Unit 4: Hydrogen Atom, Central Force Dynamics & Hydrogenic Orbitals
Rigorous quantum mechanical formulation of central force dynamics and one-electron atomic systems. Separation of center-of-mass and relative coordinates in spherical coordinates, angular momentum eigenvalues and spherical harmonics, radial Schrödinger equation solution via confluent hypergeometric functions and Associated Laguerre polynomials, hydrogenic orbital structures, radial probability distributions, spin-orbit coupling, fine structure relativistic corrections, and magnetic Zeeman effects.
§4.1 Central Force Hamiltonian & Spherical Coordinate Reduction
The hydrogen atom consists of a nucleus of charge \(+Z e\) (mass \(M\)) and an electron of charge \(-e\) (mass \(m_e\)) interacting via the attractive Coulomb electrostatic potential.
Separation of Center-of-Mass and Relative Coordinates
The total classical and quantum Hamiltonian for the two-body system is:
By introducing the center-of-mass vector \(\mathbf{R} = \frac{M \mathbf{r}_N + m_e \mathbf{r}_e}{M + m_e}\) with total mass \(M_{\text{tot}} = M + m_e\), and the relative position vector \(\mathbf{r} = \mathbf{r}_e - \mathbf{r}_N\) with reduced mass:
the kinetic energy operator factors exactly into:
where the Coulomb potential \(V(r) = -\frac{Z e^2}{4\pi \varepsilon_0 r}\) depends strictly on the scalar radial distance \(r = |\mathbf{r}|\), establishing central force symmetry. The center-of-mass motion describes a free particle of mass \(M_{\text{tot}}\), while the internal electronic dynamics are governed by the relative Hamiltonian:
Spherical Polar Coordinate Representation
In spherical polar coordinates \((r, \theta, \phi)\), where \(x = r \sin\theta \cos\phi\), \(y = r \sin\theta \sin\phi\), \(z = r \cos\theta\), the Laplacian operator transforms to:
Notice that the angular derivative operators are precisely related to the square of the orbital angular momentum operator \(\hat{L}^2\):
Consequently, the relative Hamiltonian is written compactly as:
Because \(\hat{H}\), \(\hat{L}^2\), and \(\hat{L}_z\) mutually commute:
they possess a complete simultaneous orthonormal eigenbasis. The full stationary wavefunctions \(\psi(r, \theta, \phi)\) separate into a radial function and angular Spherical Harmonics:
Substituting this separable product and utilizing \(\hat{L}^2 Y_l^m(\theta, \phi) = \hbar^2 l(l + 1) Y_l^m(\theta, \phi)\), the angular dependence cancels identically, yielding the Radial Schrödinger Equation:
The term \(\frac{\hbar^2 l(l+1)}{2\mu r^2}\) represents a repulsive centrifugal barrier arising from orbital angular momentum.
§4.2 Associated Laguerre Polynomials & Exact Hydrogenic Eigenfunctions
To solve the radial equation analytically for bound states (\(E < 0\)), we define the reduced radial function \(u(r) = r R(r)\).
Dimensionless Radial Equation
Substituting \(u(r)\) converts the equation into:
We introduce the parameter \(\kappa = \sqrt{-2\mu E}/\hbar\) and the dimensionless variable:
In terms of \(\rho\), the differential equation becomes:
where \(\lambda = \frac{Z e^2}{4\pi \varepsilon_0 \hbar} \sqrt{\frac{\mu}{-2 E}}\).
Asymptotic Analysis
1. As \(\rho \rightarrow \infty\): The equation reduces to \(\frac{d^2 u}{d\rho^2} - \frac{1}{4} u = 0\), whose physically well-behaved normalizable solution is \(u(\rho) \sim e^{-\rho / 2}\).
2. As \(\rho \rightarrow 0\): The dominant term is \(\frac{d^2 u}{d\rho^2} - \frac{l(l+1)}{\rho^2} u = 0\). The ansatz \(u(\rho) \sim \rho^s\) requires \(s(s - 1) = l(l + 1)\), which has roots \(s = l + 1\) and \(s = -l\). Since \(u(0)\) must vanish for \(R(0)\) to remain finite, we retain \(s = l + 1\). Thus, \(u(\rho) \sim \rho^{l+1}\), which implies \(R(\rho) \sim \rho^l\).
Series Expansion & Associated Laguerre Polynomials
Factoring out the asymptotic behavior, we set:
Substituting into the radial equation yields Kummer's confluent hypergeometric equation for \(v(\rho)\):
For the power series expansion \(v(\rho) = \sum_{k=0}^\infty a_k \rho^k\) to terminate into a polynomial of degree \(n_r\) (preventing \(v(\rho)\) from diverging as \(e^\rho\) at large distances), the numerator in the recurrence relation must vanish:
Defining the principal quantum number \(n = n_r + l + 1\), we have \(\lambda = n\), where \(n \in \{1, 2, 3, \dots\}\) and \(l \in \{0, 1, \dots, n-1\}\). The polynomial solutions are proportional to the Associated Laguerre Polynomials \(L_{n-l-1}^{2l+1}(\rho)\):
Normalized Hydrogenic Radial Wavefunctions
The fully normalized radial eigenfunction is:
where \(a_\mu = \frac{4\pi \varepsilon_0 \hbar^2}{\mu e^2} \approx a_0 = 0.529177 \text{ \AA}\) is the modified Bohr radius and \(\rho = \frac{2 Z r}{n a_\mu}\). Explicit formulas for the lowest states:
- 1s (\(n=1, l=0\)): \(R_{10}(r) = 2 \left(\frac{Z}{a_0}\right)^{3/2} e^{-Z r / a_0}\)
- 2s (\(n=2, l=0\)): \(R_{20}(r) = \frac{1}{\sqrt{2}} \left(\frac{Z}{a_0}\right)^{3/2} \left(1 - \frac{Z r}{2 a_0}\right) e^{-Z r / 2 a_0}\)
- 2p (\(n=2, l=1\)): \(R_{21}(r) = \frac{1}{2\sqrt{6}} \left(\frac{Z}{a_0}\right)^{3/2} \left(\frac{Z r}{a_0}\right) e^{-Z r / 2 a_0}\)
- 3s (\(n=3, l=0\)): \(R_{30}(r) = \frac{2}{81\sqrt{3}} \left(\frac{Z}{a_0}\right)^{3/2} \left(27 - 18\frac{Zr}{a_0} + 2\frac{Z^2 r^2}{a_0^2}\right) e^{-Zr/3a_0}\)
§4.3 Hydrogenic Energy Spectrum, Degeneracy & The Rydberg Formula
From the polynomial termination condition \(\lambda = n\), the bound-state energy eigenvalues are obtained directly:
where \(R_y = \frac{\mu e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2} \approx 13.60569\text{ eV} = 1\text{ Ry} = \frac{1}{2} E_h\) (where \(E_h = 27.2114\text{ eV}\) is 1 Hartree).
Energy Levels and Quantum Numbers
The electronic energy depends strictly on the principal quantum number \(n\) and is independent of \(l\) and \(m\). This degeneracy is remarkable:
1. Principal quantum number \(n\): \(n \in \{1, 2, 3, \dots\}\) governs overall orbital scale and energy.
2. Azimuthal (orbital) quantum number \(l\): \(l \in \{0, 1, 2, \dots, n - 1\}\) designates orbital angular momentum magnitude \(|\mathbf{L}| = \hbar \sqrt{l(l+1)}\).
3. Magnetic quantum number \(m\): \(m \in \{-l, -l+1, \dots, +l\}\) governs the spatial projection \(L_z = m\hbar\).
Degeneracy Analysis
For a given \(n\), the orbital angular momentum can take \(n\) values (\(l = 0, 1, \dots, n-1\)). For each \(l\), there are \(2l + 1\) distinct \(m\) projections. The total spatial orbital degeneracy \(g_n\) is:
Including electron spin degeneracy (\(m_s = \pm 1/2\)), the total state degeneracy is \(2 n^2\):
- \(n = 1\): \(1^2 = 1\) orbital (\(1s\)) \(\rightarrow\) 2 quantum states
- \(n = 2\): \(2^2 = 4\) orbitals (\(2s, 2p_x, 2p_y, 2p_z\)) \(\rightarrow\) 8 quantum states
- \(n = 3\): \(3^2 = 9\) orbitals (\(3s, 3p (3), 3d (5)\)) \(\rightarrow\) 18 quantum states
The accidental degeneracy across different \(l\) values for a given \(n\) is a consequence of the higher dynamical symmetry group \(SO(4)\) generated by the conservation of the Laplace-Runge-Lenz vector in a pure \(1/r\) Coulomb potential.
The Rydberg Formula for Optical Transitions
A radiative transition between an upper state \(n_2\) and a lower state \(n_1\) involves photon emission or absorption with photon energy:
The transition wavenumber \(\tilde{\nu} = 1/\lambda\) is given by the general Rydberg formula:
where \(R_\infty = \frac{m_e e^4}{8 \varepsilon_0^2 h^3 c} \approx 109737.316\text{ cm}^{-1}\), and \(R_H = R_\infty \frac{M_p}{M_p + m_e} \approx 109677.583\text{ cm}^{-1}\) for atomic hydrogen. The classic spectral series of hydrogen:
- Lyman Series (\(n_1 = 1, n_2 \ge 2\)): Ultraviolet range (\(121.6\text{ nm} \rightarrow 91.2\text{ nm}\))
- Balmer Series (\(n_1 = 2, n_2 \ge 3\)): Visible range (\(656.3\text{ nm} \text{ [H}\alpha\text{]} \rightarrow 364.6\text{ nm}\))
- Paschen Series (\(n_1 = 3, n_2 \ge 4\)): Near-infrared range (\(1875\text{ nm} \rightarrow 820.4\text{ nm}\))
- Brackett Series (\(n_1 = 4\)) and Pfund Series (\(n_1 = 5\)): Mid-infrared range.
§4.4 Radial Probability Distribution Functions: Shell Structure & Radial Nodes
The physical probability of locating the electron inside an infinitesimal volume element \(d\tau = r^2 \sin\theta \, dr \, d\theta \, d\phi\) is:
The Radial Distribution Function \(P(r)\)
To find the probability of finding the electron at a radial distance between \(r\) and \(r + dr\) regardless of angle, we integrate over all angles \(\theta \in [0, \pi]\) and \(\phi \in [0, 2\pi]\):
Because the Spherical Harmonics are normalized (\(\int |Y_l^m|^2 d\Omega = 1\)), we obtain the Radial Probability Density:
Notice the crucial difference:
- The probability density per unit volume at the nucleus (\(r = 0\)) is \([R_{n0}(0)]^2 > 0\) for all \(s\)-orbitals (\(l = 0\)).
- The radial probability density \(P(r)\) vanishes at the nucleus: \(P_{n l}(0) = 0^2 [R_{n l}(0)]^2 = 0\) for all states because the spherical shell volume \(4\pi r^2 dr\) shrinks to zero as \(r \rightarrow 0\).
Most Probable Radius vs Mean Radius \(\langle r \rangle\)
For the ground state of hydrogen (\(1s, Z = 1\)):
To find the most probable radius \(r_{\text{mp}}\), we maximize \(P_{10}(r)\):
The most probable distance of the electron from the proton in the ground state matches the historical first Bohr radius! In contrast, the expectation value (average distance) \(\langle r \rangle\) is:
Because the exponential tail extends outward, \(\langle r \rangle > r_{\text{mp}}\). In general:
Node Topology
The total number of nodes in an atomic orbital wavefunction is \(n - 1\):
1. Radial nodes: Points where \(R_{n l}(r) = 0\) (excluding \(r = 0\) and \(r = \infty\)). The number of radial nodes is:
2. Angular nodal surfaces: Conical or planar surfaces where \(Y_l^m(\theta, \phi) = 0\). The number of angular nodes is:
3. Total nodes:
Examples:
- \(1s\): \(n=1, l=0 \implies 0\) radial nodes, 0 angular nodes.
- \(2s\): \(n=2, l=0 \implies 1\) radial node (\(r = 2 a_0 / Z\)), 0 angular nodes.
- \(2p\): \(n=2, l=1 \implies 0\) radial nodes, 1 angular nodal plane (e.g., \(xy\)-plane for \(p_z\)).
- \(3s\): \(n=3, l=0 \implies 2\) radial nodes, 0 angular nodes.
- \(3d\): \(n=3, l=2 \implies 0\) radial nodes, 2 angular nodal surfaces.
§4.5 Angular Wavefunctions, Real Spherical Harmonics & Orbital Shapes
The angular parts of hydrogenic wavefunctions are the Spherical Harmonics \(Y_l^m(\theta, \phi)\), normalized over the unit sphere:
with \(Y_l^{-m}(\theta, \phi) = (-1)^m [Y_l^m(\theta, \phi)]^*\).
Real Orbitals in Chemistry
In physical chemistry, complex orbitals with \(m = \pm 1, \pm 2\) are generally replaced by real linear combinations that point along Cartesian coordinate axes. Since the Hamiltonian is degenerate in \(m\), any linear combination of degenerate eigenfunctions is also an exact eigenfunction.
\(p\)-Orbitals (\(l = 1\)):
- \(p_z = Y_1^0 = \sqrt{\frac{3}{4\pi}} \cos\theta = \sqrt{\frac{3}{4\pi}} \frac{z}{r}\)
- \(p_x = \frac{1}{\sqrt{2}} (-Y_1^1 + Y_1^{-1}) = \sqrt{\frac{3}{4\pi}} \sin\theta \cos\phi = \sqrt{\frac{3}{4\pi}} \frac{x}{r}\)
- \(p_y = \frac{i}{\sqrt{2}} (Y_1^1 + Y_1^{-1}) = \sqrt{\frac{3}{4\pi}} \sin\theta \sin\phi = \sqrt{\frac{3}{4\pi}} \frac{y}{r}\)
Each \(p\)-orbital consists of two lobes with opposite signs (phases) separated by an angular nodal plane through the nucleus:
- \(p_z\) has nodal plane \(z = 0\) (\(xy\)-plane).
- \(p_x\) has nodal plane \(x = 0\) (\(yz\)-plane).
- \(p_y\) has nodal plane \(y = 0\) (\(xz\)-plane).
\(d\)-Orbitals (\(l = 2\)):
There are five real \(d\)-orbitals formed from \(l = 2\) Spherical Harmonics:
- \(d_{z^2} = Y_2^0 = \sqrt{\frac{5}{16\pi}} (3\cos^2\theta - 1) = \sqrt{\frac{5}{16\pi}} \frac{3z^2 - r^2}{r^2}\) (two conical nodes at \(\cos\theta = \pm 1/\sqrt{3} \approx 54.74^\circ\))
- \(d_{xz} = \frac{1}{\sqrt{2}} (-Y_2^1 + Y_2^{-1}) = \sqrt{\frac{15}{4\pi}} \sin\theta \cos\theta \cos\phi = \sqrt{\frac{15}{4\pi}} \frac{xz}{r^2}\)
- \(d_{yz} = \frac{i}{\sqrt{2}} (Y_2^1 + Y_2^{-1}) = \sqrt{\frac{15}{4\pi}} \sin\theta \cos\theta \sin\phi = \sqrt{\frac{15}{4\pi}} \frac{yz}{r^2}\)
- \(d_{xy} = \frac{i}{\sqrt{2}} (-Y_2^2 + Y_2^{-2}) = \sqrt{\frac{15}{4\pi}} \sin^2\theta \sin\phi \cos\phi = \sqrt{\frac{15}{4\pi}} \frac{xy}{r^2}\)
- \(d_{x^2-y^2} = \frac{1}{\sqrt{2}} (Y_2^2 + Y_2^{-2}) = \sqrt{\frac{15}{16\pi}} \sin^2\theta \cos(2\phi) = \sqrt{\frac{15}{16\pi}} \frac{x^2 - y^2}{r^2}\)
Orbital Boundary Surfaces
In molecular chemistry, orbital shapes are conventionally visualized as 3D isosurfaces encompassing a specified probability (typically 90% or 95% of total electron density):
The spatial directional orientation of \(p\) and \(d\) orbitals governs chemical valence, hybridization (\(sp, sp^2, sp^3, d^2sp^3\)), crystal field splitting in transition metal complexes, and stereochemistry.
§4.6 Spin-Orbit Coupling, Fine Structure & Relativistic Dirac Corrections
In high-resolution spectroscopy, hydrogenic spectral lines reveal small splittings called fine structure, with energy differences of order \(\alpha^2 E_n \sim 10^{-4}\text{ eV}\), where \(\alpha = \frac{e^2}{4\pi\varepsilon_0 \hbar c} \approx \frac{1}{137.036}\) is the fine structure constant.
The Spin-Orbit Interaction Hamiltonian
From the rest frame of the orbiting electron, the positively charged nucleus circulates with velocity \(-\mathbf{v}\), creating an effective internal magnetic field \(\mathbf{B}_{\text{int}}\):
The electron possesses an intrinsic magnetic dipole moment \(\boldsymbol{\mu}_s = -g_s \frac{e}{2 m_e} \mathbf{S} \approx -\frac{e}{m_e} \mathbf{S}\). Incorporating the Thomas precession factor of \(1/2\) (due to the accelerating non-inertial reference frame of the electron), the spin-orbit Hamiltonian is:
Total Angular Momentum Coupling
Define total angular momentum \(\hat{\mathbf{J}} = \hat{\mathbf{L}} + \hat{\mathbf{S}}\). Squaring both sides:
The eigenvalues of \(\hat{\mathbf{L}} \cdot \hat{\mathbf{S}}\) in the coupled representation \(|j, m_j, l, s\rangle\) are:
where for a single electron \(s = 1/2\), so \(j = l + 1/2\) or \(j = l - 1/2\) (for \(l > 0\)).
Relativistic Corrections and Total Fine Structure
The full relativistic correction to order \(\alpha^2\) comprises three terms:
1. Relativistic kinetic energy correction: \(\hat{H}_{\text{rel}} = -\frac{\hat{p}^4}{8 m_e^3 c^2}\)
2. Spin-orbit coupling: \(\hat{H}_{\text{SO}}\)
3. Darwin term (for \(s\)-states, \(l=0\)): \(\hat{H}_D = \frac{\pi \hbar^2 Z e^2}{2 m_e^2 c^2 (4\pi \varepsilon_0)} \delta^3(\mathbf{r})\)
Remarkably, Paul Dirac's relativistic wave equation yields the exact combined energy formula:
States with the same principal quantum number \(n\) and the same total angular momentum \(j\) have identical energies in Dirac theory:
- The \(2s_{1/2}\) (\(n=2, l=0, j=1/2\)) and \(2p_{1/2}\) (\(n=2, l=1, j=1/2\)) levels are strictly degenerate in the Dirac equation.
- In 1947, Willis Lamb and Robert Retherford discovered that \(2s_{1/2}\) lies approximately \(1057.8\text{ MHz}\) above \(2p_{1/2}\). This Lamb shift arises from quantum electrodynamic (QED) vacuum fluctuations of the electromagnetic field and electron self-energy.
§4.7 Zeeman Effect: Normal vs Anomalous Splitting in Magnetic Fields
When an atom is placed in an external uniform magnetic field \(\mathbf{B} = B \hat{\mathbf{z}}\), its spectral lines split into closely spaced components.
Magnetic Interaction Hamiltonian
The total magnetic dipole moment of an atomic electron is:
where \(\mu_B = \frac{e \hbar}{2 m_e} \approx 9.274 \times 10^{-24}\text{ J}\cdot\text{T}^{-1}\) is the Bohr magneton, and \(g_e \approx 2.002319\) is the electron spin gyromagnetic \(g\)-factor (taken as \(2\) to first order). The Zeeman Hamiltonian is:
Normal Zeeman Effect (Zero Spin, \(S = 0\))
In singlet states where total spin \(S = 0\), \(\hat{\mathbf{J}} = \hat{\mathbf{L}}\), and the interaction Hamiltonian reduces to:
Under optical dipole selection rules \(\Delta m_l = 0, \pm 1\), any spectral transition of frequency \(\nu_0\) splits into exactly three equally spaced lines:
The central line (\(\Delta m = 0\), \(\pi\)-component) is linearly polarized parallel to \(\mathbf{B}\), while the side lines (\(\Delta m = \pm 1\), \(\sigma\)-components) are circularly polarized perpendicular to \(\mathbf{B}\).
Anomalous Zeeman Effect (Non-Zero Spin, Weak Field)
When spin is non-zero and the external magnetic field is weak compared to internal spin-orbit coupling (\(B \ll B_{\text{int}} \sim 10\text{ T}\)), \(j\) and \(m_j\) remain good quantum numbers. In first-order perturbation theory, the Zeeman energy shift is:
where \(g_J\) is the Landé \(g\)-factor:
Values of \(g_J\):
- For pure orbital state (\(s=0\)): \(g_J = 1\)
- For pure spin state (\(l=0, j=1/2, s=1/2\)): \(g_J = 2\)
- For \(^2P_{1/2}\) (\(l=1, s=1/2, j=1/2\)): \(g_J = 1 + \frac{3/4 + 3/4 - 2}{2 \times 3/4} = 1 + \frac{-1/2}{3/2} = 1 - \frac{1}{3} = \frac{2}{3}\)
- For \(^2P_{3/2}\) (\(l=1, s=1/2, j=3/2\)): \(g_J = 1 + \frac{15/4 + 3/4 - 2}{2 \times 15/4} = 1 + \frac{5/2}{15/2} = 1 + \frac{1}{3} = \frac{4}{3}\)
Because different states possess different \(g_J\) factors, transitions split into complex multiplet patterns with more than 3 lines (the anomalous Zeeman effect).
Strong-Field Paschen-Back Limit
When the magnetic field is very strong (\(B \gg B_{\text{int}}\)), the external field decouples \(\mathbf{L}\) and \(\mathbf{S}\). Both \(\hat{L}_z\) and \(\hat{S}_z\) become independently conserved:
With electric dipole selection rules \(\Delta m_s = 0\) and \(\Delta m_l = 0, \pm 1\), the spectrum collapses back into a triplet resembling the normal Zeeman effect.
§4.8 The Stark Effect: Linear vs Quadratic Shifts in Static Electric Fields
When an atom is exposed to an external uniform electrostatic field \(\mathbf{E}_{\text{ext}} = \mathcal{E} \hat{\mathbf{z}}\), its spectral lines shift and split—the Stark Effect.
The Stark Interaction Hamiltonian
The electric dipole moment of an atomic electron is \(\mathbf{d} = -e \mathbf{r}\). The electrostatic perturbation Hamiltonian is:
Quadratic Stark Effect (Non-Degenerate Ground State)
For the ground state of hydrogen (\(1s\), \(n = 1\)), the first-order energy correction is:
This vanishes identically because \(|\psi_{1s}|^2\) is symmetric under spatial inversion (\(z \rightarrow -z\)) while \(z\) is antisymmetric (Laporte parity selection rule). The lowest non-vanishing shift occurs at second order:
where \(\alpha_{\text{pol}}\) is the static atomic dipole polarizability. For ground-state hydrogen:
The energy shift is strictly proportional to \(\mathcal{E}^2\) (Quadratic Stark Effect).
Linear Stark Effect (Degenerate Excited States)
For the first excited state (\(n = 2\)), there are 4 degenerate orbitals: \(2s, 2p_0, 2p_{+1}, 2p_{-1}\). Because the perturbation \(\hat{H}' = e \mathcal{E} z\) preserves cylindrical symmetry around \(z\), \([\hat{H}', \hat{L}_z] = 0\). Matrix elements between states with different \(m\) vanish:
- \(2p_{+1}\) (\(m = +1\)) and \(2p_{-1}\) (\(m = -1\)) have no partners with the same \(m\) and different parity, so their first-order shifts are zero.
- The \(2s\) (\(m = 0\)) and \(2p_z\) (\(m = 0\)) states possess different parities and couple strongly:
The secular determinant within the \(\{2s, 2p_z\}\) subspace is:
The degenerate \(n = 2\) level splits linearly with field strength (Linear Stark Effect):
- \(E_+^{(1)} = +3 e a_0 \mathcal{E}\) (eigenstate \(\frac{1}{\sqrt{2}}(2s - 2p_z)\), dipole pointing against field)
- \(E_0^{(1)} = 0\) (two-fold degenerate: \(2p_x, 2p_y\))
- \(E_-^{(1)} = -3 e a_0 \mathcal{E}\) (eigenstate \(\frac{1}{\sqrt{2}}(2s + 2p_z)\), dipole pointing along field)
The linear Stark effect is unique to hydrogenic systems due to their accidental \(l\)-degeneracy.
## Advanced Mathematical Supplement: SO(4) Symmetry & Runge-Lenz Invariant
Dynamical SO(4) Symmetry of the Hydrogen Atom & The Laplace-Runge-Lenz Vector
The accidental degeneracy of hydrogenic energy levels with respect to orbital angular momentum \(l\) (where \(E_n\) depends strictly on \(n\)) is not an accident. It is the direct consequence of an underlying four-dimensional rotational symmetry \(\mathbf{SO}(4)\).
The Quantum Laplace-Runge-Lenz Operator
In classical Keplerian planetary orbits, the Laplace-Runge-Lenz vector points from the focus along the major axis toward perihelion, remaining strictly constant in time. In quantum mechanics, the Hermitian Laplace-Runge-Lenz operator is:
Evaluating the commutator with the Coulomb Hamiltonian \(\hat{H} = \frac{\hat{\mathbf{p}}^2}{2\mu} - \frac{Z e^2}{4\pi\varepsilon_0 r}\):
Because both \(\hat{\mathbf{L}}\) and \(\hat{\mathbf{M}}\) commute with \(\hat{H}\), the hydrogen atom possesses six independent conserved continuous generators.
The SO(4) Lie Algebra
For bound states (\(E < 0\)), define the scaled vector:
The commutation relations between \(\hat{\mathbf{L}}\) and \(\hat{\mathbf{K}}\) are:
This is precisely the Lie algebra of the four-dimensional orthogonal rotation group \(\mathbf{SO}(4)\). By defining two decoupled commuting angular momentum vectors:
they satisfy two independent \(\mathfrak{su}(2)\) algebras:
Because \(\hat{\mathbf{L}} \cdot \hat{\mathbf{K}} = 0\), the Casimirs are equal: \(\hat{\mathbf{I}}_1^2 = \hat{\mathbf{I}}_2^2 = j(j + 1)\hbar^2\). The total Casimir relates directly to the Hamiltonian:
Setting \(2j + 1 = n\), the bound-state energy eigenvalues are obtained algebraically without ever solving differential equations:
The state degeneracy is \((2j + 1)^2 = n^2\), matching the observed spatial orbital degeneracy.
## Research Monograph: Relativistic Dirac Equation & Quantum Electrodynamic Corrections
Dirac's Relativistic Wave Mechanics
Paul Dirac sought a first-order wave equation that is linear in both time and space derivatives to ensure positive-definite probability densities while satisfying relativistic invariance \(E^2 = p^2 c^2 + m_0^2 c^4\):
where \(\Psi\) is a 4-component spinor (bispinor), and \(\boldsymbol{\alpha} = (\alpha_x, \alpha_y, \alpha_z)\) and \(\beta\) are \(4 \times 4\) Dirac matrices:
where \(\sigma_i\) are the \(2 \times 2\) Pauli spin matrices.
Consequences of the Dirac Equation
1. Intrinsic Spin \(s = 1/2\) and \(g = 2\): Electron spin and the gyromagnetic ratio \(g_s = 2\) emerge automatically from relativistic invariance without empirical introduction!
2. Antimatter (The Positron): Negative energy solutions with \(E < -m_e c^2\) led to Dirac's hole theory, predicting the existence of the positron, experimentally discovered by Carl Anderson in 1932.
3. Zitterbewegung: Rapid trembling motion of the electron position operator at frequency \(\omega \sim 2 m_e c^2 / \hbar \approx 10^{21}\text{ Hz}\) over a Compton wavelength \(\hbar / m_e c \approx 3.86 \times 10^{-13}\text{ m}\).
Quantum Electrodynamic (QED) Radiative Corrections
While Dirac's equation predicts exact degeneracy for \(2s_{1/2}\) and \(2p_{1/2}\), Quantum Electrodynamics (QED) accounts for interaction with the quantized electromagnetic vacuum:
1. Electron Self-Energy & Vacuum Fluctuations: Zero-point fluctuations of the vacuum electric field jiggle the bound electron, smearing its interaction with the nuclear Coulomb potential. Because the \(s\)-electron spends time at the nucleus while \(p\)-electrons do not, this shifts the \(2s_{1/2}\) level upward by \(+1010\text{ MHz}\).
2. Vacuum Polarization: Virtual electron-positron pairs in the vacuum shield the bare nuclear charge at distances \(r \gg \hbar / m_e c\). At very short distances, the electron penetrates the screening cloud, feeling a slightly stronger effective attraction, which lowers the \(2s_{1/2}\) level by \(-27\text{ MHz}\).
The net QED shift (\(+1057.8\text{ MHz}\)) matches the experimental Lamb shift with ten-figure accuracy, representing the pinnacle of modern theoretical physics.
Worked Problems & Step-by-Step Quantum Derivations
Multi-step solved problems covering Planck distribution, photoelectric kinetics, Compton shift, de Broglie wavelengths, uncertainty relations, and Hermitian operator commutation algebra.
For a one-electron hydrogenic atom with nuclear charge \(Z\):
- Derive the expression for the most probable radius \(r_{\text{mp}}\) of the electron in the \(1s\) state.
- Evaluate the expectation value \(\langle r \rangle\) in the \(1s\) state and compare it to \(r_{\text{mp}}\).
- Derive the two local maxima and the radial node of the radial probability distribution \(P_{2s}(r)\) for the \(2s\) state, and determine which peak represents the primary outer electron shell.
Comprehensive Multi-Step Solution:
Step 1: Most Probable Radius of the \(1s\) State
The radial wavefunction for the \(1s\) orbital is:
The radial probability distribution function is:
To find the extremum, differentiate \(P_{1s}(r)\) with respect to \(r\) and set to zero:
Factoring common terms:
For \(r > 0\), the non-trivial solution is:
For neutral hydrogen (\(Z = 1\)), \(r_{\text{mp}} = a_0 \approx 0.529177\text{ \AA}\).
Step 2: Expectation Value \(\langle r \rangle\) of the \(1s\) State
The quantum expectation value is:
Using the standard definite integral \(\int_0^\infty x^n e^{-a x} dx = \frac{n!}{a^{n+1}}\) with \(n = 3\) and \(a = \frac{2Z}{a_0}\):
Multiplying by the prefactor:
The ratio is \(\frac{\langle r \rangle}{r_{\text{mp}}} = \frac{1.5 a_0 / Z}{a_0 / Z} = 1.5\). The expectation value is 50% larger than the most probable distance because of the long asymmetric exponential tail at large radii.
Step 3: Radial Probability Distribution and Peaks of the \(2s\) State
The radial wavefunction for the \(2s\) state is:
The radial node occurs where \(R_{20}(r) = 0\):
For \(Z = 1\), \(r_{\text{node}} = 2 a_0\). The radial probability distribution is:
Let \(x = \frac{Zr}{a_0}\). Then \(P_{2s}(x) \propto x^2 (1 - x/2)^2 e^{-x} = \frac{1}{4} x^2 (2 - x)^2 e^{-x} = \frac{1}{4} (2x - x^2)^2 e^{-x}\). Differentiating with respect to \(x\):
The roots are:
- \(x = 0\) (minimum at origin)
- \(x = 2\) (node minimum, \(P_{2s} = 0\))
- Solving \(x^2 - 6x + 4 = 0\):
- Inner maximum: \(x_1 = 3 - \sqrt{5} \approx 0.764 \implies r_1 \approx 0.764 \frac{a_0}{Z}\)
- Outer maximum: \(x_2 = 3 + \sqrt{5} \approx 5.236 \implies r_2 \approx 5.236 \frac{a_0}{Z}\)
Comparing peak heights:
- Inner peak at \(x_1 \approx 0.764\): \((2(0.764) - 0.764^2)^2 e^{-0.764} \approx (1.528 - 0.584)^2 (0.466) \approx (0.891)(0.466) \approx 0.415\)
- Outer peak at \(x_2 \approx 5.236\): \((2(5.236) - 5.236^2)^2 e^{-5.236} \approx (10.472 - 27.416)^2 e^{-5.236} \approx (-16.944)^2 (0.00532) \approx 287.1 \times 0.00532 \approx 1.528\)
The outer maximum at \(r \approx 5.236 a_0 / Z\) is nearly 4 times larger than the inner peak. The inner peak corresponds to core electron penetration, while the outer peak defines the valence shell.
For the ground state (\(1s\)) of atomic hydrogen (\(Z = 1\)):
- Calculate the expectation value of inverse radial distance \(\langle r^{-1} \rangle\).
- Compute the expectation value of the Coulomb potential energy \(\langle V \rangle\).
- Using the known ground state energy \(E_1 = -13.606\text{ eV}\), calculate the expectation value of kinetic energy \(\langle T \rangle\) and verify the quantum mechanical virial theorem \(\langle T \rangle = -\frac{1}{2} \langle V \rangle\).
Comprehensive Multi-Step Solution:
Step 1: Calculation of \(\langle r^{-1} \rangle\)
The ground state radial wavefunction is \(R_{10}(r) = 2 a_0^{-3/2} e^{-r/a_0}\). The expectation value of \(r^{-1}\) is:
Using \(\int_0^\infty x e^{-a x} dx = \frac{1}{a^2}\) with \(a = \frac{2}{a_0}\):
Therefore:
Notice that \(\langle r^{-1} \rangle = \frac{1}{a_0} \ne \frac{1}{\langle r \rangle} = \frac{1}{1.5 a_0} = \frac{2}{3 a_0}\).
Step 2: Expectation Value of Coulomb Potential Energy \(\langle V \rangle\)
The Coulomb potential energy operator is \(V(r) = -\frac{e^2}{4\pi \varepsilon_0 r}\). Taking the expectation value:
Recall the definition of the Bohr radius: \(a_0 = \frac{4\pi \varepsilon_0 \hbar^2}{m_e e^2} \implies \frac{e^2}{4\pi \varepsilon_0 a_0} = \frac{e^4 m_e}{(4\pi \varepsilon_0)^2 \hbar^2} = 2 R_y = 27.2114\text{ eV}\). Thus:
Step 3: Verification of the Virial Theorem
The total Hamiltonian is \(\hat{H} = \hat{T} + \hat{V}\). The total energy expectation value in the ground state is:
Solving for kinetic energy:
Now evaluate the virial relation:
The quantum virial theorem \(\langle T \rangle = -\frac{1}{2} \langle V \rangle\) holds with exact mathematical precision.
For hydrogenic ions \(\text{He}^+\) (\(Z = 2\)) and \(\text{Li}^{2+}\) (\(Z = 3\)):
- Determine the energy (in eV) of the first three principal levels (\(n = 1, 2, 3\)) and calculate the degeneracy of each level including electron spin.
- Calculate the wavelength (in nm) of the transition corresponding to the analog of the Balmer-\(\alpha\) line (\(n = 3 \rightarrow 2\)) for \(\text{He}^+\) and for \(\text{Li}^{2+}\).
- Determine the minimum photon energy required to fully ionize \(\text{Li}^{2+}\) from its \(n = 2\) excited state.
Comprehensive Multi-Step Solution:
Step 1: Energy Levels and Degeneracies
The hydrogenic energy levels are given by:
The state degeneracy including spin is \(g_n = 2 n^2\).
- For \(n = 1\): \(g_1 = 2(1)^2 = 2\)
- For \(n = 2\): \(g_2 = 2(2)^2 = 8\)
- For \(n = 3\): \(g_3 = 2(3)^2 = 18\)
Calculated energies:
1. For \(\text{He}^+\) (\(Z = 2, Z^2 = 4\)):
- \(E_1 = -4 \times 13.6057 = -54.423\text{ eV}\)
- \(E_2 = -4 \times \frac{13.6057}{4} = -13.606\text{ eV}\)
- \(E_3 = -4 \times \frac{13.6057}{9} = -6.047\text{ eV}\)
2. For \(\text{Li}^{2+}\) (\(Z = 3, Z^2 = 9\)):
- \(E_1 = -9 \times 13.6057 = -122.451\text{ eV}\)
- \(E_2 = -9 \times \frac{13.6057}{4} = -30.613\text{ eV}\)
- \(E_3 = -9 \times \frac{13.6057}{9} = -13.606\text{ eV}\)
Step 2: Transition Wavelengths for \(n = 3 \rightarrow 2\)
The Rydberg formula gives:
For atomic hydrogen (\(Z = 1\)):
Because \(\lambda \propto \frac{1}{Z^2}\):
1. For \(\text{He}^+\) (\(Z = 2\)):
2. For \(\text{Li}^{2+}\) (\(Z = 3\)):
Step 3: Ionization Energy of \(\text{Li}^{2+}\) from \(n = 2\)
Ionization corresponds to exciting the electron from \(n = 2\) to \(n = \infty\) (\(E_\infty = 0\)):
The corresponding threshold ionization wavelength is:
Given the radial wavefunctions for \(n = 3\) states in hydrogen (\(Z = 1\)):
where \(\rho_3 = \frac{2 r}{3 a_0}\):
- Determine the exact radial positions of the nodes for the \(3s\) orbital in units of \(a_0\).
- Determine the radial node position for the \(3p\) orbital.
- Show that the \(3d\) orbital has zero radial nodes, and explain the chemical significance of orbital penetration near the nucleus.
Comprehensive Multi-Step Solution:
Step 1: Radial Nodes of the \(3s\) Orbital
The number of radial nodes for \(3s\) (\(n = 3, l = 0\)) is \(n - l - 1 = 3 - 0 - 1 = 2\). Setting \(R_{3s}(r) = 0\) requires the quadratic factor to vanish:
Solving via the quadratic formula:
Using \(\sqrt{3} \approx 1.73205\):
- \(\rho_{3, 1} = \frac{9 - 5.19615}{2} = \frac{3.80385}{2} \approx 1.9019\)
- \(\rho_{3, 2} = \frac{9 + 5.19615}{2} = \frac{14.19615}{2} \approx 7.0981\)
Recall \(\rho_3 = \frac{2 r}{3 a_0} \implies r = \frac{3 a_0}{2} \rho_3\):
1. First (inner) node:
2. Second (outer) node:
Step 2: Radial Node of the \(3p\) Orbital
For \(3p\) (\(n = 3, l = 1\)), the number of radial nodes is \(n - l - 1 = 3 - 1 - 1 = 1\). Setting \(R_{3p}(r) = 0\):
Since \(\rho_3 = 0\) represents the origin (\(r = 0\)), the radial node occurs at:
Thus, the single radial node of \(3p\) is located at exactly \(r = 9 a_0\).
Step 3: Radial Structure of the \(3d\) Orbital and Penetration Significance
For \(3d\) (\(n = 3, l = 2\)), \(n - l - 1 = 3 - 2 - 1 = 0\). The radial wavefunction is \(R_{3d}(r) \propto \rho_3^2 e^{-\rho_3 / 2}\), which has no zeros for any finite \(r > 0\). Thus, it has zero radial nodes.
Chemical Significance of Penetration: In multi-electron atoms, the effective potential experienced by an electron deviates from \(1/r\) due to electron shielding.
- As \(r \rightarrow 0\), \(R_{3s}(r) \rightarrow \text{const} \ne 0\). The \(3s\) electron penetrates inside the inner core shells (\(1s, 2s, 2p\)) and experiences a large unshielded nuclear charge \(Z_{\text{eff}}\).
- The \(3p\) electron has \(R \propto r^1\) and experiences a centrifugal barrier \(\frac{2\hbar^2}{2\mu r^2}\), reducing core penetration.
- The \(3d\) electron has \(R \propto r^2\) and experiences a severe centrifugal barrier \(\frac{6\hbar^2}{2\mu r^2}\), virtually excluding it from the core region.
Consequently, in multi-electron atoms, energy ordering splits by \(l\):
which explains the structure of the periodic table and the Aufbau order \(4s < 3d\).
- Write the explicit analytical expressions for the hydrogenic \(1s\) and \(2s\) wavefunctions \(\psi_{100}(r)\) and \(\psi_{200}(r)\).
- Evaluate the overlap integral \(S_{1s, 2s} = \int \psi_{100}^*(\mathbf{r}) \psi_{200}(\mathbf{r}) d\tau\) analytically over all space.
- Verify that the eigenfunctions are strictly orthogonal.
Comprehensive Multi-Step Solution:
Step 1: Analytical Wavefunction Expressions
Because \(l = 0\) and \(m = 0\), the Spherical Harmonic is a constant: \(Y_0^0(\theta, \phi) = \frac{1}{\sqrt{4\pi}}\). The full spatial wavefunctions are:
Step 2: Overlap Integral Evaluation
The volume element in spherical coordinates is \(d\tau = r^2 \sin\theta \, dr \, d\theta \, d\phi\). Since \(\psi_{1s}\) and \(\psi_{2s}\) are spherically symmetric (independent of \(\theta\) and \(\phi\)):
Thus:
Substitute the radial functions:
The integral becomes:
Split into two integrals:
Step 3: Verification of Orthogonality
Subtracting \(I_2\) from \(I_1\):
Therefore:
The \(1s\) and \(2s\) states are strictly orthogonal as required by the Hermitian property of \(\hat{H}\) for distinct energy eigenvalues (\(E_1 \ne E_2\)).
For the \(2p\) state of atomic hydrogen (\(n = 2, l = 1, s = 1/2\)):
- Determine the allowed values of the total angular momentum quantum number \(j\) and designate the spectral terms in standard Russell-Saunders notation \(^{2s+1}L_j\).
- Given that \(\langle r^{-3} \rangle_{2p} = \frac{1}{24 a_0^3}\), calculate the spin-orbit energy shift \(\Delta E_{\text{SO}}\) for both \(j\) levels.
- Calculate the energy splitting \(\delta E = E(2p_{3/2}) - E(2p_{1/2})\) in eV and in wavenumber (\(\text{cm}^{-1}\)).
Comprehensive Multi-Step Solution:
Step 1: Angular Momentum Coupling and Term Symbols
For \(l = 1\) and \(s = 1/2\):
With multiplicity \(2s + 1 = 2(1/2) + 1 = 2\) and letter symbol \(P\) for \(l = 1\):
- For \(j = 3/2\): \(^2P_{3/2}\) (degeneracy \(2j + 1 = 4\))
- For \(j = 1/2\): \(^2P_{1/2}\) (degeneracy \(2j + 1 = 2\))
Step 2: Spin-Orbit Energy Shifts
The spin-orbit operator is:
The expectation value of \(\hat{\mathbf{L}} \cdot \hat{\mathbf{S}}\) is:
For \(l = 1, s = 1/2\): \(l(l+1) = 2\) and \(s(s+1) = 3/4\), so \(l(l+1) + s(s+1) = 11/4\).
- For \(j = 3/2\): \(j(j+1) = \frac{3}{2} \times \frac{5}{2} = \frac{15}{4}\)
- For \(j = 1/2\): \(j(j+1) = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4}\)
The spin-orbit energy shift is:
Recall \(\alpha = \frac{e^2}{4\pi \varepsilon_0 \hbar c}\) and \(a_0 = \frac{4\pi \varepsilon_0 \hbar^2}{m_e e^2}\). Then:
Substituting \(\langle r^{-3} \rangle_{2p} = \frac{1}{24 a_0^3}\):
With \(|E_1| = 13.6057\text{ eV}\) and \(\alpha \approx \frac{1}{137.036}\):
Step 3: Fine Structure Splitting \(\delta E\)
The energy splitting between the two states is:
Converting to wavenumber:
In frequency: \(\Delta \nu = c \Delta \tilde{\nu} = (2.998 \times 10^{10}\text{ cm/s})(0.365\text{ cm}^{-1}) \approx 10.95\text{ GHz}\). This \(10.95\text{ GHz}\) doublet splitting in the Balmer-\(\alpha\) line was one of the earliest experimental triumphs of relativistic quantum mechanics.
The famous yellow sodium doublet (D-lines) arises from the transition \(3p \rightarrow 3s\):
- \(D_1\): \(^2P_{1/2} \rightarrow ^2S_{1/2}\) (\(\lambda = 589.6\text{ nm}\))
- \(D_2\): \(^2P_{3/2} \rightarrow ^2S_{1/2}\) (\(\lambda = 589.0\text{ nm}\))
In a uniform magnetic field of \(B = 1.50\text{ Tesla}\):
- Calculate the Landé \(g\)-factors for the three states involved: \(^2S_{1/2}\), \(^2P_{1/2}\), and \(^2P_{3/2}\).
- Determine the Zeeman sub-level shifts \(\Delta E(m_j)\) for each state.
- Using electric dipole selection rules \(\Delta m_j = 0, \pm 1\), determine the number of Zeeman transition components and their frequency shifts from line center for both \(D_1\) and \(D_2\).
Comprehensive Multi-Step Solution:
Step 1: Calculation of Landé \(g\)-Factors
The Landé formula is:
For a single valence electron, \(s = 1/2 \implies s(s+1) = 3/4\).
1. Ground state \(3^2S_{1/2}\) (\(l = 0, j = 1/2\)):
2. Excited state \(3^2P_{1/2}\) (\(l = 1, j = 1/2\)):
3. Excited state \(3^2P_{3/2}\) (\(l = 1, j = 3/2\)):
Step 2: Energy Shifts \(\Delta E(m_j) = g_J \mu_B B m_j\)
Evaluate the unit energy \(\mu_B B\):
In frequency units:
Sub-level shifts:
1. \(^2S_{1/2}\) (\(g = 2\)):
- \(m_j = +1/2 \implies \Delta E = 2 \times (+1/2) \mu_B B = +1 \mu_B B\)
- \(m_j = -1/2 \implies \Delta E = 2 \times (-1/2) \mu_B B = -1 \mu_B B\)
2. \(^2P_{1/2}\) (\(g = 2/3\)):
- \(m_j = +1/2 \implies \Delta E = \frac{2}{3} (+1/2) \mu_B B = +\frac{1}{3} \mu_B B\)
- \(m_j = -1/2 \implies \Delta E = \frac{2}{3} (-1/2) \mu_B B = -\frac{1}{3} \mu_B B\)
3. \(^2P_{3/2}\) (\(g = 4/3\)):
- \(m_j = +3/2 \implies \Delta E = \frac{4}{3} (+3/2) \mu_B B = +2 \mu_B B\)
- \(m_j = +1/2 \implies \Delta E = \frac{4}{3} (+1/2) \mu_B B = +\frac{2}{3} \mu_B B\)
- \(m_j = -1/2 \implies \Delta E = \frac{4}{3} (-1/2) \mu_B B = -\frac{2}{3} \mu_B B\)
- \(m_j = -3/2 \implies \Delta E = \frac{4}{3} (-3/2) \mu_B B = -2 \mu_B B\)
Step 3: Transition Splittings Under Selection Rule \(\Delta m_j = m_j' - m_j'' = 0, \pm 1\)
1. \(D_1\) Line (\(^2P_{1/2} \rightarrow ^2S_{1/2}\)):
- Upper \(m_j' = +1/2 \rightarrow\) Lower \(+1/2\) (\(\Delta m_j = 0\)): \(\Delta E_{\text{trans}} = +1/3 - 1 = -2/3 \mu_B B\) (\(\pi\)-line)
- Upper \(m_j' = -1/2 \rightarrow\) Lower \(-1/2\) (\(\Delta m_j = 0\)): \(\Delta E_{\text{trans}} = -1/3 - (-1) = +2/3 \mu_B B\) (\(\pi\)-line)
- Upper \(m_j' = +1/2 \rightarrow\) Lower \(-1/2\) (\(\Delta m_j = +1\)): \(\Delta E_{\text{trans}} = +1/3 - (-1) = +4/3 \mu_B B\) (\(\sigma\)-line)
- Upper \(m_j' = -1/2 \rightarrow\) Lower \(+1/2\) (\(\Delta m_j = -1\)): \(\Delta E_{\text{trans}} = -1/3 - 1 = -4/3 \mu_B B\) (\(\sigma\)-line)
The \(D_1\) line splits into 4 distinct components at \(\pm \frac{2}{3} \delta\nu_0\) and \(\pm \frac{4}{3} \delta\nu_0\) (\(\pm 14.0\text{ GHz}\) and \(\pm 28.0\text{ GHz}\)).
2. \(D_2\) Line (\(^2P_{3/2} \rightarrow ^2S_{1/2}\)):
- \(\Delta m_j = 0\):
- \(+1/2 \rightarrow +1/2\): \(+2/3 - 1 = -1/3 \mu_B B\)
- \(-1/2 \rightarrow -1/2\): \(-2/3 - (-1) = +1/3 \mu_B B\)
- \(\Delta m_j = +1\):
- \(+3/2 \rightarrow +1/2\): \(+2 - 1 = +1 \mu_B B\)
- \(+1/2 \rightarrow -1/2\): \(+2/3 - (-1) = +5/3 \mu_B B\)
- \(\Delta m_j = -1\):
- \(-1/2 \rightarrow +1/2\): \(-2/3 - 1 = -5/3 \mu_B B\)
- \(-3/2 \rightarrow -1/2\): \(-2 - (-1) = -1 \mu_B B\)
The \(D_2\) line splits into 6 distinct components at \(\pm \frac{1}{3} \delta\nu_0\), \(\pm 1 \delta\nu_0\), and \(\pm \frac{5}{3} \delta\nu_0\) (\(\pm 7.0\text{ GHz}\), \(\pm 21.0\text{ GHz}\), and \(\pm 35.0\text{ GHz}\)).
Consider the \(n = 2\) excited state of atomic hydrogen in an external electric field \(\mathcal{E} = 1.00 \times 10^7\text{ V/m}\) along the \(z\)-axis.
- The perturbation matrix element between \(2s\) and \(2p_z\) is \(H'_{12} = \langle 2s | e \mathcal{E} z | 2p_z \rangle = -3 e a_0 \mathcal{E}\). Evaluate \(H'_{12}\) in electron volts.
- Calculate the energy splitting \(\Delta E_{\text{Stark}}\) between the upper and lower Stark levels.
- Determine the electric dipole moment of the perturbed eigenstates in units of Debye.
Comprehensive Multi-Step Solution:
Step 1: Matrix Element Evaluation
Given \(a_0 = 5.29177 \times 10^{-11}\text{ m}\), \(e = 1.60218 \times 10^{-19}\text{ C}\), and \(\mathcal{E} = 1.00 \times 10^7\text{ V/m}\):
In Joules:
In electron volts:
Step 2: Energy Splitting \(\Delta E_{\text{Stark}}\)
The secular equation within the \(\{2s, 2p_z\}\) subspace yields eigenvalues:
The total splitting between the upper and lower levels is:
Converting to frequency:
Step 3: Induced Electric Dipole Moment
The normalized eigenstates are:
The electric dipole moment operator is \(\hat{d}_z = -e \hat{z}\). The permanent dipole moment of state \(|\psi_\pm\rangle\) is:
Evaluate numerically:
Using \(1\text{ Debye (D)} \approx 3.33564 \times 10^{-30}\text{ C}\cdot\text{m}\):
Each mixed \(s-p\) hybrid state possesses a gigantic permanent electric dipole moment of \(\pm 7.63\text{ D}\) (larger than the dipole moment of water, which is \(1.85\text{ D}\)), aligning parallel or antiparallel to the field.
For a nucleus with non-spherical charge distribution and electric quadrupole moment \(Q\) interacting with an atomic electron:
- Define the electric field gradient (EFG) tensor component \(q_{z z} = \frac{\partial^2 V}{\partial z^2}\) at the nucleus.
- Given that for a hydrogenic orbital with quantum numbers \(n, l\):
Show that \(q_{z z} = 0\) for all \(s\)-orbitals (\(l = 0\)).
- For a \(2p_z\) electron in a hydrogenic ion of charge \(Z\), where \(\langle r^{-3} \rangle_{2p} = \frac{Z^3}{24 a_0^3}\), calculate \(q_{z z}\) in terms of fundamental constants.
Comprehensive Multi-Step Solution:
Step 1: Definition of Electric Field Gradient (EFG)
The electrostatic potential produced at the nucleus (\(\mathbf{r} = 0\)) by an electron at position \(\mathbf{r}\) is:
The Electric Field Gradient (EFG) tensor \(V_{i j}\) is the second spatial derivative of the electrostatic potential:
In traceless principal axis form, the principal EFG component is:
Step 2: Vanishing of EFG for \(s\)-Orbitals (\(l = 0\))
For any \(s\)-orbital, the angular wavefunction is spherically symmetric:
The angular integration over the sphere is:
Evaluate the integral:
Because the angular factor \(3\cos^2\theta - 1 = 2 P_2(\cos\theta)\) is orthogonal to \(P_0(\cos\theta) = 1\), the integral vanishes identically:
All \(s\)-electrons produce exactly zero electric field gradient at the nucleus, meaning they cannot induce nuclear quadrupole splitting!
Step 3: EFG for \(2p_z\) Electron (\(l = 1, m_l = 0\))
For a \(2p_z\) orbital, \(l = 1\) and \(m_l = 0\): Evaluate the angular prefactor:
Substitute into the formula with \(\langle r^{-3} \rangle_{2p} = \frac{Z^3}{24 a_0^3}\):
Including the electrostatic constant \(\frac{1}{4\pi\varepsilon_0}\):
This non-zero electric field gradient couples to the nuclear electric quadrupole moment \(e Q\), generating nuclear quadrupole resonance (NQR) and hyperfine quadrupole splittings in high-resolution NMR and microwave spectra.