Density Matrix Formalism

Quantum mechanics hands you a state vector $|\psi\rangle$ and statistical mechanics hands you an ensemble of possibilities weighted by probabilities; the density matrix fuses the two into one object. The state vector is the language of perfect knowledge — every amplitude and phase specified, a pure state. But real atomic ensembles almost never live there. A thermal vapour is a Boltzmann-weighted mixture of energy eigenstates; an atom that has just spontaneously emitted is entangled with a photon it can no longer see, and alone occupies a mixed state; collisions scramble phases in ways no wavefunction tracks. Von Neumann built the density operator $\rho$ precisely to carry both kinds of uncertainty at once — the quantum uncertainty inside a single $|\psi\rangle$ , and the classical uncertainty of not knowing which $|\psi_i\rangle$ you have. Written out, $\rho=\sum_i p_i\,|\psi_i\rangle\langle\psi_i|$ with $p_i\ge 0$ and $\sum_i p_i=1$ , its matrix elements split into two distinct jobs. The diagonal entries $\rho_{nn}$ are the populations — probabilities of finding the atom in level $n$ . The off-diagonal entries $\rho_{nm}$ are the coherences, whose size and phase measure how well-defined the superposition between $n$ and $m$ is. Coherences are the source of the oscillating atomic dipole that drives absorption and stimulated emission; erase them and the atom radiates only through population differences, and the Einstein rate-equation picture drops out. Dynamics obey the Liouville–von Neumann equation $\mathrm{i}\hbar\,\dot\rho=[H,\rho]$ , and once relaxation is added this becomes the optical Bloch equations — the master language wherever coherence, statistical mixing, and dissipation coexist: laser spectroscopy, coherent control, quantum optics, and quantum information.

Key takeaways

The density operator $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$ generalizes the state vector to statistical mixtures, subsystems of entangled states, and ensembles no wavefunction can capture.
Its diagonal elements $\rho_{nn}$ are populations (probabilities); its off-diagonal elements $\rho_{nm}$ are coherences, encoding the phase relations behind interference, Rabi flopping, and dipole radiation.
The equation of motion is $\mathrm{i}\hbar\,\dot\rho=[H,\rho]$ ; adding relaxation yields the optical Bloch equations governing absorption, saturation, and fluorescence.
Purity $\mathrm{Tr}(\rho^2)$ diagnoses the state — $1$ pure, toward $1/d$ maximally mixed — and every observable follows from a single trace, $\langle A\rangle=\mathrm{Tr}(\rho A)$ .
A two-level state maps onto the Bloch sphere: populations set the polar angle, coherences the azimuth, and relaxation drags the tip inward toward equilibrium.
Key idea: the density operator packages populations (diagonal, probabilities) and coherences (off-diagonal, phase relations) into one object $\rho$ whose Hermiticity, unit trace, and positivity guarantee sensible probabilities, and whose evolution $\mathrm{i}\hbar\dot\rho=[H,\rho]$ plus relaxation reproduces all of quantum optics — from Rabi flopping to the Einstein rate equations.

Where this lives today

The density matrix is the daily language of atomic, molecular, and optical physics. Every magneto-optical trap and laser-cooling stage is designed by solving the optical Bloch equations: the steady-state $\rho_{ee}^{ss}=\tfrac12\,s/(1+s)$ sets the scattering force, its ceiling $\Gamma/2$ caps the deceleration, and the coherences set the sub-Doppler features that carried cooling below the Doppler limit. In atomic clocks and quantum computers the same $\rho$ tracks the qubit, and the coherence time $T_2=1/\gamma$ is the number the field fights to extend — every dynamical-decoupling sequence and magnetic shield is an attempt to keep the off-diagonal elements alive longer. Saturated-absorption spectroscopy reads the steady-state coherence directly as a Lorentzian susceptibility, and the reduced density matrix — a subsystem after its environment is traced away — is the working model of decoherence that sets error rates for quantum information.

The physical picture

Think of $\rho$ as a small square ledger. Down the diagonal it records how much of the atom sits in each level — ordinary probabilities that add to one. Off the diagonal it records how tightly the phases between levels are locked: a perfectly phased superposition fills those cells to maximum magnitude, a randomized ensemble leaves them at zero. Both the classical statistician's question ("which level?") and the interferometer's question ("what phase relation?") are answered by reading different cells of the same matrix — which is what makes $\rho$ irreducible.

Geometrically, a two-level system maps onto the Bloch sphere. A pure state is a unit arrow touching the surface: populations fix its polar angle, coherences fix its azimuth. Statistical mixing shortens the arrow so it retreats inside the ball, and a maximally mixed state sits at the centre with no direction at all. Relaxation is friction: longitudinal decay $\Gamma$ pulls the polar component toward the ground state as atoms emit, transverse decay $\gamma$ collapses the azimuthal component as coherences dephase. The purity $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\mathbf s|^2)$ is just how far the arrow reaches — the single number telling pure ( $|\mathbf s|=1$ ) from mixed.

Figure

The Bloch sphere: populations set the polar angle, coherences the azimuth. A pure state reaches the surface at vector length 1 with Tr(ρ²) = 1, a mixed state sits inside at length 0.71 with Tr(ρ²) = 0.75, the maximally mixed state at the centre with Tr(ρ²) = 0.5. Γ drags the tip toward the ground state, γ collapses it toward the axis

Why introduce $\rho$ when $|\psi\rangle$ already works for isolated systems? Because open systems and subsystems generically have no wavefunction. Trace out the photon after spontaneous emission and the atom alone is genuinely mixed — no $|\psi\rangle$ in Hilbert space reproduces its statistics. Here the density matrix is not a convenience but the only correct description. And every observable still emerges from one operation, $\langle A\rangle=\mathrm{Tr}(\rho A)$ , a trace that performs the quantum expectation inside each pure component and the classical average across components in a single stroke.

The core equations

The starting point is the definition. Instead of a single ket, the state is a probability-weighted sum of projectors onto the pure states $|\psi_i\rangle$ that might have been prepared; the kets need not be orthogonal, which lets the same $\rho$ be decomposed many ways.

\rho=\sum_i p_i\,|\psi_i\rangle\langle\psi_i|,\qquad p_i\ge 0,\ \sum_i p_i=1 \qquad (1)

where: $\rho$ — density operator; $p_i$ — classical probability of having prepared $|\psi_i\rangle$ ; $|\psi_i\rangle$ — normalized pure states (not necessarily orthogonal). A pure state is the special case where one $p_i=1$ ; any spread of the $p_i$ over distinct states is genuine statistical mixing. Three properties make $\rho$ a legitimate description of probabilities; a matrix failing any of them is not a physical state. It is Hermitian (real eigenvalues), unit-trace (total probability conserved), and positive semidefinite (no measurement returns a negative probability).

\rho^\dagger=\rho,\qquad \mathrm{Tr}\,\rho=1,\qquad \langle\phi|\rho|\phi\rangle\ge 0\ \ \forall\,|\phi\rangle \qquad (2)

where: $\rho^\dagger=\rho$ — Hermiticity; $\mathrm{Tr}\,\rho=1$ — probability conserved; $\langle\phi|\rho|\phi\rangle\ge0$ — positivity. Positivity is the sharpest constraint: it caps every coherence at $|\rho_{nm}|\le\sqrt{\rho_{nn}\rho_{mm}}$ , so phase locking can never exceed what the populations permit. Every measurable quantity is extracted the same way — a single trace of the operator against $\rho$ — replacing both the quantum expectation of a pure state and the classical average over the ensemble at once.

\langle A\rangle=\mathrm{Tr}(\rho A)=\sum_{n,m}\rho_{nm}A_{mn} \qquad (3)

where: $A$ — any observable; $\rho_{nm}=\langle n|\rho|m\rangle$ in the basis $\{|n\rangle\}$ ; $A_{mn}=\langle m|A|n\rangle$ . Because coherences $\rho_{nm}$ multiply off-diagonal operator elements, an off-diagonal observable like the dipole moment reads out precisely the coherences, not the populations. The state's evolution is fixed by the Hamiltonian through a commutator — the density-operator sibling of the Schrödinger equation. Note the ordering $[H,\rho]$ , opposite in sign to the Heisenberg equation for operators.

\mathrm{i}\hbar\,\frac{\partial\rho}{\partial t}=[H,\rho] \qquad (4)

where: $H$ — Hamiltonian; $[H,\rho]=H\rho-\rho H$ . Coherent evolution never changes the eigenvalues of $\rho$ , so purity is conserved: a closed system that starts pure stays pure. Mixedness enters only through relaxation or through tracing out a partner. The single number separating pure from mixed is the purity. It equals one exactly when $\rho$ is a projector onto one state and falls as probability spreads across eigenstates, bottoming out at $1/d$ in $d$ dimensions.

\mathrm{Tr}(\rho^2)=\sum_k\lambda_k^2\le 1,\qquad \mathrm{Tr}(\rho^2)=1\ \Leftrightarrow\ \text{pure} \qquad (5)

where: $\lambda_k$ — eigenvalues of $\rho$ ( $\sum_k\lambda_k=1$ ). Purity is basis-independent, so it is a genuine physical property of the state and the cleanest experimental handle on decoherence. For a two-level system the whole state collapses onto a single three-vector. Writing $\rho$ with the identity and the Pauli matrices exposes the geometry: the Bloch vector $\mathbf s$ carries everything, its length is the purity, and its components split into a population part and a coherence part.

\rho=\tfrac12\big(\mathbb{1}+\mathbf{s}\cdot\boldsymbol{\sigma}\big),\qquad |\mathbf s|\le 1,\qquad \mathrm{Tr}(\rho^2)=\tfrac12\big(1+|\mathbf s|^2\big) \qquad (6)

where: $\mathbf s=(s_x,s_y,s_z)$ — Bloch vector; $\boldsymbol\sigma$ — Pauli matrices; $s_z=\rho_{ee}-\rho_{gg}$ — inversion; $\rho_{eg}=\tfrac12(s_x-\mathrm i s_y)$ — coherence. $|\mathbf s|=1$ pure (surface), $|\mathbf s|<1$ mixed (interior), $\mathbf s=0$ maximally mixed (centre). Finally, the operation that manufactures mixed states from pure ones: the partial trace. Summing over one half of an entangled pair leaves the other in a mixed state even though the pair as a whole is pure — the formal statement that a subsystem of an entangled state has no wavefunction of its own.

\rho_A=\mathrm{Tr}_B(\rho_{AB}),\qquad \rho_{AB}\ \text{entangled}\ \Rightarrow\ \mathrm{Tr}(\rho_A^2)<1 \qquad (7)

where: $\mathrm{Tr}_B$ — partial trace over subsystem $B$ (e.g. the emitted photon); $\rho_A$ — reduced state of $A$ (the atom alone). Entanglement with an unobserved partner is a source of mixedness distinct from classical ignorance — and, for spontaneous emission, the physical one.

Derivation AThe Liouville von Neumann equation of motion

The goal is to earn equation (4) rather than assert it: to show that if every pure member of the ensemble obeys the Schrödinger equation, the density matrix built from them obeys $\mathrm{i}\hbar\dot\rho=[H,\rho]$ , and to see where the commutator — with its sign opposite to Heisenberg's — comes from. This matters because the sign is a frequent error, and because the derivation shows that coherent evolution conserves purity, so mixedness can only arise from relaxation or from tracing out a partner.

Step 1. Write the Schrödinger equation for one ensemble member and its conjugate: $\mathrm{i}\hbar\,\partial_t|\psi_i\rangle=H|\psi_i\rangle$ and $-\mathrm{i}\hbar\,\partial_t\langle\psi_i|=\langle\psi_i|H$ . The weights $p_i$ are fixed at $t=0$ and do not evolve. Why: the ket and bra pick up opposite signs of $\mathrm i\hbar\,\partial_t$ , and that opposition becomes the commutator.

Step 2. Differentiate $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$ by the product rule: $\mathrm{i}\hbar\,\partial_t\rho=\sum_i p_i\big[(\mathrm{i}\hbar\partial_t|\psi_i\rangle)\langle\psi_i|+|\psi_i\rangle(\mathrm{i}\hbar\partial_t\langle\psi_i|)\big]$ . Why: only the states carry time dependence, so the derivative acts term by term and leaves the weights untouched.

Step 3. Substitute Step 1. The first bracket gives $H|\psi_i\rangle\langle\psi_i|$ , the second $-|\psi_i\rangle\langle\psi_i|H$ ; summing, $\mathrm{i}\hbar\,\partial_t\rho=H\rho-\rho H=[H,\rho]$ . Why: each bracketed sum is exactly $\rho$ , so the equation closes on $\rho$ itself — autonomous, needing no reference to individual $|\psi_i\rangle$ .

Step 4. Contrast the Heisenberg equation $\mathrm{i}\hbar\,\dot A=[A,H]$ , whose commutator is reversed: $\rho$ is a state built from kets, not a Heisenberg operator, so its generator sits on the left. Why: the sign is the difference between a coherence rotating the right way on the Bloch sphere and the wrong way.

Go deeper · Derivation B · 1 minWhy a mixed state has no wavefunction the reduced density matrix after emission

Derivation BWhy a mixed state has no wavefunction the reduced density matrix after emission

The aim is the most convincing reason the density matrix is indispensable: after an atom spontaneously emits, it is genuinely mixed, and no wavefunction reproduces its statistics, even though the atom plus photon together remain perfectly pure. This is not classical ignorance about which state was prepared; it is entanglement with a partner we chose not to look at. The calculation is short and exact, and it is the microscopic origin of decoherence, of why the diagonal Einstein picture is valid, and of the coherence-decay term $\gamma$ .

Step 2. Form the joint operator $\rho_{AB}=|\Psi\rangle\langle\Psi|$ , still pure: $\mathrm{Tr}(\rho_{AB}^2)=1$ . Why: nothing irreversible has happened to the whole — the lost information is only about which part we keep.

Step 3. Trace over the field, $\rho_{\text{atom}}=\sum_{n=0,1}\langle n|\rho_{AB}|n\rangle$ . The cross terms carry $\langle 0|1_k\rangle=0$ , since zero- and one-photon states are orthogonal. Why: those cross terms are the atomic coherences $\alpha\beta^*|e\rangle\langle g|$ — the trace kills them because the photon's which-path record is perfect.

Step 4. What survives is diagonal, $\rho_{\text{atom}}=|\alpha|^2|e\rangle\langle e|+|\beta|^2|g\rangle\langle g|$ , purity $|\alpha|^4+|\beta|^4$ . For $|\alpha|^2=|\beta|^2=\tfrac12$ that is $\tfrac12<1$ : a mixed state no $|\psi\rangle$ can match. Why: had the field states been identical ( $\langle0|1_k\rangle=1$ , no entanglement), the coherence would survive and $\rho$ factor into a pure atomic state — so entanglement, not ignorance, mixed the atom.

Go deeper · Derivation C · 2 minThe optical Bloch equations and steady state saturation

Derivation CThe optical Bloch equations and steady state saturation

The payoff: take a two-level atom driven by a near-resonant laser, add spontaneous emission and dephasing as relaxation, and read off the steady state. This is where the formalism becomes the working tool of laser spectroscopy — it yields the Lorentzian absorption line, the saturation of a strong transition, and, in the fast-dephasing limit, the Einstein rate equations exactly. Resonance fluorescence, saturated-absorption spectroscopy, and laser cooling all rest on the two coupled equations below.

Step 1. Split $H=H_0+H_{\text{int}}$ with dipole coupling $H_{\text{int}}=-\mathbf d\cdot\mathbf E$ , and define the Rabi frequency $\Omega=\langle e|\mathbf d\cdot\boldsymbol\varepsilon|g\rangle E_0/\hbar$ and detuning $\delta=\omega-\omega_0$ . Move to the rotating frame, make the rotating-wave approximation, and use the slow coherence $\tilde\rho_{eg}=\rho_{eg}e^{\mathrm i\omega t}$ . Why: $\Omega$ measures how hard the field shakes the dipole and $\delta$ how far off resonance; the rotating frame removes the fast optical oscillation.

Step 2. Evaluate $\mathrm i\hbar\dot\rho=[H,\rho]$ in the rotating frame; the commutator couples populations and coherences through $\Omega$ : $\dot\rho_{ee}=\tfrac{\mathrm i\Omega}{2}(\tilde\rho_{eg}-\tilde\rho_{ge})$ and $\dot{\tilde\rho}_{eg}=\mathrm i\delta\,\tilde\rho_{eg}+\tfrac{\mathrm i\Omega}{2}(\rho_{ee}-\rho_{gg})$ . Why: driving does not change populations directly — it first builds a coherence, which then feeds the excited population, so coherence is the engine of absorption.

Step 3. Add relaxation: population decays at $\Gamma=1/\tau$ , coherence at $\gamma=\Gamma/2+\gamma_{\text{ph}}$ ( $\gamma_{\text{ph}}$ = collisional dephasing). The optical Bloch equations follow: $\dot\rho_{ee}=-\Gamma\rho_{ee}+\tfrac{\mathrm i\Omega}{2}(\tilde\rho_{eg}-\tilde\rho_{ge})$ and $\dot{\tilde\rho}_{eg}=-(\gamma-\mathrm i\delta)\tilde\rho_{eg}+\tfrac{\mathrm i\Omega}{2}(\rho_{ee}-\rho_{gg})$ . Why: for pure radiative decay the coherence dies at half the population rate because amplitude decays half as fast as probability; collisions add to $\gamma$ without touching $\Gamma$ .

Step 4. Set $\dot{\tilde\rho}_{eg}=0$ : $\tilde\rho_{eg}^{ss}=\tfrac{\mathrm i\Omega}{2}(\rho_{gg}-\rho_{ee})/(\gamma-\mathrm i\delta)$ . Its absorptive part $\propto\gamma/(\gamma^2+\delta^2)$ is a Lorentzian of full width $2\gamma$ in detuning — the natural line shape read off the coherence. Why: the atom responds like a damped driven oscillator, and the density matrix delivers that Lorentzian without leaving the two-level picture.

Step 5. Feed back to get $\rho_{ee}^{ss}=\tfrac12\,s/(1+s)$ with $s=(\Omega^2/2)/(\delta^2+\gamma^2)$ . As $s\to\infty$ , $\rho_{ee}\to\tfrac12$ : populations equalize and net absorption stops. Why: this ceiling makes fluorescence and laser cooling finite — an atom scatters at most $\Gamma/2$ photons per second however hard you drive it.

Step 6. Check the classical limit: for weak driving and fast dephasing ( $\Omega\to0$ , $\gamma$ large) eliminate the coherence adiabatically to leave $\dot\rho_{ee}=-\Gamma\rho_{ee}+B\,u(\omega)(\rho_{gg}-\rho_{ee})$ — the Einstein rate equation, $B$ carried by $\Omega^2$ and $A$ by $\Gamma$ . Why: this is the seam where coherent quantum optics rejoins the rate-equation picture, showing the Einstein coefficients as a limit of the density matrix, not a separate postulate.

Limiting cases

The formalism interpolates smoothly between familiar regimes as the coherences and the mixing are dialled:

Case	Result	Meaning
Pure state	$\mathrm{Tr}(\rho^2)=1,\ \rho^2=\rho$	$\rho=\vert\psi\rangle\langle\psi\vert$ is a projector; all statistics come from one wavefunction — the ordinary Schrödinger limit.
Maximally mixed	$\rho=\tfrac1d\mathbb 1,\ \mathrm{Tr}(\rho^2)=\tfrac1d$	every basis state equally likely, all coherences zero, no preferred Bloch direction — the fully thermalized $T\to\infty$ limit.
Coherences damped ( $\gamma\to\infty$ )	$\dot\rho_{nn}\to$ Einstein rate equations	adiabatically eliminate $\rho_{nm}$ ; only populations evolve, recovering the $B$ -coefficient absorption and $A$ -coefficient spontaneous rates.
Thermal equilibrium	$\rho=e^{-H/k_{\mathrm B}T}/\mathrm{Tr}\,e^{-H/k_{\mathrm B}T}$	diagonal in the energy basis, Boltzmann-weighted populations, exactly zero coherence — the canonical density matrix of statistical mechanics.
Strong driving ( $s\to\infty$ )	$\rho_{ee}^{ss}\to\tfrac12$	populations saturate, scattering rate ceils at $\Gamma/2$ .

Common trapA mixed state is the same thing as a coherent superposition

It is not, and conflating the two is the most common error with $\rho$ . The pure superposition $\tfrac1{\sqrt2}(|1\rangle+|2\rangle)$ has $\rho_{11}=\rho_{22}=\rho_{12}=\rho_{21}=\tfrac12$ — full coherences, $\mathrm{Tr}(\rho^2)=1$ . The 50/50 mixture has the identical diagonal but $\rho_{12}=\rho_{21}=0$ and $\mathrm{Tr}(\rho^2)=\tfrac12$ . Same populations, utterly different physics: only the superposition carries a macroscopic oscillating dipole and can interfere or Rabi-flop; the mixture is inert. Never infer the state from populations alone — the coherences, invisible to a population measurement, decide whether the atom radiates.

Worked examples

1 · The thermal excited-state population of the sodium D line (a confidence builder)

A sodium vapour at $T=300$ K sits in thermal equilibrium. The D transition is at $\lambda=589$ nm. What fraction of atoms is in the excited state, and what does that say about the coherences?

Step 1. The transition frequency is $\nu=c/\lambda=(2.998\times10^8)/(589\times10^{-9})=5.09\times10^{14}$ Hz, so $E=h\nu=3.37\times10^{-19}$ J $=2.11$ eV — an optical energy of a couple of electron-volts.

Step 2. Compare the thermal energy $k_{\mathrm B}T=(1.381\times10^{-23})(300)=4.14\times10^{-21}$ J $=0.0259$ eV. The ratio $E/k_{\mathrm B}T=2.11/0.0259=81.4$ — eighty-one thermal quanta.

Step 3. The Boltzmann-weighted excited fraction is $\rho_{ee}\approx e^{-81.4}=\mathbf{4.4\times10^{-36}}$ . For all practical purposes $\rho_{gg}=1$ , $\rho_{ee}\approx0$ , purity $\mathrm{Tr}(\rho^2)\approx1$ — essentially the pure ground state.

Step 4. The lesson is off-diagonal. A thermal density matrix is diagonal in the energy basis: its coherences $\rho_{eg}$ are exactly zero, so an undriven vapour carries no optical dipole and cannot emit coherently — to make it radiate you must drive it, building the coherence $\rho_{eg}$ that equilibrium forbids. Hence the next example needs a laser.

2 · Saturating the sodium D₂ line with a resonant laser

The Na D₂ line has a natural linewidth $\Gamma=2\pi\times9.79$ MHz (radiative lifetime $\tau=1/\Gamma\approx16.2$ ns). Drive it exactly on resonance ( $\delta=0$ ) with pure radiative damping ( $\gamma=\Gamma/2$ ) at Rabi frequency $\Omega=\Gamma$ . Find the steady-state excited population and the photon scattering rate, and the maximum possible scattering rate.

Step 1. Convert: $\Gamma=2\pi\times9.79\times10^6=6.15\times10^7\ \mathrm{s^{-1}}$ , so $\tau=1/\Gamma\approx16.2$ ns.

Step 2. On resonance $s=(\Omega^2/2)/(\Gamma/2)^2=2\Omega^2/\Gamma^2$ ; with $\Omega=\Gamma$ , $s=2$ .

Step 3. The excited population is $\rho_{ee}^{ss}=\tfrac12\,s/(1+s)=\tfrac12\cdot\tfrac{2}{3}=\mathbf{0.333}$ — one atom in three excited at any instant.

Step 4. Each excited atom decays at $\Gamma$ , so $R=\Gamma\rho_{ee}^{ss}=(6.15\times10^7)/3=\mathbf{2.05\times10^7\ photons/s}$ , about 20 million per second.

Step 5. As the drive $\to\infty$ , $\rho_{ee}^{ss}\to\tfrac12$ and $R_{\max}=\Gamma/2=\mathbf{3.08\times10^7\ photons/s}$ . However intense the laser, a two-level atom cannot scatter faster than $\Gamma/2\approx3.1\times10^7$ photons per second — the ceiling that limits the force in laser cooling.

3 · Purity and the Bloch vector of a partially coherent state

An atom is described by populations $\rho_{11}=0.75$ , $\rho_{22}=0.25$ and a real coherence $\rho_{12}=\rho_{21}=0.25$ . Is this a valid state, and how mixed is it?

Step 1. Check positivity. The coherence is capped at $|\rho_{12}|\le\sqrt{\rho_{11}\rho_{22}}=\sqrt{0.75\times0.25}=0.433$ . Since $0.25<0.433$ , the state is physical.

Step 2. Compute the purity directly, $\mathrm{Tr}(\rho^2)=\rho_{11}^2+\rho_{22}^2+2\rho_{12}^2=0.5625+0.0625+2(0.0625)=\mathbf{0.75}$ . Below one, so the state is mixed.

Step 3. Cross-check via the Bloch vector: $s_z=\rho_{22}-\rho_{11}=-0.5$ , $s_x=2\rho_{12}=0.5$ , giving $|\mathbf s|=\sqrt{0.25+0.25}=0.707$ . Then $\mathrm{Tr}(\rho^2)=\tfrac12(1+|\mathbf s|^2)=\tfrac12(1.5)=0.75$ — the two routes agree. The state is genuinely mixed, $\mathrm{Tr}(\rho^2)=0.75$ , with a Bloch vector of length $0.71$ reaching partway to the surface.

Physics of the portrait

The density matrix is where quantum amplitudes and classical probabilities were first made to live in one equation — and the person who built that equation was a mathematician working out what a quantum measurement really is.

Origins · 1 minJohn von Neumann and how this was discovered

John von Neumann

1903–1957 · the density operator and the foundations of quantum statistics

John von Neumann introduced the density operator in 1927, in the same run of papers that cast quantum mechanics into the rigorous language of Hilbert space. He wanted a single object to describe a system known only statistically — a beam of atoms with a spread of spins, a subsystem of a larger correlated whole — and found it in $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$ , with the rule $\langle A\rangle=\mathrm{Tr}(\rho A)$ turning any observable into a trace. He proved the equation of motion $\mathrm{i}\hbar\dot\rho=[H,\rho]$ that bears his and Liouville's names, and defined the entropy $S=-k_{\mathrm B}\mathrm{Tr}(\rho\ln\rho)$ measuring mixedness — zero for a pure state, maximal for the fully thermalized one. Lev Landau reached the density matrix independently that same year, studying how a subsystem loses its wavefunction — the very mechanism behind the reduced density matrix here. Von Neumann's 1932 Mathematische Grundlagen der Quantenmechanik systematized the whole apparatus, and it is still the framework in which decoherence, quantum information, and the optical Bloch equations are phrased. The formalism reduces to ordinary wavefunctions exactly when the state is pure, and takes over everywhere a wavefunction cannot go.

History

1927

— Landau introduces the density matrix to describe a subsystem that has lost its own wavefunction

1927

— von Neumann formulates the density operator, the rule ⟨A⟩ = Tr(ρA), and the equation iħρ̇ = [H,ρ]

1932

— von Neumann's Mathematische Grundlagen systematizes ρ, purity Tr(ρ²), and the entropy S = −k Tr(ρ ln ρ)

1946

— Bloch writes phenomenological relaxation equations with a magnetization vector and the times T₁, T₂

1957

— Feynman, Vernon, and Hellwarth map any two-level system onto a Bloch vector on a sphere

Going further

References. Foot, Atomic Physics (2005), Ch. 7 — density matrix and optical Bloch equations · Loudon, The Quantum Theory of Light (3rd ed., 2000), Ch. 2 and 6 — density operator and resonance fluorescence · Cohen-Tannoudji, Diu and Laloë, Quantum Mechanics, Vol. 1, Complement E_III — the density operator · Demtröder, Laser Spectroscopy, Vol. 1 — coherence and relaxation in spectroscopy · Sakurai and Napolitano, Modern Quantum Mechanics — statistical mixtures and the density operator · Blum, Density Matrix Theory and Applications (3rd ed., 2012).

Provenance. Definitions and properties of $\rho$ follow Cohen-Tannoudji Complement E_III and Sakurai; the Liouville–von Neumann derivation and the optical Bloch equations follow Foot Ch. 7 and Loudon Ch. 6. A verifier should reconfirm: the sign of $[H,\rho]$ against Heisenberg's $[A,H]$ ; the steady state $\rho_{ee}^{ss}=\tfrac12\,s/(1+s)$ with $s=(\Omega^2/2)/(\delta^2+\gamma^2)$ ; the relation $\gamma=\Gamma/2$ for pure radiative decay; and the purity bound $\mathrm{Tr}(\rho^2)\le1$ with equality iff pure. The worked numbers are standard: Na D at 589 nm gives $E/k_{\mathrm B}T=81.4$ and $\rho_{ee}\approx4.4\times10^{-36}$ at 300 K; Na D₂ with $\Gamma=2\pi\times9.79$ MHz ( $\tau\approx16.2$ ns) gives $R_{\max}=\Gamma/2=3.08\times10^7\ \mathrm{s^{-1}}$ ; the partial-coherence example has $\mathrm{Tr}(\rho^2)=0.75$ , $|\mathbf s|=0.71$ .

Builds on: Electric Dipole Approximation · Transition Dipole Moment · Einstein A and B Coefficients. Leads to: Stimulated Emission · Spontaneous Emission · Lorentzian Line Shape · Radiative Lifetime and Natural Linewidth.

Ask the tutor: Ask me to walk you through building the optical Bloch equations for your own atom and reading its saturation and linewidth straight off the steady-state density matrix.

What is the difference between a pure state and a mixed state?

A pure state is one you know completely as a single wavefunction, so $\rho=|\psi\rangle\langle\psi|$ and $\mathrm{Tr}(\rho^2)=1$ ; a mixed state is a probabilistic ensemble of several wavefunctions, $\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|$ with more than one nonzero $p_i$ and $\mathrm{Tr}(\rho^2)<1$ . The crucial subtlety: a coherent superposition like $\tfrac1{\sqrt2}(|1\rangle+|2\rangle)$ is pure, not mixed — it is one definite wavefunction. Mixedness means genuine classical uncertainty about which wavefunction you have, or entanglement with something you traced away, and only mixedness lowers the purity below one.

Are the coherences the same as a superposition?

They are the fingerprint of one. In a coherent superposition the off-diagonal elements $\rho_{12}$ are nonzero and carry a definite phase; in a statistical mixture of the same two levels they are exactly zero even though the populations are identical. So a nonzero coherence is precisely what tells you a superposition, not a mixture, is present. Physically the coherence is the amplitude of the oscillating atomic dipole — what absorbs, emits stimulated radiation, and interferes. Kill the coherences and the atom still has populations but behaves like a classical probabilistic object with no wave character.

Why does the density matrix use a trace for every measurement?

Because the trace does two averages in one operation: inside each pure component it computes the quantum expectation $\langle\psi_i|A|\psi_i\rangle$ , and across components it weights by the classical probabilities $p_i$ — exactly the two kinds of averaging a mixed state demands. Algebraically $\mathrm{Tr}(\rho A)=\sum_{n,m}\rho_{nm}A_{mn}$ , which shows a second virtue: off-diagonal observables like the dipole moment pick out the coherences, while diagonal observables like energy pick out the populations. The trace is also basis-independent, so the prediction never depends on how you wrote the matrix.

How does the density matrix recover the Einstein rate equations?

Through the fast-dephasing limit of the optical Bloch equations. When $\gamma$ is large, $\rho_{eg}$ relaxes to its quasi-steady value almost instantly and can be eliminated adiabatically — set $\dot{\tilde\rho}_{eg}=0$ and substitute back. What remains are closed equations for populations alone, $\dot\rho_{ee}=-\Gamma\rho_{ee}+B\,u(\omega)(\rho_{gg}-\rho_{ee})$ , with a stimulated rate $\propto\Omega^2\gamma/(\delta^2+\gamma^2)$ (the $B$ coefficient) and a spontaneous rate $\Gamma$ (the $A$ coefficient). The Einstein picture is thus not a separate theory but the coherence-free corner of the density matrix, valid whenever phases scramble faster than populations change.

What sets the coherence decay rate γ, and why is it Γ over 2?

Two things add into $\gamma$ . First, spontaneous emission: since the coherence is an amplitude and the excited population is a probability (amplitude squared), the amplitude decays at half the rate, so radiative decay alone gives $\gamma=\Gamma/2$ . Second, pure dephasing $\gamma_{\text{ph}}$ from collisions or field fluctuations, which randomize the phase without changing populations, adds directly: $\gamma=\Gamma/2+\gamma_{\text{ph}}$ . In magnetic-resonance language these are the two relaxation times, $T_1=1/\Gamma$ (longitudinal) and $T_2=1/\gamma$ (transverse), with $T_2\le2T_1$ ; pure dephasing is what pushes $T_2$ below its radiative ceiling.

Can a system start pure and become mixed on its own?

Not under closed, unitary evolution — the Liouville–von Neumann equation conserves the eigenvalues of $\rho$ , hence its purity: a truly isolated state that starts pure stays pure forever. Mixedness arises only two ways: you add relaxation by hand (shorthand for coupling to a reservoir), or you entangle with another system and trace it out. Spontaneous emission is the textbook case — the atom entangles with the photon field and, considered alone, becomes mixed even though atom-plus-field remains pure. So decoherence is never spontaneous in a closed universe; it is always the shadow of a correlation with something you stopped tracking.