Comment on “A note on the infimum problem of Hilbert space effects” [J. Math. Phys. 47, 102103 (2006)]

Let $H$ be a (complex) Hilbert space and $B(H)$ the $C*$-algebra of bounded linear operators in $H$⁠. The cone of bounded non-negative operators in $H$ is denoted by $B(H)+$⁠. The order relation ⩽ is the natural one: Given $A,B∊B(H)+$⁠, $A⩽B$ if and only if $⟨Ah,h⟩⩽⟨Bh,h⟩$ for all $h∊H$⁠.

Given $A,B∊B(H)+$⁠, the parallel sum of $A$ and $B$ is the bounded non-negative operator denoted by $A:B$ and defined (e.g., see Ref. 6) by

Parallel addition is separately nondecreasing and $0⩽A:B⩽A,B$⁠.

The shorted operator of $A$ by $B$ is the bounded non-negative operator denoted by $[A]B$ and defined (e.g., see, Ref. 1) by

where the strong operator limit exists because $(nA):B⩽B$ and $(nA):B⩽((n+1)A):B$ for all $n∊N$⁠. Note that $[A]B⩽B$ but $[A]B$ may not be comparable with $A$⁠.

Another formula to calculate $[A]B$ is available. Namely, let $PA,B$ denote the orthogonal projection of $H$ onto the closure of the subspace ${h∊H∣B1∕2h∊Ran(A1∕2)}$⁠, where $RanC$ denotes the range of the operator $C$⁠.

Theorem 1: (Reference 5). If $A,B∊B(H)+$ then $[A]B=B1∕2PA,BB1∕2$⁠.

Given $A,B∊B(H)+$⁠, let $A∧B$ denote the infimum of $A$ and $B$⁠, if it exists, and defined as the greatest lower bound of the set ${A,B}$ in the partially ordered set $B(H)+$⁠, that is, $A∧B⩽A,B$ and, if $C∊B(H)+$ is such that $C⩽A,B$⁠, then $C⩽A∧B$⁠. Questions as characterizations of the existence and computation of the infimum operator are related to the lattice properties of quantum effects, see Refs. 3 and 4. The most comprehensive answer is the following.

Theorem 2: (Reference 2). Let $A,B∊B(H)+$⁠. Then $A∧B$ exists if and only if the shorted operators $[A]B$ and $[B]A$ are comparable, that is, either $[A]B⩽[B]A$ or $[B]A⩽[A]B$⁠. In this case, $A∧B=min{[A]B,[B]A}$⁠.

In this short note we show that the two main results of Ref. 7 can be easily derived from Theorems 2 and 1. The first one is the positive answer to a conjecture in Ref. 3.

Theorem 3: (Reference 7). Let $A,B∊B(H)+$ be injective and at least one of them invertible. Then $A∧B$ exists if and only if $A$ and $B$ are comparable.

Proof: Assume that $A$ is injective and $B$ is invertible. Then $B1∕2$ is invertible and $Ran(B1∕2)=H$⁠. Thus, ${h∊H∣A1∕2h∊Ran(B1∕2)}=H$ and hence $PB,A=I$⁠. In particular, by Theorem 1 we have $[B]A=A$⁠.

On the other hand,

and then taking into account that $A1∕2$ is injective and hence, that it has dense range, it follows that $PA,B=I$⁠. Therefore, $[A]B=B$⁠. The statement follows now from Theorem 2.◼

For $C∊B(H)$ we denote by $σ(C)$ the spectrum of the operator $A$⁠. The latter main result of Ref. 7 is the following.

Theorem 4: (Reference 7). Let $B∊B(H)+$⁠. Then $I∧B$ exists if and only if either $σ(B)⊆{0}∪[1,‖B‖]$ or $σ(B)⊆[0,1]$⁠.

Proof: Because $Ker(B)$ is reducing both the identity operator $I$ and $B$⁠, without loss of generality we can assume that $B$ is injective. In this case, the statement to be proven becomes as follows: if $B∊B(H)+$ is injective then $I∧B$ exists if and only if either $σ(B)⊆[0,1]$ or $σ(B)⊆[1,‖B‖]$⁠. Elementary spectral theory shows that $σ(B)⊆[0,1]$ is equivalent with $B⩽I$⁠, while $σ(B)⊆[1,‖B‖]$ is equivalent with $I⩽B$⁠. Thus, the statement to be proven is a particular case of Theorem 3.◼