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## Question

Suppose I have a $D$-dimensional density matrix $\rho_0$

$\rho_0^\dagger = \rho_0 \quad, \quad \mathrm{Tr} \rho_0 = 1 \quad, \quad \rho_0 > 0,$

with a known spectrum $\{\lambda_i^0\}$ and von Neumann entropy

$H_0 = - \sum_{i=1}^D \lambda_i^0 \ln \lambda_i^0$.

Now we look at the perturbed density matrix $\rho = \rho_0 + \sigma$, where $\sigma$ need not be positive. Suppose we have a bound on the size of the Hilbert-Schmidt norm of the perturbation

$|| \sigma ||_{\mathrm{HS}} = || \rho - \rho_0 ||_{\mathrm{HS}} \le \epsilon$

where

$|| \sigma ||_{\mathrm{HS}}^2 = \sum_k \sum_{k'} | \sigma_{k,k'} |^2 = \sum_k \sum_{k'} | \langle e_k , \sigma \; e_{k'} \rangle |^2$

for any basis $e_k$.

What bound can we put on the perturbation in entropy

$\Delta H = |H - H_0|$

in terms of both $\epsilon$ and the spectrum $\{\lambda_i^0\}$?

## Prior Art

To demonstrate the continuity of the entropy, Fannes established an upper bound on the entropy perturbation in terms of the trace norm

$T = \frac{1}{2}|| \sigma ||_1 = \frac{1}{2} \sum_k \sum_{k'} | \sigma_{k,k'} | = \frac{1}{2}\sum_k \sum_{k'} | \langle e_k , \sigma \; e_{k'} \rangle |$.

Importantly, it was for two arbitrary density matricies, in the sense that the bound did not depend on a known spectrum of $\rho_0$ (just on $T$ and $D$). This was subsequently improved to the optimal inequality by Audenaert:

$|H - H_0| < T \; \log (D-1) + H_2\; [T,1-T],$

where

$H_2\; [T,1-T] = -T \; \log T - (1-T) \log (1-T)$

is the binary entropy. (See [Wikipedia][1].)

However, both Fannes and Audenaert's proofs involve breaking the perturbation into positive and negative parts

$\sigma = \sigma_+ - \sigma_- ,$

where $\sigma_+, \sigma_- > 0$. (Actually, Audenaert first reduces the problem to classical probability distributions, and then breaks the probability perturbations into positive and negative parts, which is the same thing.) As far as I can tell, this is only useful when working with a 1-norm, not a 2-norm, so the two proofs don't offer me much guidance. In addition, neither takes advantage of the fact that we're working from a known matrix $\rho_0$; they only depend on the trace distance $T$ and the dimension $D$.

Now, one can just naively use with worst-case bound between the 1-norm the 2-norm

$T = \frac{1}{2}|| \sigma ||_1 \le \frac{1}{2} \sqrt{D} || \sigma || _{\mathrm{HS}}$