Reduced dynamics and the Kraus representation

Units are SI with ħ explicit; nothing here is relativistic, so no metric signature enters (the course uses (+, −, −, −) where four-vectors appear). Start S and E uncorrelated, ρ_S ⊗ ρ_E, with ρ_E = Σⱼ pⱼ|eⱼ⟩⟨eⱼ| in its eigenbasis, and let one unitary U act on both. Quantum Mechanics II's partial trace, Tr_E X = Σₖ ⟨eₖ|X|eₖ⟩, gives ρ_S(t) = Σⱼₖ pⱼ ⟨eₖ|U|eⱼ⟩ ρ_S ⟨eⱼ|U†|eₖ⟩. Each ⟨eₖ|U|eⱼ⟩ is a partial matrix element, an operator on S alone, so ρ_S(t) = Σⱼₖ Kⱼₖ ρ_S Kⱼₖ† with Kⱼₖ = √pⱼ ⟨eₖ|U|eⱼ⟩, derived. Unitarity supplies the one constraint: Σⱼₖ Kⱼₖ†Kⱼₖ = Σⱼ pⱼ ⟨eⱼ|U†(Σₖ|eₖ⟩⟨eₖ|)U|eⱼ⟩ = 1_S, so the trace survives for every input. Checks: each K is dimensionless, as an amplitude must be; with no coupling, U = U_S ⊗ U_E, the sum collapses to U_S ρ_S U_S†, the closed-system limit; and a pure environment gives d_E operators, a mixed one up to d_E × rank ρ_E.

Let an atom S with levels |e⟩, |g⟩ trade its excitation with a two-level environment E through H = ħg(σ₊τ₋ + σ₋τ₊), in the interaction picture on resonance, with τ₊, τ₋ the ladder operators of E, g in s⁻¹ and θ = gt a dimensionless angle. |gg⟩ and |ee⟩ do not move; on the pair |eg⟩, |ge⟩, H is ħg times a swap, so U = cos θ − i sin θ × swap there. With E in |g⟩, K₀ = ⟨g|U|g⟩ = |g⟩⟨g| + cos θ|e⟩⟨e| and K₁ = ⟨e|U|g⟩ = −i sin θ σ₋, complete because cos²θ + sin²θ = 1. They give Pₑ → cos²θ Pₑ and ρ_eg → cos θ ρ_eg. If E starts excited with probability p, K₀ and K₁ take a factor √(1 − p) and two more, √p(cos θ|g⟩⟨g| + |e⟩⟨e|) and −i√p sin θ σ₊, join them, adding p sin²θ to Pₑ. This is not a toy: a circular Rydberg atom crossing a cavity whose mode holds no photon is this model, with the mode's |0⟩ and |1⟩ as E. Brune and co-workers (1996) measured Ω₀/2π = 47 kHz, so g = Ω₀/2 = 1.48 × 10⁵ s⁻¹ and the swap completes at 10.6 μs, about 20 times shorter than the 220 μs photon lifetime (both quoted).

Any equation iħ dρ/dt = [H_eff, ρ] with H_eff Hermitian keeps the purity: d Tr ρ²/dt = 2 Tr(ρ dρ/dt) = −(2i/ħ) Tr(ρH_effρ − ρρH_eff) = 0, by the cyclic property of the trace. The reduced state does not. For S pure at Bloch polar angle α, measured from |e⟩, and E excited with probability p, the populations and coherence above give Tr ρ_S² = ½[1 + (cos²θ cos α + (2p − 1) sin²θ)² + cos²θ sin²α], derived. An excited atom and E in |g⟩ give ½(1 + cos²2θ): 1 at θ = 0, then ½ at θ = π/4, where the pair (|eg⟩ − i|ge⟩)/√2 is maximally entangled and the atom alone maximally mixed, then 1 again at π/2, with the excitation wholly in E. For the Rydberg atom the minimum falls at 5.3 μs. Purity coming back is not itself memory, since decay into a large reservoir also ends in the pure |g⟩; composition is the test: Λ(π/4) applied twice leaves Pₑ = cos⁴(π/4) = ¼ where Λ(π/2) leaves 0, because the second application hands the atom a fresh environment while the real one still holds half the excitation. Topic 14.03 recovers a composition law only where the environment forgets fast.

Nothing forced the basis |eₖ⟩ used in the trace. Trace instead over |fₗ⟩ = Σₖ |eₖ⟩Wₖₗ with W unitary, and for a pure environment the new operators are Lₗ = Σₖ W*ₖₗ Kₖ. Then Σₗ Lₗ ρ Lₗ† = Σₖₖ' (Σₗ W*ₖₗ Wₖ'ₗ) Kₖ ρ Kₖ'† = Σₖ Kₖ ρ Kₖ†, because the rows of a unitary are orthonormal. Kraus sets related by a unitary therefore give one map, derived; the converse, that any two sets for the same map are so related once the shorter is padded with zero operators, is quoted. Random π phase kicks, ρ → ½ρ + ½σ_zρσ_z, have K₀ = 1/√2 and K₁ = σ_z/√2; the Hadamard W turns them into (1 ± σ_z)/2, which are |e⟩⟨e| and |g⟩⟨g|, an energy measurement whose result nobody reads. Kick and unread measurement are one channel. The number of operators depends on the dilation, but a d-level system never needs more than d² (quoted), so the four of the exchange model, for p strictly between 0 and 1, are already as many as any qubit channel requires.

A map is positive if it sends every density matrix to one, and completely positive if 1_A ⊗ Λ does so for every partner A it leaves alone. Physics demands the second, because S may be entangled with a distant qubit, and acting on S must still leave the pair in a state. Every Kraus map qualifies, derived in a line: (1 ⊗ K)σ(1 ⊗ K)† is positive whenever σ is, and a sum of positive operators is positive. Now take the transpose T: ρ → ρᵀ in the basis |e⟩, |g⟩. It is positive and keeps the trace, since ρᵀ has the eigenvalues of ρ; on the Bloch ball it is the reflection y → −y, which no rotation produces. Apply it to the second qubit of |Ψ⁻⟩ = (|eg⟩ − |ge⟩)/√2. Partial transposition, |ab⟩⟨cd| → |ad⟩⟨cb|, moves the coherences −½|eg⟩⟨ge| and −½|ge⟩⟨eg| to −½|ee⟩⟨gg| and −½|gg⟩⟨ee|, and the result has eigenvalues ½, ½, ½ and −½, shown. The −½ belongs to (|ee⟩ + |gg⟩)/√2: a probability of −½. Transposing ρ is complex conjugation, what an antiunitary does: a symmetry of a whole closed system, like time reversal, never a process on a part.

Stinespring's theorem runs the derivation backwards and is quoted: every completely positive, trace-preserving map on a d-level system is the reduced dynamics of some unitary acting on S and a pure environment of at most d² levels. Its first step can be checked. Given Kraus operators Kₖ, the map V|ψ⟩ = Σₖ Kₖ|ψ⟩ ⊗ |k⟩ obeys V†V = Σₖ Kₖ†Kₖ = 1, so V is an isometry; extending it to a unitary is the part we quote. Operator sums, completely positive trace-preserving maps and unitary dilations are therefore one class. The derivation needed a product state at t = 0 and fails without one. Put S and E in (|gg⟩ + |ee⟩)/√2 and apply a CNOT with E as control: the pair ends in |g⟩ ⊗ (|g⟩ + |e⟩)/√2, so S is pure. Start instead from the product of two maximally mixed qubits, with the same ρ_S = 1/2, half the identity: the CNOT leaves ρ_S = 1/2. One initial ρ_S, two final ones, so no map ρ_S(0) → ρ_S(t) exists. That is why topic 14.03 starts the atom and the field as a product.