Let $T$ be a bounded operator. Then, the operators $\left\lvert T \right\rvert:=\sqrt{T^\*T}$ and $\left\lvert T^\* \right\rvert:=\sqrt{TT^\*}$ are well-defined.


Is there a way to write 
$$(\left\lvert T \right\rvert -i)^{-1} -(\left\lvert T^\* \right\rvert -i)^{-1}$$ 
in terms of operators that are more accessible from $T$ than the absolute values?


The problem is that taking the square root in operator theory is a very non-explicit process. Or is there a way to simplify this difference if the explicit structure of $T$ is well-known?
Of course, I thought about the resolvent identity, but 
$$(\left\lvert T \right\rvert -i)^{-1} (\left\lvert T^\* \right\rvert-\left\lvert T \right\rvert) (\left\lvert T^\* \right\rvert -i)^{-1}$$ looks even more difficult.



Resolvent difference of absolute values!
I think the answer is "no". Consider the operator $\bar{T} = \left[\matrix{0&0\cr T&0}\right]$ acting on $H \oplus H$. Then $|\bar{T}| = \left[\matrix{|T|&0\cr 0&0}\right]$ and $|\bar{T}^\*| = \left[\matrix{0&0\cr 0&|T^\*|}\right]$. So $$(|\bar{T}| - i)^{-1} - (|\bar{T}^\*| - i)^{-1} = \left[\matrix{(|T| - i)^{-1} - i&0\cr 0& (|T^\*| - i)^{-1} + i}\right].$$
To me this says that evaluating your expression is in general no easier than evaluating $(|T| - i)^{-1}$ and $(|T^\*| - i)^{-1}$ separately, and it's hard to see how there could be any easier way to do this than by taking the absolute values in the usual manner.


