Hydrogen Spectrum

Purpose

For an electron of mass $m_e$ bound to a nucleus of charge $+Ze$ and mass $M_N$, separate the center-of-mass motion and use the reduced mass

$$\mu = \frac{m_eM_N}{m_e+M_N}.$$

The relative-coordinate Hamiltonian is

$$H = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{4\pi\epsilon_0r}.$$

Its nonrelativistic bound-state energies are

$$E_n = -\frac{\mu Z^2e^4} {2(4\pi\epsilon_0)^2\hbar^2} \frac{1}{n^2}, \qquad n=1,2,3,\ldots.$$

Equivalently, with the fine-structure constant

$$\alpha_{\rm fs} = \frac{e^2}{4\pi\epsilon_0\hbar c},$$

the spectrum is

$$E_n = -\frac{\mu c^2(Z\alpha_{\rm fs})^2}{2n^2}.$$

The energy zero is the separated-particle threshold. The negative levels form an infinite sequence accumulating at $E=0$ from below.

The canonical solution is at Hydrogen Atom. This card focuses on spectral calculations; radial functions have their own canonical page at Radial Wavefunctions.

At a glance

Quantity	Formula
Reduced mass	$\mu=m_eM_N/(m_e+M_N)$
Reduced-mass Bohr scale	$a_\mu=4\pi\epsilon_0\hbar^2/(\mu e^2)$
Charge-$Z$ length scale	$a_Z=a_\mu/Z$
Reduced-mass Rydberg energy	$\operatorname{Ry}_\mu=\mu c^2\alpha_{\rm fs}^2/2$
Bound energy	$E_n=-Z^2\operatorname{Ry}_\mu/n^2$
Spatial shell degeneracy	$g_n=n^2$
Ionization from shell $n$	$E_{\rm ion}(n)=Z^2\operatorname{Ry}_\mu/n^2$
Emission energy	$h\nu=Z^2\operatorname{Ry}_\mu(1/n_f^2-1/n_i^2)$
Coulomb virial relation	$\langle T\rangle_n=-E_n$, $\langle V\rangle_n=2E_n$

For ordinary hydrogen, the leading value is approximately

$$E_n\simeq-\frac{13.6\,\mathrm{eV}}{n^2}.$$

The quoted number includes neither the full hierarchy of precision corrections nor a claim that every real-hydrogen state is exactly degenerate.

Natural Coulomb scales

Define

$$a_\mu = \frac{4\pi\epsilon_0\hbar^2}{\mu e^2} = \frac{\hbar}{\mu c\alpha_{\rm fs}}.$$

For nuclear charge $Z$,

$$a_Z = \frac{a_\mu}{Z} = \frac{\hbar}{\mu cZ\alpha_{\rm fs}}.$$

The corresponding energy is

$$\operatorname{Ry}_\mu = \frac{\hbar^2}{2\mu a_\mu^2} = \frac{\mu e^4} {2(4\pi\epsilon_0)^2\hbar^2} = \frac12\mu c^2\alpha_{\rm fs}^2.$$

Then

$$E_n = -\frac{Z^2\operatorname{Ry}_\mu}{n^2}.$$

Lengths scale approximately as $1/Z$ and binding energies as $Z^2$ when the reduced mass is held fixed. Increasing the reduced mass makes the system smaller and more tightly bound.

Quantum numbers

The spatial bound states are labeled by

$$n=1,2,3,\ldots,$$ $$\ell=0,1,\ldots,n-1,$$ $$m=-\ell,-\ell+1,\ldots,\ell.$$

They satisfy

$$H\lvert n\ell m\rangle = E_n\lvert n\ell m\rangle,$$ $$L^2\lvert n\ell m\rangle = \hbar^2\ell(\ell+1) \lvert n\ell m\rangle,$$ $$L_z\lvert n\ell m\rangle = \hbar m\lvert n\ell m\rangle.$$

The integer $m$ in the last formula is a magnetic quantum number and is not the particle mass. The radial node number is

$$n_r=n-\ell-1.$$

The parity is

$$\Pi=(-1)^\ell.$$

The full spatial wavefunction factors as

$$\psi_{n\ell m}(r,\theta,\phi) = R_{n\ell}(r)Y_\ell^m(\theta,\phi).$$

Degeneracy

In the spinless Schrödinger-Coulomb model, the energy depends only on $n$. For one shell,

$$g_n^{\rm spatial} = \sum_{\ell=0}^{n-1}(2\ell+1) = n^2.$$

If electron spin is included but spin-dependent interactions are omitted, the count doubles:

$$g_n^{\rm spatial+spin}=2n^2.$$

The degeneracy among different $m$ values at fixed $\ell$ follows from rotational invariance and occurs for any central potential. The additional degeneracy among different $\ell$ values is special to the Coulomb problem and reflects its hidden symmetry. See Degeneracy of the Hydrogen Atom.

This ideal count is not the multiplicity of one unresolved line in every real experiment. Fine structure, the Lamb shift, hyperfine structure, isotope effects, finite nuclear size, and external fields split or reorganize the levels.

Transition energies

For emission from $n_i$ to $n_f$ with $n_i\gt n_f$,

$$h\nu = E_{n_i}-E_{n_f} \text{ in magnitude}$$

becomes

$$h\nu = Z^2\operatorname{Ry}_\mu \left( \frac{1}{n_f^2}- \frac{1}{n_i^2} \right).$$

Equivalently,

$$\frac{1}{\lambda} = R_\mu Z^2 \left( \frac{1}{n_f^2}- \frac{1}{n_i^2} \right),$$

where

$$R_\mu = \frac{\operatorname{Ry}_\mu}{hc}$$

is the reduced-mass spectroscopic Rydberg constant.

The spectral series are classified by $n_f$:

Series	Final principal quantum number
Lyman	$n_f=1$
Balmer	$n_f=2$
Paschen	$n_f=3$
Brackett	$n_f=4$

The gross energy formula determines line positions before finer splittings. It does not determine line strengths. Dipole selection rules and transition matrix elements depend on angular and radial wavefunctions.

Ionization and series limits

The continuum threshold is

$$E=0.$$

Ionizing a state in shell $n$ requires

$$E_{\rm ion}(n) = -E_n = \frac{Z^2\operatorname{Ry}_\mu}{n^2}.$$

For a fixed lower shell $n_f$, the emission series approaches its limit as $n_i\to\infty$:

$$h\nu_{\rm limit} = \frac{Z^2\operatorname{Ry}_\mu}{n_f^2}.$$

The infinite bound sequence has no highest bound state. Its levels become more closely spaced and more spatially extended near threshold.

Reduced-mass correction

Replacing the reduced mass by the electron mass is the infinite-nuclear-mass approximation. Their ratio is

$$\frac{\mu}{m_e} = \frac{M_N}{m_e+M_N} = \frac{1}{1+m_e/M_N}.$$

Therefore

$$E_n = \frac{\mu}{m_e} E_n^{(M_N\to\infty)}.$$

Finite nuclear mass reduces the magnitude of the binding energy and increases the Bohr length by the reciprocal factor. The correction is small for hydrogen but is essential in isotope comparisons and precision spectroscopy.

Center-of-mass kinetic energy is separate:

$$E_{\rm total} = \frac{P_{\rm CM}^2}{2(m_e+M_N)} +E_n.$$

The quoted atomic levels normally refer to the internal energy $E_n$.

Hydrogenic-ion scaling

For one-electron ions,

$$E_n^{(Z)} = -\frac{\mu_Z}{m_e} \frac{Z^2\operatorname{Ry}_\infty}{n^2},$$

where $\operatorname{Ry}_\infty$ uses the electron mass and an infinitely heavy nucleus. Ignoring small reduced-mass differences gives

Ion	$Z$	Ground energy	Length scale
$\mathrm H$	1	$-13.6\,\mathrm{eV}$	$a_0$
$\mathrm{He}^+$	2	$-54.4\,\mathrm{eV}$	$a_0/2$
$\mathrm{Li}^{2+}$	3	$-122.4\,\mathrm{eV}$	$a_0/3$

Neutral helium is not hydrogenic because it has two electrons. The one- electron ion $\mathrm{He}^+$ is hydrogenic.

Virial and momentum checks

For a $1/r$ potential, the virial theorem gives

$$2\langle T\rangle = -\langle V\rangle.$$

Since $E_n=\langle T\rangle+\langle V\rangle$,

$$\langle T\rangle_{n\ell m} = -E_n,$$ $$\langle V\rangle_{n\ell m} = 2E_n.$$

Consequently,

$$\langle p^2\rangle_{n\ell m} = 2\mu\langle T\rangle = -2\mu E_n = \frac{\mu^2c^2(Z\alpha_{\rm fs})^2}{n^2}.$$

The root-mean-square speed scale is

$$v_{\rm rms} = \frac{\sqrt{\langle p^2\rangle}}{\mu} = \frac{Z\alpha_{\rm fs}c}{n}.$$

This makes the nonrelativistic condition transparent: low-lying states require $Z\alpha_{\rm fs}\ll1$ for a parametrically small velocity scale.

The Hellmann–Feynman theorem or radial integrals also give

$$\left\langle\frac1r\right\rangle_{n\ell m} = \frac{Z}{a_\mu n^2} = \frac{1}{a_Zn^2}.$$

These expectation values are independent of $\ell$ in the ideal Coulomb problem, although many other radial moments are not.

Scope and corrections

The displayed spectrum assumes:

• a nonrelativistic Schrödinger equation;
• a point nucleus and exact $-Ze^2/(4\pi\epsilon_0r)$ interaction;
• no spin-orbit, relativistic kinetic, or Darwin corrections;
• no radiative Lamb shift;
• no hyperfine interaction or finite nuclear size;
• no external electric or magnetic field;
• one electron and no electron-electron interaction.

The leading relativistic scale grows with powers of $Z\alpha_{\rm fs}$. At large $Z$, a Dirac and finite-nuclear-size treatment becomes essential. For precision hydrogen, even small recoil and QED effects matter because the experimental resolution is far finer than the gross binding scale.

Units and constants

Quantity	SI units
$a_\mu$, $a_Z$	m
$\operatorname{Ry}_\mu$, $E_n$	J or eV
$R_\mu$	m$^{-1}$
$\mu$, $m_e$, $M_N$	kg
$\alpha_{\rm fs}$, $Z$, $n$	dimensionless

The formula uses SI Coulomb normalization. In Gaussian or natural-unit conventions, factors of $4\pi\epsilon_0$, $c$, and $\hbar$ move or are set to one. Translate the definition of $\alpha_{\rm fs}$ first.

Common mistakes

• Using $m_e$ when reduced-mass accuracy is required.
• Calling $-13.6\,\mathrm{eV}$ the exact energy of every real hydrogen ground state.
• Applying the hydrogen value to a hydrogenic ion without the $Z^2$ factor.
• Scaling orbital lengths as $1/Z^2$ rather than $1/Z$.
• Forgetting that the energy zero is the ionization threshold.
• Counting $n=0$ as a principal shell.
• Allowing $\ell=n$ or $\lvert m\rvert\gt\ell$.
• Calling the full $n^2$ degeneracy a consequence of rotations alone.
• Doubling the degeneracy for spin while simultaneously claiming spin interactions are included.
• Using the energy formula to infer line strengths without matrix elements and selection rules.
• Calling neutral helium hydrogenic.
• Applying the nonrelativistic point-nucleus model at high $Z$ without a validity check.

Related formulas

• Hydrogenic Ions
• Radial Wavefunctions
• Coulomb Potential
• Hydrogen Atom Angular Structure
• Atomic Orbitals
• Continuum States of the Coulomb Problem
• Hydrogen as Atomic Prototype
• Hydrogen Atom Model Card

Exercises

1. Count one shell

Show that the $n$th shell has $n^2$ spatial states.

Solution

For fixed $n$, the allowed orbital quantum numbers are $\ell=0,1,\ldots,n-1$. Each $\ell$ has $2\ell+1$ magnetic substates, so

$$\begin{aligned} g_n &= \sum_{\ell=0}^{n-1}(2\ell+1)\\ &= 2\frac{n(n-1)}{2}+n\\ &= n^2. \end{aligned}$$

Including spin while neglecting spin-dependent interactions doubles this to $2n^2$.

2. Lyman-alpha energy

Find the gross transition energy for hydrogen emission from $n_i=2$ to $n_f=1$.

Solution

For $Z=1$,

$$h\nu = \operatorname{Ry}_\mu \left( 1-\frac14 \right) = \frac34\operatorname{Ry}_\mu.$$

Using $\operatorname{Ry}_\mu\simeq13.6\,\mathrm{eV}$ gives

$$h\nu\simeq10.2\,\mathrm{eV}.$$

This is the gross nonrelativistic line energy. Fine and Lamb structure split the real spectral line at higher resolution.

3. Reduced-mass direction

Does finite nuclear mass increase or decrease the magnitude of the binding energy relative to the infinite-mass nucleus?

Solution

The reduced mass obeys

$$\mu = \frac{m_eM_N}{m_e+M_N} \lt m_e.$$

Since $\lvert E_n\rvert$ is proportional to $\mu$, finite nuclear mass decreases the binding magnitude. It also increases the length scale because $a_\mu$ is proportional to $1/\mu$.

4. Virial check

Use the Coulomb virial theorem to express $\langle T\rangle$, $\langle V\rangle$, and $\langle p^2\rangle$ in terms of $E_n$.

Solution

For $V\propto r^{-1}$,

$$2\langle T\rangle=-\langle V\rangle.$$

Together with $E_n=\langle T\rangle+\langle V\rangle$, this gives

$$\langle T\rangle=-E_n, \qquad \langle V\rangle=2E_n.$$

Finally,

$$\langle p^2\rangle = 2\mu\langle T\rangle = -2\mu E_n.$$

Because $E_n\lt0$, both $\langle T\rangle$ and $\langle p^2\rangle$ are positive.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018, Ch. 4.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994, Ch. 13.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1, Wiley, 1977.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• H. A. Bethe and E. E. Salpeter, Quantum Mechanics of One- and Two-Electron Atoms, Springer, 1957.