Hydrogen Atom Bound-State Spectrum

Derivation

Hydrogen Atom Bound-State Spectrum

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Statement

Starting from the radial Schrödinger equation for a single electron of reduced mass \(\mu\) bound in the Coulomb potential \(V(r)=-\,e^2/4\pi\epsilon_0 r\), we show that normalisable (\(E<0\)) solutions exist only for a discrete set of energies \(E_n=-E_{\mathrm R}/n^2\) with principal quantum number \(n=1,2,3,\dots\), where \(E_{\mathrm R}=\mu e^4/2(4\pi\epsilon_0)^2\hbar^2\) is the Rydberg energy, and that the corresponding radial wavefunctions are \(R_{nl}(r)\propto (r/na_0)^{l}\,e^{-r/na_0}\,L_{n-l-1}^{2l+1}(2r/na_0)\), the associated Laguerre functions, with \(l=0,1,\dots,n-1\).

Why it matters

This is the only realistic three-dimensional atom for which the bound-state problem is solved in closed form, and it fixes the numbers — \(13.6\ \mathrm{eV}\), the Bohr radius \(a_0\), the Lyman and Balmer lines — against which all of atomic physics is calibrated. The exact spectrum is the zeroth-order platform onto which fine structure, hyperfine structure, the Lamb shift and external-field perturbations are all built.

Its structure is also generic: the interplay of a centrifugal barrier, an attractive tail and a normalisability condition that quantises energy recurs in every central-potential bound-state problem. Hydrogen is where a physicist first sees quantisation emerge as a boundary-condition requirement rather than as a postulate.

Assumptions

Non-relativistic electron.Drop it and the Schrödinger equation is replaced by the Dirac equation; the \(n^2\) degeneracy in \(l,\,j\) splits into fine structure of order \(\alpha^2 E_{\mathrm R}\).

Point, static, spinless Coulomb source.A finite nuclear size, nuclear motion beyond the two-body reduction, or magnetic (spin–spin) coupling shifts levels (hyperfine, isotope, volume shifts) and removes exact solvability.

Pure \(1/r\) potential.Any deviation (screening in multi-electron ions, a \(1/r^2\) or short-range core term) breaks the accidental \(l\)-degeneracy and the closed-form Laguerre solution.

Bound states only, \(E<0\).For \(E\ge 0\) the spectrum is continuous (scattering states); the termination argument below does not apply and one obtains Coulomb wavefunctions instead of polynomials.

Reduced-mass two-body reduction already performed.If one uses the bare electron mass \(m_e\) instead of \(\mu=m_e m_p/(m_e+m_p)\), every level is off by the factor \(m_e/\mu\approx 1.00054\), which is visible in precision spectroscopy and in the H/D isotope shift.

Derivation

\[ -\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2}+\left[-\frac{e^2}{4\pi\epsilon_0\,r}+\frac{\hbar^2\,l(l+1)}{2\mu\,r^2}\right]u=E\,u \]

The prior separation result gives the radial equation for \(u(r)=rR(r)\); substituting the Coulomb potential yields this one-dimensional equation on \(r\in[0,\infty)\) with an effective potential (Coulomb well plus centrifugal barrier). A

\[ E<0,\qquad \kappa\equiv\frac{\sqrt{-2\mu E}}{\hbar},\qquad E=-\frac{\hbar^2\kappa^2}{2\mu} \]

We seek bound states, for which \(E<0\); defining the positive constant \(\kappa\) fixes the length scale \(1/\kappa\) of the exponential decay and makes the following substitutions dimensionless. A

\[ \rho\equiv\kappa r,\qquad \rho_0\equiv\frac{\mu e^2}{2\pi\epsilon_0\hbar^2\kappa},\qquad \frac{d^2u}{d\rho^2}=\left[1-\frac{\rho_0}{\rho}+\frac{l(l+1)}{\rho^2}\right]u \]

Rescaling \(r\to\rho=\kappa r\) and dividing through by \(E=-\hbar^2\kappa^2/2\mu\) removes all dimensional constants except the single dimensionless coupling \(\rho_0\); this is the standard nondimensional Coulomb radial equation. B

\[ \rho\to\infty:\quad \frac{d^2u}{d\rho^2}\approx u\ \Rightarrow\ u\sim e^{-\rho} \]

At large \(\rho\) the \(1/\rho\) and \(1/\rho^2\) terms vanish; of the two exponential solutions \(e^{\pm\rho}\), only \(e^{-\rho}\) is normalisable, so we peel it off. B

\[ \rho\to 0:\quad \frac{d^2u}{d\rho^2}\approx\frac{l(l+1)}{\rho^2}u\ \Rightarrow\ u\sim\rho^{\,l+1} \]

Near the origin the centrifugal term dominates; the indicial roots are \(\rho^{l+1}\) and \(\rho^{-l}\), and only \(\rho^{l+1}\) keeps \(u(0)=0\) and \(R=u/r\) finite. B

\[ u(\rho)=\rho^{\,l+1}e^{-\rho}\,v(\rho)\ \Rightarrow\ \rho\,v''+2(l+1-\rho)\,v'+\big[\rho_0-2(l+1)\big]v=0 \]

Factoring out both asymptotic behaviours leaves a well-behaved function \(v(\rho)\); direct differentiation and substitution into Step 3 gives this Laguerre-type ODE for \(v\), regular at the origin. B

\[ v(\rho)=\sum_{j=0}^{\infty}c_j\,\rho^{\,j}\ \Rightarrow\ c_{j+1}=\frac{2(j+l+1)-\rho_0}{(j+1)(j+2l+2)}\,c_j \]

A Frobenius power series inserted into Step 6 gives a two-term recurrence by matching each power of \(\rho\); it determines every coefficient from \(c_0\). C

\[ \text{if it did not terminate: } \frac{c_{j+1}}{c_j}\to\frac{2}{j}\ \Rightarrow\ v\sim e^{2\rho}\ \Rightarrow\ u\sim e^{+\rho}\ \text{(non-normalisable)} \]

\[ \therefore\ \exists\,j_{\max}=N\ \text{with}\ 2(N+l+1)=\rho_0 \]

An infinite series grows like \(e^{2\rho}\), overwhelming \(e^{-\rho}\); normalisability forces the series to terminate, i.e. the numerator must vanish at some non-negative integer \(N\) (the radial quantum number). C

\[ n\equiv N+l+1\in\{1,2,3,\dots\}\ \Rightarrow\ \rho_0=2n,\qquad l=0,1,\dots,n-1 \]

Defining the principal quantum number \(n\) collects the termination condition into \(\rho_0=2n\); since \(N\ge 0\) and \(l\ge 0\), the allowed angular momenta at fixed \(n\) are \(l\le n-1\). A

\[ \rho_0=\frac{\mu e^2}{2\pi\epsilon_0\hbar^2\kappa}=2n\ \Rightarrow\ \kappa=\frac{1}{n a_0},\qquad a_0\equiv\frac{4\pi\epsilon_0\hbar^2}{\mu e^2} \]

\[ E_n=-\frac{\hbar^2\kappa^2}{2\mu}=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}\,\frac{1}{n^2} \]

Solving \(\rho_0=2n\) for \(\kappa\) identifies the Bohr radius \(a_0\), and inserting \(\kappa=1/na_0\) into \(E=-\hbar^2\kappa^2/2\mu\) delivers the quantised spectrum. B

\[ v(\rho)\propto L_{\,n-l-1}^{\,2l+1}(2\rho),\qquad R_{nl}(r)=\sqrt{\left(\frac{2}{na_0}\right)^{\!3}\frac{(n-l-1)!}{2n\,(n+l)!}}\;\left(\frac{2r}{na_0}\right)^{\!l}e^{-r/na_0}\,L_{\,n-l-1}^{\,2l+1}\!\left(\frac{2r}{na_0}\right) \]

The terminating polynomial solving Step 6 with \(\rho_0=2n\) is, by definition, the associated Laguerre polynomial \(L_{n-l-1}^{2l+1}\); imposing \(\int_0^\infty|R|^2 r^2\,dr=1\) with the Laguerre orthogonality integral fixes the normalisation. C

Result

\[ E_n=-\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}\,\frac{1}{n^2}=-\frac{E_{\mathrm R}}{n^2},\qquad E_{\mathrm R}=\tfrac12\alpha^2\mu c^2\approx 13.606\ \mathrm{eV},\qquad n=1,2,3,\dots \]

\[ R_{nl}(r)=\sqrt{\left(\frac{2}{na_0}\right)^{\!3}\frac{(n-l-1)!}{2n\,(n+l)!}}\;\left(\frac{2r}{na_0}\right)^{\!l}e^{-r/na_0}\,L_{\,n-l-1}^{\,2l+1}\!\left(\frac{2r}{na_0}\right) \]

Reading. The energy depends only on \(n\), so at fixed \(n\) the \(n^2\) states with \(l=0,\dots,n-1\) and \(m=-l,\dots,l\) are degenerate (\(2n^2\) including spin). The \(l\)-independence is the "accidental" Coulomb degeneracy, tied to a hidden \(SO(4)\) symmetry (the conserved Runge–Lenz vector). The radial function has \(n-l-1\) nodes, decays on the scale \(na_0\), and vanishes as \(r^{l}\) at the origin; only \(l=0\) states have nonzero amplitude at the nucleus.

Units check. \(e^2/4\pi\epsilon_0\) has units \(\mathrm{J\,m}\), so \([e^2/4\pi\epsilon_0]^2=\mathrm{J^2 m^2}\); dividing by \(\hbar^2\;(\mathrm{J^2 s^2})\) gives \(\mathrm{m^2\,s^{-2}}\), and multiplying by \(\mu\;(\mathrm{kg})\) yields \(\mathrm{kg\,m^2\,s^{-2}}=\mathrm J\). The form \(E_{\mathrm R}=\tfrac12\alpha^2\mu c^2\) is manifestly an energy with \(\alpha\) dimensionless. \(a_0=4\pi\epsilon_0\hbar^2/\mu e^2\) has units \((\mathrm{C^2 J^{-1} m^{-1}})(\mathrm{J^2 s^2})/(\mathrm{kg\,C^2})=\mathrm m\).

Limiting cases

\(n\to\infty\): \(E_n\to 0^-\) with spacing \(\Delta E\sim 2E_{\mathrm R}/n^3\), the Rydberg limit merging into the ionisation continuum — infinitely many bound states because the Coulomb tail is long-ranged.
Nuclear charge \(Ze\) (hydrogen-like ion): replace \(e^2\to Ze^2\), giving \(E_n=-Z^2E_{\mathrm R}/n^2\) and \(a_0\to a_0/Z\); levels scale as \(Z^2\), orbitals shrink as \(1/Z\).
\(m_p\to\infty\): \(\mu\to m_e\), recovering the infinite-mass Rydberg \(R_\infty\); finite \(m_p\) lowers \(E_{\mathrm R}\) by the factor \(\mu/m_e\approx 0.99946\).
Circular states \(l=n-1\): the Laguerre polynomial is a constant, \(R_{nl}\propto r^{\,n-1}e^{-r/na_0}\) with no radial nodes — the closest quantum analogue of a Bohr circular orbit.

Breaks when

Relativistic / fine-structure regime. Terms of order \((v/c)^2\sim\alpha^2\) — relativistic kinetic correction, spin–orbit coupling, Darwin term — split the exact \(l\)-degeneracy by \(\sim\alpha^2 E_{\mathrm R}\approx 10^{-4}\,\mathrm{eV}\); the Schrödinger spectrum is then only leading order.
Continuum, \(E\ge 0\). The series-termination argument (Step 8) fails, energies are not quantised, and one gets oscillatory Coulomb scattering wavefunctions rather than Laguerre polynomials.
Strong external fields. A field comparable to the internal field (\(\sim 5\times 10^{11}\,\mathrm{V/m}\), or \(B\sim 10^5\,\mathrm T\)) makes the Stark/Zeeman terms non-perturbative; the \(1/r\) symmetry is broken and states mix or ionise.
Many electrons. For \(Z\ge 2\) with more than one electron, electron–electron repulsion screens the nucleus, destroys the pure \(1/r\) form and the accidental degeneracy, and no closed-form spectrum exists.

Failure modes

Confusing \(n\), \(l\), and the radial index. Writing the Laguerre order as \(L_n^l\) instead of \(L_{n-l-1}^{2l+1}\): the upper index is \(2l+1\), the degree (node count) is \(n-l-1\).
Using \(m_e\) instead of the reduced mass \(\mu\). Harmless to two significant figures but wrong at the \(5\times10^{-4}\) level, exactly where the H–D isotope shift lives.
Forgetting the \(u=rR\) substitution. Applying the flat-space radial boundary condition to \(R\) rather than \(u\) mishandles the origin and gives the wrong indicial root.
Dropping the \(\rho^{-l}\) root by hand-waving. It must be excluded by normalisability/finiteness of \(R\) at \(r=0\), not merely asserted.
Assuming \(l\) can equal \(n\). The constraint \(l\le n-1\) follows from \(N=n-l-1\ge 0\); "2d" or "1p" states do not exist.
Quoting \(2n\) degeneracy instead of \(n^2\) (or \(2n^2\) with spin). Summing \(\sum_{l=0}^{n-1}(2l+1)=n^2\), not \(2n\).

Discussion

The deepest structural point is that quantisation here is a boundary-condition phenomenon. Nothing in the differential equation is discrete; the integers \(n\) appear solely because a generic solution blows up at infinity and only a measure-zero set of energies produces the terminating polynomial that decays. This is the same mechanism as the particle in a box or the harmonic oscillator, and it is worth internalising that "energy is quantised" is a theorem about normalisable solutions, not an axiom.

The exact \(l\)-independence of \(E_n\) is special to the \(1/r\) potential. In any other central potential the energy depends on both \(n\) and \(l\); the Coulomb (and isotropic-oscillator) cases are singled out by an extra conserved quantity — the Laplace–Runge–Lenz vector — which enlarges the rotational \(SO(3)\) symmetry to \(SO(4)\) for bound states. The "accidental" degeneracy is therefore not accidental at all: it is the fingerprint of a dynamical symmetry, and Pauli exploited exactly this to obtain the spectrum algebraically before Schrödinger's wave treatment.

Physically, each level balances two competing energies. Confining the electron to a scale \(a\) costs kinetic energy \(\sim\hbar^2/2\mu a^2\) (uncertainty principle) while the Coulomb attraction gains \(\sim -e^2/4\pi\epsilon_0 a\); minimising the sum gives \(a\sim a_0\) and \(E\sim -E_{\mathrm R}\), reproducing the exact result up to factors. The virial theorem sharpens this: for a \(1/r\) potential \(\langle T\rangle=-E_n\) and \(\langle V\rangle=2E_n\), so the atom's size and binding are locked together.

Modern precision reveals what this derivation omits. The measured \(2s_{1/2}\)–\(2p_{1/2}\) Lamb shift (\(\sim 1057\ \mathrm{MHz}\)) is a QED radiative effect absent from both Schrödinger and Dirac theory; the \(21\ \mathrm{cm}\) line is the ground-state hyperfine splitting from the proton spin; and the proton charge radius enters the \(s\)-states through the finite-size (Darwin-like) shift, the quantity at the heart of the "proton radius puzzle" in muonic hydrogen. The clean \(-E_{\mathrm R}/n^2\) is thus the first term of a long, well-controlled expansion.

Common misconceptions. The electron is not orbiting; \(|R_{nl}(r)|^2 r^2\) is a stationary probability density with no classical trajectory, and the "Bohr radius" is a probability scale, not an orbit radius. The degeneracy in \(m\) is ordinary rotational symmetry, but the degeneracy in \(l\) is the special \(SO(4)\) one and is lifted by any departure from a pure \(1/r\) law. Finally, \(n\) is not "the number of nodes" — the radial node count is \(n-l-1\), and the total (radial plus angular) node count is \(n-1\).

Worked examples

Example 1 — Lyman-\(\alpha\) wavelength (\(n=2\to 1\)).

\[ E_n=-\frac{E_{\mathrm R}}{n^2},\qquad E_{\mathrm R}=13.606\ \mathrm{eV} \]

Use the derived spectrum; energies in eV. A

\[ \Delta E=E_2-E_1=-E_{\mathrm R}\!\left(\frac{1}{4}-\frac{1}{1}\right)=\frac{3}{4}E_{\mathrm R} \]

The emitted photon carries the level difference. A

\[ \Delta E=\tfrac34(13.606\ \mathrm{eV})=10.20\ \mathrm{eV} \]

Insert numbers. A

\[ \lambda=\frac{hc}{\Delta E}=\frac{1239.84\ \mathrm{eV\,nm}}{10.20\ \mathrm{eV}} \]

Photon energy–wavelength relation with \(hc=1239.84\ \mathrm{eV\,nm}\). A

\[ \lambda\approx 121.6\ \mathrm{nm} \]

Reading. The Lyman-\(\alpha\) line, in the far ultraviolet — the strongest hydrogen emission line and a workhorse of astrophysics. Units check. \(\mathrm{eV\,nm}/\mathrm{eV}=\mathrm{nm}\).

Example 2 — Rydberg energy from fundamental constants.

\[ E_{\mathrm R}=\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2}=\tfrac12\,\alpha^2\,\mu c^2 \]

Rewrite the derived constant using the fine-structure constant \(\alpha=e^2/4\pi\epsilon_0\hbar c\). B

\[ \alpha=\frac{1}{137.036},\qquad \mu c^2\approx m_e c^2=0.5110\times 10^{6}\ \mathrm{eV} \]

Insert the measured fine-structure constant and electron rest energy (infinite-mass limit). A

\[ E_{\mathrm R}=\tfrac12\left(\frac{1}{137.036}\right)^{2}(0.5110\times10^{6}\ \mathrm{eV})=\tfrac12(5.325\times10^{-5})(5.110\times10^{5}\ \mathrm{eV}) \]

\(\alpha^2=5.325\times10^{-5}\); multiply through. A

\[ E_{\mathrm R}=\tfrac12(27.21\ \mathrm{eV})=13.61\ \mathrm{eV} \]

Finite proton mass would multiply by \(\mu/m_e=0.99946\), giving \(13.598\ \mathrm{eV}\). A

\[ E_{\mathrm R}\approx 13.6\ \mathrm{eV}\quad(\text{ionisation energy of ground-state H}) \]

Reading. Binding hydrogen's electron takes \(13.6\ \mathrm{eV}\); the compact form \(\tfrac12\alpha^2 m_e c^2\) shows binding is a fraction \(\sim\alpha^2\) of the rest energy, i.e. the atom is weakly relativistic. Units check. \(\alpha\) dimensionless, \(\mu c^2\) an energy, so \(E_{\mathrm R}\) is an energy.

Problems

Compute the wavelength of the Balmer-\(\alpha\) (H\(\alpha\), \(n=3\to2\)) line.
Solution
\(\Delta E=E_{\mathrm R}(1/4-1/9)=13.606\times 5/36=1.889\ \mathrm{eV}\). Then \(\lambda=hc/\Delta E=1239.84/1.889=656.3\ \mathrm{nm}\) — the red line of the Balmer series (visible).
How many distinct quantum states (including spin) share the energy \(E_3\)? List the allowed \(l\).
Solution
For \(n=3\), \(l=0,1,2\) with \((2l+1)\) values of \(m\): \(1+3+5=9=n^2\) orbital states. Doubling for spin gives \(2n^2=18\) states.
Find the ionisation energy of hydrogen prepared in the first excited state (\(n=2\)).
Solution
Ionisation requires reaching \(E=0\): \(-E_2=E_{\mathrm R}/4=13.606/4=3.40\ \mathrm{eV}\).
State the number of radial nodes of the \(3s\), \(3p\) and \(3d\) states, and the total node count of each.
Solution
Radial nodes \(=n-l-1\): \(3s\ (l=0)\to2\); \(3p\ (l=1)\to1\); \(3d\ (l=2)\to0\). The total (radial + angular) node count is \(n-1=2\) for all three: \(3s=2+0\), \(3p=1+1\), \(3d=0+2\).
Using \(\langle 1/r\rangle_{nl}=1/(n^2 a_0)\), evaluate \(\langle V\rangle\) for the ground state and verify the virial relation \(\langle V\rangle=2E_n\).
Solution
\(\langle V\rangle=-\dfrac{e^2}{4\pi\epsilon_0}\Big\langle\dfrac1r\Big\rangle=-\dfrac{e^2}{4\pi\epsilon_0 a_0}\) for \(n=1\). Since \(a_0=4\pi\epsilon_0\hbar^2/\mu e^2\), \(\dfrac{e^2}{4\pi\epsilon_0 a_0}=\dfrac{\mu e^4}{(4\pi\epsilon_0)^2\hbar^2}=2E_{\mathrm R}=27.2\ \mathrm{eV}\). Thus \(\langle V\rangle=-27.2\ \mathrm{eV}=2E_1\) since \(E_1=-13.6\ \mathrm{eV}\); consistent with the virial theorem, and \(\langle T\rangle=E_1-\langle V\rangle=+13.6\ \mathrm{eV}=-E_1\).