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Derivation

The Density Matrix and von Neumann Equation

D-364 Home PU-401 Threads chance · matter Depends on schrodinger-equation, tensor-product-hilbert-space
Statement

For a quantum system whose state is known only statistically — a mixed state realised as an ensemble \(\{p_i,\,|\psi_i\rangle\}\) with \(p_i\ge 0\) and \(\sum_i p_i = 1\) — the density operator \(\hat{\rho}=\sum_i p_i\,|\psi_i\rangle\langle\psi_i|\) encodes all measurable information. We prove that \(\hat{\rho}\) is Hermitian, positive semi-definite, and of unit trace; that under unitary Schrödinger evolution it obeys the von Neumann equation \(i\hbar\,\partial_t\hat{\rho}=[\hat{H},\hat{\rho}]\); and that the state of a subsystem \(A\) of a composite \(\mathcal{H}_A\otimes\mathcal{H}_B\) is the reduced density matrix \(\hat{\rho}_A=\operatorname{Tr}_B\hat{\rho}\), the unique operator reproducing every local expectation value.

Why it matters

A single ket \(|\psi\rangle\) describes only a pure state — maximal knowledge. The moment a system entangles with anything we do not track (a measuring apparatus, a thermal bath, a partner particle sent to a distant lab), no ket describes it, yet it still has definite statistics. The density matrix is the object that survives this loss of information: it is the true state space of quantum mechanics, of which pure states are merely the extreme points.

It is also the language in which quantum statistical mechanics, decoherence, open-system dynamics, and quantum information are actually written. The thermal state \(\hat{\rho}\propto e^{-\beta\hat{H}}\), the qubit Bloch ball, entanglement entropy, and the Lindblad master equation all begin here. On the chance thread it is where classical probability and quantum amplitude coexist; on the matter thread it is how a piece of the world behaves once the rest is traced away.

Assumptions
The ensemble probabilities are classical.If the \(p_i\) were themselves amplitudes to be added coherently, we would have a pure superposition, not a mixture; dropping this collapses the whole distinction the density matrix exists to make. Closed-system evolution is unitary, generated by a Hermitian \(\hat{H}\).If \(\hat{H}\) is non-Hermitian or the evolution is non-unitary, the derivation of \(i\hbar\,\partial_t\hat{\rho}=[\hat{H},\hat{\rho}]\) fails and trace and positivity are no longer preserved — one needs the Lindblad generalisation instead. The composite Hilbert space is a genuine tensor product \(\mathcal{H}_A\otimes\mathcal{H}_B\).Partial trace is defined relative to a factorisation; without a fixed tensor structure (e.g. for indistinguishable particles on (anti)symmetrised spaces) "the subsystem" is ambiguous and \(\operatorname{Tr}_B\) is not well defined. Observables of \(A\) act as \(\hat{O}_A\otimes\hat{\mathbf{1}}_B\).The reduced density matrix is defined by reproducing exactly these local expectation values; if the measurement couples nontrivially to \(B\), \(\hat{\rho}_A\) alone no longer suffices.
Derivation
1
\[ \hat{\rho}\;=\;\sum_i p_i\,|\psi_i\rangle\langle\psi_i|,\qquad p_i\ge 0,\quad \sum_i p_i = 1 . \]
Definition: build the operator whose \(i\)-th term is the pure-state projector \(|\psi_i\rangle\langle\psi_i|\) weighted by the classical probability of preparing \(|\psi_i\rangle\). The \(|\psi_i\rangle\) are normalised but need not be orthogonal. A
2
\[ \hat{\rho}^{\dagger}=\sum_i p_i\left(|\psi_i\rangle\langle\psi_i|\right)^{\dagger}=\sum_i p_i\,|\psi_i\rangle\langle\psi_i|=\hat{\rho}. \]
Hermiticity. Each projector is self-adjoint and the \(p_i\) are real, so the sum is self-adjoint; hence \(\hat{\rho}\) has real eigenvalues and an orthonormal eigenbasis (spectral theorem). A
3
\[ \langle\phi|\hat{\rho}|\phi\rangle=\sum_i p_i\,\langle\phi|\psi_i\rangle\langle\psi_i|\phi\rangle=\sum_i p_i\,\bigl|\langle\phi|\psi_i\rangle\bigr|^2\;\ge\;0\quad\forall\,|\phi\rangle. \]
Positive semi-definiteness. Each term is a non-negative probability times a squared modulus; therefore every eigenvalue of \(\hat{\rho}\) lies in \([0,1]\). A
4
\[ \operatorname{Tr}\hat{\rho}=\sum_i p_i\operatorname{Tr}\!\left(|\psi_i\rangle\langle\psi_i|\right)=\sum_i p_i\,\langle\psi_i|\psi_i\rangle=\sum_i p_i=1. \]
Unit trace, using \(\operatorname{Tr}(|\psi_i\rangle\langle\psi_i|)=\langle\psi_i|\psi_i\rangle=1\) and normalisation of the ensemble weights. Steps 2–4 are exactly the defining properties of a density operator. A
5
\[ \langle\hat{A}\rangle=\sum_i p_i\,\langle\psi_i|\hat{A}|\psi_i\rangle=\sum_i p_i\sum_n\langle\psi_i|n\rangle\langle n|\hat{A}|\psi_i\rangle. \]
Expectation values. Average the quantum expectation in each pure state over the classical ensemble, then insert a resolution of identity \(\sum_n|n\rangle\langle n|=\hat{\mathbf 1}\). A
6
\[ \langle\hat{A}\rangle=\sum_n\langle n|\hat{A}\Bigl(\sum_i p_i|\psi_i\rangle\langle\psi_i|\Bigr)|n\rangle=\sum_n\langle n|\hat{A}\hat{\rho}|n\rangle=\operatorname{Tr}\!\left(\hat{\rho}\hat{A}\right). \]
Regroup the sum over \(i\) into \(\hat{\rho}\), then read off the trace. Every measurable prediction depends on the ensemble only through \(\hat{\rho}\) — distinct ensembles giving the same \(\hat{\rho}\) are physically indistinguishable. B
7
\[ i\hbar\,\partial_t|\psi_i(t)\rangle=\hat{H}|\psi_i(t)\rangle,\qquad -i\hbar\,\partial_t\langle\psi_i(t)|=\langle\psi_i(t)|\hat{H}. \]
Dynamics. Each ensemble member evolves by the Schrödinger equation (assumed prior result); the bra equation is its adjoint, using \(\hat{H}^{\dagger}=\hat{H}\). The classical weights \(p_i\) are fixed labels and do not evolve. A
8
\[ i\hbar\,\partial_t\hat{\rho}=\sum_i p_i\Bigl[\bigl(i\hbar\,\partial_t|\psi_i\rangle\bigr)\langle\psi_i|+|\psi_i\rangle\bigl(i\hbar\,\partial_t\langle\psi_i|\bigr)\Bigr]. \]
Differentiate \(\hat{\rho}=\sum_i p_i|\psi_i\rangle\langle\psi_i|\) with the product rule and multiply through by \(i\hbar\). B
9
\[ i\hbar\,\partial_t\hat{\rho}=\sum_i p_i\Bigl[\hat{H}|\psi_i\rangle\langle\psi_i|-|\psi_i\rangle\langle\psi_i|\hat{H}\Bigr]=\hat{H}\hat{\rho}-\hat{\rho}\hat{H}=[\hat{H},\hat{\rho}]. \]
Substitute step 7 (note the bra term carries \(-\hat{H}\)), then pull \(\hat{H}\) outside the \(i\)-sum on each side to reconstruct \(\hat{\rho}\). This is the von Neumann equation. B
10
\[ \hat{\rho}(t)=\hat{U}(t)\,\hat{\rho}(0)\,\hat{U}^{\dagger}(t),\qquad \hat{U}(t)=e^{-i\hat{H}t/\hbar}. \]
Formal solution for time-independent \(\hat{H}\): substitute and check \(i\hbar\,\partial_t\hat{U}\hat{\rho}\hat{U}^{\dagger}=\hat{H}\hat{U}\hat{\rho}\hat{U}^{\dagger}-\hat{U}\hat{\rho}\hat{U}^{\dagger}\hat{H}=[\hat{H},\hat{\rho}]\). Unitary conjugation preserves \(\hat{\rho}^{\dagger}=\hat{\rho}\), \(\operatorname{Tr}\hat{\rho}=1\), and the spectrum, so the eigenvalues \(p_i\) are constants of motion. C
11
\[ \langle\hat{O}_A\otimes\hat{\mathbf 1}_B\rangle=\operatorname{Tr}_{AB}\!\Bigl[(\hat{O}_A\otimes\hat{\mathbf 1}_B)\,\hat{\rho}\Bigr]=\sum_{a a' b}\langle a b|(\hat{O}_A\otimes\hat{\mathbf 1}_B)|a' b\rangle\,\langle a' b|\hat{\rho}|a b\rangle. \]
Reduced state, setup. Demand an operator \(\hat{\rho}_A\) on \(\mathcal{H}_A\) alone that reproduces every local expectation value. Write the full trace in a product basis \(\{|a\rangle\otimes|b\rangle\}\) with \(\hat{O}_A\) acting only on the \(A\) index. C
12
\[ =\sum_{a a'}\langle a|\hat{O}_A|a'\rangle\underbrace{\sum_b\langle a' b|\hat{\rho}|a b\rangle}_{\displaystyle \langle a'|\hat{\rho}_A|a\rangle}=\operatorname{Tr}_A\!\left(\hat{O}_A\,\hat{\rho}_A\right),\qquad \hat{\rho}_A\equiv\operatorname{Tr}_B\hat{\rho}. \]
The \(\hat{\mathbf 1}_B\) forces the \(B\) indices equal, so the sum over \(b\) is a partial trace. Matching to \(\operatorname{Tr}_A(\hat{O}_A\hat{\rho}_A)\) for all \(\hat{O}_A\) uniquely defines \(\hat{\rho}_A=\operatorname{Tr}_B\hat{\rho}\); it inherits Hermiticity, positivity, and unit trace. C
Result
\[ \hat{\rho}=\sum_i p_i|\psi_i\rangle\langle\psi_i|,\qquad \langle\hat{A}\rangle=\operatorname{Tr}(\hat{\rho}\hat{A}),\qquad i\hbar\,\partial_t\hat{\rho}=[\hat{H},\hat{\rho}],\qquad \hat{\rho}_A=\operatorname{Tr}_B\hat{\rho}. \]

Reading. The density operator is the state of any quantum system for which we lack a wavefunction. Its three defining properties — Hermitian, positive, unit trace — encode "these are genuine probabilities." Expectation values are traces against it; it evolves by the commutator with \(\hat{H}\) (the operator analogue of the classical Liouville equation, with an extra minus sign relative to Heisenberg observables); and the accessible state of a part of a larger system is obtained by tracing out the rest. Purity is measured by \(\operatorname{Tr}\hat{\rho}^2\le 1\), with equality iff the state is pure.

Units check. \(|\psi_i\rangle\langle\psi_i|\) and the \(p_i\) are dimensionless, so \(\hat{\rho}\) and \(\operatorname{Tr}(\hat{\rho}\hat{A})\) carry the dimensions of \(\hat{A}\) alone — correct. In the von Neumann equation \([\hat{H},\hat{\rho}]\) has units of energy and \(i\hbar\,\partial_t\hat{\rho}\) has units of (J·s)(1/s) = J, so both sides are energies. \(\hat{\rho}_A\) remains dimensionless with \(\operatorname{Tr}_A\hat{\rho}_A=1\).

Limiting cases
  • Pure state: one \(p_i=1\), so \(\hat{\rho}=|\psi\rangle\langle\psi|\) is a rank-one projector with \(\hat{\rho}^2=\hat{\rho}\) and \(\operatorname{Tr}\hat{\rho}^2=1\); the von Neumann equation reduces to the Schrödinger equation.
  • Maximally mixed: \(\hat{\rho}=\hat{\mathbf 1}/d\) in \(d\) dimensions gives \([\hat{H},\hat{\rho}]=0\) — a stationary state of total ignorance, \(\operatorname{Tr}\hat{\rho}^2=1/d\).
  • Product state: \(\hat{\rho}=\hat{\rho}_A\otimes\hat{\rho}_B\) (no correlations) yields \(\operatorname{Tr}_B\hat{\rho}=\hat{\rho}_A\) exactly, with no loss of purity on tracing.
  • Thermal equilibrium: \(\hat{\rho}=e^{-\beta\hat{H}}/Z\) commutes with \(\hat{H}\), hence \(\partial_t\hat{\rho}=0\); the diagonal weights are Boltzmann factors.
  • Energy eigenbasis: if \([\hat{\rho},\hat{H}]=0\) the populations \(\rho_{nn}\) are constant while coherences oscillate as \(\rho_{mn}(t)=\rho_{mn}(0)e^{-i(E_m-E_n)t/\hbar}\).
Breaks when
  • Open systems / decoherence. When the system exchanges information with an unmonitored environment, its reduced dynamics are non-unitary: \(i\hbar\,\partial_t\hat{\rho}=[\hat{H},\hat{\rho}]\) is replaced by a Lindblad master equation with dissipator terms, and \(\operatorname{Tr}\hat{\rho}^2\) decreases. The von Neumann equation holds only for the closed total system.
  • No tensor factorisation. For identical particles the physical Hilbert space is (anti)symmetrised, not a clean \(\mathcal{H}_A\otimes\mathcal{H}_B\); "trace out particle 2" is ambiguous and \(\hat{\rho}_A\) is not uniquely defined without specifying a mode/orbital partition.
  • Non-Hermitian or time-dependent effective Hamiltonians. Effective \(\hat{H}\) with complex eigenvalues (e.g. modelling decay) break trace preservation; the naive commutator equation then leaks probability and must be handled with care.
  • Explicit measurement/collapse. A projective measurement updates \(\hat{\rho}\mapsto \sum_k \hat{P}_k\hat{\rho}\hat{P}_k\) (or a conditioned single \(\hat{P}_k\hat{\rho}\hat{P}_k/p_k\)); this discontinuous, non-unitary jump is not generated by any \([\hat{H},\hat{\rho}]\).
Failure modes
  • Confusing a mixture with a superposition. \(\tfrac12(|0\rangle\langle0|+|1\rangle\langle1|)\) (diagonal, no coherences) is not the state \(|{+}\rangle\langle{+}|\) with \(|{+}\rangle=\tfrac{1}{\sqrt2}(|0\rangle+|1\rangle)\) (off-diagonal terms present). They give different statistics for \(\hat{\sigma}_x\).
  • Sign error in the equation of motion. Writing \(i\hbar\,\partial_t\hat{\rho}=[\hat{\rho},\hat{H}]\) flips the sign; the density matrix is not an observable, so it does not obey the Heisenberg equation \(i\hbar\,\dot{\hat A}=[\hat A,\hat H]\).
  • Tracing the wrong factor. Computing \(\operatorname{Tr}_A\) when the subsystem of interest is \(A\) — you must trace out the complement \(B\).
  • Assuming \(\hat{\rho}_A\) is pure because \(\hat{\rho}_{AB}\) is. For an entangled pure total state the reduced state is generically mixed; \(\operatorname{Tr}\hat{\rho}_A^2<1\) is the signature of entanglement.
  • Treating off-diagonal elements as probabilities. Only the diagonal \(\rho_{nn}\ge0\) are populations; the coherences \(\rho_{mn}\) are complex and constrained by positivity \(|\rho_{mn}|^2\le\rho_{mm}\rho_{nn}\).
  • Forgetting basis-dependence of "diagonal." Whether a state looks mixed-diagonal depends on basis; purity \(\operatorname{Tr}\hat{\rho}^2\) is the basis-independent invariant.
Discussion

The density matrix resolves a genuine conceptual gap: standard quantum mechanics assigns a ket to a system, but a subsystem of an entangled pair has no ket, and neither does a system whose preparation involved a classical coin flip. Both situations produce the same kind of object — a positive, unit-trace operator — yet they are physically different. "Proper" mixtures arise from classical ignorance about which \(|\psi_i\rangle\) was prepared; "improper" mixtures arise from entanglement with an inaccessible partner. Crucially, no local experiment can tell them apart, which is exactly why \(\hat{\rho}\) — not the ensemble list — is the physical state.

The von Neumann equation sits in a precise analogy with classical statistical mechanics. The classical phase-space density obeys the Liouville equation \(\partial_t f=\{H,f\}\) with the Poisson bracket; quantisation \(\{\,,\,\}\to \frac{1}{i\hbar}[\,,\,]\) turns this into \(\partial_t\hat{\rho}=\frac{1}{i\hbar}[\hat{H},\hat{\rho}]\). Both say the same thing: a distribution flows rigidly along the trajectories generated by \(H\), so its "shape" (entropy, eigenvalue spectrum) is conserved under closed evolution. Von Neumann entropy \(S=-\operatorname{Tr}(\hat{\rho}\ln\hat{\rho})\) is therefore constant in time for a closed system — irreversibility must come from coarse-graining or from tracing out an environment, never from the unitary dynamics itself.

The partial trace is where the arrow of decoherence enters. Take an entangled pure state \(|\Psi\rangle=\sum_k c_k|a_k\rangle|b_k\rangle\) (its Schmidt form); then \(\hat{\rho}_A=\sum_k|c_k|^2|a_k\rangle\langle a_k|\) is diagonal and mixed. The information that made the total state pure lives entirely in the \(A\)–\(B\) correlations, which the local observer cannot see. Decoherence is precisely this process running dynamically: as a system entangles with ever more environmental degrees of freedom, its reduced state loses coherences in the pointer basis and comes to look classical — without any collapse postulate, purely from unitary evolution of the whole plus a partial trace.

At the deepest level the density matrix is what makes quantum mechanics a probabilistic theory with a well-defined state space. Gleason's theorem shows that on a Hilbert space of dimension \(\ge3\), any consistent probability assignment to measurement outcomes is given by \(p=\operatorname{Tr}(\hat{\rho}\hat{P})\) for some density operator — the Born rule and the density matrix are forced, not postulated. The set of density operators is convex, with pure states as its extreme points; mixed states are the interior. This convex geometry is the arena of quantum information: entanglement measures, channel capacities, and the distinguishability of states are all statements about distances and boundaries in the space of \(\hat{\rho}\). Common misconceptions: that \(\hat{\rho}\) is "just a bookkeeping device for classical uncertainty" — it also captures irreducibly quantum coherence; that a diagonal \(\hat{\rho}\) is always classical — diagonality is basis-dependent; and that the reduced state's mixedness reflects our ignorance — for an entangled pure state it is objective and observer-independent.

Worked examples
1
Ensemble vs. superposition for a spin-\(\tfrac12\): compare (i) the equal mixture \(\hat{\rho}_{\text{mix}}=\tfrac12|{\uparrow}\rangle\langle{\uparrow}|+\tfrac12|{\downarrow}\rangle\langle{\downarrow}|\) with (ii) the pure state \(|{\rightarrow}\rangle=\tfrac{1}{\sqrt2}(|{\uparrow}\rangle+|{\downarrow}\rangle)\), by predicting \(\langle\hat{S}_x\rangle\) with \(\hat{S}_x=\tfrac{\hbar}{2}\hat{\sigma}_x\).
Setup: write both as \(2\times2\) matrices in the \(\{|{\uparrow}\rangle,|{\downarrow}\rangle\}\) basis. Symbols first, numbers after. A
2
\[ \hat{\rho}_{\text{mix}}=\begin{pmatrix}\tfrac12&0\\[2pt]0&\tfrac12\end{pmatrix},\qquad \hat{\rho}_{\text{pure}}=|{\rightarrow}\rangle\langle{\rightarrow}|=\begin{pmatrix}\tfrac12&\tfrac12\\[2pt]\tfrac12&\tfrac12\end{pmatrix},\qquad \hat{\sigma}_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}. \]
The mixture has zero coherences; the pure state has equal off-diagonal coherences \(\tfrac12\) — this is the entire physical difference. A
3
\[ \langle\hat{S}_x\rangle_{\text{mix}}=\tfrac{\hbar}{2}\operatorname{Tr}(\hat{\rho}_{\text{mix}}\hat{\sigma}_x)=\tfrac{\hbar}{2}\operatorname{Tr}\!\begin{pmatrix}0&\tfrac12\\ \tfrac12&0\end{pmatrix}=0, \qquad \langle\hat{S}_x\rangle_{\text{pure}}=\tfrac{\hbar}{2}\operatorname{Tr}\!\begin{pmatrix}\tfrac12&\tfrac12\\ \tfrac12&\tfrac12\end{pmatrix}=\tfrac{\hbar}{2}. \]
Trace against \(\hat{\sigma}_x\). Purity confirms the distinction: \(\operatorname{Tr}\hat{\rho}_{\text{mix}}^2=\tfrac12\) (mixed), \(\operatorname{Tr}\hat{\rho}_{\text{pure}}^2=1\) (pure). B
\[ \langle\hat{S}_x\rangle_{\text{mix}}=0,\qquad \langle\hat{S}_x\rangle_{\text{pure}}=+\tfrac{\hbar}{2}=+0.527\times10^{-34}\ \text{J·s}. \]

Reading. Both states are "half up, half down" for a \(\hat{S}_z\) measurement, but only the coherent superposition points along \(+x\). The off-diagonal elements of \(\hat{\rho}\) are physically real, measurable quantities — not bookkeeping.

1
Reduced state of a Bell pair: for \(|\Phi^{+}\rangle=\tfrac{1}{\sqrt2}\bigl(|{\uparrow}\rangle_A|{\uparrow}\rangle_B+|{\downarrow}\rangle_A|{\downarrow}\rangle_B\bigr)\), find \(\hat{\rho}_A=\operatorname{Tr}_B|\Phi^{+}\rangle\langle\Phi^{+}|\) and its purity.
Setup: form the full \(4\times4\) projector, then trace over \(B\) by summing the diagonal \(B\)-blocks. B
2
\[ \hat{\rho}_A=\sum_{b\in\{\uparrow,\downarrow\}}\langle b|_B\,|\Phi^{+}\rangle\langle\Phi^{+}|\,|b\rangle_B =\tfrac12|{\uparrow}\rangle_A\langle{\uparrow}|+\tfrac12|{\downarrow}\rangle_A\langle{\downarrow}|. \]
Only terms with matching \(B\) labels survive the partial trace; the cross terms \(|{\uparrow}{\uparrow}\rangle\langle{\downarrow}{\downarrow}|\) have orthogonal \(B\) states and vanish. B
3
\[ \hat{\rho}_A=\tfrac12\hat{\mathbf 1},\qquad \operatorname{Tr}\hat{\rho}_A^2=\left(\tfrac12\right)^2+\left(\tfrac12\right)^2=\tfrac12,\qquad S_A=-\operatorname{Tr}(\hat{\rho}_A\ln\hat{\rho}_A)=\ln 2 . \]
The reduced state is maximally mixed even though the global state is pure — the hallmark of maximal entanglement. Entanglement entropy \(S_A=\ln 2 = 1\) bit. C
\[ \hat{\rho}_A=\tfrac12\hat{\mathbf 1},\qquad \operatorname{Tr}\hat{\rho}_A^2=\tfrac12,\qquad S_A=\ln2\approx0.693. \]

Reading. A local observer of one Bell qubit sees perfect randomness with no preferred axis. All correlations that made \(|\Phi^{+}\rangle\) pure are stored non-locally in the \(A\)–\(B\) link, invisible to either party alone. This is the operational content of "entanglement."

Problems
  1. Show that for any density operator \(\operatorname{Tr}\hat{\rho}^2\le 1\), with equality if and only if \(\hat{\rho}\) is pure.
    Solution Diagonalise \(\hat{\rho}=\sum_k \lambda_k|k\rangle\langle k|\) with \(\lambda_k\ge0\), \(\sum_k\lambda_k=1\). Then \(\operatorname{Tr}\hat{\rho}^2=\sum_k\lambda_k^2\). Since \(0\le\lambda_k\le1\), \(\lambda_k^2\le\lambda_k\), so \(\sum_k\lambda_k^2\le\sum_k\lambda_k=1\). Equality requires \(\lambda_k^2=\lambda_k\) for every \(k\), i.e. each \(\lambda_k\in\{0,1\}\); combined with \(\sum_k\lambda_k=1\) this forces exactly one eigenvalue equal to 1, so \(\hat{\rho}=|k\rangle\langle k|\) is a rank-one projector — a pure state. \(\blacksquare\)
  2. A qubit is in the state \(\hat{\rho}=\tfrac12(\hat{\mathbf 1}+\vec r\cdot\vec{\hat\sigma})\) with Bloch vector \(\vec r=(0.6,\,0,\,0.0)\). Compute \(\langle\hat{\sigma}_x\rangle\), the purity, and state whether \(\hat{\rho}\) is a valid state.
    Solution For the Bloch form \(\langle\hat{\sigma}_i\rangle=\operatorname{Tr}(\hat{\rho}\hat{\sigma}_i)=r_i\), so \(\langle\hat{\sigma}_x\rangle=0.6\). Purity: \(\operatorname{Tr}\hat{\rho}^2=\tfrac12(1+|\vec r|^2)=\tfrac12(1+0.36)=0.68\). Validity requires \(|\vec r|\le1\): here \(|\vec r|=0.6\le1\), so \(\hat{\rho}\) is a valid mixed state (eigenvalues \(\tfrac12(1\pm0.6)=0.8,\ 0.2\ge0\)). It sits inside the Bloch ball, hence mixed since \(0.68<1\).
  3. A two-level atom has \(\hat{H}=\tfrac{\hbar\omega_0}{2}\hat{\sigma}_z\) and initial state \(\hat{\rho}(0)=\tfrac12(\hat{\mathbf 1}+\hat{\sigma}_x)\). Solve the von Neumann equation for \(\hat{\rho}(t)\) and give \(\langle\hat{\sigma}_x(t)\rangle\).
    Solution \(\hat{\rho}(t)=\hat U\hat{\rho}(0)\hat U^\dagger\) with \(\hat U=e^{-i\hat H t/\hbar}=e^{-i\omega_0 t\hat\sigma_z/2}=\operatorname{diag}(e^{-i\omega_0 t/2},e^{+i\omega_0 t/2})\). Now \(\hat{\rho}(0)=\tfrac12\begin{pmatrix}1&1\\1&1\end{pmatrix}\). Conjugating multiplies the off-diagonal \((1,2)\) element by \(e^{-i\omega_0 t}\) and \((2,1)\) by \(e^{+i\omega_0 t}\): \(\hat{\rho}(t)=\tfrac12\begin{pmatrix}1&e^{-i\omega_0 t}\\ e^{+i\omega_0 t}&1\end{pmatrix}=\tfrac12(\hat{\mathbf 1}+\cos\omega_0 t\,\hat\sigma_x+\sin\omega_0 t\,\hat\sigma_y)\). Hence \(\langle\hat{\sigma}_x(t)\rangle=\operatorname{Tr}(\hat{\rho}\hat\sigma_x)=\cos\omega_0 t\): the coherence precesses (Larmor precession in the equatorial plane) while populations stay fixed.
  4. For the pure product state \(|\Psi\rangle=|{\uparrow}\rangle_A\otimes\tfrac{1}{\sqrt2}(|{\uparrow}\rangle_B+|{\downarrow}\rangle_B)\), compute \(\hat{\rho}_A\) and \(\hat{\rho}_B\). Are they pure? Contrast with the Bell-pair worked example.
    Solution Because the state factorises, \(\hat{\rho}=\hat{\rho}_A\otimes\hat{\rho}_B\) with \(\hat{\rho}_A=|{\uparrow}\rangle_A\langle{\uparrow}|\) and \(\hat{\rho}_B=|{+}\rangle_B\langle{+}|\), \(|{+}\rangle=\tfrac1{\sqrt2}(|{\uparrow}\rangle+|{\downarrow}\rangle)\). Partial trace: \(\operatorname{Tr}_B\hat{\rho}=(\operatorname{Tr}\hat{\rho}_B)\hat{\rho}_A=\hat{\rho}_A\) since \(\operatorname{Tr}\hat{\rho}_B=1\); similarly \(\hat{\rho}_B\). Both are rank-one projectors: \(\operatorname{Tr}\hat{\rho}_A^2=\operatorname{Tr}\hat{\rho}_B^2=1\), pure. Contrast: a product pure state gives pure reduced states (entropy 0), whereas the Bell pair gave maximally mixed reduced states (entropy \(\ln2\)). Reduced-state mixedness is the diagnostic separating product from entangled.
  5. A harmonic oscillator (frequency \(\omega\), \(\hbar\omega=0.10\ \text{eV}\)) is in thermal equilibrium at \(T=300\ \text{K}\). Write \(\hat{\rho}\) in the energy eigenbasis, and compute the ground-state population \(\rho_{00}\). (Use \(k_B T\approx 0.0259\ \text{eV}\).)
    Solution Thermal state \(\hat{\rho}=e^{-\beta\hat H}/Z\), \(\beta=1/k_BT\), diagonal in the number basis: \(\rho_{nn}=e^{-\beta E_n}/Z\) with \(E_n=\hbar\omega(n+\tfrac12)\). The half-quantum shifts cancel in the ratio, so \(\rho_{nn}=(1-e^{-\beta\hbar\omega})e^{-n\beta\hbar\omega}\) (geometric series \(Z_{\text{rel}}=1/(1-e^{-\beta\hbar\omega})\)). Numbers: \(\beta\hbar\omega=\hbar\omega/k_BT=0.10/0.0259=3.86\); \(e^{-3.86}=0.0210\). Ground-state population \(\rho_{00}=1-e^{-\beta\hbar\omega}=1-0.0210=0.979\). So about \(97.9\%\) of the population sits in the ground state — the mode is nearly frozen, as expected when \(\hbar\omega\gg k_BT\).