Stationary & Time-Dependent Perturbation Theory

1. The Rayleigh-Schrödinger Perturbation Series

Consider a quantum system described by a Hamiltonian $\hat{H}$ partitioned into an unperturbed Hermitian part $\hat{H}_0$ with known complete orthonormal eigenstates $\{|n^{(0)}\rangle\}$ and energies $\{E_n^{(0)}\}$, and a perturbing potential $\hat{V}$ modulated by a continuous bookkeeping parameter $\lambda \in [0, 1]$:

$$\hat{H} = \hat{H}_0 + \lambda \hat{V}$$ $$\hat{H}_0 |n^{(0)}\rangle = E_n^{(0)} |n^{(0)}\rangle, \quad \langle m^{(0)} | n^{(0)} \rangle = \delta_{mn}$$

For a non-degenerate unperturbed energy level ($E_m^{(0)} \ne E_n^{(0)}$ for all $m \ne n$), the exact energy eigenvalues $E_n$ and state kets $|n\rangle$ can be expanded as power series in $\lambda$:

$$E_n = E_n^{(0)} + \lambda E_n^{(1)} + \lambda^2 E_n^{(2)} + \lambda^3 E_n^{(3)} + \dots$$ $$|n\rangle = |n^{(0)}\rangle + \lambda |n^{(1)}\rangle + \lambda^2 |n^{(2)}\rangle + \dots$$

Imposing intermediate normalization $\langle n^{(0)} | n \rangle = 1$, all higher-order corrections are strictly orthogonal to the unperturbed state:

$$\langle n^{(0)} | n^{(k)}\rangle = 0 \quad \forall k \ge 1$$

2. First- and Second-Order Perturbation Equations

Substituting the expansions into the time-independent Schrödinger equation $(\hat{H}_0 + \lambda \hat{V})|n\rangle = E_n |n\rangle$ and collecting like powers of $\lambda$:

$$\mathcal{O}(\lambda^0): \quad (\hat{H}_0 - E_n^{(0)})|n^{(0)}\rangle = 0$$ $$\mathcal{O}(\lambda^1): \quad (\hat{H}_0 - E_n^{(0)})|n^{(1)}\rangle = (E_n^{(1)} - \hat{V})|n^{(0)}\rangle$$ $$\mathcal{O}(\lambda^2): \quad (\hat{H}_0 - E_n^{(0)})|n^{(2)}\rangle = E_n^{(2)}|n^{(0)}\rangle + (E_n^{(1)} - \hat{V})|n^{(1)}\rangle$$

Projecting the first-order equation onto $\langle n^{(0)}|$:

$$\langle n^{(0)} | (\hat{H}_0 - E_n^{(0)}) | n^{(1)}\rangle = 0 = \langle n^{(0)} | (E_n^{(1)} - \hat{V}) | n^{(0)}\rangle \implies E_n^{(1)} = \langle n^{(0)} | \hat{V} | n^{(0)}\rangle$$

The first-order energy shift is the expectation value of the perturbing potential in the unperturbed state.

To determine $|n^{(1)}\rangle$, we project onto an orthogonal unperturbed state $\langle m^{(0)}|$ ($m \ne n$):

$$\langle m^{(0)} | (\hat{H}_0 - E_n^{(0)}) | n^{(1)}\rangle = (E_m^{(0)} - E_n^{(0)})\langle m^{(0)} | n^{(1)}\rangle = -\langle m^{(0)} | \hat{V} | n^{(0)}\rangle$$ $$\langle m^{(0)} | n^{(1)}\rangle = \frac{\langle m^{(0)} | \hat{V} | n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}} \implies |n^{(1)}\rangle = \sum_{m \ne n} \frac{\langle m^{(0)} | \hat{V} | n^{(0)}\rangle}{E_n^{(0)} - E_m^{(0)}} |m^{(0)}\rangle$$

Projecting the second-order equation onto $\langle n^{(0)}|$:

$$E_n^{(2)} = \langle n^{(0)} | \hat{V} | n^{(1)}\rangle = \sum_{m \ne n} \frac{|\langle m^{(0)} | \hat{V} | n^{(0)}\rangle|^2}{E_n^{(0)} - E_m^{(0)}}$$

Key Physical Insight: For the ground state ($n = 0$), the denominator $E_0^{(0)} - E_m^{(0)} < 0$ is strictly negative for all $m > 0$. Hence, the second-order energy shift of the ground state is always non-positive ($E_0^{(2)} \le 0$). The ground state energy is depressed by any perturbation that couples it to excited states.

1. Breakdown of the Non-Degenerate Formula

When the unperturbed eigenvalue $E_n^{(0)}$ is $g$-fold degenerate, there exist $g$ orthonormal eigenstates $\{|n_1^{(0)}\rangle, |n_2^{(0)}\rangle, \dots, |n_g^{(0)}\rangle\}$ possessing the identical energy:

$$\hat{H}_0 |n_i^{(0)}\rangle = E_n^{(0)} |n_i^{(0)}\rangle \quad (i = 1, \dots, g)$$

In the non-degenerate formula for $|n^{(1)}\rangle$, energy denominators of the form $E_n^{(0)} - E_m^{(0)}$ vanish for states within the degenerate subspace, causing formal divergences.

2. The Secular Equation in the Degenerate Subspace

To lift the ambiguity, we must construct the correct zero-order linear combinations $|\psi_\alpha^{(0)}\rangle$ within the $g$-dimensional degenerate eigenspace:

$$|\psi_\alpha^{(0)}\rangle = \sum_{j=1}^g c_{\alpha j} |n_j^{(0)}\rangle$$

Writing the first-order Schrödinger equation for $|\psi_\alpha\rangle$:

$$(\hat{H}_0 - E_n^{(0)})|\psi_\alpha^{(1)}\rangle = (E_\alpha^{(1)} - \hat{V})|\psi_\alpha^{(0)}\rangle$$

Projecting onto any basis bra $\langle n_i^{(0)}|$ belonging to the degenerate subspace:

$$\langle n_i^{(0)} | (\hat{H}_0 - E_n^{(0)}) | \psi_\alpha^{(1)}\rangle = 0 = \langle n_i^{(0)} | (E_\alpha^{(1)} - \hat{V}) \sum_{j=1}^g c_{\alpha j} |n_j^{(0)}\rangle$$ $$\sum_{j=1}^g \left( \langle n_i^{(0)} | \hat{V} | n_j^{(0)}\rangle - E_\alpha^{(1)} \delta_{ij} \right) c_{\alpha j} = 0$$

Defining matrix elements $V_{ij} \equiv \langle n_i^{(0)} | \hat{V} | n_j^{(0)}\rangle$, non-trivial coefficients $c_{\alpha j}$ exist if and only if the secular determinant vanishes:

$$\det\left( V_{ij} - E^{(1)} \delta_{ij} \right) = 0$$ $$\begin{vmatrix} V_{11} - E^{(1)} & V_{12} & \cdots & V_{1g} \\ V_{21} & V_{22} - E^{(1)} & \cdots & V_{2g} \\ \vdots & \vdots & \ddots & \vdots \\ V_{g1} & V_{g2} & \cdots & V_{gg} - E^{(1)} \end{vmatrix} = 0$$

The roots $E_\alpha^{(1)}$ ($\alpha = 1, \dots, g$) yield the first-order energy splittings. If all $g$ roots are distinct, the perturbation completely lifts the degeneracy. If a symmetry operator $\hat{A}$ commutes with both $\hat{H}_0$ and $\hat{V}$, choosing the unperturbed basis to be simultaneous eigenstates of $\hat{A}$ automatically diagonalizes the matrix $V_{ij}$ ($V_{ij} \propto \delta_{ij}$), avoiding the need to solve the full determinant.

1. Physical Origin of Fine Structure

The non-relativistic Bohr/Schrödinger theory of hydrogen yields energy levels $E_n = -\frac{13.6\text{ eV}}{n^2}$ that depend solely on the principal quantum number $n$, exhibiting an $n^2$-fold degeneracy. High-resolution spectroscopy reveals that these spectral lines are split by fine structure, originating from three distinct relativistic corrections of order $\alpha^4 m c^2 \approx 10^{-4} E_n$:

$$\hat{H}_{\text{FS}} = \hat{H}_{\text{rel}} + \hat{H}_{\text{SO}} + \hat{H}_{\text{Darwin}}$$

2. The Relativistic Mass-Kinetic Correction

Expanding the relativistic energy-momentum relation $E = \sqrt{p^2 c^2 + m^2 c^4} = mc^2 + \frac{p^2}{2m} - \frac{p^4}{8m^3 c^2} + \dots$:

$$\hat{H}_{\text{rel}} = -\frac{\hat{p}^4}{8m^3 c^2} = -\frac{1}{2mc^2}\left(\frac{\hat{p}^2}{2m}\right)^2 = -\frac{1}{2mc^2}\left(\hat{H}_0 - V(r)\right)^2$$

Evaluating the expectation value in hydrogenic state $|n, l, m_l\rangle$:

$$\Delta E_{\text{rel}} = \langle \hat{H}_{\text{rel}} \rangle = -\frac{E_n^2}{2mc^2}\left( \frac{4n}{l + 1/2} - 3 \right)$$

3. Spin-Orbit Coupling and Thomas Precession

An electron moving with velocity $\vec{v}$ through the central Coulomb field $\vec{E} = -\frac{1}{e}\frac{dV}{dr}\frac{\vec{r}}{r}$ experiences in its rest frame an effective magnetic field:

$$\vec{B}' \approx -\frac{\vec{v} \times \vec{E}}{c} = \frac{1}{m e c r}\frac{dV}{dr}(\vec{r} \times \vec{p}) = \frac{1}{m e c r}\frac{dV}{dr}\vec{L}$$

Coupling to the electron's intrinsic magnetic dipole moment $\vec{\mu}_s = -g_s \frac{e}{2m}\vec{S}$ (with $g_s = 2$), and including the kinematic relativistic Thomas precession factor of $1/2$ arising from the rotating non-inertial reference frame:

$$\hat{H}_{\text{SO}} = \frac{1}{2m^2 c^2}\frac{1}{r}\frac{dV}{dr}(\vec{L}\cdot\vec{S}) = \frac{e^2}{8\pi\varepsilon_0 m^2 c^2 r^3}(\vec{L}\cdot\vec{S})$$

In the coupled angular momentum basis $|j, m_j, l, s=1/2\rangle$ where $\vec{J} = \vec{L} + \vec{S}$:

$$\vec{J}^2 = \vec{L}^2 + \vec{S}^2 + 2\vec{L}\cdot\vec{S} \implies \vec{L}\cdot\vec{S} = \frac{\hbar^2}{2}\left(j(j+1) - l(l+1) - 3/4\right)$$

For $l > 0$, using $\langle r^{-3} \rangle = \frac{1}{a_0^3 n^3 l(l+1/2)(l+1)}$:

$$\Delta E_{\text{SO}} = \frac{E_n^2}{mc^2}\frac{j(j+1) - l(l+1) - 3/4}{l(l+1/2)(l+1)}$$

4. The Darwin Term and Total Fine Structure Formula

The Darwin term arises from relativistic quantum Zitterbewegung (rapid trembling motion of the electron over the Compton wavelength $\lambda_c \sim \hbar/mc$), smearing out the Coulomb potential at the origin:

$$\hat{H}_{\text{Darwin}} = \frac{\pi\hbar^2}{2m^2 c^2}\left(\frac{e^2}{4\pi\varepsilon_0}\right)\delta^3(\vec{r})$$

Since only $s$-orbitals ($l=0$) have non-vanishing probability density at the nucleus ($|\psi(0)|^2 = \frac{1}{\pi a_0^3 n^3}$):

$$\Delta E_{\text{Darwin}} = \frac{E_n^2}{mc^2} n \quad (\text{for } l = 0)$$

Remarkably, when all three corrections are summed, the $l$-dependent terms algebraically collapse into a single unified formula depending exclusively on $n$ and $j$:

$$\Delta E_{\text{FS}} = \Delta E_{\text{rel}} + \Delta E_{\text{SO}} + \Delta E_{\text{Darwin}} = \frac{E_n \alpha^2}{n}\left( \frac{1}{j + 1/2} - \frac{3}{4n} \right)$$

Energy levels with identical total angular momentum $j$ remain degenerate (e.g., $2s_{1/2}$ and $2p_{1/2}$), a degeneracy lifted only by quantum electrodynamic vacuum fluctuations (the Lamb shift $\approx 1057\text{ MHz}$).

1. The Stark Effect: Atoms in an Electric Field

When an atom is placed in a uniform external electric field $\vec{\mathcal{E}} = \mathcal{E}\hat{z}$, the electrostatic perturbation is:

$$\hat{V}_{\text{Stark}} = -q\vec{\mathcal{E}}\cdot\vec{r} = e\mathcal{E}\hat{z} = e\mathcal{E}r\cos\theta$$

Since $\hat{z}$ is odd under spatial parity ($\hat{\Pi}^\dagger \hat{z} \hat{\Pi} = -\hat{z}$), the diagonal expectation value vanishes for any state of definite parity:

$$E_n^{(1)} = e\mathcal{E}\langle n^{(0)} | \hat{z} | n^{(0)} \rangle = 0$$

Consequently, non-degenerate atomic states (such as the Hydrogen $1s$ ground state) exhibit only a quadratic Stark effect:

$$\Delta E_{1s} = \sum_{n \ne 1} \frac{|\langle n | e\mathcal{E}z | 1s \rangle|^2}{E_{1s}^{(0)} - E_n^{(0)}} = -\frac{1}{2}\alpha_{\text{pol}}\mathcal{E}^2 = -\frac{9}{4}(4\pi\varepsilon_0)a_0^3 \mathcal{E}^2$$

where $\alpha_{\text{pol}} = \frac{9}{2}(4\pi\varepsilon_0)a_0^3$ is the atomic polarizability of ground state hydrogen.

In the degenerate excited states of hydrogen (such as $n=2$, with states $|2s\rangle, |2p_0\rangle, |2p_1\rangle, |2p_{-1}\rangle$), parity-opposite states $|2s\rangle$ and $|2p_0\rangle$ have non-vanishing dipole coupling:

$$\langle 2s | e\mathcal{E}z | 2p_0 \rangle = -3 e a_0 \mathcal{E}$$

Diagonalizing the $2\times 2$ submatrix yields the linear Stark effect:

$$\Delta E = \pm 3 e a_0 \mathcal{E}$$

The energy shifts linearly with the applied electric field, forming permanent electric dipole moments.

2. The Zeeman Effect: Atoms in a Magnetic Field

For an atom placed in a uniform magnetic field $\vec{B} = B\hat{z}$, the magnetic dipole interaction Hamiltonian is:

$$\hat{H}_B = -\vec{\mu}\cdot\vec{B} = \frac{\mu_B}{\hbar}(\hat{L}_z + g_s \hat{S}_z)B = \frac{\mu_B B}{\hbar}(\hat{J}_z + \hat{S}_z)$$

where $\mu_B = \frac{e\hbar}{2m_e} \approx 9.274 \times 10^{-24}\text{ J/T}$ is the Bohr magneton and $g_s \approx 2$.

Two distinct physical regimes arise depending on the ratio of magnetic energy to the fine-structure splitting:

1. Weak Field Regime (Zeeman Effect, $\mu_B B \ll \Delta E_{\text{FS}}$): Fine structure dominates; $\vec{J}$ is a good quantum number. In the coupled basis $|j, m_j, l, s\rangle$, using the projection theorem: $$\Delta E_Z = \langle j, m_j | \hat{H}_B | j, m_j \rangle = g_J \mu_B B m_j$$ where $g_J$ is the Landé $g$-factor: $$g_J = 1 + \frac{j(j+1) + s(s+1) - l(l+1)}{2j(j+1)}$$
2. Strong Field Regime (Paschen-Back Effect, $\mu_B B \gg \Delta E_{\text{FS}}$): The magnetic field uncouples $\vec{L}$ and $\vec{S}$. States are described by the uncoupled basis $|l, m_l, s, m_s\rangle$: $$\Delta E_{PB} = \mu_B B(m_l + 2m_s)$$

1. Transition Amplitudes in the Interaction Picture

Consider a system with unperturbed Hamiltonian $\hat{H}_0$ subject to a time-dependent perturbation $\hat{V}(t)$ turned on at $t = 0$. In the Schrödinger picture:

$$|\psi(t)\rangle = \sum_n c_n(t) e^{-i E_n^{(0)} t/\hbar} |n^{(0)}\rangle$$

Substituting into $i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle = (\hat{H}_0 + \hat{V}(t))|\psi(t)\rangle$ yields the exact set of coupled differential equations for the transition amplitudes:

$$i\hbar \frac{dc_f(t)}{dt} = \sum_k \langle f^{(0)} | \hat{V}(t) | k^{(0)} \rangle e^{i\omega_{fk}t} c_k(t)$$

where $\omega_{fk} \equiv \frac{E_f^{(0)} - E_k^{(0)}}{\hbar}$ is the Bohr angular frequency of the transition.

2. First-Order Transition Amplitude

Assuming the system is initially prepared in state $|i^{(0)}\rangle$ at $t = 0$, so $c_k(0) = \delta_{ki}$:

$$c_f^{(1)}(t) = -\frac{i}{\hbar}\int_0^t \langle f^{(0)} | \hat{V}(t') | i^{(0)} \rangle e^{i\omega_{fi}t'}\,dt'$$

The transition probability from state $|i\rangle$ to state $|f\rangle$ is:

$$\mathcal{P}_{i\to f}(t) = |c_f^{(1)}(t)|^2 = \frac{1}{\hbar^2}\left|\int_0^t V_{fi}(t')e^{i\omega_{fi}t'}\,dt'\right|^2$$

3. Constant Perturbation Switched on at $t=0$

For a constant potential $\hat{V}(t) = \hat{V}$ for $t \ge 0$:

$$c_f^{(1)}(t) = -\frac{i}{\hbar}V_{fi}\int_0^t e^{i\omega_{fi}t'}\,dt' = -\frac{i}{\hbar}V_{fi} \frac{e^{i\omega_{fi}t} - 1}{i\omega_{fi}} = -\frac{V_{fi}}{\hbar\omega_{fi}} e^{i\omega_{fi}t/2} 2i\sin\left(\frac{\omega_{fi}t}{2}\right)$$ $$\mathcal{P}_{i\to f}(t) = \frac{4|V_{fi}|^2}{\hbar^2\omega_{fi}^2}\sin^2\left(\frac{\omega_{fi}t}{2}\right) = \frac{|V_{fi}|^2}{\hbar^2} t^2 \left[\frac{\sin(\omega_{fi}t/2)}{\omega_{fi}t/2}\right]^2$$

The function exhibits a sharp central diffraction peak with height proportional to $t^2$ and width $\Delta\omega \sim 2\pi/t$, establishing the time-energy uncertainty relation $\Delta E \Delta t \sim 2\pi\hbar$.

1. Transition into a Continuum of Final States

When the final state $|f\rangle$ belongs to a dense or continuous spectrum of states with density of states $\rho(E_f)$ (the number of states per unit energy interval $dE$), the total transition probability is the integral over all continuum final states:

$$\mathcal{P}_{\text{total}}(t) = \int \mathcal{P}_{i\to f}(t) \rho(E_f)\,dE_f = \int \frac{4|V_{fi}|^2}{\hbar^2}\frac{\sin^2[(\omega_{fi}t)/2]}{\omega_{fi}^2} \rho(E_f)\,dE_f$$

Using the representation of the Dirac delta distribution:

$$\lim_{t\to\infty} \frac{\sin^2(\alpha t)}{\pi t \alpha^2} = \delta(\alpha) \implies \lim_{t\to\infty} \frac{4\sin^2\left(\frac{(E_f - E_i)t}{2\hbar}\right)}{\frac{(E_f - E_i)^2}{\hbar^2} t} = 2\pi\hbar \delta(E_f - E_i)$$

2. Fermi's Golden Rule for Constant and Harmonic Perturbations

Differentiating with respect to time yields the constant steady-state transition rate $W_{i\to f} \equiv \frac{d\mathcal{P}}{dt}$:

$$W_{i\to f} = \frac{2\pi}{\hbar}|\langle f | \hat{V} | i \rangle|^2 \rho(E_f) \delta(E_f - E_i)$$

For a periodic harmonic perturbation $\hat{V}(t) = \hat{F} e^{-i\omega t} + \hat{F}^\dagger e^{i\omega t}$:

$$W_{i\to f} = \frac{2\pi}{\hbar}|\langle f | \hat{F} | i \rangle|^2 \rho(E_f) \delta(E_f - E_i - \hbar\omega) + \frac{2\pi}{\hbar}|\langle f | \hat{F}^\dagger | i \rangle|^2 \rho(E_f) \delta(E_f - E_i + \hbar\omega)$$

The first term describes stimulated absorption (absorbing a photon of energy $\hbar\omega = E_f - E_i$), while the second describes stimulated emission (emitting a photon of energy $\hbar\omega = E_i - E_f$).

1. The Electric Dipole Approximation

In the semiclassical theory of radiation, an atom interacts with an electromagnetic plane wave vector potential $\vec{A}(\vec{r}, t) = A_0 \hat{\epsilon} \cos(\vec{k}\cdot\vec{r} - \omega t)$. Because the optical wavelength $\lambda \approx 500\text{ nm}$ is vastly larger than the atomic radius $a_0 \approx 0.05\text{ nm}$, the spatial phase factor across the atom is nearly constant:

$$e^{i\vec{k}\cdot\vec{r}} = 1 + i\vec{k}\cdot\vec{r} - \frac{1}{2}(\vec{k}\cdot\vec{r})^2 + \dots \approx 1$$

This is the electric dipole (E1) approximation. The interaction Hamiltonian reduces to the electric dipole interaction:

$$\hat{H}_{\text{int}} = -\hat{\vec{d}}\cdot\vec{\mathcal{E}}(t) = e\hat{\vec{r}}\cdot\vec{\mathcal{E}}(t)$$

2. Einstein A and B Coefficients and Spontaneous Emission

The rate of stimulated absorption is $W_{\text{abs}} = B_{12} u(\omega)$, where $u(\omega)$ is the spectral radiation energy density. The spontaneous emission rate $A_{21}$ (occurring in the absence of external radiation) is related to the stimulated coefficient by Planck's blackbody law:

$$A_{21} = \frac{\hbar\omega^3}{\pi^2 c^3} B_{21} = \frac{\omega^3 |\vec{d}_{fi}|^2}{3\pi\varepsilon_0 \hbar c^3}$$

The radiative lifetime of the excited state is $\tau = 1 / A_{21}$.

3. Electric Dipole Selection Rules

The transition dipole matrix element $\langle n', l', m' | \hat{\vec{r}} | n, l, m \rangle$ evaluates to zero unless the initial and final quantum numbers satisfy specific selection rules:

1. Parity Selection Rule: States must have opposite parity $\pi_i \pi_f = -1$.
2. Orbital Angular Momentum: $\Delta l = l' - l = \pm 1$. Direct transitions $l \to l$ (such as $2s \to 1s$) are strictly forbidden in E1 radiation.
3. Magnetic Quantum Number: $\Delta m = m' - m = 0$ (for $\pi$-polarized light parallel to $z$), and $\Delta m = \pm 1$ (for $\sigma^\pm$-circularly polarized light in the $xy$-plane).