Smallest eigenvalue of sum of two unbounded operators

Question

Suppose $T,S:D(\mathcal H)\to \mathcal H$ are two unbounded operators with discrete spectrum consisting eigenvalues $0<\lambda_1(T)\leq\lambda_2(T)\leq\dots$ and $0<\lambda_1(S)\leq\lambda_2(S)\leq\dots$. Moreover assume that $T+S$ also has discrete spectrum and that

Q. How large $\lambda_1(T+S)$ can be compared to $\lambda_1(T)$ and $\lambda_1(S)$?

Answer 1

I answer this question, which I think implies the answer to your question:

Suppose $T,S$ are possibly unbounded Hermitian operators in a Hilbert space with discrete spectrum and least eigenvalues $0$. If $T+S$ also has discrete spectrum, how large can it's least eigenvalue be?

In the case of $2 \times 2$ matrices you can solve the problem explicitly.

For general operators you can solve the Rayleigh-Schrodinger equation, which is non-linear and involves calculating the inverse of one of the operators (With discrete spectrum, calculating the inverse not a problem.).

If you want an upper bound you can construct approximants to the eigenvalue by simply choosing an arbitrary sequence of vectors which spans the Hilbert space and minimizing $\frac{(\psi , (T+S) \psi}{(\psi,\psi)}$ in the subspace spanned by the first $n$ vectors.

In special cases there may be explicit solutions to the eigenvalue problem . In one case ($x^4$ anharmonic oscillator) there are approximants from below.