# Clarification about inverses in sandwiched Renyi divergence

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?

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 $$\sigma^{-1} = \sum_i \lambda_i^{-1} |v_i\rangle \langle v_i|$$.