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 $


$ \|\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], $


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

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}} $

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}} $


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

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 $\delta$).


1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.