# Pseudoinverse of a quantum state

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 $$\sigma^{-1}$$?

• $\rho \leq \lambda \sigma$ 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 $|+\rangle\langle+| \leq \lambda |0\rangle\langle 0|$. – user13507 Oct 30 '20 at 16:52
• @user13507, ah okay! Makes sense then! I can accept your comment as an answer if you move it – user1936752 Oct 30 '20 at 16:57

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 $$D_{\max}(\rho \| \sigma) = \log \| \sigma^{-1/2} \rho \sigma^{-1/2} \|_\infty$$.
• I believe the necessary condition is $\mathrm{supp}(\rho) \subseteq \mathrm{supp}(\sigma)$. – Artemy Oct 30 '20 at 18:45