关于无界自伴算子张量和$A \otimes 1+ 1 \otimes B$离散谱中特征值$\lambda_1^A + \lambda_1^B$有限重数的判定疑问

Hey everyone, I've been digging into problems involving unbounded self-adjoint operators on Hilbert spaces lately, and there's a specific question I'm stuck on—hoping someone here can lend a hand or point me in the right direction.

Here's the setup I'm working with:

• Let $A, B$ be unbounded self-adjoint operators on Hilbert spaces $\mathcal{H}_1, \mathcal{H}_2$ respectively, both with non-empty discrete spectra.
• Specifically, the infima of their spectra $\inf , \sigma(A) = \lambda_1^A$ and $\inf , \sigma(B) = \lambda_1^B$ are isolated eigenvalues at the bottom of each spectrum, each with finite multiplicity. Also, the lower bounds of their essential spectra satisfy $\inf , \sigma_{ess}(A) > \inf , \sigma(A)$ and $\inf , \sigma_{ess}(B) > \inf , \sigma(B)$.

I've defined the operator $A \otimes 1 + 1 \otimes B$ in the classical sense—i.e., as the closure of the operator acting on the algebraic tensor product of the domains of $A$ and $B$. I already know that $\sigma(A \otimes 1 + 1 \otimes B)$ is the closure of $\sigma(A) + \sigma(B)$; in particular, $\lambda_1^A + \lambda_1^B$ is still an isolated eigenvalue, since the lower bound of the continuous spectrum is $\min(\lambda_1^A + \inf , \sigma_{ess}(B), \lambda_1^B + \inf , \sigma_{ess}(A))$, which is strictly greater than $\lambda_1^A + \lambda_1^B$.

Now here's my core question: What guarantees that $\lambda_1^A + \lambda_1^B$ still has finite multiplicity?

I've tried a few angles to work this out:

• Trying to find an explicit form of the spectral family for $A \otimes 1 + 1 \otimes B$
• Using Weyl's criterion to argue by contradiction that this eigenvalue isn't part of the essential spectrum

But I haven't been able to piece together a solid argument to reach a conclusion. Does anyone have a clue, or know of relevant results I can lean on?

备注：内容来源于stack exchange，提问作者Hugo