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Additive entropy definition does not imply monotonicity

Example

Let $s \in R$ be a fixed additive term, let $H_{i n} = C$ with its standard inner product and basis $(1)$ , and let $H_{o u t} = C^{2}$ with its standard inner product and orthonormal basis $(f_{0}, f_{1})$ . In the ordered product basis $(1 \otimes f_{0}, 1 \otimes f_{1})$ , set the joint density matrices $ρ_{0} = (\begin{array}{cc} 1 / 2 & 0 \\ 0 & 1 / 2 \end{array})$ and $ρ_{1} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ . On the ordered two-point set ${0, 1}$ define $G (t) : = s + S_{o u t} (ρ_{t}; H_{i n}, H_{o u t})$ . Then

$G (1) = G (0) - k_{B} \log 2 < G (0) .$

This is a finite matrix comparison for an additive definition; the labels $0, 1$ do not describe a physical evolution.

Example references

Finite-regulator outside density and entropy

Density operator

Ordered tensor registers, their inner product, and wire permutations

The natural logarithm as the inverse of the exponential function

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$

The exponential function is strictly increasing

Watrous, The Theory of Quantum Information, Chapter 5, section 5.2.1

Preskill, Quantum Shannon Theory, Chapter 10, section 10.2

Facts & Assumptions

Given: The one-dimensional input space $C$ , the two-dimensional output space $C^{2}$ , a fixed real $s$ , and $k_{B} > 0$ .

[F1]

Density operator: on a nonzero finite-dimensional complex inner-product space, a density operator is non-negative and has trace one.

[F2]

Finite-regulator outside density and entropy: the outside density is the input-factor partial trace, and outside entropy is $- k_{B}$ times the finite natural-log spectral sum, with $0 \log 0 = 0$ .

[F3]

Ordered tensor registers, their inner product, and wire permutations: the ordered tensor product of orthonormal factor bases is an orthonormal basis in the displayed factor order.

[F4]

The natural logarithm as the inverse of the exponential function: $\log$ is the inverse of $\exp$ on the positive reals.

[F5]

The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ : for every real $x$ , $\exp (x) > 0$ and $\exp (- x) = 1 / \exp (x)$ .

[F6]

The real exponential function and the number $e$ by a power series: the real exponential is defined by its power series; at input $0$ , every positive-degree term vanishes and the zero-degree term is $1$ , so $\exp (0) = 1$ .

[F7]

The exponential function is strictly increasing: $\exp$ is strictly increasing and therefore injective.

Verification

1.1givenF1F3algebra

Both displayed matrices are self-adjoint, non-negative, and trace one in the ordered product basis from [F3], so they are density operators by [F1].

2.1givenF2F3step 1.1algebra

Since $H_{i n}$ is one-dimensional, the partial trace in [F2] has one term and returns $ρ_{t}$ . The eigenvalues of $ρ_{0}$ are $(1 / 2, 1 / 2)$ and those of $ρ_{1}$ are $(1, 0)$ .

3.1givenF2F4F5F6F7step 2.1algebra

By [F4]–[F7], $\log 1 = 0$ , $\log (1 / 2) = - \log 2$ , and $\log 2 > 0$ : the first follows from [F6], the second from [F5] and injectivity in [F7], and the third from $1 = \exp (0) < 2 = \exp (\log 2)$ by [F6] and [F7]. Therefore [F2] gives $S_{o u t} (ρ_{0}) = k_{B} \log 2$ and $S_{o u t} (ρ_{1}) = 0$ .

4.1givenstep 3.1algebra∎

The fixed term $s$ cancels, so $G (1) - G (0) = - k_{B} \log 2 < 0$ by $k_{B} > 0$ and step 3.1. Thus the additive definition alone does not force a nondecreasing value on these ordered data.

Additive entropy definition does not imply monotonicity

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources