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.

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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.

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  • $\begingroup$ okay, if nobody in the next few days disagrees with you, I will accept your answer. Thank you for this insight. $\endgroup$ – gipom Mar 20 '17 at 0:00

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