Hydrogenlike Atoms

What you will learn

The hydrogen atom is the simplest atom for which the Schrödinger equation can be solved exactly. In contrast, helium, though it has only one additional electron, cannot be solved exactly due to the complexity introduced by electron-electron interactions.
Solving the Schrödinger equation for the hydrogen atom leverages the fact that the Coulomb potential from the nucleus is isotropic: it is radially symmetric and depends solely on the distance from the nucleus. This symmetry leads to degeneracies in the energy levels.
While the resulting energy eigenfunctions (atomic orbitals) are not necessarily isotropic, their dependence on angular coordinates arises fundamentally from the isotropy of the underlying potential.
Because angular momentum is conserved, energy eigenvalues can be classified by two angular momentum quantum numbers, $l$ and $m$ (both integers), which quantize the magnitude and projection of angular momentum.
Atomic orbitals derived for hydrogen are also foundational in describing the structure of multi-electron atoms and molecules.

Schrödinger equation for hydrogenlike atoms¶

Consider one electron and one nucleus with charge $Ze$ , where $e$ is the magnitude of the electron charge ( $1.6021773 \times 10^{-19}$ C) and $Z$ is the atomic number. Examples of such systems are H, $He^+$ , and $Li^{2+}$ .
We can reduce the two-body problem to a one-body problem of an electron with reduced mass moving with respect to a fixed nucleus. Since the electron is much lighter compared to the nucleus, we can use the electron mass instead of the reduced mass, $\mu \approx m_e$ .
We have a problem of one particle moving in a symmetric potential field in 3D. We expect to get 3 quantum numbers and anticipate some degeneracies due to this radial symmetry.
Kinetic energy operator in 3D is the same as for the particle in a box in 3D:

\frac{\hbar^2}{2m_e}\nabla^2

(1)

Potential energy operator consists of one Coulomb term encoding the electrostatic attraction between nucleus and electron:

V = - \frac{Ze^2}{4\pi\epsilon_0 r}

(2)

Here $m_e$ is the electron mass and $\epsilon_0$ is the vacuum permittivity.

H-atom in the spherical coordinate system¶

The Schrödinger equation for the H atom is a problem in full three dimensions with kinetic and potential energy terms. We also expect to get infinitely many eigenfunctions and eigenvalues since we have a bound electron.

{\left[ -\frac{\hbar^2}{2m_e}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0 r}\right]\psi(r,\theta,\phi) = E\psi(r,\theta,\phi)}

(3)

Because of the spherical symmetry of the Coulomb potential, it is convenient to work in spherical coordinates:

\nabla^2 = \nabla_r^2 + \frac{1}{r^2}\nabla_{\theta, \phi}^2

(4)

Note that the Coulomb potential term above depends only on $r$ (and not on $\theta$ or $\phi$ ). The Laplacian can be written in terms of the angular momentum operator $\hat{L}$ :

\nabla^2 = \nabla_r^2 - \frac{1}{r^2}\frac{\hat{L}^2}{\hbar^2}

(5)

By substituting this in and multiplying both sides by $2m_e r^2$ , we get:

\left[\frac{-\hbar^2}{2m_e}\nabla^2_r - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hat{L}^2}{2m_e r^2}\right]\psi_i(r,\theta,\phi) = E\psi_i(r,\theta,\phi)

(6)

Since the operator can be split into $r$ -dependent and angle-dependent parts, the solution can be written as a product of the radial and angular parts.

Separation of variables¶

{\psi_i(r,\theta,\phi) = R_{nl}(r)Y_l^m(\theta,\phi)}

(7)

The $R_{nl}$ is called the radial wavefunction, which we need to obtain.
The $Y_l^m$ are spherical harmonics, which are eigenfunctions of angular momentum $\hat{L}^2$ as discussed earlier. Hence this part is already solved.

\hat{L}^2 Y_{lm} = \hbar^2 l(l+1) Y_{lm}

(8)

Plugging $R_{nl}(r)Y_l^m(\theta,\phi)$ into the Schrödinger equation, applying the angular momentum operator, and cancelling the spherical harmonics from both sides, we end up with the radial part:

{\left[ -\frac{\hbar^2}{2m_e}\left(\frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r}\right) - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{l(l+1)\hbar^2}{2m_e r^2}\right] R_{nl}(r) = E_{nl}R_{nl}(r)}

(9)

We find that the electron is moving in an effective potential generated by the attractive Coulomb interaction and the repulsive orbital kinetic energy.

V_{eff} = - \frac{Ze^2}{4\pi\epsilon_0 r} + \frac{l(l+1)\hbar^2}{2m_e r^2}

(10)

The first (repulsive) term grows more rapidly with $1/r$ than the Coulomb potential, so it dominates at small distances if $l \neq 0$ . Both terms approach zero for large values of $r$ . The resultant potential is repulsive at short distances for $l > 0$ and is more repulsive the greater the value of $l$ . The net result of this repulsive centrifugal potential is to force electrons in orbitals with $l > 0$ on average farther from the nucleus than $l = 0$ electrons.

Effective radial potential for the hydrogen atom — Fig.1 Effective radial potential $V_{eff}(r)$ for the hydrogen atom. The attractive Coulomb term and the repulsive centrifugal term combine to push higher- $l$ electrons farther from the nucleus.

Radial wavefunctions¶

The radial eigenfunctions $R_{nl}$ can be written as (the derivations are lengthy but standard):

R_{nl}(r) = \rho^l e^{-\rho/2}{L_{n-l-1}^{2l+1}(\rho)}

(11)

Bohr radius: $a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2}$
Dimensionless distance defined via the ratio to the Bohr radius: $\rho = \frac{2Zr}{na_0}$
Laguerre polynomials: $L_{n-l-1}^{2l+1}(\rho)$

Examples of the radial wavefunctions for hydrogenlike atoms¶

Orbital	$n$	$l$	$R_{nl}$
1s	1	0	$2\left(\frac{Z}{a_0}\right)^{3/2}e^{-\rho/2}$
2s	2	0	$\frac{1}{2\sqrt{2}}\left(\frac{Z}{a_0}\right)^{3/2}(2 - \rho)e^{-\rho/2}$
2p	2	1	$\frac{1}{2\sqrt{6}}\left(\frac{Z}{a_0}\right)^{3/2}\rho e^{-\rho/2}$
3s	3	0	$\frac{1}{9\sqrt{3}}\left(\frac{Z}{a_0}\right)^{3/2}(6 - 6\rho - \rho^2)e^{-\rho/2}$
3p	3	1	$\frac{1}{9\sqrt{6}}\left(\frac{Z}{a_0}\right)^{3/2}(4 - \rho)\rho e^{-\rho/2}$
3d	3	2	$\frac{1}{9\sqrt{30}}\left(\frac{Z}{a_0}\right)^{3/2}\rho^2 e^{-\rho/2}$

Full quantum solution of the H-atom¶

The complete solution of the H-atom problem is provided by writing down all eigenfunctions and eigenvalues of the Hamiltonian:

\hat{H} |n, l, m_l\rangle = E_n |n, l, m_l\rangle

(12)

We notice that while the wavefunction depends on three quantum numbers (coming from quantization of the radial and angular coordinates), the energy depends only on the principal quantum number $n$ . This is another example of energetic degeneracy that is due to the special spherical symmetry of the H-atom.

Energy level diagram of the hydrogen atom — Fig.2 Energy levels of the hydrogen atom. All states with the same principal quantum number $n$ share the same energy, an accidental degeneracy of the Coulomb problem.

Spectrum of the H atom¶

The equation for the hydrogen atom energy can be expressed in wavenumber units ( $m^{-1}$ ; usually $cm^{-1}$ is used):

{\tilde{E}_n = \frac{E_n}{hc} = \frac{E_n}{2\pi\hbar c} = -\overbrace{\frac{m_e e^4}{4\pi c(4\pi\epsilon_0)^2\hbar^3}}^{\equiv R} \times\frac{Z^2}{n^2}}

(16)

where $R$ is the Rydberg constant and we have assumed that the nucleus has infinite mass. To be exact, the Rydberg constant depends on the nuclear mass, but this difference is very small. For the H atom, $R_H = 1.096\,775\,856 \times 10^7$ $m^{-1}$ .
The H-atom energy expression can be used to calculate the differences between energy levels:

\Delta\tilde{v}_{n_1,n_2} = \tilde{E}_{n_2} - \tilde{E}_{n_1} = -\frac{R_H Z^2}{n_2^2} + \frac{R_H Z^2}{n_1^2} = R_H Z^2\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)

(17)

The Lyman series is obtained with $n_1 = 1$ , the Balmer series with $n_1 = 2$ , and the Paschen series with $n_1 = 3$ . The ionization energy is given by:

{E_i = R_H Z^2\left(\frac{1}{1^2} - \frac{1}{\infty}\right)}

(18)

For a ground-state hydrogen atom (i.e., $n = 1$ ), the above equation gives a value of 109678 cm $^{-1}$ = 13.6057 eV. Note that the larger the nuclear charge $Z$ is, the larger the binding energy is.

Balmer series of the hydrogen spectrum — Fig.3 The Balmer series of the hydrogen atom: transitions ending on $n_1 = 2$ that give rise to the visible emission lines of hydrogen.

Quantum numbers $n$ , $l$ , and $m$ ¶

The quantum numbers in hydrogenlike atoms take on the following values, dictated by the solution of the Schrödinger equation with boundary conditions imposed on the respective radial and angular parts.

For historical reasons, the following letters are used to express the value of $l$ :

l = 0, 1, 2, 3, ... \quad \Leftrightarrow \quad \text{symbol} = s, p, d, f, ...

(23)

For $n = 1$ we have only one wavefunction: $R_{10}(r)Y_0^0(\theta,\phi)$ . This state is usually labeled $1s$ , where 1 indicates the shell number ( $n$ ) and $s$ corresponds to orbital angular momentum $l$ being zero.
For $n = 2$ , we have several possibilities: $l = 0$ or $l = 1$ . The former is labeled $2s$ . The latter is the $2p$ state and consists of three degenerate states (for example, $2p_x$ , $2p_y$ , $2p_z$ or $2p_{+1}$ , $2p_0$ , $2p_{-1}$ ). In the latter notation the values for $m$ are indicated as subscripts.
There is one more quantum number that has not been discussed yet: the spin quantum number, $m_s = \pm 1/2$ .
For one-electron systems this can take values $\pm\frac{1}{2}$ (discussed in more detail later). In the absence of magnetic fields the spin levels are degenerate, and therefore the total degeneracy of the levels is $2n^2$ .

Table of wavefunctions in Cartesian coordinates¶

The full hydrogenlike wavefunctions, written in terms of $\sigma = \frac{Zr}{a_0}$ :

$n$	$l$	$m$	Wavefunction expressed in $\sigma = \frac{Zr}{a_0}$
1	0	0	$\psi_{1s} = \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}e^{-\sigma}$
2	0	0	$\psi_{2s} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}(2 - \sigma)e^{-\sigma/2}$
2	1	0	$\psi_{2p_z} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-\sigma/2}\cos(\theta)$
2	1	$\pm 1$	$\psi_{2p_x} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-\sigma/2}\sin(\theta)\cos(\phi)$
			$\psi_{2p_y} = \frac{1}{4\sqrt{2\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma e^{-\sigma/2}\sin(\theta)\sin(\phi)$
3	0	0	$\psi_{3s} = \frac{1}{81\sqrt{3\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(27 - 18\sigma + 2\sigma^2\right)e^{-\sigma/3}$
3	1	0	$\psi_{3p_z} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(6 - \sigma\right)\sigma e^{-\sigma/3}\cos(\theta)$
3	1	$\pm 1$	$\psi_{3p_x} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(6 - \sigma\right)\sigma e^{-\sigma/3}\sin(\theta)\cos(\phi)$
			$\psi_{3p_y} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\left(6 - \sigma\right)\sigma e^{-\sigma/3}\sin(\theta)\sin(\phi)$
3	2	0	$\psi_{3d_{z^2}} = \frac{1}{81\sqrt{6\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}\left(3\cos^2(\theta) - 1\right)$
3	2	$\pm 1$	$\psi_{3d_{xz}} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}\sin(\theta)\cos(\theta)\cos(\phi)$
			$\psi_{3d_{yz}} = \frac{\sqrt{2}}{81\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}\sin(\theta)\cos(\theta)\sin(\phi)$
3	2	$\pm 2$	$\psi_{3d_{x^2-y^2}} = \frac{1}{81\sqrt{3\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}\sin^2(\theta)\cos(2\phi)$
			$\psi_{3d_{xy}} = \frac{1}{81\sqrt{3\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\sigma^2 e^{-\sigma/3}\sin^2(\theta)\sin(2\phi)$

Computing with wavefunctions¶

We can use the wavefunctions from the table above to calculate various averages by using the definition of an average in quantum mechanics, $\langle f(r) \rangle = \langle \psi |f(r)|\psi\rangle$ :

\langle r\rangle_{n,l,m} = \frac{a_0 n^2}{Z}\left( 1 + \frac{1}{2}\left[1 - \frac{l(l+1)}{n^2}\right]\right)

(24)

\langle r^2\rangle_{n,l,m} = \frac{a_0^2 n^4}{Z^2}\left( 1 + \frac{3}{2}\left[1 - \frac{l(l+1) - 1/3}{n^2}\right]\right)

(25)

\langle r^{-1}\rangle_{n,l,m} = \frac{Z}{a_0 n^2}

(26)

\langle r^{-2}\rangle_{n,l,m} = \frac{Z^2}{a_0^2 n^3 (l+1/2)}

(27)

\langle r^{-3}\rangle_{n,l,m} = \frac{Z^3}{a_0^3 n^3 (l+1/2)(l+1)}

(28)