Exercise 9

from Example Sheet 4

Exercise 9.

Continuity of the von Neumann entropy (Fannes’ inequality): Suppose \(\rho, \sigma \in \mathcal{D}( \mathcal{H})\) are states such that their trace distance \(D(\rho, \sigma)\) satisfies the bound \[ 2 D(\rho, \sigma) = \|\rho - \sigma\|_1 \le 1/e. \] Then \[ |S(\rho) - S(\sigma)| \le \|\rho - \sigma\|_1 \log d + \eta\left( \|\rho - \sigma\|_1\right), \](1) where \(d = \dim \mathcal{H}\), and \(\eta(x) := - x \log x.\)

Let us prove this theorem in steps:

  1. Let \(r_1 \ge r_2 \ge \ldots \ge r_d\) be the eigenvalues of \(\rho\) arranged in non-increasing order, and let \(s_1 \ge s_2 \ge \ldots \ge s_d\) be the eigenvalues of \(\sigma\) arranged in non-increasing order. Then prove that: \[ \|\rho - \sigma\|_1 \ge \sum_{j=1}^d |r_j - s_j| \](2)
  2. Check (using elementary calculus) that if \(|r-s| \le 1/2\), then \[ |\eta(r) - \eta(s)| \le \eta(|r-s|), \] where \(\eta(x) := - x \log x.\)
  3. Use Step 2 and the triangle inequality to prove that \[ |S(\rho) - S(\sigma)|\le \sum_j \eta(|r_j - s_j|). \]
  4. Let \(\varepsilon_j := |r_j - s_j|\), \(\forall\, j=1,2,\ldots, d\), and \(\varepsilon:= \sum_j \varepsilon_j\). Let \(\lambda_j := \varepsilon_j/\varepsilon\) and note that \(\{\lambda_j\}\) forms a probability distribution. Use this fact and Step 3 to prove that \[ |S(\rho) - S(\sigma)|\le \varepsilon\log d + \eta(\varepsilon). \]
  5. Note that \(\eta(\varepsilon)\) is a monotonically increasing function of \(\varepsilon\) for \(0 \le \varepsilon\le 1/e\). Use this to finally arrive at the statement (1).
  1. I could not find a short proof of this fact (without assuming something that directly implies it). This subquestion is then nonexaminable; see e.g. the excellent book Matrix Analysis by Bhatia1 for several (more involved) proofs.

  2. Without loss of generality, take \(r > s\) and set \(\delta := r-s\). Then \[\begin{aligned} \eta(s+\delta) - \eta(s) - \eta(\delta) &= - (s+\delta)\log (s+\delta) + s \log s + \delta\log(\delta) \\ &= - s\log (s+\delta) - \delta \log(s+\delta) + s \log s + \delta\log(\delta)\\ &= s\log\frac{s}{s+\delta} + \delta \log \frac{\delta}{s+\delta} \\ &\leq 0 \end{aligned}\] since both \(\frac{s}{s+\delta}\leq 1\) and \(\frac{\delta}{s+\delta} \leq 1\). Thus, \[ \eta(r) - \eta(s) \leq \eta(r-s). \] On the other hand, set \[ f(\delta) := \eta(s) - \eta(s+\delta) - \eta(\delta). \] It remains to show that \(f(\delta)\leq 0\) for \(\delta \leq \frac{1}{2}\) (and \(r=s+\delta\leq 1\)), which implies \[ \eta(s) - \eta(s+\delta) \leq \eta(\delta) \] meaning that we have proven \[ |\eta(r) - \eta(s)|\leq \eta(|r-s|). \]

    First, \(f(0) = 0\), and \[ f'(\delta) = - \eta'(s+\delta) -\eta'(\delta) \] while \[ \eta'(x) = \frac{\mathrm{d}}{\mathrm{d}x}(- x \log x) = - \log x - \frac{1}{\ln 2}, \] so \[ f'(\delta) = \log ( s + \delta) + \frac{1}{\ln 2} + \log (\delta) + \frac{1}{\ln 2} \] and \[ f''(\delta) = \frac{1}{\ln 2} \left[ \frac{1}{s+\delta} + \frac{1}{\delta}\right] > 0 \] for \(\delta > 0\). Thus, \(f\) is convex. We wish to prove \(f(\delta)\leq 0\) for \(\delta \in [0, \min(1-s, \frac{1}{2})]\), and by convexity, we only need to check the boundaries (since \(f\) will be less than either boundary point between the two). We have immediately \(f(0)=0\). Next, consider \(s \geq \frac{1}{2}\). Then let’s check \(g(s) := f(1-s) = \eta(s) - \eta(1-s) \leq 0\) for \(s\geq \frac{1}{2}\). We can see at \(s= \frac{1}{2}\), \(g(s)=0\), and \(g'(s) = -\log(s)+\log(1-s) \leq 0\) since \(1-s \leq s\). Thus, \(g\) is decreasing for all \(s\in [ \frac{1}{2},1]\), proving indeed, \(g(s)\leq 0\) on that interval.

    Lastly, we need to check when \(s < \frac{1}{2}\) that \(f(\frac{1}{2}) \leq 0\). In this case, \[ h(s) := f(\frac{1}{2} ) = \eta(s) - \eta(s+\tfrac{1}{2}) - \eta(\tfrac{1}{2}). \] Then \(h(0) = - 2 \eta(\frac{1}{2})\leq 0\), and \(h'(s) = - \log(s) + \log(s + \frac{1}{2}) > 0\), so \(h\) is increasing. Thus, \(h\) takes a maximum at the end of the interval we wish to test, at \(s= \frac{1}{2}\). But here, \(h(\tfrac{1}{2}) = \eta(\tfrac{1}{2} ) - \eta(1) - \eta(\tfrac{1}{2}) = 0\). Thus, \(f(\tfrac{1}{2}) \leq 0\) for \(s\in [0, \tfrac{1}{2}]\).

  3. We have that \(S(\rho) = \sum_j \eta(r_j)\) and \(S(\sigma) = \sum_j \eta(s_j)\). Therefore, \[\begin{aligned} |S(\rho) - S(\sigma)| &= \left| \sum_j \eta(r_j) - \eta(s_j)\right| \\ &\leq \sum_j |\eta(r_j) - \eta(s_j)| \\ &\leq \sum_j \eta(|r_j-s_j|). \end{aligned}\]
  4. Note that \[ \eta(xy) = - xy \log (xy) = - xy (\log x + \log y) = - xy \log x - xy \log y = y \eta(x) + x \eta(y) \] Thus, \(\eta(\varepsilon_j) = \eta(\varepsilon\lambda_j) = \varepsilon\eta(\lambda_j) + \lambda_j\eta(\varepsilon)\), and \[\begin{aligned} |S(\rho) - S(\sigma)| &\leq \sum_j \eta(\varepsilon_j) \\ &= \varepsilon\sum_j \eta(\lambda_j) + \eta(\varepsilon) \\ &= \eta(\varepsilon) + H(\{ \lambda_j\}_{j})\\ &\leq \eta(\varepsilon)+ \varepsilon\log d \end{aligned}\] using that \(\{ \lambda_j\}_{j}\) is a probability distribution, and the entropy of any probability distribution with \(d\) elements is bounded by \(\log d\).
  5. By step 1, \(\varepsilon\leq \|\rho-\sigma\|_1\). Since \(\varepsilon\mapsto \eta(\varepsilon)\) is monotonically increasing on the range \([0,1/e]\), if \(\|\rho-\sigma\|_1\leq 1/e\), then \[\begin{aligned} |S(\rho) - S(\sigma)| &\leq\eta(\varepsilon) + \varepsilon\log d \\ &\leq \eta(\|\rho-\sigma\|_1) + \|\rho-\sigma\|_1 \log d \end{aligned}\] as desired.
  1. DOI: 10.1007/978-1-4612-0653-8. In particular, equation IV.62 of that book is a restatement of this question.