Let $A$ and $B$ be two unbounded self-adjoint operators. From [this](https://mathoverflow.net/questions/139249/perturbation-of-unbounded-self-adjoint-operators) mathoverflow post, for instance, we know that $A + B$ is self-adjoint on $\mathcal{D}(A) \cap \mathcal{D}(B)$ if $A$ and $B$ are commuting and positive operators (Putnam's book is cited as a reference there). My question is: assume $A$ is positive and $A$ and $B$ commute. $B$ is not positive though, but $B$ is a relatively bounded perturbation of $A$. Could we still say $A + B$ is self-adjoint?



Sum of two unbounded self-adjoint operators
This will work if you take the assumption that $A,B$ commute in a sufficiently strong sense (commuting resolvents would be enough). Then no extra assumption is needed.


There is a version of the spectral theorem that says that there is a projection valued measure that represents both $A$ and $B$:
$$
A = \int s\, dE(s,t) , \quad B = \int t\, dE(s,t)
$$
See [here.](https://books.google.com/books?id=2d7zZ5_9izEC&pg=PA101&dq=spectral%20theorem%20commuting%20normal%20operators&hl=en&sa=X&ei=JLEuVfDfFszEsAWl1YBw&ved=0CB4Q6AEwAA#v=onepage&q=spectral%20theorem%20commuting%20normal%20operators&f=false)
Clearly this makes $\int (s+t)\, dE(s,t)$ self-adjoint on a suitable domain. Note, however, that this domain is not necessarily $D(A)\cap D(B)$, as simple examples show (multiplication by $|x|$ and $x$).


