# What is hydrogen-like

1. Jul 24, 2014

### Greg Bernhardt

Definition/Summary

A hydrogen-like or hydrogenic atom is a two-body system where the particles are bound by an electrostatic interaction. Examples of such systems include the hydrogen atom, singly ionized helium, and positronium.

Equations

The Hamiltonian of the system is given by
$$\hat{H} = \frac{\hat{p}^2}{2\mu} - \frac{Ze^2}{r}$$ where $\mu = \frac{m_1 m_2}{m_1+m_2}$ is the reduced mass of the two particles, $e$ is the elementary charge, and $Ze$ is the charge of the nucleus.

The energy eigenstates are labelled by three quantum number: the principal quantum number, $n$; the orbital angular momentum or azimuthal quantum number, $l$; and the magnetic quantum number, $m_l$. All three are integers, subject to the following conditions:
\begin{align*}
n &= 1, 2, \ldots \\
l &= 0, 1, \ldots, n-1 \\
m_l &= -l, -l+1, \ldots, 0, \ldots, l-1, l
\end{align*}
The wave functions have the form $\psi_{nlm}(r,\theta,\phi) = R_{nl}(r) Y_{lm}(\theta,\phi)$ where $Y_{lm}(\theta,\phi)$ denotes the spherical harmonics. The radial function is given by
$$R_{nl}(r) = \sqrt{\left(\frac{2}{na}\right)^3 \frac{(n-l-1)!}{2n ((n+l)!)^3}} e^{-r/na} \left(\frac{2r}{na}\right)^l L^{2l+1}_{n-l-1}\left(\frac{2r}{na}\right)$$ where $L_n^k(x)$ is an associated Laguerre polynomial and $a = \frac{\hbar^2}{Ze^2\mu}$. The energy of a state is $E_n = -\frac{Z^2e^4\mu}{2n^2\hbar^2}$.

For hydrogen, the natural length scale is the Bohr radius $a_0 = \frac{\hbar^2}{m_e e^2} = 0.0529\text{ nm}$, where $m_e$ is the mass of the electron. The energies are given by
$$E_n = -\frac{m_e e^4}{2\hbar^2}\frac{1}{n^2} = -\frac{13.6\text{ eV}}{n^2}.$$

Extended explanation

The Schrodinger equation with the electrostatic potential can be solved using the technique of separation of variables. The solution is assumed to have the form $\psi(r,\theta,\phi) = R(r) \Theta(\theta) \Phi(\phi)$ and substituted into the Schrodinger equation, which results in individual differential equations for $R(r)$, $\Theta(\theta)$, and $\Phi(\phi)$. Because of the spherical symmetry of the potential, the solutions for the angular parts turn out to be the spherical harmonics $Y_{lm}(\theta, \phi)$.

The radial function $R(r)$ satisfies the differential equation
$$\frac{1}{R}\frac{d}{dr}\left(r^2 \frac{dR}{dr}\right) + \frac{2\mu}{\hbar^2}(Er^2+Ze^2r)=l(l+1).$$ Solutions to the eigenvalue equation which satisfy physical boundary conditions turn out to contain a parameter $n$ which can only assume positive integer values.

Because the energy of the eigenstates depends only $n$, there is a $2n^2$-fold degeneracy. The factor of two arises because of the electron's spin. In actual atoms and ions, perturbations due to magnetic interactions, like the spin-orbit interaction, lift the degeneracy, and the energy of an eigenstate generally depends on $l$ and $m_l$ as well.

It is possible to derive the energies using an elegant algebraic approach. The angular momentum and Laplace-Runge-Lenz operators form a Lie algebra called SO(4), which can be decomposed into SO(3)*SO(3) â€” operators that behave much like angular momentum.

The derivation of a hydrogen-like atom's energy starts with the commutation relations for the angular momentum and the Laplace-Runge-Lenz vectors:
\begin{align*}
[L_i,L_j] &= i\hbar \epsilon_{ijk} L_k \\
[L_i,A_j] &= i\hbar \epsilon_{ijk} A_k \\
[A_i,L_j] &= i\hbar \epsilon_{ijk} A_k \\
[A_i,A_j] &= i\hbar (-2\mu H) \epsilon_{ijk} L_k
\end{align*} However, the magnitude of the quantum-mechanical LRL vector differs from that of the classical one:
$$\vec{A}^2 = 2\mu H (\vec{L}^2 + \hbar^2) + (Z\mu e^2)^2$$ though it is still orthogonal to the angular momentum: $\vec{L}\cdot \vec{A} = 0$.

Since we are concerned with bound states, the eigenvalues of $\hat{H}$ will be less than zero, and we can safely set $H = - \frac{P^2}{2\mu}$ where P is real. Setting $\vec{A} = P \vec{D}$, we find commutators
\begin{align*}
[L_i,L_j] &= i\hbar \epsilon_{ijk} L_k \\
[L_i,D_j] &= i\hbar \epsilon_{ijk} D_k \\
[D_i,L_j] &= i\hbar \epsilon_{ijk} D_k \\
[D_i,D_j] &= i\hbar \epsilon_{ijk} L_k
\end{align*} and
$$2\mu H (\vec{D}^2 + \vec{L}^2 + \hbar^2) + (Z\mu e^2)^2 = 0.$$ The SO(4) symmetry is now rather evident.

Now we decompose into SO(3)*SO(3). Let $\vec{L} = \vec{J}_1 + \vec{J}_2$ and $\vec{D} = \vec{J}_1 - \vec{J}_2$, where the $\vec{J}$'s have the commutation relations of angular momentum. Orthogonality requires $\vec{J}_1$ and $\vec{J}_2$ have the same total "angular momentum", $j$; therefore, we have
$$2\mu H (4j(j+1) + 1)\hbar^2 + (Z\mu e^2)^2 = 0,$$ yielding energies
$$E = - \frac{(Z \mu e^2)^2}{2(2j+1)^2 \hbar^2}.$$ This implies that the principal quantum number $n = 2j+1$, with $j$ not restricted to integer values, and that the total degeneracy is $n^2$ (neglecting the electron's spin).

Interestingly, it is possible to solve for a hydrogen-like atom's wave function by finding eigenvalues of $\hat{L}_z$ and $\hat{A}_z$ instead of eigenvalues of $\vec{L}^2$ and $\hat{L}_z$, though one must use cylindrical parabolic coordinates instead of spherical coordinates.

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