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The Fannes–Audenaert inequality is a mathematical bound on the difference between the von Neumann entropies of two density matrices as a function of their trace distance. It was proved by Koenraad M. R. Audenaert in 2007[1] as an optimal refinement of Mark Fannes' original inequality, which was published in 1973.[2]
Statement of inequality
For any two density matrices and of dimensions ,
where
is the (Shannon) entropy of the probability distribution ,
is the (von Neumann) entropy of a matrix with eigenvalues , and
is the trace distance between the two matrices. Note that the base for the logarithm is arbitrary, so long as the same base is used on both sides of the inequality.
Audenaert also proved that—given only the trace distance T and the dimension d—this is the optimal bound. He did this by directly exhibiting a pair of matrices which saturate the bound for any values of T and d. The matrices (which are diagonal in the same basis, i.e. they commute) are
Fannes' inequality and Audenaert's refinement
The original inequality proved by Fannes was
when . He also proved the weaker inequality
which can be used for larger T.
Fannes proved this inequality as a means to prove the continuityTemplate:Disambiguation needed of the von Neumann entropy, which did not require an optimal bound. The proof is very compact, and can be found in the textbook by Nielsen and Chuang.[3] Audenaert's proof of the optimal inequality, on the other hand, is significantly more complicated.
References
- ↑ Koenraad M. R. Audenaert, "A sharp continuity estimate for the von Neumann entropy", J. Phys. A: Math. Theor. 40 8127 (2007). Preprint: arXiv:quant-ph/0610146.
- ↑ M. Fannes, "A continuity property of the entropy density for spin lattice systems ", Communications in Mathematical Physics 31 291–294 (1973).
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