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CHAPTER 06

# Density matrices and subsystems

Learning goals. Represent mixtures, calculate partial traces, test physical density matrices, and use entropy and fidelity with explicit conventions.

## 6.1 The density operator

For a pure state, define ρ=|ψ⟩⟨ψ|\rho=|\psi\rangle\langle\psi|. If |ψ⟩=α|0⟩+β|1⟩|\psi\rangle=\alpha|0\rangle+\beta|1\rangle, then

ρ=(|α|2 α β*β α*|β|2).\rho=\begin{pmatrix} |\alpha|^2&\alpha\beta^*\\ \beta\alpha^*&|\beta|^2 \end{pmatrix}.

For a preparation choosing states |ψ j⟩|\psi_j\rangle with classical probabilities p j p_j,

ρ=∑j p j|ψ j⟩⟨ψ j|.\rho=\sum_jp_j|\psi_j\rangle\langle\psi_j|.

This is an average of outer products, not an average ket. A valid density operator is Hermitian, positive semidefinite, and trace one. Positivity means ⟨v|ρ|v⟩≥0\langle v|\rho|v\rangle\ge0 for all vectors. The eigenvalues are probabilities in an eigenstate mixture.

The Born rule and unitary evolution become

p(m)=Tr(P m ρ),ρ↦U ρ U†.p(m)=\operatorname{Tr}(P_m\rho),\qquad \rho\longmapsto U\rho U^\dagger.

The trace identity Tr(A B)=Tr(B A)\operatorname{Tr}(AB)=\operatorname{Tr}(BA) explains why the formula agrees with ⟨ψ|P m|ψ⟩\langle\psi|P_m|\psi\rangle for a pure state.

## 6.2 Coherence is basis dependent

Compare

ρ+=|+⟩⟨+|=1 2(1 1 1 1),ρ m i x=1 2|0⟩⟨0|+1 2|1⟩⟨1|=1 2(1 0 0 1).\rho_+=|+\rangle\langle+|=\frac12\begin{pmatrix}1&1\\1&1\end{pmatrix}, \qquad \rho_\mathrm{mix}=\frac12|0\rangle\langle0|+\frac12|1\rangle\langle1| =\frac12\begin{pmatrix}1&0\\0&1\end{pmatrix}.

Both have Z probabilities 1/2,1/2 1/2,1/2. But Tr(|+⟩⟨+|ρ+)=1\operatorname{Tr}(|+\rangle\langle+|\rho_+)=1, while the same expression for the mixture is 1/2 1/2. The off-diagonal terms in the Z representation encode coherence between the Z alternatives.

The mixture has many ensemble decompositions: I/2=(|+⟩⟨+|+|−⟩⟨−|)/2 I/2=(|+\rangle\langle+|+|-\rangle\langle-|)/2 as well. No local measurement can reveal which of these preparation stories occurred if only the density operator is available. A density operator summarizes the statistics, not a unique hidden list of constituent pure states [4].

## 6.3 Bloch ball and purity

Every qubit density operator has the form

ρ=I+x X+y Y+z Z 2=1 2(1+z x−i y x+i y 1−z).\rho=\frac{I+xX+yY+zZ}{2} =\frac12\begin{pmatrix}1+z&x-iy\\x+iy&1-z\end{pmatrix}.

Its eigenvalues are (1±r)/2(1\pm r)/2, where r=x 2+y 2+z 2 r=\sqrt{x^2+y^2+z^2}. Positivity requires r≤1 r\le1. Pure states lie on the sphere; mixed states lie inside the ball. The center is I/2 I/2, which is not a pure state with zero amplitudes.

Purity is

Tr(ρ 2)=1+r 2 2.\operatorname{Tr}(\rho^2)=\frac{1+r^2}{2}.

It equals one exactly for pure states. For a general D D-dimensional system it lies between 1/D 1/D and one. Purity alone does not characterize every aspect of a state’s usefulness.

## 6.4 Partial trace: forgetting access, not measuring a hidden value

Let |Ψ⟩=∑a,b c a b|a⟩A|b⟩B|\Psi\rangle=\sum_{a,b}c_{ab}|a\rangle_A|b\rangle_B. The state available to A is

ρ A=Tr B|Ψ⟩⟨Ψ|,(ρ A)a a′=∑b c a b c a′b*.\rho_A=\operatorname{Tr}_B|\Psi\rangle\langle\Psi|, \qquad (\rho_A)_{aa'}=\sum_bc_{ab}c_{a'b}^*.

Derive this by expanding the joint outer product:

ρ A B=∑a,b,a′,b′c a b c a′b′*|a⟩⟨a′|⊗|b⟩⟨b′|.\rho_{AB}=\sum_{a,b,a',b'}c_{ab}c_{a'b'}^* |a\rangle\langle a'|\otimes|b\rangle\langle b'|.

The trace of |b⟩⟨b′||b\rangle\langle b'| is δ b b′\delta_{bb'}, which eliminates terms with different environment indices. We sum matching indices, not all entries of the matrix.

For a Bell pair,

ρ A B=1 2(|00⟩⟨00|+|00⟩⟨11|+|11⟩⟨00|+|11⟩⟨11|).\rho_{AB}=\tfrac12( |00\rangle\langle00|+|00\rangle\langle11| +|11\rangle\langle00|+|11\rangle\langle11|).

The two cross terms vanish under Tr B\operatorname{Tr}_B because Tr(|0⟩⟨1|)=0\operatorname{Tr}(|0\rangle\langle1|)=0. Hence ρ A=I/2\rho_A=I/2. The joint state is pure, but each subsystem is mixed. A local mixed state need not reflect ordinary ignorance about an independently existing local pure state.

Worked partial trace. For |χ⟩=(|00⟩+2 i|01⟩+2|10⟩)/3|\chi\rangle=(|00\rangle+2i|01\rangle+2|10\rangle)/3,

ρ A=1 9(5 2 2 4),ρ B=1 9(5−2 i 2 i 4).\rho_A=\frac19\begin{pmatrix}5&2\\2&4\end{pmatrix},\qquad \rho_B=\frac19\begin{pmatrix}5&-2i\\2i&4\end{pmatrix}.

For example (ρ A)01=c 00 c 10*+c 01 c 11*=(1/3)(2/3)+(2 i/3)(0)=2/9(\rho_A)_{01}=c_{00}c_{10}^*+c_{01}c_{11}^* =(1/3)(2/3)+(2i/3)(0)=2/9. Their purities both equal (25+4+4+16)/81=49/81(25+4+4+16)/81=49/81, as required for the reductions of a bipartite pure state.

Laboratory L07 — Entanglement and partial trace. Edit all four complex amplitudes of two qubits. Inspect both reduced matrices, their Bloch vectors, purity, entropy, and pure-state concurrence. Compare a Bell state with a product having the same Z distribution on each subsystem.

## 6.5 Schmidt coefficients and entropy

The singular-value decomposition of the coefficient matrix gives the Schmidt form

|Ψ⟩=∑j λ j|u j⟩A|v j⟩B,λ j≥0,∑j λ j=1.|\Psi\rangle=\sum_j\sqrt{\lambda_j}|u_j\rangle_A|v_j\rangle_B, \qquad \lambda_j\ge0,\quad\sum_j\lambda_j=1.

The λ j\lambda_j are the nonzero eigenvalues of both reduced density matrices. One nonzero Schmidt coefficient means a product state. More than one means entanglement for a pure joint state.

Von Neumann entropy is S(ρ)=−Tr(ρ log 2 ρ)=−∑j λ j log 2 λ j S(\rho)=-\operatorname{Tr}(\rho\log_2\rho) =-\sum_j\lambda_j\log_2\lambda_j. For a bipartite pure state, S(ρ A)=S(ρ B)S(\rho_A)=S(\rho_B) quantifies entanglement. For a mixed joint state, local entropy alone is not an entanglement measure. A classically correlated mixture of 00 and 11 also has locally maximally mixed states.

## 6.6 Distinguishability and generalized measurements

Our fidelity convention is the squared convention:

F(ρ,σ)=(Tr ρ σ ρ)2.F(\rho,\sigma)= \left(\operatorname{Tr}\sqrt{\sqrt\rho\,\sigma\sqrt\rho}\right)^2.

For two pure states it is |⟨ψ|ϕ⟩|2|\langle\psi|\phi\rangle|^2. Some papers use its square root; compare definitions before comparing numbers. Trace distance is D(ρ,σ)=1 2∥ρ−σ∥1 D(\rho,\sigma)=\tfrac12\|\rho-\sigma\|_1, where the trace norm sums singular values. It bounds the difference in probability assigned to any measurement event.

A generalized measurement, or POVM, consists of positive operators E m E_m summing to I, with p(m)=Tr(E m ρ)p(m)=\operatorname{Tr}(E_m\rho). Effects specify outcome probabilities, but not the full postmeasurement state. For that one needs measurement operators M m M_m or an instrument, with E m=M m†M m E_m=M_m^\dagger M_m and conditioned state M m ρ M m†/p(m)M_m\rho M_m^\dagger/p(m). Several instruments can realize the same POVM.

## 6.7 Exercises

6.1. Find ρ\rho for the recurring single-qubit example.

Show solution / guidance

ρ=1 13(4−6 i 6 i 9)\rho=\frac1{13}\begin{pmatrix}4&-6i\\6i&9\end{pmatrix}. Its determinant is (36−36)/169=0(36-36)/169=0, trace one, and purity one.

6.2. Is (1/2 3/4 3/4 1/2)\begin{pmatrix}1/2&3/4\\3/4&1/2\end{pmatrix} physical?

Show solution / guidance

No. It is Hermitian with trace one, but eigenvalues are 5/4,−1/4 5/4,-1/4. Positivity is a separate requirement.

6.3. Reduce 3/4|00⟩+1/4|11⟩\sqrt{3/4}|00\rangle+\sqrt{1/4}|11\rangle.

Show solution / guidance

Each reduction is diag(3/4,1/4)\operatorname{diag}(3/4,1/4), purity 5/8 5/8, and entropy h 2(1/4)≈0.8113 h_2(1/4)\approx0.8113 bits. The off-diagonal joint terms vanish under the trace.

6.4. Explain why a locally maximally mixed state does not prove joint entanglement.

Show solution / guidance

Both the Bell state and the separable mixture (|00⟩⟨00|+|11⟩⟨11|)/2(|00\rangle\langle00|+|11\rangle\langle11|)/2 have reduction I/2 I/2. Joint coherences or correlations in additional bases are needed.

6.5. Calculate fidelity of |0⟩|0\rangle with |+⟩|+\rangle, and of |0⟩|0\rangle with I/2 I/2.

Show solution / guidance

Both equal 1/2 1/2 in the squared convention. Fidelity with a pure target is ⟨0|ρ|0⟩\langle0|\rho|0\rangle.

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