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Chapter 6 • Theory & Derivations

The Hydrogen Atom & Central Force Potentials

Three-dimensional quantum mechanics in spherical coordinates: central force reduction, orbital angular momentum algebra and spherical harmonics $Y_l^m(\theta,\phi)$, the radial Schrödinger equation, associated Laguerre polynomials, quantum numbers $(n, l, m)$, energy degeneracies, and radial probability distributions.

§6.1 Schrödinger Equation in Spherical Coordinates

Central Potential and Reduced Mass Reduction

The hydrogen atom consists of two interacting particles: a proton ($m_p$, position $\mathbf{r}_p$) and an electron ($m_e$, position $\mathbf{r}_e$) interacting via the central Coulomb potential:

$$V(r) = -\frac{e^2}{4\pi \epsilon_0 r}, \quad r = |\mathbf{r}_e - \mathbf{r}_p|$$

Transforming to center-of-mass coordinates $\mathbf{R}$ and relative coordinates $\mathbf{r}$, the center-of-mass motion separates into a free particle equation, while the relative motion is described by an effective single-particle equation with reduced mass $\mu$:

$$\mu = \frac{m_e m_p}{m_e + m_p} \approx m_e \left( 1 - \frac{m_e}{m_p} \right) \approx 0.99945 m_e$$

In spherical coordinates $(r, \theta, \phi)$, where $x = r\sin\theta\cos\phi$, $y = r\sin\theta\sin\phi$, $z = r\cos\theta$, the Laplacian $\nabla^2$ is:

$$\nabla^2 = \frac{1}{r^2}\frac{\partial}{\partial r}\left( r^2 \frac{\partial}{\partial r} \right) + \frac{1}{r^2 \sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta} \right) + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2}{\partial \phi^2}$$

The Time-Independent Schrödinger Equation becomes:

$$-\frac{\hbar^2}{2\mu}\nabla^2 \psi(r,\theta,\phi) + V(r)\psi(r,\theta,\phi) = E\psi(r,\theta,\phi)$$

§6.2 Orbital Angular Momentum and Spherical Harmonics

Angular Momentum Operators

Classical angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$ translates into quantum mechanical differential operators:

$$\hat{L}_x = -i\hbar \left( y\frac{\partial}{\partial z} - z\frac{\partial}{\partial y} \right), \quad \hat{L}_y = -i\hbar \left( z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \right), \quad \hat{L}_z = -i\hbar \left( x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x} \right)$$

Fundamental commutation relations:

$$[\hat{L}_x, \hat{L}_y] = i\hbar \hat{L}_z, \quad [\hat{L}_y, \hat{L}_z] = i\hbar \hat{L}_x, \quad [\hat{L}_z, \hat{L}_x] = i\hbar \hat{L}_y$$

The total angular momentum operator $\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$ commutes with each individual component:

$$[\hat{L}^2, \hat{L}_z] = 0$$

In spherical coordinates:

$$\hat{L}_z = -i\hbar \frac{\partial}{\partial \phi}$$
$$\hat{L}^2 = -\hbar^2 \left[ \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\left( \sin\theta \frac{\partial}{\partial \theta} \right) + \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} \right]$$
Separation of Variables

Factoring the wavefunction into radial and angular components:

$$\psi(r,\theta,\phi) = R(r) Y(\theta,\phi)$$

The angular functions $Y_l^m(\theta,\phi)$ are the Spherical Harmonics, simultaneous eigenfunctions of $\hat{L}^2$ and $\hat{L}_z$:

$$\hat{L}^2 Y_l^m(\theta,\phi) = l(l+1)\hbar^2 Y_l^m(\theta,\phi), \quad l \in \{0, 1, 2, \dots\}$$
$$\hat{L}_z Y_l^m(\theta,\phi) = m_l \hbar Y_l^m(\theta,\phi), \quad m_l \in \{-l, -l+1, \dots, +l\}$$

Explicitly, $Y_l^m(\theta,\phi) = \sqrt{\frac{(2l+1)}{4\pi}\frac{(l-m)!}{(l+m)!}} P_l^m(\cos\theta) e^{i m \phi}$, where $P_l^m$ are Associated Legendre polynomials.

§6.3 The Radial Equation and Energy Eigenvalues

Substituting the angular eigenvalue $l(l+1)\hbar^2$ into the full Schrödinger equation leaves the Radial Differential Equation:

$$\frac{1}{r^2}\frac{d}{dr}\left( r^2 \frac{dR}{dr} \right) + \frac{2\mu}{\hbar^2}\left[ E - V(r) - \frac{l(l+1)\hbar^2}{2\mu r^2} \right]R = 0$$

The effective potential includes an outward centrifugal barrier:

$$V_{\text{eff}}(r) = -\frac{e^2}{4\pi \epsilon_0 r} + \frac{l(l+1)\hbar^2}{2\mu r^2}$$
Bound State Solutions ($E < 0$) and Quantized Energies

Solving the radial equation via power series around $r=0$ and extracting the asymptotic behavior at $r \to \infty$ yields solutions in terms of Associated Laguerre Polynomials $L_{n-l-1}^{2l+1}$:

$$R_{nl}(r) = -\sqrt{\left(\frac{2}{n a_0}\right)^3 \frac{(n-l-1)!}{2n[(n+l)!]^3}} e^{-r/n a_0} \left( \frac{2r}{n a_0} \right)^l L_{n-l-1}^{2l+1}\left( \frac{2r}{n a_0} \right)$$

The solutions are normalizable if and only if the principal quantum number $n$ satisfies:

$$n = 1, 2, 3, \dots \qquad \text{with } l \in \{0, 1, 2, \dots, n-1\}$$

This gives the Bohr Energy Levels:

$$E_n = -\frac{\mu e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} \frac{1}{n^2} = -\frac{13.6 \text{ eV}}{n^2}$$
Degeneracy

The energy depends exclusively on $n$. For a given $n$:

  • $l$ ranges from $0$ to $n-1$ ($n$ distinct orbital angular momenta).
  • For each $l$, $m_l$ ranges from $-l$ to $+l$ ($2l+1$ values).

Total orbital degeneracy:

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

Including the two electron spin states ($m_s = \pm 1/2$), the total degeneracy is $2n^2$.

Radial Probability Density $P(r)$

The probability of finding the electron between radius $r$ and $r+dr$ integrated over all angles is:

$$P(r) dr = r^2 |R_{nl}(r)|^2 dr$$

For the ground state ($1s$: $n=1, l=0$):

$$R_{10}(r) = \frac{2}{a_0^{3/2}} e^{-r/a_0} \implies P(r) = \frac{4}{a_0^3} r^2 e^{-2r/a_0}$$

The maximum of $P(r)$ occurs at $\frac{dP}{dr} = 0 \implies r_{\text{max}} = a_0 = 0.529 \text{ Å}$, matching the Bohr radius.

See Simulation 6.1 below to visualize radial distribution curves $P(r)$ and 2D quantum electron cloud slices for $1s, 2s, 2p,$ and $3d$ orbitals.

📝 Chapter Worked Examples & Exercises

Complete derivations & analytical proofs
Hard Example 6.1: Expectation Values and <1/r> for Hydrogen Ground State

Using the normalized hydrogen ground-state wavefunction $\psi_{100}(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$, calculate:\n(a) The expectation value of the electron-nuclear distance $\langle r \rangle$.\n(b) The expectation value $\langle 1/r \rangle$.\n(c) The average potential energy $\langle V \rangle$ and kinetic energy $\langle T \rangle$, verifying the quantum Virial theorem for Coulomb potentials ($2\langle T \rangle + \langle V \rangle = 0$).

Step 1: Calculate and <1/r>
$$\langle r \rangle = \int_0^\infty r P(r) dr = \frac{4}{a_0^3} \int_0^\infty r^3 e^{-2r/a_0} dr $$ $$\text{Using } \int_0^\infty x^n e^{-a x} dx = \frac{n!}{a^{n+1}} \implies \langle r \rangle = \frac{4}{a_0^3} \frac{3!}{(2/a_0)^4} = \frac{4 \times 6}{16} a_0 = \frac{3}{2} a_0 $$ $$\langle 1/r \rangle = \frac{4}{a_0^3} \int_0^\infty r e^{-2r/a_0} dr = \frac{4}{a_0^3} \frac{1!}{(2/a_0)^2} = \frac{1}{a_0}$$

Notice that $\langle r \rangle = 1.5 a_0$, whereas the most probable distance is $r_{\text{mp}} = a_0$. The radial probability distribution has a long exponential tail extending outward.

Step 2: Verify the Coulomb Virial Theorem
$$\langle V \rangle = -\frac{e^2}{4\pi \epsilon_0} \left\langle \frac{1}{r} \right\rangle = -\frac{e^2}{4\pi \epsilon_0 a_0} = 2 E_1 = -27.2 \text{ eV} $$ $$\langle T \rangle = E_1 - \langle V \rangle = -13.6 \text{ eV} - (-27.2 \text{ eV}) = +13.6 \text{ eV} = -E_1 $$ $$2\langle T \rangle + \langle V \rangle = 2(+13.6) + (-27.2) = 0$$

For any homogeneous potential of degree $k$ ($V \propto r^k$), the Virial theorem states $2\langle T \rangle = k\langle V \rangle$. For the Coulomb potential, $k = -1$, yielding $2\langle T \rangle = -\langle V \rangle$, exactly confirmed.

An electron is accelerated from rest through an electrostatic potential difference of $V = 150 \text{ V}$.\n(a) Determine its de Broglie wavelength using non-relativistic mechanics.\n(b) At what accelerating potential does the relativistic correction to the de Broglie wavelength exceed $1\%$?
Step 1: Non-relativistic Calculation
$$K = e V = 150 \text{ eV} = 150 \times 1.602 \times 10^{-19} \text{ J} = 2.403 \times 10^{-17} \text{ J} $$ $$p = \sqrt{2 m_e K} = \sqrt{2(9.109 \times 10^{-31} \text{ kg})(2.403 \times 10^{-17} \text{ J})} = 6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s} $$ $$\lambda = \frac{h}{p} = \frac{6.626 \times 10^{-34} \text{ J}\cdot\text{s}}{6.617 \times 10^{-24} \text{ kg}\cdot\text{m/s}} = 1.001 \times 10^{-10} \text{ m} = 1.001 \text{ Å}$$
For quick calculations, note that $\lambda = \sqrt{\frac{150}{V}} \text{ Å}$. For $V = 150 \text{ V}$, $\lambda = 1.00 \text{ Å}$, which corresponds to typical atomic crystal lattice spacings.
Step 2: Relativistic Condition
$$E^2 = p^2 c^2 + m_0^2 c^4 \implies p = \frac{1}{c}\sqrt{K(K + 2m_0 c^2)} $$ $$\lambda_{\text{rel}} = \frac{h c}{\sqrt{K(K + 2m_0 c^2)}} = \frac{\lambda_{\text{class}}}{\sqrt{1 + \frac{K}{2m_0 c^2}}} \approx \lambda_{\text{class}}\left(1 - \frac{K}{4 m_0 c^2}\right) $$ $$\frac{\Delta \lambda}{\lambda} \approx \frac{K}{4 m_0 c^2} \ge 0.01 \implies K \ge 0.04 m_0 c^2 = 0.04 (511 \text{ keV}) \approx 20.44 \text{ keV}$$
When accelerating potentials exceed roughly $20 \text{ kV}$ (typical in transmission electron microscopes), relativistic momentum corrections become necessary.