proof of Theorem 3.10 (Barnum-Knill) on pretty-good measurements in John Watrous' book

Question

The near-optimality result for pretty-good measurements as given in Barnum & Knill's original paper holds only for a commuting ensemble of states, Theorem 2 in

https://arxiv.org/pdf/quant-ph/0004088.pdf

Theorem 3.10 in Watrous' book makes no such assumptions on the ensemble so it seems like a generalization of Barnum & Knill's paper.

However, I'm unable to see how equation 3.55 is obtained in the proof of this theorem.

This is in Chapter 3 of the book: https://cs.uwaterloo.ca/~watrous/TQI/TQI.3.pdf

Basically, we are given

$\rho = \sum_{a\in\Sigma} \eta (a)$,

which implies $\textrm{im}(\eta(a))\subseteq \textrm{im}(\rho)$. (I understand this bit.)

Equation 3.55 reads

$\langle \nu(a),\eta(a)\rangle = \langle \rho^{1/4}\nu(a)\rho^{1/4}, (\rho^+)^{1/4} \eta(a) (\rho^+)^{1/4} \rangle.~~~~*$

Now, the right-hand side is equal to

$\textrm{tr}\big(\nu(a)(\rho^+\rho)^{1/4} \eta(a)(\rho^+\rho)^{1/4}\big)$,

since $\rho$ commutes with its pseudo-inverse $\rho^+$.

I'm wondering how this gives the equality in equation $*$ since $\rho^+ \rho \neq \mathbb{1}$.

Answer 1

As also mentioned in this other answer of mine, that identity relies on the following general statement: given any pair of Hermitian matrices, $A,B$, and an Hermitian operator $C$ such that $\operatorname{im}(B)\subseteq \operatorname{im}(C)$, we have $$\langle A,B\rangle = \langle CAC,C^+ BC^+\rangle.$$ To see this, note that the RHS equals $\operatorname{tr}(ACC^+BC^+ C)$, and from the definition of pseudoinverse you have $CC^+=C^+ C=\Pi_{\operatorname{im}(C)}$, that is, the projection onto the support of $C$ (image and support are identical for Hermitian operators). Thus the RHS equals $$\operatorname{tr}(A \Pi_{\operatorname{im}(C)} B \Pi_{\operatorname{im}(C)}).$$ But if $\operatorname{im}(B)\subseteq \operatorname{im}(C)$, then $\Pi_{\operatorname{im}(C)} B \Pi_{\operatorname{im}(C)}=B$, hence the conclusion.

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To connect this with the statement at hand, you need only observe that $\rho^{1/4}$ has the same support as $\rho$, and because $\rho=\sum_a \eta(a)$ we have $\operatorname{im}(\rho)\supseteq \operatorname{im}(\eta(a))$ for all $a$.