Clarification about inverses in sandwiched Renyi divergence

Question

The sandwiched Renyi divergence is defined as in
$$ \tilde{D}_\alpha(\rho\|\sigma):=\frac{1}{\alpha−1}\log tr[(\sigma^{\frac{1−\alpha}{2\alpha}}\rho \sigma^{\frac{1−\alpha}{2 \alpha }})^\alpha] $$

The divergence measure takes on finite values when $\rho, \sigma$ are not orthogonal to each other and is $\infty$ otherwise.

If we take $\alpha > 1$, the above expression involves fractional powers of $\sigma^{-1}$. Since non-orthogonality (or overlap of support of $\rho, \sigma$) is important, how should we consider the inverse of $\sigma$? Will the generalized inverse of a matrix suffice?

Answer 1

Firstly, the sandwiched divergence can be infinite even when $\rho$ and $\sigma$ are not orthogonal. For example, consider $\rho = \frac{|0\rangle \langle 0| + |1\rangle\langle 1|}{2}$ and $\sigma = |0\rangle \langle 0|$. We can think of this as the limiting case of $\sigma_\epsilon = (1-\epsilon) |0\rangle \langle 0| + \epsilon |1 \rangle \langle 1|$. Computing the divergence with $\sigma_\epsilon$ we don't have any issues because everything is full rank and we find $$ D_\alpha(\rho\|\sigma_\epsilon) = \frac{1}{\alpha-1}\log \left( 2^{-\alpha}\left((1-\epsilon)^{1-\alpha} + \epsilon^{1-\alpha}\right) \right) $$ which tends to $+\infty$ as $\epsilon \to 0$ when $\alpha > 1$.

The correct condition for finiteness is that for every vector $|v\rangle$ such that $\sigma |v\rangle = 0$ we must also have $\rho |v \rangle =0$. In other words the kernel of $\sigma$ is contained in the kernel of $\rho$. So for example if we were to swap the roles of $\rho$ and $\sigma$ in the previous example, so $\sigma = \frac{|0\rangle \langle 0| + |1\rangle\langle 1|}{2}$ and $\rho = |0\rangle \langle 0|$, then this satisfies our condition for finiteness. Moreover we can see it as a limiting case of $\rho_\epsilon = (1-\epsilon) |0\rangle \langle 0| + \epsilon |1 \rangle \langle 1|$ which gives $$ D_\alpha(\rho_\epsilon\|\sigma) = \frac{1}{\alpha-1}\log \left( 2^{\alpha-1}\left((1-\epsilon)^{\alpha} + \epsilon^{\alpha}\right) \right) $$ which tends to $1$ as $\epsilon \to 0$ whenever $\alpha > 1$ which also agrees with a direct computation of $D_\alpha(\rho\|\sigma)$.

To answer your question, yes you can take the pseudoinverse which for Hermitian operators can be computed readily via the spectral decomposition. If $\sigma = \sum_i \lambda_i |v_i\rangle \langle v_i|$ is the spectral decomposition of $\sigma$ then $\sigma^{-1} = \sum_i \lambda_i^{-1} |v_i\rangle \langle v_i|$.

Answer 2

This quantity was defined independently in https://arxiv.org/abs/1306.1586, where many of its properties were explored therein.

To answer your question, the quantity can be defined in the general case as

\begin{align} \widetilde{D}_\alpha(\rho\Vert\sigma) & := \sup_{\varepsilon > 0} \frac{1}{\alpha-1} \ln\operatorname{Tr}\left[\left(\sigma(\varepsilon)^{\frac{1-\alpha}{2\alpha}} \rho \sigma(\varepsilon)^{\frac{1-\alpha}{2\alpha}}\right)^\alpha\right]\\ & = \lim_{\varepsilon \to 0^+} \frac{1}{\alpha-1} \ln\operatorname{Tr}\left[\left(\sigma(\varepsilon)^{\frac{1-\alpha}{2\alpha}} \rho \sigma(\varepsilon)^{\frac{1-\alpha}{2\alpha}}\right)^\alpha\right] \end{align}

where $\sigma(\varepsilon):=\sigma+\varepsilon I$. Since $\sigma(\varepsilon)$ is invertible, the inverse is a true inverse.

It follows from the reasoning in the proof of Proposition 7.29 on page 363 of https://arxiv.org/abs/2011.04672v2 that

$$ \widetilde{D}_\alpha(\rho\Vert\sigma) = \begin{cases} \frac{1}{\alpha-1} \ln\operatorname{Tr}\left[\left(\sigma^{\frac{1-\alpha}{2\alpha}} \rho \sigma^{\frac{1-\alpha}{2\alpha}}\right)^\alpha\right] & \text{if } \alpha \in (0,1) \vee (\alpha> 1 \wedge \operatorname{supp}(\rho) \subseteq \operatorname{supp}(\sigma) )\\ +\infty & \text{if } \alpha > 1 \wedge \operatorname{supp}(\rho) \not \subseteq \operatorname{supp}(\sigma) \end{cases} $$

The expression with the inverse here is understood as generalized inverse. Alternatively, since the condition $\operatorname{supp}(\rho) \subseteq \operatorname{supp}(\sigma) $ holds, without loss of generality one can take the support of $\sigma$ to be the whole Hilbert space.