Let $A, B$ be unbounded self-adjoint operators on Hilbert spaces $\mathcal{H_1}, \mathcal{H_2}$, with both non-empty discrete spectra. Let us say, for instance, $\inf \, \sigma(A) = \lambda_1^A$ and $\inf \, \sigma(B) = \lambda_1^B$ are isolated eigenvalues with finite multiplicity at the bottom of the spectrum, so the lower bounds of the essential spectra of both operators check out $\inf \, \sigma_{ess}(A) > \inf \, \sigma(A) $ and $\inf \, \sigma_{ess}(B) > \inf \, \sigma(B)$.
Now let us define $A \otimes 1 + 1 \otimes B$ classically, i.e. as the closure of this operator on the algebraic tensor product of their domains. We know that $\sigma(A \otimes 1 + 1 \otimes B)$ is the closure of $\sigma(A)+ \sigma(B)$, in particular $\lambda_1^A + \lambda_1^B$ will still be an isolated eigenvalue since the lower bound of the continuous spectrum will be at $\min(\lambda_1^A + \inf \, \sigma_{ess}(B), \lambda_1^B + \inf \, \sigma_{ess}(A))$. Now here is my question: what guarantees me that $\lambda_1^A + \lambda_1^B$ is still of finite multiplicity? Does someone have a clue or a reference? I thought of different things: seeking for an explicit form of the spectral family of $\sigma(A \otimes 1 + 1 \otimes B)$ is the closure of $\sigma(A)+ \sigma(B)$, proving by absurd that it's not in the essential spectrum using Weyl criterion, but I don't manage to have the good argument to conclude.