Is it true that $|r_i-s_i| \le 1/2$ for all $i$, where $r_i$ and $s_i$ are the eigenvalues of density matrices $\rho$ and $\sigma$?

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?

• I don't think so. They clearly stated $d$ is the dimension of the Hilbert space. May 17 at 23:40
• – glS
May 18 at 6:21
• @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 $\left|\eta(r)-\eta(s)\right| \le \eta(\left|r-s\right|)$. May 18 at 12:03

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 $$\frac{1}{\mathrm{e}}$$.

• 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 $T(\rho, \sigma)$. May 18 at 1:10
• @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 May 18 at 9:11
• 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. May 18 at 10:51
• 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'. May 18 at 12:40
• @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 May 19 at 13:26