The max-relative entropy between two states is defined as $D_{\max}(\rho\|\sigma) = \log\lambda$, where $\lambda$ is the smallest real number that satisfies $\rho\leq \lambda\sigma$, where $A\leq B$ is used to denote that $B-A$ is positive semidefinite.

An alternative way to express the max relative entropy is

$$D_{\max}(\rho\|\sigma) = \|\sigma^{-1/2}\rho\sigma^{-1/2}\|_{\infty},$$

where $\|\cdot\|_\infty$ is the operator norm which essentially picks out the largest eigenvalue. I see that the essential idea is

\begin{align} \rho &\leq \lambda\sigma \\ \sigma^{-1/2}\rho\sigma^{-1/2}&\leq \lambda I \end{align}

Choosing the smallest possible $\lambda$ results in equality and hence one recovers $D_{\max}(\rho\|\sigma)$ this way.

I assumed that $\sigma^{-1}$ here is obtained by

  1. Diagonalizing $\sigma$
  2. Taking the reciprocal of all nonzero eigenvalues and leaving zero eigenvalues as they are
  3. Undiagonalizing $\sigma$ again.

However, this doesn't make sense because to me because when the support of $\rho$ is bigger than the support of $\sigma$, $D_{\max}(\rho\|\sigma) = \infty$. However, it looks like $\|\sigma^{-1/2}\rho\sigma^{1/2}\|_\infty$ can never be infinity.

So how does one obtain $\sigma^{-1}$?

  • 1
    $\begingroup$ $\rho \leq \lambda \sigma$ is only equivalent to $\sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda I$ when $\sigma$ is invertible (or at least when $\mathrm{supp}(\rho) \subset \mathrm{supp}(\sigma)$ so you can restrict to $\mathrm{supp}(\sigma)$), otherwise it is not true. Consider something like $|+\rangle\langle+| \leq \lambda |0\rangle\langle 0|$. $\endgroup$
    – user13507
    Oct 30, 2020 at 16:52
  • $\begingroup$ @user13507, ah okay! Makes sense then! I can accept your comment as an answer if you move it $\endgroup$ Oct 30, 2020 at 16:57

1 Answer 1


There is a problem in the derivation you presented, since $\rho \leq \lambda \sigma$ is only equivalent to $\sigma^{-1/2} \rho \sigma^{-1/2} \leq \lambda I$ when $\sigma$ is invertible (or at least when $\mathrm{supp}(\rho) \subset \mathrm{supp}(\sigma)$, so that you can restrict the space to $\mathrm{supp}(\sigma)$ instead of the whole Hilbert space). The property is not true otherwise: consider e.g. $|+\rangle\langle+| \leq \lambda |0\rangle\langle 0|$ as a counterexample.

All in all, only when the condition $\mathrm{supp}(\rho) \subset \mathrm{supp}(\sigma)$ is satisfied can we conclude that $D_{\max}(\rho \| \sigma) = \log \| \sigma^{-1/2} \rho \sigma^{-1/2} \|_\infty$.

  • 1
    $\begingroup$ I believe the necessary condition is $\mathrm{supp}(\rho) \subseteq \mathrm{supp}(\sigma)$. $\endgroup$
    – Artemy
    Oct 30, 2020 at 18:45

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