Most of the quantum mechanics problems you solve first are time-independent — the Hamiltonian doesn’t change, so the system just sits in stationary states or evolves in a trivial way.
But real quantum systems get disturbed constantly. An atom gets hit by light. A nucleus decays. An electron absorbs a photon and jumps to a higher energy level.
Time-dependent perturbation theory (TDPT) is the tool we use to handle these situations. The setup is:
- We have a known, time-independent “unperturbed” Hamiltonian \(H_0\) whose eigenstates we already understand.
- We add a small, time-dependent perturbation \(H'(t)\).
- We want to calculate the probability that the system, starting in one state, transitions to another.
The logic runs through the whole unit: start with the general formalism, then first-order transitions, then Fermi’s Golden Rule, then the specific case of light-matter interaction, and finally spontaneous emission and Einstein coefficients.
1. Time-Dependent Perturbation Theory (The General Framework)
The Hamiltonian
We write the total Hamiltonian as:
$$
H(t) = H_0 + H'(t)
$$
where:
- \(H_0\) is time-independent, with known eigenstates satisfying \(H_0 \phi_n = E_n \phi_n\).
- \(H'(t)\) is the time-dependent perturbation. “Small” means we can treat it as a correction.
Expanding the wavefunction
At any time \(t\), we can expand the state \(\Psi(t)\) in the basis of the unperturbed eigenstates, but allowing the coefficients to vary with time:
$$
\Psi(t) = \sum_n c_n(t) \, \phi_n \, e^{-iE_n t/\hbar}
$$
Why the exponential factor? It’s standard practice to factor out the known time dependence from \(H_0\). What remains — the \(c_n(t)\) — captures the effect of the perturbation. If \(H’ = 0\), all \(c_n(t)\) are constant.
The quantity \(|c_n(t)|^2\) is the probability of finding the system in state \(n\) at time \(t\).
The equation for the coefficients
Plug this expansion into the time-dependent Schrödinger equation \(i\hbar \partial \Psi/\partial t = H \Psi\). After working through the algebra (the key derivation you should know), you get:
$$
i\hbar \frac{dc_m}{dt} = \sum_n c_n(t) \, \langle \phi_m | H'(t) | \phi_n \rangle \, e^{i\omega_{mn} t}
$$
where \(\omega_{mn} = (E_m – E_n)/\hbar\).
This is the central equation of TDPT. It tells you how probability flows between different states over time.
What controls transitions
The matrix element \(\langle \phi_m | H'(t) | \phi_n \rangle\) is the key quantity:
- If this matrix element is zero for a particular pair of states, no direct transition between them occurs (at least in first order — more on that later).
- If it’s nonzero, the perturbation can drive the system from \(n\) to \(m\).
The exponential factor \(e^{i\omega_{mn} t}\) matters too. When the perturbation oscillates at a frequency that matches \(\omega_{mn}\), you get a resonance — transitions become much more likely. That’s where absorption and emission come from.
2. First-Order Perturbation Theory
The basic assumption
We assume the perturbation \(H'(t)\) is weak enough that the system mostly stays in its initial state. That means we can approximate:
- The system starts in some initial state \(i\) at \(t = 0\), so \(c_i(0) = 1\) and \(c_{n \neq i}(0) = 0\).
- For short times or weak perturbations, the coefficients on the right side of the TDPT equation remain approximately \(c_i(t) \approx 1\) and all others \(\approx 0\).
This is the first-order approximation — we only keep terms linear in \(H’\).
The transition amplitude
Starting from the general TDPT equation:
$$
i\hbar \frac{dc_f}{dt} = \sum_n c_n(t) \, \langle \phi_f | H'(t) | \phi_n \rangle \, e^{i\omega_{fn} t}
$$
If we keep only the initial state \(i\) on the right-hand side (since \(c_i \approx 1\) and others are negligible), we get:
$$
i\hbar \frac{dc_f}{dt} \approx \langle \phi_f | H'(t) | \phi_i \rangle \, e^{i\omega_{fi} t}
$$
Integrate from \(0\) to \(t\):
$$ c_f^{(1)}(t) = \frac{1}{i\hbar} \int_0^t \langle \phi_f | H'(t’) | \phi_i \rangle \, e^{i\omega_{fi} t’} \, dt’ $$
This is the first-order transition amplitude from state \(i\) to state \(f\).
The transition probability
The probability that the system has made the transition by time \(t\) is:
$$ P_{i \rightarrow f}(t) = |c_f^{(1)}(t)|^2 $$
What to notice
- The matrix element \(\langle f | H'(t) | i \rangle\) appears linearly. If it’s zero, the first-order probability is zero (you’d need to go to second order).
- The probability grows with time — but only valid while \(P \ll 1\) (so the approximation \(c_i \approx 1\) still holds).
- The integral tells you which final states are accessible. If \(H'(t)\) oscillates at a certain frequency, the integral becomes large only when that frequency matches \(\omega_{fi}\).
A concrete example: constant perturbation
If \(H'(t)\) is constant (turned on suddenly at \(t=0\) and left constant), then \(\langle f|H’|i \rangle\) comes out of the integral. For \(f \neq i\):
$$ c_f^{(1)}(t) = \frac{\langle f|H’|i \rangle}{i\hbar} \int_0^t e^{i\omega_{fi} t’} \, dt’ = \frac{\langle f|H’|i \rangle}{\hbar \omega_{fi}} \left( e^{i\omega_{fi} t} – 1 \right) $$
The probability oscillates:
$$ P_{i \rightarrow f}(t) = \frac{4|\langle f|H’|i \rangle|^2}{\hbar^2 \omega_{fi}^2} \, \sin^2\left( \frac{\omega_{fi} t}{2} \right) $$
This oscillates in time rather than growing linearly — a signature of a constant perturbation. For very short times, \(P \propto t^2\).
3. Fermi’s Golden Rule
The problem with first-order results
The first-order transition probability we derived works fine for a single final state. But in many real situations — like a decay into a continuum of states — we don’t care about one specific final state. We care about how many final states are available and how fast the transition happens.
Decay rates, scattering cross sections, and absorption coefficients all involve transitions into a dense set of final states.
The setup
We have:
- A perturbation \(H'(t)\) that is constant in time (or turned on adiabatically).
- A set of final states with energies \(E_f\) that form a continuum.
- We want the transition rate \(W\) (probability per unit time) from an initial state \(i\) to any final state in that continuum.
The derivation
Starting from the constant perturbation result:
$$
P_{i \rightarrow f}(t) = \frac{4|\langle f|H’|i \rangle|^2}{\hbar^2 \omega_{fi}^2} \, \sin^2\left( \frac{\omega_{fi} t}{2} \right)
$$
Write \(\omega_{fi} = (E_f – E_i)/\hbar\). For large \(t\), the function \(\sin^2(xt)/(\pi x^2 t)\) approaches a delta function \(\delta(x)\). More precisely:
$$ \frac{\sin^2(\omega_{fi} t/2)}{(\omega_{fi}/2)^2} \approx 2\pi t \, \delta(E_f – E_i)/\hbar $$
Summing over final states and converting to an integral using the density of states \(\rho(E_f)\):
$$ W = \frac{2\pi}{\hbar} \, |\langle f|H’|i \rangle|^2 \, \rho(E_f) $$
where the matrix element is evaluated at \(E_f = E_i\).
The formula — Fermi’s Golden Rule
$$
\fbox{$W = \frac{2\pi}{\hbar} \, |\langle f| H’ | i \rangle|^2 \, \rho(E_f)$}
$$
What each term means
| Term | Meaning |
|---|---|
| \(W\) | Transition rate (probability per unit time) |
| \(\langle f|H’|i \rangle\) | Matrix element of the perturbation between initial and final states |
| \(\rho(E_f)\) | Density of final states — number of states per unit energy interval at \(E_f = E_i\) |
| \(2\pi/\hbar\) | A constant that makes the units work out |
Key physical insights
Energy conservation — The delta function \(\delta(E_f – E_i)\) is baked into the derivation. Transitions only happen when the final state energy equals the initial state energy (for a time-independent perturbation).
Rate is constant — Unlike \(P(t)\) which grows then oscillates, \(W\) is constant in time. This matches exponential decay laws: \(P(t) \propto e^{-Wt}\).
Two ways to increase the rate — Either strengthen the coupling (larger matrix element) or have more final states available (larger \(\rho(E_f)\)).
When is it valid?
- The perturbation must be weak (first-order approximation holds).
- The observation time \(t\) must be long enough that the energy-conserving delta function sharpens up, but short enough that the total probability remains small (\(Wt \ll 1\)).
- The density of states should vary slowly near \(E_f = E_i\).
A common application: constant perturbation turned on at (t=0)
Fermi’s Golden Rule tells you that after a short initial transient, the system decays at a constant rate. For example, in radioactive decay or photoemission, the number of particles in the initial state decays as:
$$
N_i(t) = N_i(0) \, e^{-Wt}
$$
where \(W\) is given by the Golden Rule.
4. Harmonic (Periodic) Perturbation
Why this matters
So far we looked at constant perturbations. But the most common physical situation is an oscillating perturbation — like an electromagnetic wave hitting an atom. Light has an electric field that oscillates as \(\cos(\omega t)\) or \(e^{-i\omega t}\).
Harmonic perturbations are the key to understanding absorption and stimulated emission.
The form of the perturbation
We write a harmonic perturbation as:
$$ H'(t) = V e^{-i\omega t} + V^\dagger e^{i\omega t} $$
where:
- \(V\) is an operator that does not depend on time.
- The two terms correspond to absorption (\(e^{-i\omega t}\)) and stimulated emission (\(e^{i\omega t}\)).
- The Hermitian conjugate \(V^\dagger\) ensures \(H'(t)\) is Hermitian (real energies).
For a real classical field like \(2V \cos(\omega t)\), this decomposes into \(V e^{-i\omega t} + V e^{i\omega t}\) (with \(V\) Hermitian).
First-order transition amplitude
Plug \(H'(t)\) into the first-order expression:
$$ c_f^{(1)}(t) = \frac{1}{i\hbar} \int_0^t \langle f | V e^{-i\omega t’} + V^\dagger e^{i\omega t’} | i \rangle \, e^{i\omega_{fi} t’} \, dt’ $$
This splits into two terms:
$$ c_f^{(1)}(t) = \frac{1}{i\hbar} \langle f | V | i \rangle \int_0^t e^{i(\omega_{fi} – \omega) t’} \, dt’ + \frac{1}{i\hbar} \langle f | V^\dagger | i \rangle \int_0^t e^{i(\omega_{fi} + \omega) t’} \, dt’ $$
The resonance condition
Each integral becomes large when its exponent is near zero:
| Term | Large when | Physical process |
|---|---|---|
| First term | \(\omega_{fi} – \omega \approx 0\) → \(E_f – E_i \approx \hbar \omega\) | (system gains energy \(\hbar \omega\)) |
| Second term | \(\omega_{fi} + \omega \approx 0\) → \(E_f – E_i \approx -\hbar \omega\) | Stimulated emission (system loses energy \(\hbar \omega\)) |
So energy is exchanged in quanta of \(\hbar \omega\).
Transition probability near resonance
For absorption (assuming \(\omega \approx \omega_{fi}\)), the first term dominates. The integral gives:
$$ c_f^{(1)}(t) = \frac{\langle f | V | i \rangle}{i\hbar} \cdot \frac{e^{i(\omega_{fi} – \omega)t} – 1}{i(\omega_{fi} – \omega)} $$
The probability is:
$$ P_{i \rightarrow f}(t) = \frac{|\langle f | V | i \rangle|^2}{\hbar^2} \cdot \frac{4 \sin^2\left( \frac{(\omega_{fi} – \omega)t}{2} \right)}{(\omega_{fi} – \omega)^2} $$
Shape of the resonance
As a function of \(\omega\), this is a peaked function:
- Maximum at \(\omega = \omega_{fi}\)
- Width \(\sim 2\pi/t\)
- Height \(\propto t^2\)
For large \(t\), the function approaches a delta function:
$$ \frac{4 \sin^2((\omega_{fi} – \omega)t/2)}{(\omega_{fi} – \omega)^2} \rightarrow 2\pi t \, \delta(\omega_{fi} – \omega) $$
Transition rate for harmonic perturbation
Using the same logic as Fermi’s Golden Rule, the absorption rate is:
$$ W = \frac{2\pi}{\hbar} \, |\langle f | V | i \rangle|^2 \, \rho(E_f) $$
valuated at \(E_f = E_i + \hbar \omega\).
This looks identical to the constant perturbation case, but with two important differences:
- The matrix element uses \(V\), not the full \(H’\).
- Energy conservation now reads \(E_f = E_i + \hbar \omega\) (absorption) or \(E_f = E_i – \hbar \omega\) (stimulated emission).
Absorption vs. stimulated emission
The two processes are symmetric. For a Hermitian \(V\) (real field like \(\cos(\omega t)\)):
$$ |\langle f | V | i \rangle|^2 = |\langle i | V | f \rangle|^2 $$
So the rates for absorption and stimulated emission are equal. This is a crucial result — Einstein used this symmetry to derive the (A) and (B) coefficients long before quantum mechanics was fully developed.
Summary table
| Process | Energy change | Resonance condition | Rate |
|---|---|---|---|
| Absorption | \(E_f = E_i + \hbar \omega\) | (\omega = \omega_{fi}\) | \(\frac{2\pi}{\hbar} |\langle f|V|i \rangle|^2 \rho(E_f)\) |
| Stimulated emission | \(E_f = E_i – \hbar \omega\) | \(\omega = \omega_{if}\) | Same as absorption |
5. Semi-Classical Theory of Radiation
What “semi-classical” means
In this approach:
- The atom is treated quantum mechanically — it has discrete energy levels, wavefunctions, etc.
- The electromagnetic field is treated classically — as a continuous wave with electric field \(\mathbf{E}(t)\).
This mixed approach works surprisingly well for absorption and stimulated emission. But it famously fails for spontaneous emission — that requires a fully quantum field.
The interaction Hamiltonian
An electron in an atom, with charge \(-e\), interacts with an applied electric field \(\mathbf{E}(t)\). The potential energy is:
$$ H'(t) = -\mathbf{d} \cdot \mathbf{E}(t) $$
where \(\mathbf{d} = -e \mathbf{r}\) is the electric dipole moment operator of the electron.
So:
$$ H'(t) = e \, \mathbf{r} \cdot \mathbf{E}(t) $$
The dipole approximation
The electric field varies in space as \(e^{i\mathbf{k} \cdot \mathbf{r}}\). But over the size of an atom (angstroms), the wavelength of visible light (hundreds of nanometers) is much larger. So \(\mathbf{k} \cdot \mathbf{r} \ll 1\).
That means we can approximate the field as uniform across the atom:
$$ \mathbf{E}(\mathbf{r}, t) \approx \mathbf{E}(t) \quad \text{(independent of position)} $$
This is the dipole approximation. It’s valid when:
$$
\lambda \gg a_0
$$
where \(\lambda\) is the wavelength and \(a_0\) is the Bohr radius (roughly \(10^{-10}\) m). For visible light, \(\lambda \sim 10^{-7}\) m, so this holds nicely.
A monochromatic field
For a plane wave oscillating at frequency \(\omega\):
$$ \mathbf{E}(t) = \mathbf{E}_0 \cos(\omega t) = \frac{\mathbf{E}_0}{2} \left( e^{-i\omega t} + e^{i\omega t} \right) $$
Then the perturbation becomes:
$$ H'(t) = \frac{e}{2} \, \mathbf{r} \cdot \mathbf{E}_0 \left( e^{-i\omega t} + e^{i\omega t} \right) $$
Comparing with the harmonic perturbation form \(H'(t) = V e^{-i\omega t} + V^\dagger e^{i\omega t}\):
$$ V = \frac{e}{2} \, \mathbf{r} \cdot \mathbf{E}_0 $$
The matrix element
The key quantity controlling transitions is:
$$ \langle f | V | i \rangle = \frac{e}{2} \, \mathbf{E}_0 \cdot \langle f | \mathbf{r} | i \rangle $$
The vector \(\langle f | \mathbf{r} | i \rangle\) is called the dipole matrix element. Its magnitude squared is proportional to the transition probability.
Selection rules
The dipole matrix element \(\langle f | \mathbf{r} | i \rangle\) is often zero for many pairs of states. When it’s zero, the transition is dipole-forbidden (though it might still occur via higher-order processes like quadrupole transitions).
For hydrogen-like atoms with central potential, \(\mathbf{r}\) is an odd parity operator (it changes sign under \(\mathbf{r} \to -\mathbf{r}\)). Since initial and final states have definite parity:
- \(\langle f | \mathbf{r} | i \rangle\) is nonzero only if the parity of \(f\) and \(i\) are opposite.
More specific selection rules:
| Quantum number | Selection rule |
|---|---|
| Orbital angular momentum \(l\) | \(\Delta l = \pm 1\) |
| Magnetic quantum number \(m\) | \(\Delta m = 0, \pm 1\) |
| Parity | Must change |
For hydrogen \(n\) (principal quantum number): no restriction — any \(\Delta n\) is allowed as long as \(\Delta l = \pm 1\).
Why \(\Delta l = \pm 1\)?
This comes from the angular momentum properties of \(\mathbf{r}\). The position operator can be written in terms of spherical harmonics with \(l=1\). So it connects a state with angular momentum \(l\) to states with \(l \pm 1\) (like adding angular momentum 1 to \(l\)).
The \(\Delta m\) rules depend on the polarization of the light:
| Polarization | (\Delta m) |
|---|---|
| Linear along \(z\) | \(0\) |
| Circular (right/left) | \(+1\) or \(-1\) |
The transition rate
Using Fermi’s Golden Rule for absorption:
$$ W = \frac{2\pi}{\hbar} \, |\langle f | V | i \rangle|^2 \, \rho(E_f) $$
Substitute \(V = (e/2) \mathbf{r} \cdot \mathbf{E}_0\):
$$ W = \frac{2\pi}{\hbar} \left( \frac{e}{2} \right)^2 |\mathbf{E}_0 \cdot \langle f | \mathbf{r} | i \rangle|^2 \, \rho(E_f) $$
For an isotropic distribution of polarizations, we average over the angle between \(\mathbf{E}_0\) and \(\langle \mathbf{r} \rangle\):
$$ |\mathbf{E}_0 \cdot \langle \mathbf{r} \rangle|^2 \to \frac{1}{3} |\mathbf{E}_0|^2 |\langle \mathbf{r} \rangle|^2 $$
What this theory cannot explain
The semi-classical theory works for absorption and stimulated emission (processes driven by an external field). But it completely fails to explain spontaneous emission — an excited atom decaying to a lower state even in the absence of any external field.
Why? Because in classical electromagnetism, if there’s no field, the interaction \(-\mathbf{d} \cdot \mathbf{E}\) is zero. Nothing causes the decay. But in reality, excited atoms do decay spontaneously.
The resolution: the electromagnetic field must be quantized. Even in the vacuum, there are zero-point fluctuations that can trigger emission. That’s what comes next.
6. Einstein Coefficients
The historical context
Before quantum mechanics was fully developed, Einstein in 1917 figured out a clever way to understand radiation processes using only thermodynamic arguments. He didn’t know about the quantum field, but he correctly deduced the relationships between absorption, stimulated emission, and spontaneous emission.
His insight: even without a full theory, the rates must balance at thermal equilibrium. That forced a relationship that turned out to be exactly right.
The three processes
Consider two energy levels of an atom: \(E_1\) (lower) and \(E_2\) (upper), with \(E_2 – E_1 = \hbar \omega\).
In the presence of radiation with energy density \(\rho(\omega)\) (energy per unit volume per unit frequency), three things can happen:
| Process | Description | Rate |
|---|---|---|
| Absorption | Atom in \(E_1\) absorbs a photon and goes to \(E_2\) | \(R_{1\to2} = B_{12} \, \rho(\omega) \, N_1\) |
| Stimulated emission | Atom in \(E_2\) is triggered by a photon to go to \(E_1\), emitting another photon | \(R_{2\to1}^{\text{stim}} = B_{21} \, \rho(\omega) \, N_2\) |
| Spontaneous emission | Atom in \(E_2\) decays to \(E_1\) on its own, no external photon needed | \(R_{2\to1}^{\text{spon}} = A_{21} \, N_2\) |
Where:
- \(N_1, N_2\) = number of atoms in level 1 and 2
- \(B_{12}, B_{21}, A_{21}\) = Einstein coefficients (constants that depend on the atom)
The Einstein coefficients defined
| Coefficient | Meaning | Units |
|---|---|---|
| \(B_{12}\) | Probability per unit time per unit energy density for absorption | \(\text{m}^3/\text{J} \cdot \text{s}^2\) |
| \(B_{21}\) | Probability per unit time per unit energy density for stimulated emission | Same as \(B_{12}\) |
| \(A_{21}\) | Probability per unit time for spontaneous emission | \(\text{s}^{-1}\) |
The key relation from equilibrium
At thermal equilibrium, the number of transitions upward must equal the number downward (detailed balance):
$$ B_{12} \, \rho(\omega) \, N_1 = B_{21} \, \rho(\omega) \, N_2 + A_{21} \, N_2 $$
The ratio \(N_2/N_1\) at temperature \(T\) is given by the Boltzmann distribution:
$$ \frac{N_2}{N_1} = \frac{g_2}{g_1} \, e^{-(E_2 – E_1)/k_B T} = \frac{g_2}{g_1} \, e^{-\hbar \omega / k_B T} $$
where \(g_1, g_2\) are the degeneracies (statistical weights) of the two levels.
Plug this into the detailed balance equation and solve for \(\rho(\omega)\):
$$ \rho(\omega) = \frac{A_{21}/B_{21}}{(g_1 B_{12}/g_2 B_{21}) e^{\hbar \omega/k_B T} – 1} $$
But we also know from Planck’s blackbody radiation law:
$$ \rho(\omega) = \frac{\hbar \omega^3}{\pi^2 c^3} \cdot \frac{1}{e^{\hbar \omega/k_B T} – 1} $$
For these two expressions to match for all \(T\), we must have:
The Einstein relations
$$
g_1 B_{12} = g_2 B_{21}
$$
$$
A_{21} = \frac{\hbar \omega^3}{\pi^2 c^3} \, B_{21}
$$
For non-degenerate levels (\(g_1 = g_2 = 1\)):
$$ B_{12} = B_{21} $$
$$ A_{21} = \frac{\hbar \omega^3}{\pi^2 c^3} \, B_{21} $$
What these relations tell us
Equality of absorption and stimulated emission rates — \(B_{12} = B_{21}\) means that for a given pair of states, the probability of absorbing a photon (from the lower state) equals the probability of being stimulated to emit a photon (from the upper state). This matches what we found in the harmonic perturbation section.
Spontaneous emission grows as (\omega^3) — \(A_{21} \propto \omega^3\) means that atoms in excited states decay much faster at higher frequencies. Blue light decays faster than red light. This makes sense: higher frequency means more vacuum modes available to emit into.
Connecting to quantum theory
From our semi-classical result for absorption, we can identify:
$$ B_{12} = \frac{\pi |\langle 2 | \mathbf{d} \cdot \hat{\epsilon} | 1 \rangle|^2}{3 \epsilon_0 \hbar^2} $$
Then using the Einstein relation \(A_{21} = (\hbar \omega^3 / \pi^2 c^3) B_{21}\):
$$ A_{21} = \frac{\omega^3 |\langle 2 | \mathbf{d} \cdot \hat{\epsilon} | 1 \rangle|^2}{3 \pi \epsilon_0 \hbar c^3} $$
This is the spontaneous emission rate — a prediction that semi-classical theory couldn’t make on its own. Einstein got the form from thermodynamics; quantum electrodynamics later derived it from first principles.
Population inversion and lasers
Notice: if \(N_2 > N_1\) (population inversion), the stimulated emission rate \(B_{21} \rho N_2\) can exceed the absorption rate \(B_{12} \rho N_1\). This means light gets amplified as it passes through the medium — the basis of laser operation.
Einstein’s coefficients are the foundation of laser physics. Without the \(A_{21}\) term, you couldn’t empty the upper state; without the \(B_{21} = B_{12}\) symmetry, you couldn’t get net gain from inversion.
Summary table
| Coefficient | Role | Proportionality |
|---|---|---|
| \(B_{12}\) | Absorption | \(|\langle f | \mathbf{d} | i \rangle|^2\) |
| \(B_{21}\) | Stimulated emission | Same as \(B_{12}\) |
| \(A_{21}\) | Spontaneous emission | \(\omega^3 B_{21}\) |
7. Spontaneous Emission (Quantum Theory)
Why semi-classical theory fails
In the semi-classical picture, the atom is quantum but the electromagnetic field is classical. The interaction is \(H'(t) = -\mathbf{d} \cdot \mathbf{E}(t)\). If there’s no external field (\(\mathbf{E} = 0\)), then \(H’ = 0\) and nothing happens. An excited atom would sit in its upper state forever.
But real atoms decay. An electron in an excited state drops to a lower state and emits a photon — even in complete darkness, in vacuum, with no external light.
This tells us: the vacuum itself is not empty. The electromagnetic field has quantum fluctuations even in its ground state. These fluctuations can trigger emission.
Quantizing the electromagnetic field
The key idea: the EM field is a collection of harmonic oscillators — one for each mode (each wave vector \(\mathbf{k}\) and polarization \(\epsilon\)). Each mode is quantized just like a quantum harmonic oscillator.
The vector potential operator (in Coulomb gauge) is:
$$ \hat{\mathbf{A}}(\mathbf{r}) = \sum_{\mathbf{k}, \lambda} \sqrt{\frac{\hbar}{2\epsilon_0 \omega V}} \left( a_{\mathbf{k},\lambda} \, \epsilon_{\lambda} e^{i\mathbf{k}\cdot\mathbf{r}} + a_{\mathbf{k},\lambda}^\dagger \, \epsilon_{\lambda}^* e^{-i\mathbf{k}\cdot\mathbf{r}} \right) $$
where:
- \(a_{\mathbf{k},\lambda}^\dagger\) = creation operator for a photon in mode \((\mathbf{k}, \lambda)\)
- \(a_{\mathbf{k},\lambda}\) = annihilation operator
- \(V\) = quantization volume (drops out of final rates)
The interaction Hamiltonian (quantum)
For an electron in an atom, the interaction with the quantized field is (in the dipole approximation):
$$ \hat{H}’ = -\frac{e}{m} \hat{\mathbf{A}} \cdot \hat{\mathbf{p}} + \frac{e^2}{2m} \hat{A}^2 $$
For spontaneous emission, the first term (linear in \(\hat{\mathbf{A}}\)) dominates. The \(\hat{A}^2\) term matters for two-photon processes and scattering.
In the dipole approximation and using the \(e\mathbf{r} \cdot \mathbf{E}\) form:
$$ \hat{H}’ = -\hat{\mathbf{d}} \cdot \hat{\mathbf{E}} $$
where \(\hat{\mathbf{d}} = -e\hat{\mathbf{r}}\) and \(\hat{\mathbf{E}}\) is the quantized electric field operator.
Spontaneous emission as a transition
We start with the atom in an excited state \(|i\rangle\) (say \(E_2\)) and no photons in the field. The initial state is:
$$ | \text{initial} \rangle = |i\rangle \otimes |0\rangle $$
where \(|0\rangle\) is the vacuum (zero photons in all modes).
The final state has the atom in a lower state \(|f\rangle\) (say \(E_1\)) and one photon emitted into a specific mode \((\mathbf{k}, \lambda)\):
$$ | \text{final} \rangle = |f\rangle \otimes |1_{\mathbf{k},\lambda}\rangle $$
Energy conservation: \(E_i = E_f + \hbar \omega_k\), so \(\omega_k = (E_i – E_f)/\hbar\).
The matrix element
The transition amplitude comes from \(\langle f, 1_{\mathbf{k},\lambda} | \hat{H}’ | i, 0 \rangle\).
Using \(\hat{\mathbf{E}} \propto i\omega (\hat{\mathbf{A}} – \hat{\mathbf{A}}^\dagger)\), the relevant term is the one with \(a_{\mathbf{k},\lambda}^\dagger\) (creation operator), since it turns \(|0\rangle\) into \(|1_{\mathbf{k},\lambda}\rangle\).
The result:
$$ \langle f, 1_{\mathbf{k},\lambda} | \hat{H}’ | i, 0 \rangle = i \sqrt{\frac{\hbar \omega}{2\epsilon_0 V}} \, \epsilon_{\lambda} \cdot \langle f | \hat{\mathbf{d}} | i \rangle $$
where \(\langle f | \hat{\mathbf{d}} | i \rangle = -e \langle f | \mathbf{r} | i \rangle\) is the dipole matrix element.
Fermi’s Golden Rule for spontaneous emission
We sum over all final states — all possible photon modes \((\mathbf{k}, \lambda)\) that satisfy energy conservation. The density of photon states in free space is:
$$ \rho(E_f) = \frac{V \omega^2}{(2\pi c)^3 \hbar} \, d\Omega $$
for photons emitted into solid angle \(d\Omega\).
Applying Fermi’s Golden Rule:
$$ W_{i\to f} = \frac{2\pi}{\hbar} \sum_{\mathbf{k},\lambda} |\langle f, 1_{\mathbf{k},\lambda} | \hat{H}’ | i, 0 \rangle|^2 \, \delta(E_i – E_f – \hbar \omega) $$
Summing over polarizations and integrating over \(\mathbf{k}\) gives:
The spontaneous emission rate
$$
\fbox{$W = \frac{\omega^3 |\langle f | \hat{\mathbf{d}} | i \rangle|^2}{3\pi \epsilon_0 \hbar c^3}$}
$$
For an average over dipole orientations and summing over final polarizations:
$$ W = \frac{4 \omega^3 |\langle f | \mathbf{d} | i \rangle|^2}{3 \epsilon_0 \hbar c^3} $$
depending on how you average. The key point: this matches exactly the (\(A_{21}\) acoefficient Einstein derived from thermodynamics.
What this tells us physically
Spontaneous emission is not “spontaneous” — it’s driven by vacuum fluctuations. The vacuum always has zero-point energy \(\frac{1}{2}\hbar\omega\) in each mode. These fluctuations give a nonzero \(\langle \mathbf{E}^2 \rangle\) even in empty space, and that triggers the decay.
The vacuum is a physical thing — it’s not just “nothing.” It can cause real effects: spontaneous emission, the Lamb shift (small energy shifts of atomic levels), and the Casimir effect (forces between conducting plates).
The rate scales as \(\omega^3\) — that means:
- Blue light decays faster than red light
- For very high frequencies (X-rays), spontaneous emission is extremely fast
- For very low frequencies (radio), spontaneous emission is negligible — stimulated processes dominate
The lifetime
The lifetime \(\tau\) of an excited state is the inverse of the total decay rate:
$$ \tau = \frac{1}{W_{\text{total}}} = \frac{1}{\sum_f W_{i \to f}} $$
where the sum runs over all lower states \(f\) that are dipole-allowed.
Typical lifetimes:
- Visible light transitions: \(\tau \sim 10^{-8}\) to \(10^{-9}\) seconds (nanoseconds)
- Forbidden transitions (weak or no dipole): \(\tau\) can be milliseconds or even seconds
Comparison: semi-classical vs. quantum theory
| Process | Semi-classical | Quantum field |
|---|---|---|
| Absorption | Works | Works |
| Stimulated emission | Works | Works |
| Spontaneous emission | Fails | Works |
| Lamb shift | Fails | Works |
| Vacuum effects | N/A | Predicts |
The full quantum theory of radiation (Quantum Electrodynamics, or QED) is one of the most precisely tested theories in physics. Spontaneous emission was its first major success.
8. Second Quantization
What is second quantization?
The name is terrible and misleading. It’s not “quantizing twice.” Here’s what it actually means:
- First quantization: Wavefunctions become operators acting on states. Particles have quantum states (position, momentum, etc.), but the number of particles is fixed.
- Second quantization: The fields themselves become operators. The number of particles can change — particles can be created and destroyed.
Second quantization is the natural language for:
- Photons (which are constantly created and absorbed)
- Many-body systems (condensed matter, nuclear physics)
- Quantum field theory
Why we need it for photons
In ordinary quantum mechanics, we have a fixed number of particles. The wavefunction \(\Psi(x_1, x_2, …, x_N)\) describes \(N\) particles, and \(N\) never changes.
But photons are different. An atom emits a photon that wasn’t there before. It absorbs a photon that ceases to exist. The number of photons is not fixed.
We need a formalism where creating and annihilating particles are the basic operations.
The harmonic oscillator reminder
Every mode of the electromagnetic field is a quantum harmonic oscillator. For a single oscillator:
| Quantity | Expression |
|---|---|
| Hamiltonian | \(\hat{H} = \hbar\omega(\hat{a}^\dagger \hat{a} + \frac{1}{2})\) |
| Energy eigenstates | \(|n\rangle\) with \(E_n = \hbar\omega(n + \frac{1}{2})\) |
| Ground state | \(|0\rangle\) (n=0 particles) |
| Excited state | \(|n\rangle\) (n particles/quanta) |
Creation and annihilation operators
For a single mode, we define:
| Operator | Action | Effect |
|---|---|---|
| \(\hat{a}\) | Annihilation | \(\hat{a}|n\rangle = \sqrt{n}|n-1\rangle\) |
| \(\hat{a}^\dagger\) | Creation | \(\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle\) |
Key properties:
$$ \hat{a}|0\rangle = 0 \quad \text{(can’t annihilate nothing)} $$
$$ \hat{a}^\dagger|0\rangle = |1\rangle \quad \text{(create one particle from vacuum)} $$
The number operator:
$$ \hat{N} = \hat{a}^\dagger \hat{a} $$
$$ \hat{N}|n\rangle = n|n\rangle $$
Commutation relations (bosons)
Photons are bosons (integer spin). For bosons, creation and annihilation operators satisfy:
$$ [\hat{a}, \hat{a}^\dagger] = \hat{a}\hat{a}^\dagger – \hat{a}^\dagger\hat{a} = 1 $$
$$ [\hat{a}, \hat{a}] = [\hat{a}^\dagger, \hat{a}^\dagger] = 0 $$
For multiple independent modes \(i, j\):
$$
[\hat{a}_i, \hat{a}_j^\dagger] = \delta_{ij}
$$
All other commutators are zero.
Fock space
The total state of a system with many photons is described by specifying how many photons are in each mode:
$$ |n_{\mathbf{k}_1,\lambda_1}, n_{\mathbf{k}_2,\lambda_2}, … \rangle $$
This is called Fock space or occupation number representation. The vacuum state \(|0,0,0,…\rangle\) is usually written simply as \(|0\rangle\).
A single-photon state in mode \((\mathbf{k}, \lambda)\) is:
$$ \hat{a}_{\mathbf{k},\lambda}^\dagger |0\rangle = |1_{\mathbf{k},\lambda}\rangle $$
A two-photon state (both in same mode) is:
$$ \frac{1}{\sqrt{2}} (\hat{a}_{\mathbf{k},\lambda}^\dagger)^2 |0\rangle = |2_{\mathbf{k},\lambda}\rangle $$
The \(\frac{1}{\sqrt{2}}\) ensures proper normalization.
The quantized EM field (again, more cleanly)
Now we can write the vector potential operator concisely:
$$ \hat{\mathbf{A}}(\mathbf{r}) = \sum_{\mathbf{k},\lambda} \sqrt{\frac{\hbar}{2\epsilon_0 \omega_k V}} \left( \hat{a}_{\mathbf{k},\lambda} \, \epsilon_{\lambda} e^{i\mathbf{k}\cdot\mathbf{r}} + \hat{a}_{\mathbf{k},\lambda}^\dagger \, \epsilon_{\lambda}^* e^{-i\mathbf{k}\cdot\mathbf{r}} \right) $$
The Hamiltonian is:
$$ \hat{H} = \sum_{\mathbf{k},\lambda} \hbar\omega_k \left( \hat{a}_{\mathbf{k},\lambda}^\dagger \hat{a}_{\mathbf{k},\lambda} + \frac{1}{2} \right) $$
The \(\frac{1}{2}\) is the zero-point energy of each mode. It’s infinite (sum over infinite modes) — a problem that gets fixed by renormalization.
For fermions (bonus)
Electrons are fermions (half-integer spin). They also need second quantization, but with different rules:
| Property | Bosons (photons) | Fermions (electrons) |
|---|---|---|
| Statistics | Bose-Einstein | Fermi-Dirac |
| Commutation | \([\hat{a}_i, \hat{a}_j^\dagger] = \delta_{ij}\) | \(\{\hat{c}_i, \hat{c}_j^\dagger\} = \delta_{ij}\) |
| Pauli exclusion | No limit on \(n\) | \(n = 0\) or \(1\) only |
| Operator action | \(\hat{a}_i^\dagger|n_i\rangle = \sqrt{n_i+1}|n_i+1\rangle\) | \(\hat{c}_i^\dagger|0\rangle = |1\rangle\), \(\hat{c}_i^\dagger|1\rangle = 0\) |
The curly braces \(\{\,,\}\) denote an anticommutator:
$$ \{\hat{A}, \hat{B}\} = \hat{A}\hat{B} + \hat{B}\hat{A} $$
Key ideas to remember
| Concept | Meaning |
|---|---|
| Second quantization | Fields → operators, particle number not fixed |
| Creation operator \(\hat{a}^\dagger\) | Adds one particle to a mode |
| Annihilation operator \(\hat{a}\) | Removes one particle from a mode |
| Fock space | States labeled by occupation numbers |
| Bosons | \([\hat{a}, \hat{a}^\dagger] = 1\), unlimited occupancy |
| Fermions | \(\{\hat{c}, \hat{c}^\dagger\} = 1\), Pauli exclusion |
| Vacuum \(|0\rangle\) | Zero particles in all modes |
| Number operator \(\hat{N}\) | Counts particles in a mode |
Why “second quantization” is confusing
Historically:
- First quantization: \(x\) and \(p\) become operators. The wavefunction is a function.
- Second quantization: The wavefunction itself becomes an operator. Fields are quantized.
A better name would be quantum field theory or occupation number representation. But the name stuck.
Most importantly: second quantization doesn’t add new physical content — it’s a more powerful mathematical framework for systems where particle number changes. For spontaneous emission, it’s the natural language because photons appear and disappear.