Hydrogen Atom

The hydrogen atom is the canonical exactly solvable three-dimensional bound-state problem. In the nonrelativistic approximation, the electron and proton interact through an attractive Coulomb potential. After separating the center-of-mass motion, the relative coordinate obeys a one-body central-potential problem.

Hydrogen is important because it is both concrete and structurally rich: it introduces principal, orbital, and magnetic quantum numbers; it connects spherical harmonics to radial special functions; and it exposes degeneracies that later corrections split.

The model here is the spinless nonrelativistic point-Coulomb Hamiltonian with reduced mass. Fine structure, hyperfine structure, the Lamb shift, nuclear size, and external fields are corrections to this starting point.

Helpful background. Central Potentials to separate rotation-invariant Hamiltonians into angular sectors.

Two-Body Reduction

Let the electron and proton masses be $m_e$ and $m_p$. The reduced mass is

$$\mu = \frac{m_em_p}{m_e+m_p}.$$

After separating the free center-of-mass motion, the relative-coordinate Hamiltonian is

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

The stationary Schrödinger equation is

$$\hat H\psi(\mathbf r)=E\psi(\mathbf r).$$

It is useful to introduce the reduced-mass Bohr radius

$$a_0 = \frac{4\pi\epsilon_0\hbar^2}{\mu e^2}.$$

If one uses the electron mass instead of $\mu$, this is the infinite-proton-mass approximation. The numerical difference is small for hydrogen but conceptually important.

Separation and Quantum Numbers

Because the potential depends only on $r$, write

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

The angular quantum numbers satisfy

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

The principal quantum number is

$$n=1,2,3,\ldots.$$

The angular meaning of the $n,\ell,m$ labels, parity, and shell multiplets is developed in Hydrogen Atom Angular Structure.

Within the spinless bound-state sector, the commuting observables $H$, $L^2$, and $L_z$ distinguish the states by $n$, $\ell$, and $m$. Their role as a sector-dependent complete set is explained in Complete Sets of Commuting Observables.

The radial equation is the Coulomb specialization of the general radial equation:

$$-\frac{\hbar^2}{2\mu}\frac{d^2u}{dr^2} + \left[ -\frac{e^2}{4\pi\epsilon_0r} + \frac{\hbar^2\ell(\ell+1)}{2\mu r^2} \right]u =Eu, \qquad u=rR.$$

Normalizable bound-state solutions exist only for the allowed combinations of $n$ and $\ell$ above. The angular part is supplied by spherical harmonics, and the radial polynomial part is supplied by Laguerre polynomials.

Why the bound energies are discrete

Let $E<0$, $\kappa=\sqrt{-2\mu E}/\hbar$, and $\rho=2\kappa r$. The regular origin behavior suggests $u=\rho^{\ell+1}e^{-\rho/2}v(\rho)$: the first factor handles the origin and the exponential handles infinity. Substitution into the radial equation leaves

$$\rho v''+(2\ell+2-\rho)v'+(\lambda-\ell-1)v=0, \qquad \lambda=\frac{1}{\kappa a_0}.$$

For $v=\sum_{j=0}^\infty c_j\rho^j$, equal powers give

$$c_{j+1}=\frac{j+\ell+1-\lambda}{(j+1)(j+2\ell+2)}c_j.$$

A generic regular series has an exponentially growing part at large $\rho$, defeating the prefactor $e^{-\rho/2}$. To obtain a normalizable state, the series must terminate at some integer $j=n_r\geq0$. Hence $\lambda=n_r+\ell+1\equiv n$, so $\kappa=1/(na_0)$. This gives both the discrete energies below and the allowed range $0\leq\ell\leq n-1$. The terminating function is a generalized Laguerre polynomial.

Energy Spectrum

The bound-state energies are

$$E_n = -\frac{\mu e^4}{2(4\pi\epsilon_0)^2\hbar^2} \frac{1}{n^2} = -\frac{\hbar^2}{2\mu a_0^2} \frac{1}{n^2}.$$

For ordinary hydrogen this is approximately

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

with the reduced-mass correction included in high-precision values.

The energy depends only on $n$, not on $\ell$ or $m$, in the ideal nonrelativistic Coulomb problem. The $m$ degeneracy follows from rotational symmetry for any central potential. The additional $\ell$ degeneracy is special to the Coulomb potential and is explained in Degeneracy of the Hydrogen Atom.

For a fixed $n$, the number of spatial bound states is

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

If spin is included without spin-dependent interactions, this doubles to $2n^2$. Fine structure, hyperfine structure, Lamb-shift physics, and external fields split this ideal degeneracy.

Radial Wavefunctions

The detailed reference for the functions in this section is Radial Wavefunctions.

With

$$\rho=\frac{2r}{na_0},$$

one common normalized convention is

$$R_{n\ell}(r) = \sqrt{ \left( \frac{2}{na_0} \right)^3 \frac{(n-\ell-1)!}{2n[(n+\ell)!]} } e^{-\rho/2} \rho^\ell L_{n-\ell-1}^{(2\ell+1)}(\rho).$$

The full wavefunction is $R_{n\ell}Y_\ell^m$. The normalization convention is

$$\int_0^\infty \lvert R_{n\ell}(r)\rvert^2r^2\,dr =1, \qquad \int_{S^2} \lvert Y_\ell^m\rvert^2\,d\Omega =1.$$

Equivalently, if $u_{n\ell}=rR_{n\ell}$, then

$$\int_0^\infty \lvert u_{n\ell}(r)\rvert^2\,dr =1.$$

The ground state has $n=1$, $\ell=0$, and $m=0$. Its radial and full wavefunctions are

$$R_{10}(r) = 2a_0^{-3/2}e^{-r/a_0}, \qquad \psi_{100}(r,\theta,\phi) = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}.$$

The wavefunction is largest at the origin, but the radial probability density includes the volume factor $r^2$:

$$P_{10}(r) = \lvert R_{10}(r)\rvert^2r^2 = \frac{4r^2}{a_0^3}e^{-2r/a_0}.$$

This radial probability density is maximal at $r=a_0$. This is one precise sense in which the Bohr radius sets the size of the ground-state atom.

Orbitals and Nodes

Hydrogenic orbitals are stationary one-electron wavefunctions, not classical electron trajectories. The labels $s,p,d,\ldots$ correspond to $\ell=0,1,2,\ldots$. Complex orbitals are natural angular-momentum eigenstates; real orbitals are particular linear combinations within a degenerate $m$ subspace. The notation, real-versus-complex basis choice, nodal surfaces, and visualization cautions are developed in Atomic Orbitals.

For fixed $n$ and $\ell$, the number of radial nodes is

$$n-\ell-1.$$

Angular nodes come from the spherical harmonics. The total nodal structure is therefore a product of radial and angular information.

Physical Interpretation

The Coulomb potential supplies an infinite ladder of bound states accumulating at $E=0$ from below. States with larger $n$ are less tightly bound and extend to larger radii. The continuum above $E=0$ describes ionized scattering states; see Continuum States of the Coulomb Problem for the positive-energy sector.

The quantum numbers have distinct meanings:

• $n$ controls the energy and overall radial scale in the ideal Coulomb problem.
• $\ell$ controls orbital angular momentum and the centrifugal barrier.
• $m$ controls the projection of orbital angular momentum along the chosen $z$ axis.

The $z$ axis is a convention unless an external field or measurement apparatus selects a direction.

Coulomb virial and momentum checks

For a stationary bound state, the virial theorem for an inverse-radius potential gives

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

Combining this with $E_n=\langle T\rangle+\langle V\rangle$ fixes the separate kinetic and potential expectations:

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

The same identities hold for a hydrogenic ion. Using its energy $E_n=-\mu c^2(Z\alpha_{\rm fs})^2/(2n^2)$ gives

$$v_{\rm rms}=\frac{\sqrt{\langle p^2\rangle}}\mu =\frac{Z\alpha_{\rm fs}c}{n}, \qquad \alpha_{\rm fs}=\frac{e^2}{4\pi\epsilon_0\hbar c}.$$

Here the speed is that of the relative coordinate, and ordinary hydrogen has $Z=1$. The scale makes the nonrelativistic condition visible: low-lying states require $Z\alpha_{\rm fs}\ll1$. These virial identities also test numerical bound states independently of an energy comparison. A computed energy can look accurate while the separately evaluated kinetic and potential terms remain unconverged.

Common Mistakes

• Using $m_e$ without noticing the reduced-mass approximation.
• Treating orbitals as paths followed by an electron.
• Forgetting the $r^2$ measure when interpreting radial probabilities.
• Assuming the ideal $n^2$ degeneracy survives all physical corrections.
• Confusing the radial wavefunction $R_{n\ell}$ with the radial probability density $\lvert R_{n\ell}\rvert^2r^2$.
• Thinking that $m$ changes the radial equation in a rotationally invariant Coulomb problem.

Exercises

1. Normalize the ground-state radial wavefunction $R_{10}(r)=2a_0^{-3/2}e^{-r/a_0}$ using the radial measure.

Solution

Compute

$$\int_0^\infty \lvert R_{10}(r)\rvert^2r^2\,dr = \frac{4}{a_0^3} \int_0^\infty r^2e^{-2r/a_0}\,dr.$$

Using

$$\int_0^\infty r^2e^{-\alpha r}\,dr = \frac{2}{\alpha^3},$$

with $\alpha=2/a_0$, the integral becomes

$$\frac{4}{a_0^3} \frac{2}{(2/a_0)^3} =1.$$

2. Show that the spatial degeneracy of a fixed principal shell is $n^2$.

Solution

For fixed $n$, the allowed orbital quantum numbers are $\ell=0,1,\ldots,n-1$. For each $\ell$, there are $2\ell+1$ allowed values of $m$. Therefore

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

3. Find the most probable radius in the ground state.

Solution

The radial probability density is

$$P_{10}(r) = \frac{4r^2}{a_0^3}e^{-2r/a_0}.$$

Its derivative is proportional to

$$\frac{d}{dr} \left( r^2e^{-2r/a_0} \right) = 2re^{-2r/a_0} \left( 1-\frac{r}{a_0} \right).$$

The critical points are $r=0$ and $r=a_0$. The nonzero maximum occurs at

$$r=a_0.$$

4. Starting from the two-particle kinetic energy, show why the internal Hamiltonian contains the reduced mass. Allow a nucleus of mass $M_N$ and charge $+Ze$; hydrogen is $M_N=m_p$, $Z=1$.

Solution

Let $M=m_e+M_N$ and choose the center-of-mass and relative variables

$$\mathbf R=\frac{m_e\mathbf r_e+M_N\mathbf r_N}{M}, \qquad \mathbf r=\mathbf r_e-\mathbf r_N,$$ $$\mathbf P=\mathbf p_e+\mathbf p_N, \qquad \mathbf p=\frac{M_N\mathbf p_e-m_e\mathbf p_N}{M}.$$

The inverse momentum relations are $\mathbf p_e=(m_e/M)\mathbf P+\mathbf p$ and $\mathbf p_N=(M_N/M)\mathbf P-\mathbf p$. Expanding the kinetic energy cancels the cross terms:

$$\frac{\mathbf p_e^2}{2m_e}+\frac{\mathbf p_N^2}{2M_N} =\frac{\mathbf P^2}{2M}+\frac{\mathbf p^2}{2\mu}, \qquad \frac1\mu=\frac1{m_e}+\frac1{M_N}.$$

Since the Coulomb potential depends only on $r$, the total Hamiltonian separates into free center-of-mass motion and

$$H_{\rm rel}=\frac{\mathbf p^2}{2\mu}-\frac{Ze^2}{4\pi\epsilon_0r}.$$

The momentum identity also holds for operators on the common domain of the kinetic terms. The internal eigenvalue $E_n$ excludes the free center-of-mass contribution $P^2/(2M)$.

5. Find the gross Lyman-alpha photon energy for hydrogen emission from $n_i=2$ to $n_f=1$, neglecting photon recoil and fine, hyperfine, and radiative splittings.

Solution

Write $E_n=-\operatorname{Ry}_\mu/n^2$, where $\operatorname{Ry}_\mu$ is the reduced-mass Rydberg energy. Then

$$h\nu=E_2-E_1=\operatorname{Ry}_\mu\left(1-\frac14\right) =\frac34\operatorname{Ry}_\mu\simeq10.2\,\mathrm{eV}.$$

The numerical estimate uses $\operatorname{Ry}_\mu\simeq13.6\,\mathrm{eV}$. It is a gross-structure energy, not a prediction of each resolved component of the observed line. The spectrum reference lists the other spectral series and their limits.

6. Use the Coulomb virial theorem to determine $\langle T\rangle$, $\langle V\rangle$, and $\langle p^2\rangle$ from a bound-state energy $E_n$. Check their signs.

Solution

From $2\langle T\rangle=-\langle V\rangle$ and $E_n=\langle T\rangle+\langle V\rangle$, one obtains

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

Because $E_n<0$, the kinetic energy and squared momentum are positive, while the attractive potential expectation is negative. The potential magnitude is twice the kinetic energy, so their sum remains negative.

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed., Pergamon, 1977.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.