one finds:

$$R_{21} (r) = \frac{6}{r} \rho^2 e^{-\rho}.$$

However, this is not normalized. To normalize, compute $\int_0^{\infty} r^2 R_{21} (r) dr$ and insert the appropriate constant prefactor to make equal to unity. One obtains the result:

$$R_{21} (r) = \sqrt{\frac{\rho}{3r}} \frac{2}{r} \rho^2 e^{-\rho}.$$

The termination condition for the series defining $v(\rho)$ puts the following restriction on $n$:

$$n = \frac{me^2}{4\pi \epsilon_0 \hbar^2} \frac{\hbar}{\sqrt{-2mE}}.$$

Since $n$ is an integer, this constrains the allowed values of the energy of the electron in hydrogen to be:

$$E = -\left(\frac{me^4}{32\hbar^2 \pi^2 \epsilon_0^2} \right) \frac{1}{n^2} = \frac{-13.6 \text{ eV}}{n^2}.$$

The index $n$ thus determines the quantized energy levels of the electron in hydrogen. In Bohr-Sommerfeld theory, the values of $n$ are called shells which the electron can occupy.

While the radial solutions to the wavefunction of the electron in the hydrogen atom determine the energy levels, the angular solutions determine the electron orbitals. Now, start with the angular equation:

Angular equation $$\frac{1}{Y}\left[\frac{1}{\sin{\theta}}\frac{\partial}{\partial \theta}\left(\sin{\theta}\frac{\partial Y}{\partial \theta}\right) + \frac{1}{\sin^2{\theta}} \frac{{\partial}^{2}{Y}}{\partial{\phi}^{2}}\right] = -\ell(\ell+1).$$

The solutions to the angular equation are well-known; they are the spherical harmonics indexed by two integers: $m$ and the separation constant $\ell$ from above:

$$Y^m_{\ell} (\theta, \phi) = \sqrt{\frac{2\ell + 1}{4\pi} \frac{(\ell - m)!}{(\ell + m)!}} P^m_{\ell} (\cos \theta) e^{im\phi},$$

where the $P^m_{\ell} (\cos \theta)$ are the Legendre polynomials, defined in the spherical harmonics wiki. The periodicity of the wavefunction of the electron in $\phi$ and $\theta$ demands that $m$ and $\ell$ are integers with $|m| \leq \ell$. These solutions contribute to the shape of atomic orbitals, since they determine the angular probability density of the electron.

Orbitals of the electron in hydrogen, giving the equiprobability surfaces for the angular wavefunctions. Top row corresponds to $\ell = 0$ and subsequent rows increase the value of $\ell$ by one while iterating over the possible values of $m$ [3].

Combining the radial and angular wavefunctions with the appropriate normalization, one can write the complete solution for the wavefunction of the electron in hydrogen:

$$\psi_{n\ell m } = R_{n \ell} (r) Y^m_{\ell} (\theta, \phi) = \sqrt{\left(\frac{2\rho}{r} \right)^3 \frac{(n- \ell - 1)!}{2n((n+\ell)!)^3}} e^{-2\rho / r} (2\rho)^{\ell} L^{2\ell + 1}_{n - \ell - 1} (2\rho) Y_{m}^{\ell} (\theta, \phi),$$

with $\rho = \frac{rme^2}{4\pi \epsilon_0 \hbar^2 n}$ from above.

One of the greatest triumphs of quantum theory is the mathematical confirmation of the spectrum of the hydrogen atom. The spectral lines of hydrogen can be measured by exciting the electron in hydrogen and observing the wavelength of the emitted radiation when the electron transitions between energy levels / shells. In transitioning between energy levels, the electron wavefunction is changing from one stationary state to another stationary state. The difference between energies in each state, i.e., the energy of the emitted radiation, is quantified by the equation:

$$E_{\gamma} = {E}_{i} - {E}_{f} = -13.6 \text{ eV} \left( \frac{1}{{n}_{i}^{2}}-\frac{1}{{n}_{f}^{2}}\right),$$

where the subscripts $i$ and $f$ denote the initial and final values of $n$, respectively. The right-hand size above is just the difference in energies between radial wavefunctions with principal quantum number $n$, as derived above.

The wavelength of the emitted photons can then be determined from the relation $E_{\gamma} = \frac{hc}{\lambda}$.

Find the wavelength of the photon emitted when the electron in a hydrogen atom falls from the $n=3$ to $n=2$ levels.

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Solution:

Set $E_{\gamma} = \frac{hc}{\lambda}$ equal to the difference in the two energy levels:

$$\frac{hc}{\lambda} = -13.6 \text{ eV} \left(\frac{1}{3^2} - \frac{1}{2^2}\right)$$

Solving for $\lambda$ yields the result:

$$\lambda = 656.4 \text{ nm}.$$

This is red light, corresponding to the so-called "H-alpha" transition in the Balmer series.

The indices $n$, $\ell$, and $m$ in the radial and angular solutions uniquely specify the spatial wavefunction of the electron in hydrogen. They are called the quantum numbers of the electron, with $n$ the principal quantum number, $\ell$ the angular, orbital, or azimuthal quantum number, and $m$ the magnetic quantum number. Roughly, $n$ specifies the energy of the electron, $\ell$ specifies the total angular momentum, and $m$ specifies the angular momentum about the $z$ axis, up to constants. Specifically,

$$\hat{L}^2 = \hbar^2 \ell (\ell + 1), \qquad \hat{L}_z = \hbar m$$

label the (squared) total angular momentum and angular momentum about the $z$ axis respectively.

A fourth quantum number, $s$, is called the spin quantum number. This number gives the spin orientation of the electron in hydrogen (or any other atom). However, the spin of the electron is not important to first order: it plays a role when there are multiple electrons, an external magnetic field, or a significant coupling with the spin of the proton.

Canonically, a shorthand notation is used in chemistry to label solutions to the Schrödinger equation for the hydrogen atom. The values $\ell = 0,1,2,3$ which most electrons in most elements occupy are labeled s,p,d, and f respectively. For example, an electron in the $4p$ orbital has $n=4$ and $p = 1$. The value of $m$ is not used often in chemistry because it often has no influence on the energy or bonding characteristics of the atom.

Compute the spatial wavefunction, total angular momentum, and angular momentum about the $z$ axis of an electron in the $2s$ state with $m = 0$.

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Solution:

The spatial wavefunction desired is $\psi_{200}$. From the general solution derived above, this is:

$$\psi_{200} (r,\theta, \phi) = \sqrt{\left(\frac{2\rho}{r} \right)^3 \frac{1}{32}} e^{-2\rho / r} L^{ 1}_{ 1} (2\rho) Y_{0}^{0} (\theta, \phi).$$

From the formula for the spherical harmonics (described in more detail in the spherical harmonics wiki), $Y^0_0 = \frac{1}{\sqrt{4\pi}}$. The desired associated Laguerre polynomial is $L^1_1 (2\rho) = 4 - 4\rho$. The full spatial wavefunction is therefore, after rearranging constants:

$$\psi_{200} (r,\theta, \phi) = \sqrt{ \frac{\rho^3}{\pi r^3}} e^{-2\rho / r} (1-\rho) .$$

Since $\ell = m = 0$, the total angular momentum and angular momentum about the $z$ axis of this electron are both zero. This corresponds to spherical symmetry of the associated wavefunction, as can be seen from the lack of angular dependence of $\psi_{200}$ above.

[1] Image from https://en.wikipedia.org/wiki/Electron under Creative Commons licensing for reuse and modification.

[2] Griffiths, David J. Introduction to Quantum Mechanics. Second Edition. Pearson: Upper Saddle River, NJ, 2006.

[3] Image from https://commons.wikimedia.org/wiki/File:Singleelectronorbitals.jpg under Creative Commons licensing for reuse and modification.