In the question
Define antilinear or antiunitary operator $Θ$ acting on the ket state and on the bra state consistently?
an answer mentions that the bra-ket $⟨β|A|α⟩$ in a proper way iff $A$ is linear and self-adjoint and that such usage in the absence of a self-adjoint operator is a shortcoming of the bra-ket notation. Why is it necessary that it be linear self-adjoint like an observable? Shouldn't the action on bra be meaningful as long as the operator has an adjoint?
The point is that the notation is ambiguous. Interpreted as $\langle \beta \lvert A \rvert \alpha \rangle = \langle \beta , A \alpha \rangle$ (where on the right I have reverted to the unambiguous math notation)$-$i.e., $A$ is understood to act to the right, then everything is fine. But the symmetric "look" of the notation suggests that $A$ can act to the left, but this is only true if $A$ is self-adjoint, because by definition, then, $\langle \beta , A \alpha \rangle = \langle A\beta , \alpha \rangle$, whereas interpreting $$ \langle \beta \lvert A \rvert \alpha \rangle = \langle \beta \rvert \left(A \rvert \alpha \rangle \right)\,, $$ then $$ \langle \beta \lvert A \rvert \alpha \rangle^* = \left( \langle \beta \rvert \left(A \rvert \alpha \rangle \right) \right)^* = \left(A \rvert \alpha \rangle \right)^{\dagger} \lvert \beta \rangle =\langle \alpha \rvert A^{\dagger}\rvert \beta\rangle\,, $$ where now we can again interpret this as matrix multiplication, i.e., $A^{\dagger}$ acts to the right. One can see how this makes a difference. For more details, see this answer and the answers in the duplicate question.
