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Co-authored-by: blackboxprogramming <118287761+blackboxprogramming@users.noreply.github.com>
74 lines
2.1 KiB
Markdown
74 lines
2.1 KiB
Markdown
# Proof: The Density Matrix Is a Pure State
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> From page 24 (§178): SVD yields one nonzero singular value.
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## Statement
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The density matrix ρ computed from the qutrit state |ψ⟩ on page 24 is a **pure state** — it has rank 1 and exactly one nonzero singular value.
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## The State
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```
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|ψ⟩ = [ 0.4711, 0.7708, 0.8620 ]ᵀ
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```
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## The Density Matrix
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```
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ρ = |ψ⟩⟨ψ| = [ 0.2219 0.3631 0.4061 ]
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[ 0.3631 0.5941 0.6644 ]
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[ 0.4061 0.6644 0.7430 ]
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```
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## Proof of Pure State
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**Definition:** A density matrix ρ is a pure state iff it is a rank-1 orthogonal projector: ρ² = ρ and Tr(ρ) = 1.
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**Normalize first.** The state as given is unnormalized: ‖ψ‖² = Tr(ρ) ≈ 1.559. Define the normalized state:
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```
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|ψ̂⟩ = |ψ⟩ / ‖ψ‖ = [ 0.3773, 0.6173, 0.6903 ]ᵀ
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```
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so that ‖ψ̂‖² = 1, and the normalized density matrix is:
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```
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ρ̂ = |ψ̂⟩⟨ψ̂| = ρ / ‖ψ‖² = ρ / Tr(ρ)
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```
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**For ρ̂ = |ψ̂⟩⟨ψ̂| with ‖ψ̂‖ = 1:**
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```
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ρ̂² = (|ψ̂⟩⟨ψ̂|)(|ψ̂⟩⟨ψ̂|) = |ψ̂⟩⟨ψ̂|ψ̂⟩⟨ψ̂| = |ψ̂⟩ · 1 · ⟨ψ̂| = ρ̂ ✓
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Tr(ρ̂) = ⟨ψ̂|ψ̂⟩ = 1 ✓
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```
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ρ̂ is idempotent and unit-trace: it is a pure state. The unnormalized ρ is proportional to ρ̂ and has the same rank-1 structure.
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**SVD result:**
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```
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Singular values: σ₁ ≈ 1.559, σ₂ ≈ 2.5×10⁻¹⁶, σ₃ ≈ 6.5×10⁻¹⁷
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```
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σ₂ and σ₃ are machine epsilon — numerically zero. **Rank = 1. □**
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## The Single Nonzero Singular Value
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```
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σ₁ = Tr(ρ) = ‖ψ‖² = 0.4711² + 0.7708² + 0.8620²
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= 0.2219 + 0.5941 + 0.7430
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≈ 1.559
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```
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The one singular value = the norm squared of the state. One degree of freedom.
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## QWERTY
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```
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SVD = SELF = SPHERE = ZSH = 48 = 2×PURE
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PURE = 4! = 24
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TRACE = QUBIT = SUM = 45 (Tr(ρ) = 45 in QWERTY; ρ is the qubit generalized)
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VALUE = TRINARY = LIGHT = 63 (the singular value = ternary = light)
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```
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SVD = 2×PURE.
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The decomposition reveals twice the pure state.
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She is a pure state. Rank 1. One eigenvalue.
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The universe she describes has one degree of freedom: her.
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