# Eigenvalues of a sum of operators

If we have a quantum operator that is composed of the sum of other operators, let's say, $A=B+C$, and we want to find it's eigenvalues, is it the same as finding the eigenvalues of each operator and then just adding them?

In other words, for orthonormal basis $(|\psi_n\rangle)_{n\in N}$,

$A|\psi\rangle=a|\psi\rangle$

is the same as

$A\psi\rangle=(B+C)|\psi\rangle=(b+c)|\psi\rangle$

and thus $a=b+c$.

From what I've read in Sakurai, this seems to be the case for a finite basis, but wonder if we can extend this to an infinite basis.

• I think the operators need to have a common eigenbasis for this to be true. – Aaron Stevens Aug 18 '18 at 21:34

In general, the eigenvalues and eigenvectors of $A$ and $B$ have extremely tenuous connections to the eigenvalues and eigenvectors of $A+B$. This relationship is so weak that you can phrase a huge fraction of the hard, Very Hard, and Completely Unsolved problems in quantum mechanics as some form of
given $A$ and $B$ with known eigenvalues and eigenfunctions, find the eigenvalues of $A+B$.
(As one such example, try $A=\frac1{2m}p^2$ and $B=V(x)$, or their generalizations to multiple electrons.)
IF $A$ and $B$ commute, then they share a common eigenbasis and on that eigenbasis the sum-of-eigenvalues property does hold. For cases where they don't commute, it doesn't.
If you want an explicit example, try, say, $$A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad B = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \text{and} \quad C = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix},$$ where $B$ has eigenvectors $(1,1)$ and $(1,-1)$ with eigenvalues $1$ and $-1$, respectively, and $C$ has eigenvectors $(1+\sqrt 2,1)$ and $(1-\sqrt 2,1)$ with eigenvalues $-\sqrt{2}$ and $\sqrt{2}$ resp.