# Smallest eigenvalue of sum of two unbounded operators

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 $$\lambda_1(S)$$?

• You could write the Question as: "Suppose $T,S:D(\mathcal H)\to D(\mathcal H)$ are two unbounded Hermetian operators with discrete spectrum and least eigenvalues $0$. If $T+S$ also has discrete spectrum, how large can it's least eigenvalue be?" (Just adding constants to $S,T$.) Also, did you mean $D(\mathcal H)\to \mathcal H$ ? – Keith McClary Nov 20 '19 at 17:52
• Also, do you know the answer for $2 \times 2$ matrices? – Keith McClary Nov 20 '19 at 18:14
• Yes, edited it. I'm so used to copy-pasting in LaTex. – BigM Nov 20 '19 at 18:29
• You have a better chance of getting an Answer if you write the Question in the simplest possible way. If you agree that the answer follows from that of the special case I have stated then you should ask that. Anyone who could answer knows that discrete eigenvalues can be arranged in ascending sequence, so you don't need to write that out. – Keith McClary Nov 22 '19 at 17:55

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.
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.