# Linear Mapping and integration

I have been reading the paper - "Introduction to Quantum Fisher Information". In section 1.2 the author talks about the linear map $\mathbb{J}_D$, which he defines as follows:

Let $D \in M_n$ be a positive invertible matrix. The linear mapping $\mathbb{J}^f_D:M_n \to M_n$ is defined by the formula

$\mathbb{J}_D^f=f(\mathbb{L}_D\mathbb{R}^{-1}_D)\mathbb{R}_D$

where $f:\mathbb{R}^+\to\mathbb{R}^+$,

$\mathbb{L}_D(X)=DX$ and $\mathbb{R}_D(X)=XD$.

Also, the author points out that the inverse mapping of $\mathbb{J}_D^f$ is the given by

$\frac{1}{f}(\mathbb{L}_D\mathbb{R}^{-1}_D)\mathbb{R}_D^{-1}$

Then he gives the following example, of which I am able to get the first part but not the inverse map.

Example 1: If $f(x)=(x+1)/2$, then

$\mathbb{J}_DB=\frac{1}{2}(DB+BD)$.

$\mathbb{J}_DB=\int_0^\infty \text{exp}(-tD/2)B\text{exp}(-tD/2)dt$

Can you tell me how does he get the required formula for the inverse map? Also, I would like to know what is the meaning of the map $\mathbb{J}_D$? What does the integral of the linear map represent? Can you also provide me some reference book where I could look for these topics?

-

One way to think of the operation ${J}_D^f$ is as follows: $M_n$ has the structure of a Hilbert space by the inner product $\langle X, Y \rangle = {\rm Tr}(X^* Y) = {\rm Tr}(Y X^*)$. Then the transformations $L_D R_{D}^{-1}$ and ${R}_D$ are positive operators on this Hilbert space. Since they commute, you may take the functional calculus $f(s) \otimes g(t) \mapsto f(L_D R_{D}^{-1}) g({R}_D)$ from the space of functions on ${\rm Sp}(L_D R_D^{-1}) \times {\rm Sp}(R_D)$ into the space of bounded operators on $M_n$ (since the spectrum ${\rm Sp}(L_D{R}_{D^{-1}})$ of $L_D$ is a finite subset of ${R}$, there is no need to worry about the regularity of $f$. Ditto for $g$). Then ${J}_D^f$ is simply the image of $f(s) \otimes t$. Since the functional calculus is an algebra homomorphism, the inverse of ${J}_D^f$ is represented by $(1/f(s)) \otimes (1/t)$, which equals $\frac{1}{f}(L_D{R}_D^{-1}){R}_D^{-1}$. When $E \subset M_n$ is the joint eigenspace for $L_D {R}_{D^{-1}}$ and ${R}_D$ with the corresponding eigenvalues $a b^{-1}$ and $b$ (i.e. $L_D$ has eigenvalue $a$ on this subspace), ${J}_D^f$ acts by $f(a/b)b$ on $E$.
About the computation of $\mathbb{J}_D$ for the case of $f(t) = \frac{t+1}{2}$, the integral formula follows from the identity $\int_0^\infty exp(-t a / 2) exp(-t b / 2) = \frac{2}{a+b} = \frac{2}{ab^{-1}+1} \frac{1}{b}$.
Edit: I made a stupid mistake in the first version, and this is the corrected version. Sorry for the change of notation from $\mathbb{L}_D$ to $L_D$, etc. I somehow couldn't make it to work.
Thanks for your reply. But I am still not clear on the computation. After applying the integral formula I have the following expression $\left(\int_0^\infty exp(-t\mathbb{L}_D/2)exp(-t\mathbb{R}_D/2) dt\right)(B)$. How does one evaluate this? –  Shishir Pandey Oct 26 '10 at 10:48
If $B$ satisfies $\mathbb{L}_D(B) = a B$ and $\mathbb{R}_D(B) = b B$, you have $exp(-t\mathbb{L}_D/2)exp(-t\mathbb{R}_D/2)(B) = exp(-ta/2)exp(-tb/2) B$ for any $t$. Its integral is, as above, $\frac{1}{f(ab^{-1})} b^{-1} B$, which equals $\frac{1}{f}(\mathbb{L}_D\mathbb{R}_D^{-1})\mathbb{R}_D^{-1}(B) = \mathbb{J}_D(B)$. For general matrix $B$, you can decompose it as a sum of the simultaneous eigenvectors for $\mathbb{L}_D$ and $\mothbb{R}_D$ because they are commuting positive operators. Functional calculus is compatible with this decomposition. –  Makoto Yamashita Oct 27 '10 at 1:50