Chapter 5 • Theory & Derivations

Unit 5: Quantum Approximation Methods: Perturbation & Variational

Comprehensive mathematical and quantum formulation of analytical approximation methods for systems lacking closed-form solutions: non-degenerate Rayleigh-Schrödinger perturbation theory through second order, degenerate perturbation theory and secular determinants, the Rayleigh-Ritz variational principle and upper bound theorems, linear variational expansions, ground-state helium atom variational optimization, time-dependent perturbation theory and Fermi's Golden Rule, and avoided crossings in two-level quantum dynamics.

§5.1 Non-Degenerate Rayleigh-Schrödinger Perturbation Theory

Exact analytical solutions to the Schrödinger equation exist only for a handful of idealized physical models (free particle, harmonic oscillator, rigid rotor, hydrogen atom). For all multi-electron atoms and molecules, approximate quantum methods are indispensable.

Rayleigh-Schrödinger Formalism

Consider an unperturbed Hamiltonian \(\hat{H}^{(0)}\) with a known complete orthonormal set of eigenfunctions \(\psi_n^{(0)}\) and non-degenerate eigenvalues \(E_n^{(0)}\):

\hat{H}^{(0)} \psi_n^{(0)} = E_n^{(0)} \psi_n^{(0)}, \quad \langle \psi_m^{(0)} | \psi_n^{(0)} \rangle = \delta_{m n}

The total Hamiltonian is perturbed by a small interaction term \(\hat{H}'\):

\hat{H} = \hat{H}^{(0)} + \lambda \hat{H}'

where \(\lambda \in [0, 1]\) is a formal dimensionless order-tracking perturbation parameter. We expand the exact wavefunction \(\psi_n\) and energy \(E_n\) in power series:

\psi_n = \psi_n^{(0)} + \lambda \psi_n^{(1)} + \lambda^2 \psi_n^{(2)} + \dots

E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \dots

Substituting into the Schrödinger equation \(\hat{H} \psi_n = E_n \psi_n\) and collecting like powers of \(\lambda\):

\(\lambda^0\): \(\hat{H}^{(0)} \psi_n^{(0)} = E_n^{(0)} \psi_n^{(0)}\)
\(\lambda^1\): \(\hat{H}^{(0)} \psi_n^{(1)} + \hat{H}' \psi_n^{(0)} = E_n^{(0)} \psi_n^{(1)} + E_n^{(1)} \psi_n^{(0)}\)
\(\lambda^2\): \(\hat{H}^{(0)} \psi_n^{(2)} + \hat{H}' \psi_n^{(1)} = E_n^{(0)} \psi_n^{(2)} + E_n^{(1)} \psi_n^{(1)} + E_n^{(2)} \psi_n^{(0)}\)

First-Order Energy Correction

Taking the inner product of the \(\lambda^1\) equation with \(\langle \psi_n^{(0)} |\):

\langle \psi_n^{(0)} | \hat{H}^{(0)} | \psi_n^{(1)} \rangle + \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle = E_n^{(0)} \langle \psi_n^{(0)} | \psi_n^{(1)} \rangle + E_n^{(1)} \langle \psi_n^{(0)} | \psi_n^{(0)} \rangle

Because \(\hat{H}^{(0)}\) is Hermitian, \(\langle \psi_n^{(0)} | \hat{H}^{(0)} | \psi_n^{(1)} \rangle = E_n^{(0)} \langle \psi_n^{(0)} | \psi_n^{(1)} \rangle\), which cancels the first term on the right:

E_n^{(1)} = \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle = H'_{n n}

The first-order energy correction is simply the expectation value of the perturbation evaluated over the unperturbed state!

First-Order Wavefunction Correction

Expand the first-order correction in the complete unperturbed basis: \(\psi_n^{(1)} = \sum_{m} c_m \psi_m^{(0)}\). Under intermediate normalization \(\langle \psi_n^{(0)} | \psi_n \rangle = 1\), \(c_n = 0\). Taking the inner product of the \(\lambda^1\) equation with \(\langle \psi_k^{(0)} |\) (\(k \ne n\)):

E_k^{(0)} c_k + \langle \psi_k^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle = E_n^{(0)} c_k + 0 \implies c_k = \frac{\langle \psi_k^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle}{E_n^{(0)} - E_k^{(0)}}

Hence, the first-order corrected wavefunction is:

\psi_n^{(1)} = \sum_{k \ne n} \frac{\langle \psi_k^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle}{E_n^{(0)} - E_k^{(0)}} \psi_k^{(0)}

Second-Order Energy Correction

Taking the inner product of the \(\lambda^2\) equation with \(\langle \psi_n^{(0)} |\):

E_n^{(2)} = \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(1)} \rangle = \sum_{k \ne n} \frac{|\langle \psi_k^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle|^2}{E_n^{(0)} - E_k^{(0)}}

Crucial Physical Insight: For the ground state (\(n = 0\)), \(E_0^{(0)} - E_k^{(0)} < 0\) for all \(k \ne 0\). Therefore, the second-order energy correction for any ground state is strictly negative:

E_0^{(2)} \le 0

The perturbation always polarizes the ground-state charge distribution to lower the system energy.

§5.2 Degenerate Perturbation Theory: Secular Determinant & Symmetry Breaking

When the unperturbed energy eigenvalue is \(g\)-fold degenerate:

E_1^{(0)} = E_2^{(0)} = \dots = E_g^{(0)} = E^{(0)}

the standard Rayleigh-Schrödinger formula breaks down because the energy denominator \(E_n^{(0)} - E_k^{(0)} \rightarrow 0\), causing catastrophic division by zero.

The Degenerate Subspace and Secular Equation

To resolve this, we work within the \(g\)-dimensional degenerate subspace spanned by \(\{\phi_1^{(0)}, \phi_2^{(0)}, \dots, \phi_g^{(0)}\}\). Any arbitrary linear combination:

\psi^{(0)} = \sum_{j=1}^g c_j \phi_j^{(0)}

is an unperturbed eigenstate with energy \(E^{(0)}\). The goal is to identify the "proper" zeroth-order states that diagonalize the perturbation. Substituting \(\psi = \psi^{(0)} + \lambda \psi^{(1)}\) into \((\hat{H}^{(0)} + \lambda \hat{H}') \psi = (E^{(0)} + \lambda E^{(1)}) \psi\) and projecting onto \(\langle \phi_i^{(0)} |\):

\sum_{j=1}^g c_j \langle \phi_i^{(0)} | \hat{H}' | \phi_j^{(0)} \rangle = E^{(1)} c_i \sum_{j=1}^g c_j \delta_{i j} = E^{(1)} c_i

Defining the perturbation matrix elements \(H'_{i j} = \langle \phi_i^{(0)} | \hat{H}' | \phi_j^{(0)} \rangle\), this yields a set of \(g\) simultaneous linear homogeneous equations:

\sum_{j=1}^g \left( H'_{i j} - E^{(1)} \delta_{i j} \right) c_j = 0 \quad (i = 1, 2, \dots, g)

For non-trivial eigenvector solutions \(\mathbf{c} \ne \mathbf{0}\), the determinant of coefficients must vanish identically. This is the Secular Determinant:

\det \left[ \mathbf{H}' - E^{(1)} \mathbf{I} \right] = \begin{vmatrix} H'_{11} - E^{(1)} & H'_{12} & \dots & H'_{1g} \\ H'_{21} & H'_{22} - E^{(1)} & \dots & H'_{2g} \\ \vdots & \vdots & \ddots & \vdots \\ H'_{g1} & H'_{g2} & \dots & H'_{gg} - E^{(1)} \end{vmatrix} = 0

Symmetry Breaking and Lifting of Degeneracy

The roots of the characteristic polynomial yield the \(g\) first-order energy shifts \(E_1^{(1)}, E_2^{(1)}, \dots, E_g^{(1)}\).

If all roots are distinct, the perturbation completely breaks the spatial symmetry and lifts the degeneracy.
If some roots remain equal, the perturbation partially removes the degeneracy.
If the off-diagonal elements vanish by symmetry (\(H'_{i j} = 0\) for \(i \ne j\)), the original basis functions \(\phi_i^{(0)}\) are already the proper zeroth-order eigenfunctions, and the shifts are simply the diagonal expectation values \(E_i^{(1)} = H'_{i i}\).

§5.3 The Rayleigh-Ritz Variational Principle: Upper Bound Proof

The variational method is the cornerstone of modern computational quantum chemistry and electronic structure theory (including Hartree-Fock and Density Functional Theory). Unlike perturbation theory, it does not require partitioning the Hamiltonian into an unperturbed part and a small perturbation.

Statement and Proof of the Variational Theorem

Let \(\hat{H}\) be a time-independent Hamiltonian with true ground-state energy \(E_0\) and exact orthonormal eigenfunctions \(\psi_k\):

\hat{H} \psi_k = E_k \psi_k, \quad E_0 \le E_1 \le E_2 \le \dots

Let \(\Phi_{\text{trial}}\) be an arbitrary normalizable trial wavefunction that satisfies the physical boundary conditions of the system. The variational energy expectation value is defined as:

E_{\text{var}}[\Phi] = \frac{\langle \Phi | \hat{H} | \Phi \rangle}{\langle \Phi | \Phi \rangle} = \frac{\int \Phi^* \hat{H} \Phi \, d\tau}{\int |\Phi|^2 d\tau}

Theorem: For any trial function \(\Phi\), the variational energy provides a rigorous upper bound to the true ground-state energy:

E_{\text{var}}[\Phi] \ge E_0

Proof: Since the exact eigenstates \(\{\psi_k\}\) form a complete orthonormal basis, any well-behaved function \(\Phi\) can be expanded as:

\Phi = \sum_{k=0}^\infty c_k \psi_k, \quad \text{where } c_k = \langle \psi_k | \Phi \rangle

Evaluate the denominator:

\langle \Phi | \Phi \rangle = \sum_{j} \sum_{k} c_j^* c_k \langle \psi_j | \psi_k \rangle = \sum_k |c_k|^2

Evaluate the numerator:

\langle \Phi | \hat{H} | \Phi \rangle = \sum_j \sum_k c_j^* c_k \langle \psi_j | \hat{H} | \psi_k \rangle = \sum_k |c_k|^2 E_k

Subtract \(E_0 \langle \Phi | \Phi \rangle\) from the numerator:

\langle \Phi | \hat{H} | \Phi \rangle - E_0 \langle \Phi | \Phi \rangle = \sum_{k=0}^\infty |c_k|^2 (E_k - E_0)

Since \(E_0\) is the absolute ground-state eigenvalue, \(E_k - E_0 \ge 0\) for all \(k \ge 0\). Furthermore, \(|c_k|^2 \ge 0\). Therefore, every term in the sum is non-negative:

\sum_{k=0}^\infty |c_k|^2 (E_k - E_0) \ge 0 \implies \langle \Phi | \hat{H} | \Phi \rangle \ge E_0 \langle \Phi | \Phi \rangle

Dividing by \(\langle \Phi | \Phi \rangle > 0\) completes the proof:

E_{\text{var}} \ge E_0 \quad \text{Q.E.D.}

The equality \(E_{\text{var}} = E_0\) holds if and only if \(\Phi = \psi_0\) is the exact ground-state eigenfunction.

Practical Optimization Procedure

We introduce variational parameters \(\{\alpha, \beta, \dots\}\) into the trial function: \(\Phi = \Phi(\mathbf{r}; \alpha, \beta)\). The optimal approximation is obtained by minimizing \(E_{\text{var}}\):

\frac{\partial E_{\text{var}}}{\partial \alpha} = 0, \quad \frac{\partial E_{\text{var}}}{\partial \beta} = 0

The closer the trial function's mathematical form matches the true physics, the closer \(E_{\text{var}}\) approaches \(E_0\).

§5.4 Linear Variational Method & The Secular Equation: Basis Set Expansions

In molecular quantum chemistry, trial wavefunctions are routinely represented as linear expansions over a set of \(K\) predetermined basis functions \(\{\chi_1, \chi_2, \dots, \chi_K\}\) (such as atomic orbitals or Gaussian basis functions):

\Phi = \sum_{i=1}^K c_i \chi_i

The expansion coefficients \(\{c_1, \dots, c_K\}\) serve as the variational parameters.

Derivation of the Matrix Secular Equation

The energy expectation value is:

E = \frac{\langle \Phi | \hat{H} | \Phi \rangle}{\langle \Phi | \Phi \rangle} = \frac{\sum_{i=1}^K \sum_{j=1}^K c_i^* c_j H_{i j}}{\sum_{i=1}^K \sum_{j=1}^K c_i^* c_j S_{i j}}

where the Hamiltonian matrix elements \(H_{i j}\) and overlap matrix elements \(S_{i j}\) are:

H_{i j} = \langle \chi_i | \hat{H} | \chi_j \rangle = \int \chi_i^* \hat{H} \chi_j \, d\tau

S_{i j} = \langle \chi_i | \chi_j \rangle = \int \chi_i^* \chi_j \, d\tau

Rearranging:

E \sum_{i=1}^K \sum_{j=1}^K c_i^* c_j S_{i j} = \sum_{i=1}^K \sum_{j=1}^K c_i^* c_j H_{i j}

To minimize \(E\) with respect to each complex coefficient \(c_k^*\), differentiate partially:

\frac{\partial E}{\partial c_k^*} \sum_{i, j} c_i^* c_j S_{i j} + E \sum_j c_j S_{k j} = \sum_j c_j H_{k j}

Setting \(\frac{\partial E}{\partial c_k^*} = 0\) yields the system of linear equations:

\sum_{j=1}^K (H_{k j} - E S_{k j}) c_j = 0 \quad (k = 1, 2, \dots, K)

In compact matrix notation:

\mathbf{H} \mathbf{c} = E \mathbf{S} \mathbf{c}

This is a generalized matrix eigenvalue problem. For non-trivial eigenvector solutions \(\mathbf{c} \ne \mathbf{0}\):

\det(\mathbf{H} - E \mathbf{S}) = 0

This determinantal equation yields \(K\) real energy roots:

E_0 \le E_1 \le E_2 \le \dots \le E_{K-1}

According to the MacDonald-Hylleraas-Undheim theorem, each root \(E_m\) provides an upper bound to the corresponding \(m\)-th excited state of the exact Hamiltonian:

E_m \ge E_m^{\text{exact}} \quad (m = 0, 1, \dots, K-1)

Expanding the basis size (\(K \rightarrow \infty\)) monotonically converges all eigenvalues downward toward the exact spectra.

§5.5 Ground-State Helium Atom via Variational Method: Effective Nuclear Charge

The helium atom contains a nucleus of charge \(Z = +2\) and two interacting electrons. It is the prototypical three-body quantum system for which no exact closed-form solution exists.

The Helium Hamiltonian

In atomic units (\(\hbar = m_e = e = 4\pi\varepsilon_0 = 1\)):

\hat{H} = -\frac{1}{2} \nabla_1^2 - \frac{1}{2} \nabla_2^2 - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}} = \hat{h}_1 + \hat{h}_2 + \frac{1}{r_{12}}

where \(r_{12} = |\mathbf{r}_1 - \mathbf{r}_2|\) is the interelectronic distance. If we completely ignore the electron-electron repulsion \(\frac{1}{r_{12}}\), the Hamiltonian separates into two independent hydrogenic ions with \(Z = 2\):

E^{(0)} = E_1(1) + E_1(2) = -\frac{Z^2}{2} - \frac{Z^2}{2} = -Z^2 = -4\text{ a.u.} = -108.8\text{ eV}

The experimentally measured ground-state energy of helium is \(E_{\text{exp}} = -2.9037\text{ a.u.} = -79.005\text{ eV}\). Ignoring electron repulsion results in an error of nearly \(30\text{ eV}\) because it assumes each electron feels the completely unshielded nuclear charge \(+2\).

First-Order Perturbation Theory Estimate

Treating \(\hat{H}' = \frac{1}{r_{12}}\) as a perturbation on the unperturbed hydrogenic product wavefunction \(\psi^{(0)} = \frac{Z^3}{\pi} e^{-Z(r_1 + r_2)}\):

E^{(1)} = \langle \psi^{(0)} | \frac{1}{r_{12}} | \psi^{(0)} \rangle = \frac{5}{8} Z\text{ a.u.}

For helium (\(Z = 2\)), \(E^{(1)} = \frac{5}{8}(2) = 1.25\text{ a.u.} = +34.01\text{ eV}\). The first-order perturbation energy is:

E_{\text{pert}} = E^{(0)} + E^{(1)} = -4.00 + 1.25 = -2.75\text{ a.u.} = -74.83\text{ eV}

While an improvement, it still exhibits an error of \(4.17\text{ eV}\).

Variational Optimization of Effective Nuclear Charge \(Z_{\text{eff}}\)

Because each electron partially shields the other from the nuclear charge, we introduce an effective nuclear charge \(Z_{\text{eff}} = \zeta < Z\) as a variational parameter:

\Phi(\mathbf{r}_1, \mathbf{r}_2; \zeta) = \frac{\zeta^3}{\pi} e^{-\zeta(r_1 + r_2)}

We re-express the Hamiltonian in terms of the effective one-electron operators:

\hat{H} = \left( -\frac{1}{2}\nabla_1^2 - \frac{\zeta}{r_1} \right) + \left( -\frac{1}{2}\nabla_2^2 - \frac{\zeta}{r_2} \right) + (\zeta - Z) \left( \frac{1}{r_1} + \frac{1}{r_2} \right) + \frac{1}{r_{12}}

Evaluating expectation values with the scaled hydrogenic functions:

Kinetic + effective potential energy: \(-\frac{\zeta^2}{2} - \frac{\zeta^2}{2} = -\zeta^2\)
Difference potential energy: \(2(\zeta - Z) \langle r^{-1} \rangle_\zeta = 2(\zeta - Z)\zeta\)
Repulsion energy: \(\langle r_{12}^{-1} \rangle_\zeta = \frac{5}{8} \zeta\)

Summing all contributions:

E_{\text{var}}(\zeta) = \zeta^2 - 2 Z \zeta + \frac{5}{8} \zeta = \zeta^2 - \left( 2 Z - \frac{5}{8} \right) \zeta

To find the minimum, set \(\frac{d E_{\text{var}}}{d\zeta} = 0\):

2\zeta - \left( 2 Z - \frac{5}{8} \right) = 0 \implies \zeta_{\text{opt}} = Z - \frac{5}{16}

For helium (\(Z = 2\)):

\zeta_{\text{opt}} = 2 - \frac{5}{16} = \frac{27}{16} = 1.6875

Each electron shields the nucleus by a screening constant \(\sigma = 5/16 \approx 0.3125\). The optimized ground-state variational energy is:

E_{\text{var}}(\zeta_{\text{opt}}) = -\left( Z - \frac{5}{16} \right)^2 = -\left( \frac{27}{16} \right)^2 = -\frac{729}{256} \approx -2.8477\text{ a.u.} = -77.49\text{ eV}

The error relative to experiment shrinks to just \(1.52\text{ eV}\) (1.9% error), demonstrating the immense power of physical intuition combined with the variational principle.

§5.6 Time-Dependent Perturbation Theory & Fermi's Golden Rule

Stationary perturbation theory calculates energy levels of time-independent systems. When systems interact with time-varying fields (such as electromagnetic radiation triggering spectroscopy), we employ Time-Dependent Perturbation Theory (TDPT).

Time-Dependent Formulation

Consider the time-dependent Schrödinger equation:

i \hbar \frac{\partial \Psi(\mathbf{r}, t)}{\partial t} = \left[ \hat{H}_0 + \hat{V}(t) \right] \Psi(\mathbf{r}, t)

Expand \(\Psi(\mathbf{r}, t)\) in the complete basis of unperturbed stationary states \(\psi_k(\mathbf{r}) e^{-i E_k t / \hbar}\):

\Psi(\mathbf{r}, t) = \sum_k c_k(t) \psi_k(\mathbf{r}) e^{-i \omega_k t}, \quad \text{where } \omega_k = \frac{E_k}{\hbar}

Substituting into the Schrödinger equation yields the exact coupled differential equations for the transition amplitudes \(c_k(t)\):

i \hbar \frac{d c_f(t)}{dt} = \sum_i c_i(t) V_{f i}(t) e^{i \omega_{f i} t}

where \(V_{f i}(t) = \langle \psi_f | \hat{V}(t) | \psi_i \rangle\) and the Bohr transition frequency is \(\omega_{f i} = \frac{E_f - E_i}{\hbar}\).

First-Order Transition Amplitude

Assuming the system starts in initial state \(|i\rangle\) at \(t = 0\) (\(c_i(0) = 1, c_f(0) = 0\) for \(f \ne i\)):

c_f^{(1)}(t) = -\frac{i}{\hbar} \int_0^t V_{f i}(t') e^{i \omega_{f i} t'} dt'

Harmonic Perturbation and Fermi's Golden Rule

For a monochromatic harmonic perturbation \(\hat{V}(t) = \hat{F} e^{-i \omega t} + \hat{F}^\dagger e^{i \omega t}\) (as in electromagnetic absorption and stimulated emission):

c_f^{(1)}(t) = -\frac{i}{\hbar} F_{f i} \int_0^t e^{i (\omega_{f i} - \omega) t'} dt' = -\frac{F_{f i}}{\hbar} \frac{e^{i(\omega_{f i} - \omega)t} - 1}{\omega_{f i} - \omega}

The transition probability \(P_{i \rightarrow f}(t) = |c_f^{(1)}(t)|^2\) is:

P_{i \rightarrow f}(t) = \frac{4 |F_{f i}|^2}{\hbar^2} \frac{\sin^2\left( \frac{\omega_{f i} - \omega}{2} t \right)}{(\omega_{f i} - \omega)^2}

As \(t \rightarrow \infty\), the diffraction function sharply peaks at the resonance frequency \(\omega = \omega_{f i}\):

\lim_{t \rightarrow \infty} \frac{\sin^2(\Delta \omega \, t / 2)}{\pi t (\Delta \omega / 2)^2} = \delta(\Delta \omega)

For transitions into a continuum of final states with density of states \(\rho(E_f)\), the constant transition rate \(W_{i \rightarrow f} = \frac{d P}{dt}\) is given by Fermi's Golden Rule:

W_{i \rightarrow f} = \frac{2\pi}{\hbar} |F_{f i}|^2 \rho(E_f)

This fundamental theorem underpins Einstein's \(B\) coefficients, photochemical excitation rates, and spectroscopic absorption cross-sections.

§5.7 Two-Level Quantum Systems, Rabi Oscillations & Avoided Energy Crossings

Two-level systems represent the quintessential model in quantum optics, magnetic resonance, and quantum information (qubits).

The Two-State Hamiltonian

In a basis of two orthonormal states \(\{|1\rangle, |2\rangle\}\), the Hamiltonian matrix is:

\mathbf{H} = \begin{pmatrix} E_1 & V \\ V^* & E_2 \end{pmatrix}

where \(E_1\) and \(E_2\) are diabatic energies and \(V\) is the coupling matrix element. Let \(\bar{E} = \frac{E_1 + E_2}{2}\) and the energy detuning \(\Delta = E_1 - E_2\). Then:

\mathbf{H} = \bar{E} \mathbf{I} + \begin{pmatrix} \Delta/2 & V \\ V^* & -\Delta/2 \end{pmatrix}

Exact Eigenvalues and Avoided Crossing

Solving the secular determinant \(\det(\mathbf{H} - \lambda \mathbf{I}) = 0\):

(E_1 - \lambda)(E_2 - \lambda) - |V|^2 = 0 \implies \lambda^2 - (E_1 + E_2)\lambda + (E_1 E_2 - |V|^2) = 0

The exact adiabatic eigenenergies are:

E_\pm = \frac{E_1 + E_2}{2} \pm \sqrt{\left( \frac{E_1 - E_2}{2} \right)^2 + |V|^2} = \bar{E} \pm \frac{1}{2} \sqrt{\Delta^2 + 4 |V|^2}

The Avoided Crossing Phenomenon (Wigner-von Neumann Non-Crossing Rule):

When the states are uncoupled (\(V = 0\)), the diagonal diabatic energies cross cleanly at resonance \(\Delta = 0\) (\(E_1 = E_2\)).
When a coupling exists (\(V \ne 0\)), the square root is strictly positive: \(\sqrt{0 + 4|V|^2} = 2 |V|\).

The adiabatic energy levels repel each other, opening an energy gap:

\Delta E_{\text{gap}} = E_+ - E_- = \sqrt{\Delta^2 + 4 |V|^2} \ge 2 |V|

The energy curves never cross; at \(\Delta = 0\), the minimum separation is precisely \(2|V|\).

Rabi Oscillations in a Resonant Field

If a two-level system is driven on resonance (\(\Delta = 0\)) by a perturbation of amplitude \(V = \hbar \Omega_R / 2\), the probability of finding the system in the excited state oscillates sinusoidally:

P_{1 \rightarrow 2}(t) = \sin^2\left( \frac{\Omega_R t}{2} \right)

where \(\Omega_R = \frac{2 |V|}{\hbar}\) is the Rabi frequency. Complete periodic population inversion occurs at intervals of \(\tau = \pi / \Omega_R\) (a \(\pi\)-pulse).

§5.8 The WKB Semiclassical Approximation & Barrier Penetration

The Wentzel-Kramers-Brillouin (WKB) approximation is a powerful semiclassical method for solving the one-dimensional Schrödinger equation when the potential energy \(V(x)\) varies slowly compared to the local de Broglie wavelength:

\left| \frac{d\lambda_{\text{dB}}}{dx} \right| \ll 1, \quad \text{where } \lambda_{\text{dB}}(x) = \frac{h}{p(x)} = \frac{h}{\sqrt{2m(E - V(x))}}

The Semiclassical Wavefunction

Substituting the ansatz \(\psi(x) = e^{i S(x) / \hbar}\) and expanding \(S(x) = S_0(x) + \hbar S_1(x) + \dots\):

1. Classically Allowed Region (\(E > V(x)\)):

The local momentum is real: \(p(x) = \sqrt{2m(E - V(x))}\).

\psi(x) \approx \frac{C}{\sqrt{p(x)}} \exp\left( \pm \frac{i}{\hbar} \int p(x) dx \right)

The probability density \(|\psi(x)|^2 \propto \frac{1}{p(x)} \propto \frac{1}{v(x)}\) is inversely proportional to velocity, matching the classical dwell time!

2. Classically Forbidden Region (\(E < V(x)\)):

The local momentum is imaginary: \(\kappa(x) = \sqrt{2m(V(x) - E)}\).

\psi(x) \approx \frac{D}{\sqrt{\kappa(x)}} \exp\left( \pm \frac{1}{\hbar} \int \kappa(x) dx \right)

Wavefunctions decay exponentially through the barrier.

Connection Formulas and Classical Turning Points

At the classical turning points where \(E = V(x)\), \(p(x) \rightarrow 0\) and \(\lambda_{\text{dB}} \rightarrow \infty\), causing the standard WKB amplitudes to diverge. By linearizing the potential near the turning point (\(V(x) \approx E + V'(x_0)(x - x_0)\)), the exact solutions are Airy functions \(\operatorname{Ai}(z)\) and \(\operatorname{Bi}(z)\). Matching Airy asymptotic expansions across turning points yields the WKB Connection Formulas, which introduce a phase shift of \(\pi/4\) at each soft turning point.

WKB Quantization Condition (Bohr-Sommerfeld Rule)

For a bound state in a potential well with two classical turning points \(a\) and \(b\):

\int_a^b p(x) dx = \left( n + \frac{1}{2} \right) \pi \hbar \quad (n = 0, 1, 2, \dots)

Remarkably, for the harmonic oscillator, the WKB quantization rule yields the exact quantum spectrum \(E_n = (n + 1/2)\hbar\omega\)!

WKB Barrier Penetration (Gamow Tunneling)

For quantum tunneling through a broad potential barrier between turning points \(x_1\) and \(x_2\), the transmission probability is:

T_{\text{WKB}} \approx \exp\left( -2 \gamma \right) = \exp\left( -\frac{2}{\hbar} \int_{x_1}^{x_2} \sqrt{2m(V(x) - E)} \, dx \right)

George Gamow used this formula in 1928 to derive the Geiger-Nuttall law for alpha decay, providing the first quantum mechanical explanation of nuclear half-lives.

## Advanced Mathematical Supplement: Dalgarno-Lewis & Hylleraas Variational Bounds

The Dalgarno-Lewis Formalism & Dual Variational Bounds in Perturbation Theory

Standard second-order perturbation theory requires evaluating an infinite sum over all bound and continuum states of the unperturbed system:

E_0^{(2)} = \sum_{k \ne 0} \frac{|\langle \psi_k^{(0)} | \hat{H}' | \psi_0^{(0)} \rangle|^2}{E_0^{(0)} - E_k^{(0)}}

In practice, evaluating the continuum integral is often mathematically intractable.

The Dalgarno-Lewis Equation

In 1955, Alexander Dalgarno and John T. Lewis proved that the infinite summation can be converted into an equivalent first-order inhomogeneous differential equation. Define a function \(F(\mathbf{r})\) such that the first-order wavefunction is:

\psi_0^{(1)}(\mathbf{r}) = F(\mathbf{r}) \psi_0^{(0)}(\mathbf{r})

Substituting into the first-order perturbation equation \((\hat{H}^{(0)} - E_0^{(0)}) \psi_0^{(1)} = -( \hat{H}' - E_0^{(1)} ) \psi_0^{(0)}\) yields the Dalgarno-Lewis differential equation:

-\frac{\hbar^2}{2m} \left[ \psi_0^{(0)} \nabla^2 F + 2 \nabla\psi_0^{(0)} \cdot \nabla F \right] = -( \hat{H}' - E_0^{(1)} ) \psi_0^{(0)}

Once \(F(\mathbf{r})\) is found, the second-order energy is given directly by a simple single integral:

E_0^{(2)} = \langle \psi_0^{(0)} | \hat{H}' | \psi_0^{(1)} \rangle = \int |\psi_0^{(0)}(\mathbf{r})|^2 F(\mathbf{r}) [ \hat{H}'(\mathbf{r}) - E_0^{(1)} ] d\mathbf{r}

no infinite sum or continuum integration required!

The Hylleraas Variational Principle for Second-Order Energy

Egil Hylleraas formulated a variational principle that provides a rigorous upper bound on the second-order perturbation energy. Define the Hylleraas functional for a trial first-order correction \(\chi\):

J[\chi] = \langle \chi | \hat{H}^{(0)} - E_0^{(0)} | \chi \rangle + 2 \text{Re} \langle \chi | \hat{H}' - E_0^{(1)} | \psi_0^{(0)} \rangle

Theorem: For any trial function \(\chi\), the functional \(J[\chi]\) satisfies:

J[\chi] \ge E_0^{(2)}

The exact second-order energy \(E_0^{(2)}\) is the global minimum of \(J[\chi]\), achieved if and only if \(\chi = \psi_0^{(1)}\). This allows variational parameters to be optimized directly to compute second-order polarizabilities and dispersion coefficients without summing over unperturbed states.

## Research Monograph: Modern Density Functional Theory & Jacob's Ladder of Functionals

The Density Functional Paradigm

In wavefunction theory, an \(N\)-electron molecule requires a wavefunction \(\Psi(\mathbf{r}_1, \dots, \mathbf{r}_N)\) depending on \(3N\) spatial and \(N\) spin coordinates—a computational bottleneck for large chemical systems. Density Functional Theory (DFT) replaces the \(3N\)-dimensional wavefunction with the three-dimensional ground-state electron density \(\rho(\mathbf{r})\).

The Hohenberg-Kohn Theorems (1964)

1. First Theorem (Existence): The ground-state electron density \(\rho(\mathbf{r})\) uniquely determines the external potential \(v_{\text{ext}}(\mathbf{r})\) (up to an additive constant), and hence completely determines the Hamiltonian, ground-state energy, and all physical properties of the system.

2. Second Theorem (Variational Principle): The exact ground-state energy is the global minimum of the universal energy functional \(E[\rho]\):

E[\rho] = F_{\text{HK}}[\rho] + \int \rho(\mathbf{r}) v_{\text{ext}}(\mathbf{r}) d\mathbf{r} \ge E_0

where the universal functional \(F_{\text{HK}}[\rho] = T[\rho] + V_{e e}[\rho]\) is completely independent of the external potential.

The Kohn-Sham Self-Consistent Scheme (1965)

Walter Kohn and Lu Jeu Sham mapped the intractable interacting system onto a fictitious non-interacting reference system possessing the exact same ground-state density:

\rho(\mathbf{r}) = \sum_{i=1}^N |\phi_i(\mathbf{r})|^2

The energy functional is partitioned into:

E_{\text{KS}}[\rho] = T_s[\rho] + J[\rho] + \int \rho(\mathbf{r}) v_{\text{ext}}(\mathbf{r}) d\mathbf{r} + E_{\text{xc}}[\rho]

where \(T_s = -\frac{1}{2} \sum \langle \phi_i | \nabla^2 | \phi_i \rangle\) is the non-interacting kinetic energy, \(J[\rho] = \frac{1}{2} \iint \frac{\rho(\mathbf{r})\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} d\mathbf{r} d\mathbf{r}'\) is the classical Coulomb repulsion, and \(E_{\text{xc}}[\rho]\) is the Exchange-Correlation Functional, which contains all quantum many-body effects.

Jacob's Ladder of Density Functionals

John Perdew organized approximations to \(E_{\text{xc}}[\rho]\) onto "Jacob's Ladder" connecting the terrestrial realm of Hartree to the heaven of chemical accuracy:

1. Rung 1: Local Density Approximation (LDA): \(E_{\text{xc}}[\rho] = \int \rho(\mathbf{r}) \varepsilon_{\text{xc}}(\rho) d\mathbf{r}\) (homogeneous electron gas).

2. Rung 2: Generalized Gradient Approximation (GGA): Depends on density and local gradient: \(E_{\text{xc}}[\rho, \nabla\rho]\) (e.g., PBE, BLYP).

3. Rung 3: Meta-GGA: Includes kinetic energy density \(\tau(\mathbf{r}) = \frac{1}{2}\sum |\nabla\phi_i|^2\) and \(\nabla^2\rho\) (e.g., SCAN, TPSS).

4. Rung 4: Hybrid Functionals: Incorporates exact Hartree-Fock exchange (e.g., B3LYP, PBE0, HSE06).

5. Rung 5: Double Hybrids: Incorporates unoccupied virtual orbitals via MP2 or random phase approximation (RPA).

Modern Grimme dispersion corrections (DFT-D3, DFT-D4) additionally capture long-range van der Waals London dispersion, making DFT the dominant workhorse of materials science and drug design.

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.

Advanced Example 5.1: First-Order Perturbation of 1D Particle in a Box by a Delta-Function Potential

Consider a particle of mass \(m\) in an infinite potential well of width \(L\) (\(V(x) = 0\) for \(0 < x < L\), \(\infty\) elsewhere). A perturbation \(\hat{H}' = \alpha \delta\left(x - \frac{L}{2}\right)\) is added at the exact center of the box.

Write the unperturbed eigenfunctions \(\psi_n^{(0)}(x)\) and energies \(E_n^{(0)}\).
Calculate the first-order energy correction \(E_n^{(1)}\) for all quantum states \(n\).
Explain why even-\(n\) states experience zero energy shift, while odd-\(n\) states experience positive shifts.

Comprehensive Multi-Step Solution:

Step 1: Unperturbed Eigenstates

The unperturbed eigenfunctions and eigenvalues for a 1D box of width \(L\) are:

\psi_n^{(0)}(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi x}{L} \right), \quad E_n^{(0)} = \frac{n^2 \pi^2 \hbar^2}{2 m L^2} \quad (n = 1, 2, 3, \dots)

Step 2: First-Order Energy Correction

The first-order perturbation energy is:

E_n^{(1)} = \langle \psi_n^{(0)} | \hat{H}' | \psi_n^{(0)} \rangle = \int_0^L [\psi_n^{(0)}(x)]^2 \alpha \delta\left(x - \frac{L}{2}\right) dx

Using the sifting property of the Dirac delta function \(\int f(x) \delta(x - x_0) dx = f(x_0)\):

E_n^{(1)} = \alpha \left[ \psi_n^{(0)}\left(\frac{L}{2}\right) \right]^2 = \alpha \left[ \sqrt{\frac{2}{L}} \sin\left( \frac{n\pi (L/2)}{L} \right) \right]^2 = \frac{2\alpha}{L} \sin^2\left( \frac{n\pi}{2} \right)

Evaluate \(\sin^2\left(\frac{n\pi}{2}\right)\) for different \(n\):

For even \(n\) (\(n = 2, 4, 6, \dots\)):

\frac{n\pi}{2} = k\pi \implies \sin(k\pi) = 0 \implies E_n^{(1)} = 0

For odd \(n\) (\(n = 1, 3, 5, \dots\)):

\frac{n\pi}{2} = \frac{\pi}{2}, \frac{3\pi}{2}, \dots \implies \sin\left(\frac{n\pi}{2}\right) = \pm 1 \implies \sin^2\left(\frac{n\pi}{2}\right) = 1 \implies E_n^{(1)} = \frac{2\alpha}{L}

Step 3: Physical Interpretation

For even-\(n\) states, the wavefunction possesses a node at the center of the box: \(\psi_n^{(0)}(L/2) = 0\). The particle has exactly zero probability density of being located at \(x = L/2\). Because the perturbation is localized strictly at that single point, the particle never encounters the perturbation, resulting in \(E_n^{(1)} = 0\).
For odd-\(n\) states, the center of the box corresponds to an antinode (maximum or minimum amplitude): \(|\psi_n^{(0)}(L/2)| = \sqrt{2/L}\). The probability density at the center is \(\frac{2}{L}\). A repulsive delta spike (\(\alpha > 0\)) raises the energy by \(\frac{2\alpha}{L}\).

Advanced Example 5.2: Second-Order Perturbation Energy Shift for Anharmonic Perturbation cx^4

A harmonic oscillator of mass \(m\) and frequency \(\omega\) is perturbed by a quartic anharmonic potential \(\hat{H}' = c \hat{x}^4\).

Express \(\hat{x}\) in terms of ladder operators \(\hat{a}\) and \(\hat{a}^\dagger\).
Calculate the first-order energy correction \(E_0^{(1)}\) for the ground state \(|0\rangle\).
Set up and evaluate the non-zero terms in the second-order sum to find \(E_0^{(2)}\).

Comprehensive Multi-Step Solution:

Step 1: Position Operator in Ladder Representation

The position operator is:

\hat{x} = \sqrt{\frac{\hbar}{2 m \omega}} (\hat{a} + \hat{a}^\dagger)

Therefore:

\hat{x}^4 = \left(\frac{\hbar}{2 m \omega}\right)^2 (\hat{a} + \hat{a}^\dagger)^4

Step 2: First-Order Ground-State Shift \(E_0^{(1)}\)

The first-order shift is:

E_0^{(1)} = c \langle 0 | \hat{x}^4 | 0 \rangle = c \left(\frac{\hbar}{2 m \omega}\right)^2 \langle 0 | (\hat{a} + \hat{a}^\dagger)^4 | 0 \rangle

Expand \((\hat{a} + \hat{a}^\dagger)^4 | 0 \rangle\). Only terms with equal numbers of creation and annihilation operators have non-zero expectation value \(\langle 0 | \dots | 0 \rangle\). Among the 16 expansion terms, only those ending in \(\hat{a}^\dagger\) can survive acting on \(|0\rangle\). Specifically, \(\hat{a} \hat{a} \hat{a}^\dagger \hat{a}^\dagger |0\rangle = \hat{a} \hat{a} (\sqrt{2}|2\rangle) = 2 |0\rangle\), and \(\hat{a} \hat{a}^\dagger \hat{a} \hat{a}^\dagger |0\rangle = 1 |0\rangle\). Total sum: \(\langle 0 | (\hat{a} + \hat{a}^\dagger)^4 | 0 \rangle = 3\). (This also follows from the Gaussian moment: \(\langle x^4 \rangle = 3 \langle x^2 \rangle^2 = 3 \left( \frac{\hbar}{2 m \omega} \right)^2\)). Thus:

E_0^{(1)} = 3 c \left(\frac{\hbar}{2 m \omega}\right)^2 = \frac{3 c \hbar^2}{4 m^2 \omega^2}

Step 3: Second-Order Shift \(E_0^{(2)}\)

The second-order perturbation formula is:

E_0^{(2)} = \sum_{k \ne 0} \frac{|\langle k | \hat{H}' | 0 \rangle|^2}{E_0^{(0)} - E_k^{(0)}}

Since \(E_k^{(0)} - E_0^{(0)} = k \hbar \omega\), the denominator is \(-k \hbar \omega\). Now examine which states \(|k\rangle\) couple to \(|0\rangle\) via \((\hat{a} + \hat{a}^\dagger)^4\):

Operating on \(|0\rangle\), \((\hat{a} + \hat{a}^\dagger)^4 |0\rangle\) contains combinations that generate only states with parity matched to \(0\), specifically \(|0\rangle\), \(|2\rangle\), and \(|4\rangle\).

1. Coupling to \(|2\rangle\):

\langle 2 | (\hat{a} + \hat{a}^\dagger)^4 | 0 \rangle = \sqrt{2} (1 + 2 + 3) \dots \implies \langle 2 | \hat{x}^4 | 0 \rangle = \left(\frac{\hbar}{2 m \omega}\right)^2 \sqrt{2} \times 6 = 6\sqrt{2} \left(\frac{\hbar}{2 m \omega}\right)^2

Energy denominator: \(E_0^{(0)} - E_2^{(0)} = -2 \hbar \omega\). Contribution: \(\frac{|6\sqrt{2}|^2}{-2 \hbar \omega} \left(\frac{\hbar}{2 m \omega}\right)^4 = \frac{72}{-2 \hbar \omega} \left(\frac{\hbar}{2 m \omega}\right)^4 = -36 \frac{1}{\hbar\omega} \left(\frac{\hbar}{2 m \omega}\right)^4\).

2. Coupling to \(|4\rangle\):

(\hat{a}^\dagger)^4 |0\rangle = \sqrt{1 \times 2 \times 3 \times 4} |4\rangle = \sqrt{24} |4\rangle

Matrix element: \(\langle 4 | \hat{x}^4 | 0 \rangle = \sqrt{24} \left(\frac{\hbar}{2 m \omega}\right)^2\). Energy denominator: \(E_0^{(0)} - E_4^{(0)} = -4 \hbar \omega\). Contribution: \(\frac{24}{-4 \hbar \omega} \left(\frac{\hbar}{2 m \omega}\right)^4 = -6 \frac{1}{\hbar\omega} \left(\frac{\hbar}{2 m \omega}\right)^4\).

Summing both non-zero terms:

E_0^{(2)} = c^2 \left( -36 - 6 \right) \frac{1}{\hbar \omega} \left(\frac{\hbar}{2 m \omega}\right)^4 = -42 \frac{c^2}{\hbar \omega} \left( \frac{\hbar^4}{16 m^4 \omega^4} \right) = -\frac{21 c^2 \hbar^3}{8 m^4 \omega^5}

Notice that \(E_0^{(2)} < 0\), verifying the general theorem that second-order ground-state energy shifts are strictly negative.

Advanced Example 5.3: Variational Ground-State Energy for 1D Harmonic Oscillator Using Gaussian Trial

Consider a 1D harmonic oscillator with Hamiltonian \(\hat{H} = -\frac{\hbar^2}{2 m} \frac{d^2}{dx^2} + \frac{1}{2} k x^2\).

Propose a normalized Gaussian trial wavefunction \(\Phi(x; \alpha) = \left( \frac{2\alpha}{\pi} \right)^{1/4} e^{-\alpha x^2}\) with variational parameter \(\alpha > 0\).
Calculate the variational energy \(E_{\text{var}}(\alpha) = \langle \Phi | \hat{H} | \Phi \rangle\) as a function of \(\alpha\).
Find the optimal value \(\alpha_{\text{opt}}\) that minimizes \(E_{\text{var}}\), evaluate the minimum energy, and compare to the exact ground-state energy \(\frac{1}{2} \hbar \omega\).

Comprehensive Multi-Step Solution:

Step 1: Trial Wavefunction and Normalization

The trial wavefunction is:

\Phi(x; \alpha) = \left( \frac{2\alpha}{\pi} \right)^{1/4} e^{-\alpha x^2}

Verify normalization:

\int_{-\infty}^\infty |\Phi(x)|^2 dx = \left( \frac{2\alpha}{\pi} \right)^{1/2} \int_{-\infty}^\infty e^{-2\alpha x^2} dx = \left( \frac{2\alpha}{\pi} \right)^{1/2} \sqrt{\frac{\pi}{2\alpha}} = 1

Step 2: Evaluation of Kinetic and Potential Energy Expectation Values

1. Kinetic Energy:

\frac{d \Phi}{dx} = -2\alpha x \Phi

\frac{d^2 \Phi}{dx^2} = (-2\alpha + 4\alpha^2 x^2) \Phi

\langle \hat{T} \rangle = -\frac{\hbar^2}{2m} \int_{-\infty}^\infty \Phi^* \frac{d^2 \Phi}{dx^2} dx = -\frac{\hbar^2}{2m} \left[ -2\alpha + 4\alpha^2 \langle x^2 \rangle \right]

Using Gaussian integral \(\langle x^2 \rangle = \frac{1}{4\alpha}\):

\langle \hat{T} \rangle = -\frac{\hbar^2}{2m} \left[ -2\alpha + 4\alpha^2 \left( \frac{1}{4\alpha} \right) \right] = -\frac{\hbar^2}{2m} [-\alpha] = \frac{\hbar^2 \alpha}{2m}

2. Potential Energy:

\langle \hat{V} \rangle = \frac{1}{2} k \langle x^2 \rangle = \frac{1}{2} k \left( \frac{1}{4\alpha} \right) = \frac{k}{8\alpha}

Summing kinetic and potential terms:

E_{\text{var}}(\alpha) = \frac{\hbar^2 \alpha}{2m} + \frac{k}{8\alpha}

Step 3: Minimization and Comparison

To minimize \(E_{\text{var}}(\alpha)\), differentiate with respect to \(\alpha\):

\frac{d E_{\text{var}}}{d\alpha} = \frac{\hbar^2}{2m} - \frac{k}{8\alpha^2} = 0 \implies \frac{k}{8\alpha^2} = \frac{\hbar^2}{2m}

\alpha^2 = \frac{2m k}{8\hbar^2} = \frac{m k}{4\hbar^2} \implies \alpha_{\text{opt}} = \frac{\sqrt{m k}}{2\hbar} = \frac{m\omega}{2\hbar}

where \(\omega = \sqrt{k/m}\). Substitute \(\alpha_{\text{opt}}\) into \(E_{\text{var}}\):

E_{\text{var}}(\alpha_{\text{opt}}) = \frac{\hbar^2}{2m} \left( \frac{m\omega}{2\hbar} \right) + \frac{m\omega^2}{8 \left( \frac{m\omega}{2\hbar} \right)} = \frac{1}{4}\hbar\omega + \frac{1}{4}\hbar\omega = \frac{1}{2} \hbar \omega

The variational result matches the exact ground-state energy \(E_0 = \frac{1}{2}\hbar\omega\) precisely! This is because the exact analytical eigenfunction of the harmonic oscillator ground state happens to be an exact Gaussian, which belongs to the trial family.

Advanced Example 5.4: Variational Calculation of Helium Ground State with Screening Parameter

For the neutral helium atom (\(Z = 2\)):

Using the scaled trial wavefunction \(\Phi(\mathbf{r}_1, \mathbf{r}_2; \zeta) = \frac{\zeta^3}{\pi} e^{-\zeta(r_1 + r_2)}\), write the expression for \(E_{\text{var}}(\zeta)\) in atomic units.
Find the optimal screening parameter \(\zeta_{\text{opt}}\) and compute the resulting ground-state energy in atomic units (Hartrees) and in eV.
Compute the first ionization energy of helium predicted by this variational calculation and compare it with the experimental value of \(24.59\text{ eV}\).

Comprehensive Multi-Step Solution:

Step 1: Variational Energy Function

In atomic units (\(1\text{ a.u.} = 27.2114\text{ eV}\)), the variational energy expectation value for the scaled trial wavefunction is:

E_{\text{var}}(\zeta) = \zeta^2 - 2 Z \zeta + \frac{5}{8} \zeta

For helium with nuclear charge \(Z = 2\):

E_{\text{var}}(\zeta) = \zeta^2 - \left( 4 - \frac{5}{8} \right) \zeta = \zeta^2 - \frac{27}{8} \zeta

Step 2: Optimal Screening Parameter and Minimum Energy

Differentiate with respect to \(\zeta\):

\frac{d E_{\text{var}}}{d\zeta} = 2\zeta - \frac{27}{8} = 0 \implies \zeta_{\text{opt}} = \frac{27}{16} = 1.6875

Substitute \(\zeta_{\text{opt}}\) back into \(E_{\text{var}}\):

E_{\text{var}}(\zeta_{\text{opt}}) = \left(\frac{27}{16}\right)^2 - \frac{27}{8}\left(\frac{27}{16}\right) = -\left(\frac{27}{16}\right)^2 = -\frac{729}{256} \approx -2.84766\text{ a.u.}

Converting to electron volts:

E_{\text{var}} = -2.84766 \times 27.2114\text{ eV} \approx -77.488\text{ eV}

Comparing with the true experimental non-relativistic value \(E_{\text{exp}} = -79.005\text{ eV}\):

\text{Error} = \frac{-77.488 - (-79.005)}{79.005} \times 100\% = \frac{1.517\text{ eV}}{79.005\text{ eV}} \times 100\% \approx 1.92\%

As guaranteed by the variational theorem, \(E_{\text{var}} > E_{\text{exact}}\) (it is an upper bound).

Step 3: First Ionization Energy of Helium

Ionization corresponds to removing one electron to produce \(\text{He}^+\) in its ground state:

\text{He} \rightarrow \text{He}^+ + e^-

The ground state of the hydrogenic ion \(\text{He}^+\) (\(Z = 2\)) has exact energy:

E(\text{He}^+) = -\frac{Z^2}{2} = -\frac{2^2}{2} = -2.000\text{ a.u.} = -54.423\text{ eV}

The predicted first ionization energy is:

I_1 = E(\text{He}^+) - E(\text{He}) = -2.000 - (-2.84766) = +0.84766\text{ a.u.}

Converting to eV:

I_1 = 0.84766 \times 27.2114\text{ eV} \approx 23.066\text{ eV}

Comparison with experimental value:

I_{1, \text{exp}} = 24.59\text{ eV}

The simple one-parameter screened variational model captures 93.8% of the experimental ionization energy, demonstrating how shielding accounts for the dominant portion of electron correlation.

Advanced Example 5.5: Degenerate Perturbation Theory for 2D Box with Asymmetric Perturbation

A particle moves in a 2D square box with potential \(V(x, y) = 0\) for \(0 < x, y < L\) and \(\infty\) elsewhere.

Identify the unperturbed energies and degeneracies of the first excited state (\(E^{(0)} = \frac{5\pi^2\hbar^2}{2 m L^2}\)).
A perturbation \(\hat{H}' = W_0\) is applied only in the quadrant \(0 < x < L/2, 0 < y < L/2\). Set up the \(2 \times 2\) secular determinant.
Solve for the first-order energy shifts \(E^{(1)}\) and the proper zeroth-order wavefunctions.

Comprehensive Multi-Step Solution:

Step 1: Unperturbed Eigenstates and Degeneracy

The unperturbed states are:

\psi_{n_x, n_y}^{(0)}(x, y) = \frac{2}{L} \sin\left(\frac{n_x \pi x}{L}\right) \sin\left(\frac{n_y \pi y}{L}\right), \quad E_{n_x, n_y}^{(0)} = \frac{\pi^2 \hbar^2}{2 m L^2} (n_x^2 + n_y^2)

For the first excited level with energy \(E^{(0)} = \frac{5\pi^2\hbar^2}{2mL^2}\), there are two degenerate states (\(g = 2\)):

State 1: \(|1\rangle = \psi_{1, 2}^{(0)}\) (\(n_x = 1, n_y = 2\))
State 2: \(|2\rangle = \psi_{2, 1}^{(0)}\) (\(n_x = 2, n_y = 1\))

Step 2: Calculation of Matrix Elements

The perturbation is \(\hat{H}' = W_0\) for \(x \in [0, L/2]\) and \(y \in [0, L/2]\), and zero elsewhere.

1. Diagonal Elements \(H'_{11}\) and \(H'_{22}\):

H'_{11} = W_0 \int_0^{L/2} \left[ \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) \right]^2 dx \int_0^{L/2} \left[ \sqrt{\frac{2}{L}} \sin\left(\frac{2\pi y}{L}\right) \right]^2 dy

Evaluate each integral:

\int_0^{L/2} \sin^2\left(\frac{\pi x}{L}\right) dx = \left[ \frac{x}{2} - \frac{L}{4\pi} \sin\left(\frac{2\pi x}{L}\right) \right]_0^{L/2} = \frac{L}{4} - 0 = \frac{L}{4}

\int_0^{L/2} \sin^2\left(\frac{2\pi y}{L}\right) dy = \left[ \frac{y}{2} - \frac{L}{8\pi} \sin\left(\frac{4\pi y}{L}\right) \right]_0^{L/2} = \frac{L}{4} - 0 = \frac{L}{4}

Thus:

H'_{11} = W_0 \left(\frac{2}{L} \times \frac{L}{4}\right) \left(\frac{2}{L} \times \frac{L}{4}\right) = W_0 \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{W_0}{4}

By symmetry, \(H'_{22} = H'_{11} = \frac{W_0}{4}\).

2. Off-Diagonal Element \(H'_{12}\):

H'_{12} = W_0 \left[ \frac{2}{L} \int_0^{L/2} \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi x}{L}\right) dx \right]^2

Using the identity \(\sin(A)\sin(B) = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\):

\sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi x}{L}\right) = \frac{1}{2} \left[ \cos\left(\frac{\pi x}{L}\right) - \cos\left(\frac{3\pi x}{L}\right) \right]

\int_0^{L/2} \sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{2\pi x}{L}\right) dx = \frac{1}{2} \left[ \frac{L}{\pi} \sin\left(\frac{\pi x}{L}\right) - \frac{L}{3\pi} \sin\left(\frac{3\pi x}{L}\right) \right]_0^{L/2}

= \frac{L}{2\pi} \left[ \sin(\pi/2) - \frac{1}{3} \sin(3\pi/2) \right] = \frac{L}{2\pi} \left[ 1 - \frac{1}{3}(-1) \right] = \frac{L}{2\pi} \left( \frac{4}{3} \right) = \frac{2 L}{3\pi}

Multiplying by the prefactor \(\frac{2}{L}\):

\frac{2}{L} \times \frac{2L}{3\pi} = \frac{4}{3\pi}

Since both \(x\) and \(y\) integrals are identical:

H'_{12} = W_0 \left( \frac{4}{3\pi} \right)^2 = \frac{16 W_0}{9\pi^2}

Step 3: Secular Equation and Eigenvalues

The secular equation is:

\begin{vmatrix} \frac{W_0}{4} - E^{(1)} & \frac{16 W_0}{9\pi^2} \\ \frac{16 W_0}{9\pi^2} & \frac{W_0}{4} - E^{(1)} \end{vmatrix} = 0

\left( \frac{W_0}{4} - E^{(1)} \right)^2 - \left( \frac{16 W_0}{9\pi^2} \right)^2 = 0 \implies E^{(1)} = \frac{W_0}{4} \pm \frac{16 W_0}{9\pi^2}

Numerical values:

\frac{1}{4} = 0.2500, \quad \frac{16}{9\pi^2} \approx \frac{16}{88.826} \approx 0.1801

\(E_+^{(1)} = (0.2500 + 0.1801) W_0 = 0.4301 W_0\)
\(E_-^{(1)} = (0.2500 - 0.1801) W_0 = 0.0699 W_0\)

The degeneracy is completely lifted. The corresponding proper zeroth-order eigenstates are the symmetric and antisymmetric linear combinations:

\psi_+^{(0)} = \frac{1}{\sqrt{2}} (\psi_{1, 2}^{(0)} + \psi_{2, 1}^{(0)}), \quad \psi_-^{(0)} = \frac{1}{\sqrt{2}} (\psi_{1, 2}^{(0)} - \psi_{2, 1}^{(0)})

Advanced Example 5.6: Transition Probability in a Sinusoidally Driven Two-Level System

A two-level quantum system has unperturbed states \(|1\rangle\) and \(|2\rangle\) with energy difference \(\hbar \omega_0 = E_2 - E_1\). At \(t = 0\), a time-dependent perturbation \(\hat{V}(t) = V_0 \cos(\omega t) (|1\rangle\langle 2| + |2\rangle\langle 1|)\) is applied.

Using first-order time-dependent perturbation theory, derive the transition probability \(P_{1 \rightarrow 2}(t)\) assuming the system starts in \(|1\rangle\) at \(t = 0\).
Identify the resonance condition and determine the transition probability on exact resonance (\(\omega = \omega_0\)).
Discuss the limitation of first-order perturbation theory at long times \(t \gg \hbar / V_0\).

Comprehensive Multi-Step Solution:

Step 1: Derivation of Transition Probability

The first-order transition amplitude is:

c_2^{(1)}(t) = -\frac{i}{\hbar} \int_0^t \langle 2 | \hat{V}(t') | 1 \rangle e^{i \omega_0 t'} dt'

Given \(\langle 2 | \hat{V}(t') | 1 \rangle = V_0 \cos(\omega t') = \frac{V_0}{2} (e^{i\omega t'} + e^{-i\omega t'})\):

c_2^{(1)}(t) = -\frac{i V_0}{2\hbar} \int_0^t \left[ e^{i(\omega_0 + \omega)t'} + e^{i(\omega_0 - \omega)t'} \right] dt'

Near resonance (\(\omega \approx \omega_0\)), the term \(e^{i(\omega_0 + \omega)t'}\) oscillates extremely rapidly at frequency \(\sim 2\omega_0\) and averages to near zero (the Rotating Wave Approximation, RWA). Retaining the dominant secular term:

c_2^{(1)}(t) \approx -\frac{i V_0}{2\hbar} \int_0^t e^{i(\omega_0 - \omega)t'} dt' = -\frac{i V_0}{2\hbar} \left[ \frac{e^{i(\omega_0 - \omega)t} - 1}{i(\omega_0 - \omega)} \right] = -\frac{V_0}{2\hbar} \frac{e^{i\Delta\omega t} - 1}{\Delta\omega}

where detuning \(\Delta\omega = \omega_0 - \omega\). The transition probability is the modulus squared:

P_{1 \rightarrow 2}(t) = |c_2^{(1)}(t)|^2 = \frac{V_0^2}{4\hbar^2} \frac{|e^{i\Delta\omega t} - 1|^2}{(\Delta\omega)^2} = \frac{V_0^2}{4\hbar^2} \frac{4 \sin^2\left(\frac{\Delta\omega t}{2}\right)}{(\Delta\omega)^2} = \frac{V_0^2}{\hbar^2} \frac{\sin^2\left(\frac{\Delta\omega t}{2}\right)}{(\Delta\omega)^2}

Step 2: Resonance Condition and Resonant Probability

Resonance occurs when the driving field frequency matches the transition frequency:

\omega = \omega_0 \implies \Delta\omega = 0

Taking the limit as \(\Delta\omega \rightarrow 0\) using \(\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1\):

\lim_{\Delta\omega \rightarrow 0} \frac{\sin^2(\Delta\omega t / 2)}{(\Delta\omega)^2} = \left(\frac{t}{2}\right)^2 = \frac{t^2}{4}

Therefore, on exact resonance:

P_{1 \rightarrow 2}^{\text{res}}(t) = \frac{V_0^2}{4 \hbar^2} t^2

The transition probability initially grows quadratically with time \(t^2\).

Step 3: Breakdown of Perturbation Theory at Long Times

First-order perturbation theory assumes the initial state population remains close to unity: \(|c_1(t)| \approx 1\). However, the formula \(P_{1 \rightarrow 2}(t) \propto t^2\) exceeds \(1\) when:

\frac{V_0^2 t^2}{4\hbar^2} > 1 \implies t > \frac{2\hbar}{V_0}

which violates probability conservation (\(P \le 1\)). At longer times, depletion of state \(|1\rangle\) cannot be neglected. The exact non-perturbative dynamics (governed by Rabi oscillations) yields:

P_{1 \rightarrow 2}^{\text{exact}}(t) = \sin^2\left( \frac{V_0 t}{2\hbar} \right)

For small arguments (\(V_0 t / 2\hbar \ll 1\)), the Taylor expansion \(\sin(x) \approx x\) yields \(\sin^2(x) \approx x^2 = \frac{V_0^2 t^2}{4\hbar^2}\), exactly matching our first-order perturbation result. First-order theory is thus valid strictly for short times \(t \ll \hbar / V_0\).

Advanced Example 5.7: Matrix Diagonalization and Avoided Crossing in a Perturbed 2-State System

Consider a two-state quantum system with diabatic Hamiltonian:

\mathbf{H}(\delta) = \begin{pmatrix} E_0 + \delta & V \\ V & E_0 - \delta \end{pmatrix}

where \(E_0 = 10.0\text{ eV}\), \(V = 0.50\text{ eV}\), and \(\delta\) is a tunable parameter from \(-2.0\text{ eV}\) to \(+2.0\text{ eV}\).

Solve for the exact adiabatic energy eigenvalues \(E_+(\delta)\) and \(E_-(\delta)\).
Calculate the energy eigenvalues and the energy gap at \(\delta = 0\), \(\delta = 1.0\text{ eV}\), and \(\delta = 2.0\text{ eV}\).
Find the normalized adiabatic eigenvectors at \(\delta = 0\) and show that the states undergo complete mixing.

Comprehensive Multi-Step Solution:

Step 1: Exact Eigenvalues

The secular equation is:

\det(\mathbf{H} - E \mathbf{I}) = \begin{vmatrix} E_0 + \delta - E & V \\ V & E_0 - \delta - E \end{vmatrix} = 0

(E_0 - E + \delta)(E_0 - E - \delta) - V^2 = 0 \implies (E_0 - E)^2 - \delta^2 - V^2 = 0

(E_0 - E)^2 = \delta^2 + V^2 \implies E_\pm = E_0 \pm \sqrt{\delta^2 + V^2}

This describes a hyperbola in the \((\delta, E)\) plane with asymptotes \(E = E_0 \pm \delta\).

Step 2: Energy Calculations and Gap Evaluation

The energy gap is:

\Delta E(\delta) = E_+(\delta) - E_-(\delta) = 2 \sqrt{\delta^2 + V^2}

Given \(E_0 = 10.0\text{ eV}\) and \(V = 0.50\text{ eV}\):

1. At \(\delta = 0\text{ eV}\) (Resonance point):

\sqrt{0^2 + 0.5^2} = 0.50\text{ eV}

\(E_+ = 10.0 + 0.50 = 10.50\text{ eV}\)
\(E_- = 10.0 - 0.50 = 9.50\text{ eV}\)
\(\Delta E_{\text{min}} = 2 V = 1.00\text{ eV}\) (Avoided crossing gap)

2. At \(\delta = 1.0\text{ eV}\):

\sqrt{1.0^2 + 0.5^2} = \sqrt{1.0 + 0.25} = \sqrt{1.25} \approx 1.118\text{ eV}

\(E_+ = 10.0 + 1.118 = 11.118\text{ eV}\)
\(E_- = 10.0 - 1.118 = 8.882\text{ eV}\)
\(\Delta E = 2 \times 1.118 = 2.236\text{ eV}\)

3. At \(\delta = 2.0\text{ eV}\):

\sqrt{2.0^2 + 0.5^2} = \sqrt{4.0 + 0.25} = \sqrt{4.25} \approx 2.062\text{ eV}

\(E_+ = 10.0 + 2.062 = 12.062\text{ eV}\)
\(E_- = 10.0 - 2.062 = 7.938\text{ eV}\)
\(\Delta E = 2 \times 2.062 = 4.123\text{ eV}\)

Step 3: Eigenvector Mixing at Resonance

At \(\delta = 0\), the Hamiltonian matrix is:

\mathbf{H}(0) = \begin{pmatrix} E_0 & V \\ V & E_0 \end{pmatrix} = \begin{pmatrix} 10.0 & 0.5 \\ 0.5 & 10.0 \end{pmatrix}

For eigenvalue \(E_+ = E_0 + V = 10.5\text{ eV}\):

\begin{pmatrix} -V & V \\ V & -V \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies c_1 = c_2

Normalized eigenvector:

|+\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt{2}} (|1\rangle + |2\rangle)

For eigenvalue \(E_- = E_0 - V = 9.5\text{ eV}\):

\begin{pmatrix} V & V \\ V & V \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \implies c_1 = -c_2

Normalized eigenvector:

|-\rangle = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \frac{1}{\sqrt{2}} (|1\rangle - |2\rangle)

At resonance (\(\delta = 0\)), each adiabatic state is an equal 50/50 quantum superposition of both diabatic states: \(P_1 = P_2 = |1/\sqrt{2}|^2 = 0.5\). This complete hybrid mixing is the quantum basis of chemical resonance and avoided crossings in non-adiabatic molecular transitions.

Advanced Example 5.8: WKB Tunneling Transmission Through a Parabolic Potential Barrier

Consider a particle of mass \(m\) and energy \(E\) encountering an inverted parabolic potential barrier:

V(x) = V_0 - \frac{1}{2} k x^2 \quad \text{for } |x| \le x_0

with \(E < V_0\).

Determine the classical turning points \(x_1\) and \(x_2\).
Using the WKB approximation, evaluate the barrier penetration integral \(\gamma = \frac{1}{\hbar} \int_{x_1}^{x_2} \sqrt{2m(V(x) - E)} \, dx\).
Derive the transmission coefficient \(T(E)\) and show that \(\ln T \propto -(V_0 - E)\).

Comprehensive Multi-Step Solution:

Step 1: Classical Turning Points

The turning points occur where \(V(x) = E\):

V_0 - \frac{1}{2} k x^2 = E \implies \frac{1}{2} k x^2 = V_0 - E \implies x^2 = \frac{2(V_0 - E)}{k}

Let \(a = \sqrt{\frac{2(V_0 - E)}{k}}\). The turning points are:

x_1 = -a, \quad x_2 = +a

Step 2: Evaluation of the WKB Barrier Integral

The WKB barrier exponent is:

\gamma = \frac{1}{\hbar} \int_{-a}^a \sqrt{2m \left( V_0 - \frac{1}{2}kx^2 - E \right)} \, dx = \frac{\sqrt{2m}}{\hbar} \int_{-a}^a \sqrt{(V_0 - E) - \frac{1}{2}kx^2} \, dx

Factor out \(\sqrt{V_0 - E}\):

\gamma = \frac{\sqrt{2m(V_0 - E)}}{\hbar} \int_{-a}^a \sqrt{1 - \frac{k x^2}{2(V_0 - E)}} \, dx = \frac{\sqrt{2m(V_0 - E)}}{\hbar} \int_{-a}^a \sqrt{1 - \frac{x^2}{a^2}} \, dx

Let \(x = a \sin\theta\):

\int_{-a}^a \sqrt{1 - \frac{x^2}{a^2}} dx = a \int_{-\pi/2}^{\pi/2} \cos^2\theta \, d\theta = a \left( \frac{\pi}{2} \right) = \frac{\pi a}{2}

Substitute \(a = \sqrt{\frac{2(V_0 - E)}{k}}\):

\gamma = \frac{\sqrt{2m(V_0 - E)}}{\hbar} \frac{\pi}{2} \sqrt{\frac{2(V_0 - E)}{k}} = \frac{\pi}{2\hbar} \sqrt{\frac{4m}{k}} (V_0 - E) = \frac{\pi}{\hbar} \sqrt{\frac{m}{k}} (V_0 - E)

Recognizing \(\omega = \sqrt{k/m}\):

\gamma = \frac{\pi (V_0 - E)}{\hbar \omega}

Step 3: Transmission Probability \(T(E)\)

In the WKB limit for \(\gamma \gg 1\):

T(E) \approx e^{-2\gamma} = \exp\left[ -\frac{2\pi (V_0 - E)}{\hbar \omega} \right]

Taking natural logarithms:

\ln T(E) = -\frac{2\pi}{\hbar\omega} (V_0 - E)

The tunneling transmission decreases exponentially with barrier height \(V_0 - E\). The exact Kemble barrier formula replaces \(e^{-2\gamma}\) with \(\frac{1}{1 + e^{2\gamma}}\), which correctly gives \(T = 1/2\) at the barrier apex (\(E = V_0\)).

Advanced Example 5.9: Hellmann-Feynman Theorem for Harmonic and Coulomb Potentials

The Hellmann-Feynman theorem states that for an exact normalized eigenstate \(\psi_\lambda\) with energy \(E(\lambda)\) governed by Hamiltonian \(\hat{H}(\lambda)\):

\frac{d E}{d\lambda} = \left\langle \psi_\lambda \left| \frac{\partial \hat{H}}{\partial \lambda} \right| \psi_\lambda \right\rangle

Prove the Hellmann-Feynman theorem using the hermiticity of \(\hat{H}\).
For the 1D harmonic oscillator with \(\hat{H} = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2} k x^2\) and \(E_n = \left(n + \frac{1}{2}\right)\hbar \sqrt{k/m}\), use the theorem with \(\lambda = k\) to find \(\langle x^2 \rangle_n\).
For the hydrogen atom with \(\hat{H} = -\frac{\hbar^2}{2m}\nabla^2 - \frac{Z e^2}{4\pi\varepsilon_0 r}\) and \(E_n = -\frac{m Z^2 e^4}{32\pi^2\varepsilon_0^2\hbar^2 n^2}\), use the theorem with \(\lambda = Z\) to evaluate \(\langle r^{-1} \rangle_{n l}\).

Comprehensive Multi-Step Solution:

Step 1: Proof of the Hellmann-Feynman Theorem

By definition:

E(\lambda) = \langle \psi(\lambda) | \hat{H}(\lambda) | \psi(\lambda) \rangle

Differentiate both sides with respect to \(\lambda\) using the product rule:

\frac{d E}{d\lambda} = \left\langle \frac{\partial \psi}{\partial \lambda} \middle| \hat{H} \middle| \psi \right\rangle + \left\langle \psi \middle| \frac{\partial \hat{H}}{\partial \lambda} \middle| \psi \right\rangle + \left\langle \psi \middle| \hat{H} \middle| \frac{\partial \psi}{\partial \lambda} \right\rangle

Because \(\hat{H}\) is Hermitian and \(\hat{H}|\psi\rangle = E|\psi\rangle\):

\left\langle \frac{\partial \psi}{\partial \lambda} \middle| \hat{H} \middle| \psi \right\rangle = E \left\langle \frac{\partial \psi}{\partial \lambda} \middle| \psi \right\rangle

\left\langle \psi \middle| \hat{H} \middle| \frac{\partial \psi}{\partial \lambda} \right\rangle = E \left\langle \psi \middle| \frac{\partial \psi}{\partial \lambda} \right\rangle

Summing both terms:

E \left[ \left\langle \frac{\partial \psi}{\partial \lambda} \middle| \psi \right\rangle + \left\langle \psi \middle| \frac{\partial \psi}{\partial \lambda} \right\rangle \right] = E \frac{d}{d\lambda} \langle \psi | \psi \rangle

Since the wavefunction is normalized for all \(\lambda\) (\(\langle \psi | \psi \rangle = 1\)), \(\frac{d}{d\lambda}(1) = 0\). Therefore, the boundary derivative terms vanish identically, leaving:

\frac{d E}{d\lambda} = \left\langle \psi \left| \frac{\partial \hat{H}}{\partial \lambda} \right| \psi \right\rangle \quad \text{Q.E.D.}

Step 2: Evaluation of \(\langle x^2 \rangle\) for the Harmonic Oscillator

Set \(\lambda = k\) (spring constant).

Differentiate the Hamiltonian:

\frac{\partial \hat{H}}{\partial k} = \frac{\partial}{\partial k} \left( -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + \frac{1}{2} k x^2 \right) = \frac{1}{2} x^2

Differentiate the exact energy \(E_n = \left(n + \frac{1}{2}\right)\hbar \sqrt{\frac{k}{m}}\):

\frac{d E_n}{dk} = \left(n + \frac{1}{2}\right)\frac{\hbar}{\sqrt{m}} \frac{1}{2\sqrt{k}} = \frac{1}{2k} \left(n + \frac{1}{2}\right)\hbar\omega = \frac{E_n}{2k}

Equating by the Hellmann-Feynman theorem:

\left\langle \frac{1}{2} x^2 \right\rangle = \frac{d E_n}{dk} = \frac{E_n}{2k} \implies \langle x^2 \rangle_n = \frac{E_n}{k} = \left(n + \frac{1}{2}\right) \frac{\hbar}{m\omega}

Notice that \(\langle V \rangle = \frac{1}{2} k \langle x^2 \rangle = \frac{1}{2} E_n\), automatically verifying the virial theorem!

Step 3: Evaluation of \(\langle r^{-1} \rangle\) for the Hydrogen Atom

Set \(\lambda = Z\) (nuclear charge).

Differentiate the Hamiltonian:

\frac{\partial \hat{H}}{\partial Z} = \frac{\partial}{\partial Z} \left( -\frac{\hbar^2}{2m}\nabla^2 - \frac{Z e^2}{4\pi\varepsilon_0 r} \right) = -\frac{e^2}{4\pi\varepsilon_0 r}

Differentiate the energy \(E_n = -Z^2 \frac{m e^4}{32\pi^2\varepsilon_0^2\hbar^2 n^2}\):

\frac{d E_n}{d Z} = -2 Z \frac{m e^4}{32\pi^2\varepsilon_0^2\hbar^2 n^2} = \frac{2 E_n}{Z}

Apply Hellmann-Feynman:

\left\langle -\frac{e^2}{4\pi\varepsilon_0 r} \right\rangle = \frac{2 E_n}{Z}

Recall \(a_0 = \frac{4\pi\varepsilon_0\hbar^2}{m e^2}\) and \(E_n = -\frac{Z^2 e^2}{8\pi\varepsilon_0 a_0 n^2}\):

-\frac{e^2}{4\pi\varepsilon_0} \langle r^{-1} \rangle = 2 \left( -\frac{Z e^2}{8\pi\varepsilon_0 a_0 n^2} \right) = -\frac{Z e^2}{4\pi\varepsilon_0 a_0 n^2}

Dividing both sides by \(-\frac{e^2}{4\pi\varepsilon_0}\):

\langle r^{-1} \rangle_{n l} = \frac{Z}{a_0 n^2}

The Hellmann-Feynman theorem evaluates complex quantum expectation values in a single line of calculus.