Proposition: A carrier-preserving readout needs a pointer

The carrier-preserving selective maps

can be implemented by:

1. rotating the carrier into the desired readout basis;

2. allocating a pointer qubit in $|0⟩$;

3. applying CNOT from carrier to pointer;

4. destructively reading the pointer;

5. reversing the basis rotation on the returned carrier.

The implementation retains one carrier but has peak quantum occupancy two.

Proof. In the rotated basis the pointer branch map on the carrier is $|b⟩⟨b|$. Conjugating by the basis rotation gives $P_{b\mid j}$. The selective map is therefore the displayed one. The pointer name is consumed by Enter; the carrier name is returned.