# Bounding the von Neumann entropy of a density matrix with the Hilbert-Schmidt norm

## 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].)

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

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

It turns out that this is sufficient for my purposes when $$\rho_0$$ is the maximally mixed matrix $$I_D / D$$, but I need a tighter bound for other $$\rho_0$$. In other words, I need a bound which depends on the spectrum of $$\rho_0$$ (growing tighter with less mixed $$\rho_0$$).

## Probably Unnecessary Details

If it matters, the density matrix $$\rho_0$$ that I am working with can be expressed as

$$\rho_0 = \eta^{\otimes N}$$

where $$\eta$$ is two-dimensional and has eigenvalues $$\{a, 1-a\}$$. This means that $$D= 2^N$$ and $$\rho_0$$ has a spectrum of the form

$$\mathrm{spec}(\rho_0) = \{a,1-a\}^{\times N} = \{a^N, a^{N-1}(1-a), \ldots, (1-a)^N \}$$.

The bound I need must decrease with $$N$$:

$$\lim_{N\to \infty} |\Delta H| = 0$$

If it does, it will almost surely decrease exponentially in $$N$$. Currently, I am able to show that the Hilbert-Schmidt norm of my perturbation falls like

$$\|\sigma\|_{\mathrm{HS}}^2 \sim [a^2 + (1-a)^2]^{(1+\delta)N}$$

for small $$\delta > 0$$. If $$a=1/2$$, then $$\rho_0$$ is maximally mixed and

$$||\sigma||_{\mathrm{HS}}^2 \sim \frac{1}{2^{(1+\delta)N}}$$

so

$$|\Delta H| \sim T \; \log(D-1) + T - T \; \log T \sim \frac{\sqrt{D} \ln{D}}{\sqrt{2^{(1+\delta) N}}} = \frac{N \ln 2}{\sqrt{2^{\delta N}}} \to 0$$.

But if $$a < 1/2$$, the bound on $$\|\sigma\|_{\mathrm{HS}}^2$$ falls more slowly with $$N$$ and the naive application of the Fannes–Audenaert inequality gives a bound on the entropy which grows with $$N$$ (for sufficiently small $$\delta$$).

I can give you a simplification of the problem and a precise answer in the limit as $\epsilon \to 0$. This precise answer also yields a good upper bound in the upward direction $H > H_0$, using the fact that entropy is a concave function. In the downward direction $H < H_0$ things are more annoying, and a good answer depends on the size of $\epsilon$ and what type of estimate you want.
There is a convex body $B$ of density matrices $\rho$ and a map from that to the much simpler convex body $S$ of unordered spectra $\vec{\lambda}$. The simplification of your problem is that the Hilbert-Schmidt metric on density matrices descends to the Euclidean metric (or $\ell^2$ metric or Hilbert-Schmidt metric) on spectra. $S$ is a simplex, in fact the quotient of a regular simplex $T$ by its isometries. Since $H(\rho)$ only depends on its spectrum, you might as well work in $S$ or $T$ rather than in $B$. In fact $T$ is exactly the convex body of classical states on $D$ configurations rather than quantum states. So the question is not really quantum at all, it is a question about the Shannon entropy $H(\vec{\lambda})$ of distributions $\vec{\lambda}$ on a set with $D$ elements.
It is easy to take the gradient of $H(\vec{\lambda})$. The differential is just $$\sum_k -(\ln(\lambda_k) + 1)d\lambda_k.$$ Since the sum of the coordinates is constant, you can drop the second term of the summand. So to maximize the deviation of $H(\vec{\lambda})$, you should push $\vec{\lambda}$ in the direction $$\delta \lambda_k = -\ln(\lambda_k)+\frac1D \sum_k \ln(\lambda_k).$$ The norm of the gradient is then an optimal upper bound as $\epsilon \to 0$. As mentioned at the beginning, if the entropy is increasing, this derivative bound holds for any $\epsilon$, because entropy is concave. In the other direction you are "falling off a cliff" instead of "climbing to the top of the dome", and I would have to understand more about what sort of bound you want. (Maybe only because I haven't grasped all of the later details of your question.)
For the uniform state (the maximally mixed state) I can conjecture a precise answer in the down direction. Then you are in the center of the simplex $T$. I conjecture that the way to decrease entropy as much as possible is to run straight for a corner, i.e., increase one $\lambda_k$ and keep the others equal.