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Finite mode cutoffs can have unbounded entropy

Statement

Fix a real number $k_{B} > 0$ . For each integer $N \geq 1$ , let $H_{i n} = C$ with its standard inner product and basis $1$ , and let $H_{o u t, N} = (C^{2})^{\otimes N}$ with the standard product inner product and product basis. Identify the product basis of $C \otimes H_{o u t, N}$ with that of $H_{o u t, N}$ . For the density operator $ρ_{N} = I_{2^{N}} / 2^{N}$ , where $I_{2^{N}}$ is the identity matrix on this $2^{N}$ -dimensional space, the outside entropy is

$S_{o u t} (ρ_{N}; C, H_{o u t, N}) = k_{B} N \log 2,$

which tends to $+ \infty$ as $N \to \infty$ .

Statement 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 addition formula $\exp (x + y) = \exp (x) \exp (y)$

The exponential function is strictly increasing

Every complete ordered field is Archimedean

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

Preskill, Quantum Shannon Theory, Chapter 10, section 10.2

Facts & Assumptions

Given: A positive real $k_{B}$ , an integer $N \geq 1$ , the one-dimensional input space $C$ , and the $N$ -factor output space $(C^{2})^{\otimes N}$ .

[F1]

Finite-regulator outside density and entropy: the outside density is obtained by summing the diagonal blocks over an orthonormal basis of the input factor, and its entropy is $- k_{B}$ times the natural-log spectral sum.

[F2]

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

[F3]

The natural logarithm as the inverse of the exponential function: $\log$ is the inverse of the real exponential on $(0, \infty)$ .

[F4]

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

[F5]

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 $\exp (0) = 1$ .

[F6]

The exponential addition formula $\exp (x + y) = \exp (x) \exp (y)$ : $\exp (x + y) = \exp (x) \exp (y)$ for real $x, y$ .

[F7]

The exponential function is strictly increasing: the real exponential is strictly increasing, hence injective.

[F8]

In every complete ordered field $F$ , for each $x \in F$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{F}$ (Every complete ordered field is Archimedean).

[F9]

Ordered tensor registers, their inner product, and wire permutations: tensoring the standard basis in each $C^{2}$ factor gives the ordered product basis of $H_{o u t, N}$ , so its dimension is $2^{N}$ .

[F10]

The library real numbers are the Cauchy-sequence quotient $R = C / N$ (The real numbers), and this real field has the least-upper-bound property, so it is a complete ordered field (The Cauchy-sequence reals have the least-upper-bound property).

Proof

1.1givenF1F2F9algebra

The matrix $ρ_{N} = I_{2^{N}} / 2^{N}$ is non-negative and has trace $2^{N} / 2^{N} = 1$ , so it is a density operator by [F2] and [F9]. Since the input factor is one-dimensional, the defining partial trace in [F1] has one summand and gives $ρ_{o u t, N} = I_{2^{N}} / 2^{N}$ .

1.2givenF3F4F6F7algebra

The spectrum of $ρ_{o u t, N}$ consists of $2^{N}$ copies of $2^{- N}$ . For $N = 1$ , $\exp (N \log 2) = 2$ by [F3]; if $\exp (N \log 2) = 2^{N}$ , [F6] gives $\exp ((N + 1) \log 2) = \exp (N \log 2) \exp (\log 2) = 2^{N + 1}$ . Induction yields $\exp (N \log 2) = 2^{N}$ for every $N \geq 1$ . By [F4], $\exp (- N \log 2) = 1 / 2^{N} = 2^{- N}$ , so [F3] and injectivity in [F7] give $\log (2^{- N}) = - N \log 2$ .

2.1givenF1F5F7step 1.2algebra

Substitution into [F1] gives $S_{o u t} = - k_{B} 2^{N} 2^{- N} \log (2^{- N}) = k_{B} N \log 2$ . Also $\log 2 > 0$ , because [F5] gives $\exp (0) = 1 < 2 = \exp (\log 2)$ and [F7] is strictly increasing.

3.1givenF8F10step 2.1algebra∎

Let $c = k_{B} \log 2 > 0$ . For any real $M$ , [F10] makes $R$ a complete ordered field, so [F8] applied to $M / c$ supplies a natural number $N_{0} \geq 1$ with $N_{0} > M / c$ . For every integer $N \geq N_{0}$ , step 2.1 gives $S_{o u t} = c N \geq c N_{0} > M$ , which is exactly convergence to $+ \infty$ .

Finite mode cutoffs can have unbounded entropy

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources