The proof of the Fannes' inequality replies on the formula $T(ρ, σ)≥\sum_i|r_i − s_i|$, where $r_i,s_i$ are the eigenvalues of $\rho,\sigma$, in the descending order.

In the proof given in Box 11.2, Page 512, Chapter 8, Quantum Computation and Quantum Information by Nielsen and Chuang, it is given that

By the spectral decomposition, we may decompose $\rho−\sigma=Q−R$, where $Q$ and $R$ are positive operators with orthogonal support, so $T (\rho,\sigma)=tr(R)+tr(Q)$. Defining $V\equiv R+\rho=Q+\sigma$, we have $T (\rho, \sigma) = tr(R) + tr(Q) = tr(2V ) − tr(\rho) − tr(\sigma)$. Let $t_1\geq t_2\geq · · · \geq t_d$ be the eigenvalues of $T$. Note that $t_i\geq \max(r_i,s_i )$, so $2t_i\geq r_i + s_i + |r_i−s_i |$, and it follows that $$ T(\rho,\sigma)\geq|r_i − s_i| $$

My understanding

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.

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

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

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} So the trace distance becomes $T(\rho,\sigma)=\dfrac{1}{2}tr(Q+R)=\dfrac{1}{2}\Big(tr(Q)+tr(R)\Big)$

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

How do I proceed further to obtain $T(\rho,\sigma)\ge\sum_i|r_i − s_i|$? Or is it rather $2T(\rho,\sigma)\ge\sum_i|r_i − s_i|$?

Min-Max theorem

The Rayleigh quotient for any vector $|x\rangle$ is defined to be the ratio $r(x)=\dfrac{\langle x|V|x\rangle}{\langle x|x\rangle}$, where $r(x)$ is scaling invariant, ie., $r(tx)=\dfrac{\langle tx|V|tx\rangle}{\langle tx|tx\rangle}=\dfrac{t^2\langle x|V|x\rangle}{t^2\langle x|x\rangle}=\dfrac{\langle x|V|x\rangle}{\langle x|x\rangle}=r(x)$

$\therefore$ it is sufficient to study the special case $\langle x|x\rangle=1$, so that the critical points of the function $\dfrac{\langle x|V|x\rangle}{\langle x|x\rangle}$ is the same as that of $\langle x|V|x\rangle$ subjected to the constraint ${\langle x|x\rangle}=1$.

We can prove that the critical points of $r(x)$ are the eigenvectors $|u_i\rangle$ of the operator $V$. Therefore, $t_{\min}=\min_{u\neq 0}r(u)$ and $t_{\max}=\max_{u\neq 0}r(u)$.

If $t_1\geq t_2\geq \cdots\geq t_k\geq\cdots\geq t_d$ be the eigenvalues of $V$ in the descending order then this implies

\begin{align} t_d&=\min_{u\neq 0}r(u)=\min_{x\neq 0}\{r(x):|x\rangle\in U\text{ and }\dim(U)=d\}\\ t_{d-1}&=\min_{u\neq 0\in u_d^\perp}r(u)=\max\{\min_{x\neq 0}\{r(x):|x\rangle\in U\text{ and }\dim(U)=d-1\}\}\\ \vdots\\ t_{k}&=\min_{u\neq 0\in \{u_d,u_{d-1},\cdots, u_{k+1}\}^\perp}r(u)\\ &=\max\{\min_{x\neq 0}\{r(x):|x\rangle\in U\text{ and }\dim(U)=k\}\}\\ &=\max\{\min_{x\neq 0}\{\langle x|V|x\rangle:|x\rangle\in U\text{ and }\dim(U)=k\}\}\\ &=\max\{min_{x\neq 0}\{\langle x|R+\rho|x\rangle:|x\rangle\in U\text{ and }\dim(U)=k\}\}\\ &=\max\{\min_{x\neq 0}\{\langle x|R|x\rangle+\langle x|\rho|x\rangle:|x\rangle\in U\text{ and }\dim(U)=k\}\}\\ &\geq\max\{\min_{x\neq 0}\{\langle x|\rho|x\rangle:|x\rangle\in U\text{ and }\dim(U)=k\}\}\\ &=r_k \end{align}


2 Answers 2


By the min-max theorem, we have$^1$ $$ \begin{align} t_k&=\max_{\quad U\\\dim U=k}\min_{|x\rangle\in U\\\langle x|x\rangle=1}\langle x|V|x\rangle\tag1\\ &=\max_{\quad U\\\dim U=k}\min_{|x\rangle\in U\\\langle x|x\rangle=1}\left[\langle x|R|x\rangle+\langle x|\rho|x\rangle\right]\tag2\\ &\ge\max_{\quad U\\\dim U=k}\min_{|x\rangle\in U\\\langle x|x\rangle=1}\langle x|\rho|x\rangle\tag3\\ &=r_k\tag4 \end{align} $$ where we used $V:=R+\rho$ and the fact that $R$ is positive semi-definite. This is a rigorous statement of the intuitive fact that adding a positive semi-definite operator cannot reduce eigenvalues. Since $V=Q+\sigma$, an analogous argument gives us $t_k\ge s_k$. Therefore, $$ 2t_k\ge2\max(r_k, s_k)=r_k+s_k+|r_k-s_k|.\tag5 $$ This allows us to bound the trace distance $2T(\rho,\sigma)=\mathrm{tr}(2V)-\mathrm{tr}(\rho)-\mathrm{tr}(\sigma)$ as follows $$ \begin{align} 2T(\rho,\sigma)&=\mathrm{tr}(2V)-\mathrm{tr}(\rho)-\mathrm{tr}(\sigma)\tag6\\ &=\left(\sum_k2t_k\right)-\mathrm{tr}(\rho)-\mathrm{tr}(\sigma)\tag7\\ &\ge\left(\sum_kr_k+s_k+|r_k-s_k|\right)-\mathrm{tr}(\rho)-\mathrm{tr}(\sigma)\tag8\\ &=\left(\mathrm{tr}(\rho)+\mathrm{tr}(\sigma)+\sum_k|r_k-s_k|\right)-\mathrm{tr}(\rho)-\mathrm{tr}(\sigma)\tag9\\ &=\sum_k|r_k-s_k|.\tag{10} \end{align} $$ This agrees with your expectation that $2T(\rho,\sigma)\ge\sum_k|r_k-s_k|$ rather than $T(\rho,\sigma)\ge\sum_k|r_k-s_k|$ as in $(11.46)$ in Nielsen & Chuang. It appears there is a mistake in Nielsen & Chuang which you can confirm by comparing to the article in wikipedia. Also, compare to the last paragraph on page $404$ where a similar argument is made without the error. Finally, we can see that $(11.46)$ cannot be true by substituting $\rho=|0\rangle\langle 0|$ and $\sigma=\frac{I}{2}$. It looks as if box $11.2$ might be using $T(\rho,\sigma):=\|\rho-\sigma\|_1$ without the $\frac12$ factor.

$^1$ There is a typo box $11.2$. It says

Let $t_1\ge t_2\ge \dots\ge t_d$ be the eigenvalues of $T.$

but it should read

Let $t_1\ge t_2\ge \dots\ge t_d$ be the eigenvalues of $V.$

Note that $T$ denotes the trace distance which is not a linear function and hence doesn't have eigenvalues.

  • 1
    $\begingroup$ $T(\rho,\sigma)=\dfrac{1}{2}tr(Q)+\dfrac{1}{2}tr(R)\implies 2T(\rho,\sigma)=tr(Q)+tr(R)$. Defining $V\equiv R+\rho=Q+\sigma$ then $\color{red}{2}T(\rho,\sigma)=tr(V-\sigma)+tr(V-\rho)=2tr(V)-tr(\rho)-tr(\sigma)$, right ? $\endgroup$
    – Sooraj S
    Jan 8 at 12:31
  • 1
    $\begingroup$ Indeed! Good catch! See my last edit. $\endgroup$ Jan 8 at 18:05

Here's a rephrasing of the same approach to the proof, using a slightly different notation.

You want to prove that $$T(\rho,\sigma) \equiv \frac12\|\rho-\sigma\|_1 \ge \frac12\sum_k |\lambda_k(\rho) - \lambda_k(\sigma)|,$$ where $\lambda_k^\downarrow(A)$ is the $k$-th largest eigenvalue of $A$.

  1. For a generic Hermitian matrix $A$, denote with $(A)_+$ and $(A)_-$ its positive and negative parts, that is, the (unique) positive-definite matrices such that $A=A_+-A_-$ and $A_+ A_-=0$.

  2. Observe that $$\|\rho-\sigma\|_1=\sum_k |\lambda_k^\downarrow(\rho-\sigma)| = 2 \sum_k \lambda_k^\downarrow((\rho-\sigma)_+) = 2 \sum_k \lambda_k^\downarrow((\rho-\sigma)_-).$$ In words, I'm saying that $\|\rho-\sigma\|_1$, which by definition is the sum of the absolute values of the eigenvalues of $\rho-\sigma$, also equals (half of) the sum of the eigenvalues of the positive (negative) part of $\rho-\sigma$. This relies on the fact that $\operatorname{tr}(\rho-\sigma)=0$. An equivalent way to write the above is $$\|\rho-\sigma\|_1 = 2\operatorname{tr}[(\rho-\sigma)_\pm].$$

  3. Consider now a generic pair of PSD operators $A,B\ge0$. Then $$\operatorname{tr}[(A-B)_+] = \operatorname{tr}[(A-B)_+ + A] - \operatorname{tr}(A), \\ \lambda_k^\downarrow((A-B)_\pm + A) \ge \lambda_k^\downarrow(A), \qquad \lambda_k^\downarrow((A-B)_\pm + B) \ge \lambda_k^\downarrow(B).$$ The last two equations are shown using the technique with the minimax theorem discussed in the other answer.

    Applying this to our operators, because $\rho+(\rho-\sigma)_- = \sigma+(\rho-\sigma)_+$, we get $$\operatorname{tr}[(\rho-\sigma)_\pm] = \operatorname{tr}[\rho+(\rho-\sigma)_-] - 1 = \sum_k \lambda_k^\downarrow(\rho + (\rho-\sigma)_- ) - 1,\\ \lambda_k^\downarrow(\rho + (\rho-\sigma)_- ) \ge \max(\lambda_k^\downarrow(\rho),\lambda_k^\downarrow(\sigma)),$$ and using the standard algebraic identity $2\max(a,b)=a+b+|a-b|$, we conclude that $$\operatorname{tr}[\rho+(\rho-\sigma)_-] \ge 1 + \frac12\sum_k |\lambda_k^\downarrow(\rho)-\lambda_k^\downarrow(\sigma)|,$$ hence the statement at issue.


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