If $\rho$ and $\sigma$ are density matrices such that the trace distance between them satisfies $T(\rho,\sigma)\leq1/e$. Then the Fannes' inequality states that $$|S(\rho)-S(\sigma)|\leq T(\rho,\sigma)\log d+\eta(T(\rho,\sigma))$$

Proof of Fannes' inequality is given in Box 11.2, Page 512, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, as

Let $r_1\geq r_2\geq\cdots\geq r_d$ be the eigenvalues of $\rho$, and Let $s_1\geq s_2\geq\cdots\geq s_d$ be the eigenvalues of $\sigma$, in descending order, where $d$ is the dimension of the Hilbert space.

Let $H=\rho-\sigma$, which is hermitian since both $\rho$ and $\sigma$ are positive semidefinite.

The spectral decomposition of $H$ can be written as, $H=\rho-\sigma=VD_HV^\dagger$.

Let $D_Q$ be the diagonal matrix that has all the positive eigenvalues of $H$, and $D_R$ contains the negation of all the negative eigenvalues of $H$, such that we can define $VD_QV^\dagger=Q$ and $VD_RV^\dagger=R$. Therefore, $$ D_H=D_Q-D_R\\ H=\rho-\sigma=VD_HV^\dagger=V(D_Q-D_R)V^\dagger=VD_QV^\dagger-VD_RV^\dagger=Q-R $$ $\implies D_Q,D_R$, and thereby $Q,R$ have orthogonal supports, since $Q$ only supports the positive eigenspace and $R$ only supports the negative eigenspace.

$\therefore D_QD_R=D_RD_Q=0\implies QR=RQ=0$

The trace distance between $\rho$ and $\sigma$ is defined as, $T(\rho,\sigma)=\dfrac{1}{2}tr|\rho-\sigma|=\dfrac{1}{2}tr\sqrt{(\rho-\sigma)^\dagger(\rho-\sigma)}$ \begin{align} (\rho-\sigma)^\dagger(\rho-\sigma)&=(Q-R)^\dagger(Q-R)=(Q^\dagger-R^\dagger)(Q-R)\\ &=Q^\dagger Q+R^\dagger R-Q^\dagger R-R^\dagger Q\\ &=QQ+RR-QR-RQ=Q^2+R^2\text{ , since }QR=RQ=0\\ &=QQ+RR+QR+RQ=(Q+R)^2\\ \sqrt{(\rho-\sigma)^\dagger(\rho-\sigma)}&=Q+R \end{align} The trace distance becomes $T(\rho,\sigma)=\dfrac{1}{2}tr(Q+R)=\dfrac{1}{2}\Big(tr(Q)+tr(R)\Big)$

So we have $\rho-\sigma=Q-R$ with $|\rho-\sigma|=Q+R$, where $Q$ and $R$ are positive operators with orthogonal supports.

Defining $V\equiv R+\rho=Q+\sigma$ obtains $2T(\rho,\sigma)=tr(Q)+tr(R)=tr(2V)-tr(\rho)-tr(\sigma)$

Applying the min-max theorem, as discussed in How to show $T(\rho,\sigma)≥\sum_i|r_i − s_i|$ with $r_i,s_i$ eigenvalues of $\rho,\sigma$?, we can prove $2t_k\ge2\max(r_k, s_k)=r_k+s_k+|r_k-s_k|$, which upper bounds the trace distance as


This much is clear!

The remaining part of the proof is as follows :

By calculus whenever $|r − s| ≤ 1/2$ it follows that $|η(r) − η(s)| ≤ η(|r − s|)$. A moment’s thought shows that $|r_i − s_i | ≤ 1/2$ for all $i$, so $$ |S(ρ) − S(σ)|=|\sum_{i}(η(r_i)−η(s_i))|≤η(|r_i−s_i|) $$ Setting $Δ≡\sum_i|r_i − s_i|$ and observing that $η(|r_i−s_i|)=Δη(|r_i−s_i|/Δ)− |r_i−s_i|\log(Δ)$, we see that $$ |S(ρ)−S(σ)|≤Δη(|r_i−s_i|/Δ)+η(Δ)≤Δ\log d+η(Δ) $$ where we applied Theorem 11.2 to obtain the second inequality. But $Δ≤T(ρ, σ)$ by $(11.46)$, so by the monotonicity of $η(·)$ on the interval $[0, 1/e]$, $$ |S(ρ) − S(σ)|≤T(ρ,σ)\log d + η(T (ρ, σ)) $$ whenever $T (ρ, σ) ≤ 1/e$, which is Fannes’ inequality


How do I prove that, whenever $|r − s| ≤ 1/2$, it follows that $|η(r) − η(s)| ≤ η(|r − s|)$?

The graph of $\eta(x)=-x\log x$ function is


In the domain $0\leq x\leq y\leq 1$ we have

\begin{align} |\eta(x)-\eta(y)|&=|-x\log x+y\log y|\\ &=|-x\log x+\frac{x}{\ln 2}+y\log y -\frac{y}{\ln 2}+\frac{y}{\ln 2}-\frac{x}{\ln 2}|\\ &\leq |-(x\log x-\frac{x}{\ln 2})+(y\log y-\frac{y}{\ln 2})|+\frac{|y-x|}{\ln 2}\\ &=|\int_x^y \log tdt|+\frac{|y-x|}{\ln 2}=-\int_x^y \log tdt+\frac{|y-x|}{\ln 2}\\ &=-\int_x^{x+(y-x)} \log tdt+\frac{|y-x|}{\ln 2}\\ &\leq -\int_0^{y-x} \log tdt+\frac{|y-x|}{\ln 2}\\ &=-\Big(t\log t-t\Big)_0^{y-x}+\frac{|y-x|}{\ln 2}\\ &=-(y-x)\log(y-x)+(y-x)+\frac{|y-x|}{\ln 2}\\ &=\eta(|y-x|)+|y-x|+\frac{|y-x|}{\ln 2} \end{align}

I am not able to find the intended inequality $|η(r) − η(s)| ≤ η(|r − s|)$.

It would be helpful if it is clarified so that the rest of the proof can be understood.

  • 2
    $\begingroup$ There's a lot of unnecessary information in the body of the question $\endgroup$
    – Rammus
    Commented Jan 9, 2023 at 18:18
  • $\begingroup$ @Rammus I tried to reformat the question. Is it fine now ? $\endgroup$
    – Sooraj S
    Commented Jan 9, 2023 at 21:09
  • $\begingroup$ I think everything from the link all the way down to "Question" should be removed since it's not relevant to the question. It does provide context which is important but that is already done by referring to box 11.2 in N&C. $\endgroup$ Commented Jan 9, 2023 at 21:14
  • 1
    $\begingroup$ @AdamZalcman Thanks for your suggestion. I posted a question on Math Stack Exchange regarding the proof : math.stackexchange.com/questions/4614700/… $\endgroup$
    – Sooraj S
    Commented Jan 10, 2023 at 15:01
  • 1
    $\begingroup$ I'm closing this question because it was already asked and answered on math, and there is not much to add about it here $\endgroup$
    – glS
    Commented Jan 10, 2023 at 23:10