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
    Commented 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 $\endgroup$ Commented Oct 30, 2020 at 16:57

2 Answers 2


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
    Commented Oct 30, 2020 at 18:45

It's worth noting that even if $\operatorname{supp}(\rho)\subseteq\operatorname{supp}(\sigma)$, which you need to have $D_{\rm max}(\rho\|\sigma)<\infty$, you don't necessarily have $\sigma$ invertible. Consider as a trivial counterexample $D_{\rm max}(\mathbb{P}_\psi\|\mathbb{P}_\psi)=0$ where $\mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|$. Then $\mathbb{P}_\psi$ isn't invertible, but the given formula still applies by simply replacing $\sigma^{-1/2}=\mathbb{P}_\psi$.

Inverse vs pseudoinverse — A more general understanding of the formula is taking $\sigma^{-1}$ to mean the pseudoinverseo of $\sigma$. This is already implicit in the question which described the procedure as involving "taking the reciprocal of all nonzero eigenvalues and leaving zero eigenvalues as they are". The leaving zero eigenvalues as they are amounts to taking the pseudoinverse instead of the proper inverse.

Definition of $D_{\rm max}$ as smallest $\log\eta$ such that — For completeness, let me also include where this formula for the max-relative entropy comes from. Let's start from the commonly adopted definition, discussed e.g. here, as $$D_{\rm max}(P\|Q) = \log \inf\{\eta\ge0: \,\, P\le \eta Q\}.$$ In words, we're saying: take the log of the smallest non-negative $\eta\in\mathbb{R}$ such that $\eta Q-P\ge0$. Being $P,Q\ge0$, this inequality is equivalent to $$\eta\langle u,Qu\rangle \ge \langle u,Pu\rangle \iff \eta \ge \frac{\langle u,Pu\rangle}{ \langle u,Qu\rangle}, \quad \forall u\in\mathbb{C}^n,$$ where I'm assuming $P,Q$ act on $\mathbb{C}^n$.

Equivalent formulation with operator norm — But also, we can exploit the fact that in general $A\le B$ with $A,B\ge0$ is equivalent to $\sqrt{B^+}A\sqrt{B^+}\le I$ with $B^+$ pseudoinverse of $B$. We can get this observing projecting the inequality onto the support of $B$, which contains the support of $A$, and within which $B$ is invertible. Then just observe that the inverse of $B$ in its support is the same as the pseudoinverse of $B$ projected on said support.

Armed with this relation, we see that $P\le \eta Q$ is equivalent to $\sqrt{Q^+} P\sqrt{Q^+}\le \eta$. But then, the smallest such $\eta$ must be the largest eigenvalue of $\sqrt{Q^+} P\sqrt{Q^+}$ (which is still positive semidefinite). We conclude that $$D_{\rm max}(P\|Q)=\log \|\sqrt{Q^+}P\sqrt{Q^+}\|_\infty.$$


Classical cases — If $P,Q\ge0$ are diagonal, then all the above collapses to the classical case. In particular, the operator norm just becomes the max over the diagonal elements, and we have $$D_{\rm max}(P\|Q) = \log \inf \{\eta\ge0: \,\, P_{ii}\le \eta Q_{ii}\forall i\} = \log \max_i \frac{P_{ii} }{Q_{ii}}.$$

Cases with $P$ pure — Let's now consider $P= \mathbb{P}_\psi$ a pure state, and $Q=\sigma$ an invertible state (there's no loss of generality here: if $\sigma$ isn't invertible, just replace the inverse with the pseudoinverse). Observe that in this case $\sigma^{-1/2}\mathbb{P}_\psi \sigma^{-1/2}$ has unit rank, and thus its operator norm equals the trace, which in turn equals $\langle \psi|\sigma^{-1}|\psi\rangle$. We then conclude that $$D_{\rm max}(\mathbb{P}_\psi\| \sigma) = \log \langle \psi|\sigma^{-1}|\psi\rangle.$$ For example, if $P=\mathbb{P}_0$ and $\sigma=\frac13\mathbb{P}_++\frac23\mathbb{P}_-$, then $D_{\rm max}(\mathbb{P}_0\|\sigma) = \log\frac94$. It's interesting to note how this corresponds to the min value of $\eta=9/4$ in the original definition, that is, to the inequality: $$\mathbb{P}_0 \le \frac94 \sigma=\frac38\begin{pmatrix}3&-1\\-1&3\end{pmatrix}.$$ Clearly in this case the inequality isn't just about the diagonal elements, but can nevertheless be verified as tight from $\det(\frac94 \sigma-\mathbb{P}_0)=0$.

On a similar note, if $\rho=\mathbb{P}_\Psi$ is bipartite, then $$D_{\rm max}(\rho\|I\otimes \sigma) = D_{\rm max}(\mathbb{P}_\Psi\|I\otimes \sigma) = \log \langle \operatorname{tr}_1(\mathbb{P}_\Psi),\sigma^{-1}\rangle. $$ If furthermore $\sigma=\rho_B=\operatorname{tr}_1(\mathbb{P}_\Psi)$, then we get $$D_{\rm max}(\mathbb{P}_\Psi\|I\otimes \operatorname{tr}_1(\mathbb{P}_\Psi)) = \log r$$ with $r$ the Schmidt rank of $|\Psi\rangle$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.