First, let me preface this statement by saying I know that there exists no (unique) self adjoint extension of the standard differential operator for the space $L_2([0,1])$.
However, when one attempts to actually prove this fact first they can go down the path of first enforcing $\psi(0) = \psi(1)$=0 as a boundary for the domain of $P$.
This actually makes the operator hermitian. However with this condition it is easily shown the adjoint of $P$ has a larger domain (namely no requirement of boundary conditions on $\psi$). Thus this operator as defined with its boundary condition can not be self adjoint.
But why can’t I simply use the operator $P = -i \frac{d}{dx}$ without any restriction on the domain? In other words, what the operator would be in the case of the real line. I understand that this wouldn’t be self adjoint, but isn’t the reason we want the operator to be self adjoint in the first place is so it admits a complete eigenbasis? However, the eigenstates of the real line operator surely span this smaller subspace. And isn’t this all we really need (that, and the eigenvalues being real which they are?).
Basically it seems we can span the whole space with generalized eigenfunctions ($\exp{(ikx)}$) using a moment operator $P$ that isn’t self adjoint. So why is this “wrong” to do?