I have found that the function $$f(z)=e^{iqz}-\frac{a}{2iq+a}e^{iq|z|}$$ satisfies $$-\partial_z^2f(z)=(q^2+a\delta(z))f(z).$$ Note this looks like an eigenvalue equation, but I am not sure if we can label it as one since $q^2+a\delta(z)$ is not constant (?), i.e., it does not belong to a field $\mathbb{K}$. Moreover, $f(z)$ is NOT an eigenfunction of $-i\partial_z$: $$-i\partial_z f(z)=qe^{iqz}-qsgn(z)\frac{a}{2iq+a}e^{iq|z|}=q\left(e^{iqz}-sgn(z)\frac{a}{2iq+a}e^{iq|z|}\right).$$
Is there any condition under which the eigenvectors of an operator, say $A^2$, are also eigenvectors of $A$, where $A$ is hermitian? For example, what happens if I try to define the operator $$D_z=\sqrt{\partial_z^2+a\delta(z)}?$$ Note that $$(-iD_z)^2 f(z)=-D_z^2 f(z)=-(\partial_z^2+a\delta(z))f(z)=q^2f(z),$$ which IS an eigenvalue equation. Let me rephrase my question: is there any circumstance under which we can assure $-iD_z f(z)=\pm q f(z)$ given $(-iD_z)^2 f(z)=q^2f(z)$? Can I even define $D_z$?!
I look for an answer in the scope of a vector space of square integrable functions. Correct me if I am wrong, but I think that in finite-dimensional spaces, we can asure that $$A^2x=\lambda^2x\implies Ax=\lambda x,$$ for $A$ hermitian via the spectral theorem. (Obviously the converse is true.) Since this theorem is also true on an infinite-dimensional Hilbert space, would it be the path to proving what I intend to?
Sorry if I have any misconceptions, and thank you for the insight!