In Nielsen and Chuang's Box 11.2: Continuity of the entropy, in the process of proving the Fannes' inequality, it says:

A moment’s thought shows that $\left|r_i − s_i\right| \le 1/2$ for all i,

The following seems to be a counter example:

Consider a 4-d Hilbert space spanned by $\{|1\rangle, |2\rangle, |3\rangle, |4\rangle\}$.

Let $\rho = |1\rangle\!\langle1|$ and $\sigma = \frac14\sum\limits_{i=1}^4(|i\rangle\!\langle i|)$.

We have $r_1=1$, $r_2=r_3=r_4=0$, and $s_1 = s_2 = s_3 = s_4 = 1/4$.

$r_1-s_1 = 1-1/4 = 3/4 > 1/2$.

It contradicts with the claims of $\left|r_i-s_i\right|<1/2$ for all $i$.

Where did I go wrong?

  • $\begingroup$ I don't think so. They clearly stated $d$ is the dimension of the Hilbert space. $\endgroup$
    – Guangliang
    May 17 at 23:40
  • $\begingroup$ see also: quantumcomputing.stackexchange.com/a/29598/55 $\endgroup$
    – glS
    May 18 at 6:21
  • $\begingroup$ @glS NC is using a unusual definition of trace distance. But it's consistent within its proof. However, no matter which definition is used, still needs $\left|\eta(r)-\eta(s)\right| \le \eta(\left|r-s\right|)$. $\endgroup$
    – Guangliang
    May 18 at 12:03

1 Answer 1


In this proof, we assume that the trace distance between $\rho$ and $\sigma$ is upper-bounded by $\frac{1}{\mathrm{e}}<\frac12$.

As mentioned by glS in the comments, the trace distance is lower-bounded by half the sum of the $\left|r_i-s_i\right|$. Suppose that for some $i$, we have $\left|r_i-s_i\right|>\frac12$. At this point, half this sum is already $\frac14$. Intuitively, since the sum of the $r_i$ and of the $s_i$ are both equal to $1$, there got to be a term where the difference between the twos will be too large for the trace distance to be less than $\frac{1}{\mathrm{e}}$. There is surely an elegant way to prove it, but I didn't find it, so here's a reasoning by induction.

To be more formal, the statement we want to prove is the following one:

Let $\left(r_i\right)_i$ and $\left(s_i\right)_i$ be two sequences of length $d$ such that:

  • $\sum_ir_i=\sum_js_i=1$
  • There is an index $j$ such that $\left|r_j-s_j\right|>\frac12$

Then $\frac12\sum_i\left|r_i-s_i\right|>\frac12$.

First, let us consider the qubit case, that is $d=2$. Let us suppose without loss of generality that $r_0=\frac12+s_0+\varepsilon$, with $0<\varepsilon\leqslant\frac12$. Then we necessarily have $r_1=\frac12-\varepsilon-s_0$, since $r_0+r_1=1$. We now replace $s_0$ by $1-s_1$ to find that $r_1=-\frac12-\varepsilon+s_1$, which finally gives us a lower-bound for the trace distance of $\frac12+\varepsilon$, which is strictly larger than $\frac12$.

Let us now assume that the proposition is true for some $d$ and let us show it for $d+1$. Since $d+1\geqslant3$, we know that there are at least two couples $\left(r_x,s_x\right)$ and $\left(r_y,s_y\right)$ such that $r_x\geqslant s_x$ and $r_y\geqslant s_y$ or the other way around. In this case, we can regroup these two couples into a single one $\left(r_x+r_y,s_x+s_y\right)$ without changing the computation of the lower-bound. These new sequences are of length $d$ and sum to $1$, and we still have the existence of an index on which the absolute value of the difference of the terms is larger than $\frac12$, we can thus use our assumption and conclude that the trace distance is necessarily lower-bounded by $\frac12$.

There is probably a (way) more elegant way to show this property, but I think that's rigorous. All in all, your counterexample doesn't apply since the trace distance between the density matrices that you're considering is equal to $\frac34$, which is larger than $\frac{1}{\mathrm{e}}$.

  • $\begingroup$ Thanks for pointing out the upper bound restriction of the trace distance. That solves the problem since $\sum_i|r_i-s_i| \le T(\rho,\sigma) \le 1/e < 1/2$. However, the last sentence of Box 11.2 says "The weaker form of Fannes' inequality for general $T(\rho, \sigma)$ follows with minor modifications." I still don't see how to work it out without the upper bound of $T(\rho, \sigma)$. $\endgroup$
    – Guangliang
    May 18 at 1:10
  • $\begingroup$ @Guangliang I've edited tha answer because it seems there's a typo in NC (it lacks a factor $\frac12$, so the proof is a bit harder). I'll edit it once again a bit later to incorporate your last question $\endgroup$ May 18 at 9:11
  • $\begingroup$ NC uses a unusual definition of trace-distance. But the usage is consistent within the proof. But the claim $|r_i-s_i|\le1/2$ is used to proof $\left|\sum_i(\eta(r_i)-\eta(s_i))\right| \le \sum_i\eta(\left|r_i-s_i\right|)$, which is needed no matter which trace distance definition is used. $\endgroup$
    – Guangliang
    May 18 at 10:51
  • $\begingroup$ Just did some calculation with my counter example. $\left|\sum_i(\eta(r_i)-\eta(s_i))\right| = 2$ and $\sum_i\eta(\left|r_i-s_i\right|) \approx 1.8113$. So the inequality part of Eq. 11.47 does not hold. To prove the weaker version of Fannes' inequality seems to need more than just 'minor modifications'. $\endgroup$
    – Guangliang
    May 18 at 12:40
  • $\begingroup$ @Guangliang I didn't spend much time on it but didn't manage to prove it. I think your best strategy is to open a new question focused on proving the general case, so that people know what the problem is about $\endgroup$ May 19 at 13:26

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