# What are the eigenvectors of the superoperator $[H,\cdot]$ with $H$ the Hamiltonian?

Let $$\{A_\alpha\}$$ be a set of hermitian operators and $$\{\Pi(\varepsilon)\}$$ a set of projectors on the (finite-dimensional) $$\varepsilon$$ subspace. Define $$A_\alpha(\Delta\varepsilon)=\sum_{\varepsilon\varepsilon'\atop{\varepsilon-\varepsilon'=\Delta\varepsilon}}\Pi(\varepsilon)A_\alpha\Pi(\varepsilon')$$ (notice how the sum is on every $$\varepsilon$$ and $$\varepsilon'$$ such that the difference of the two is $$\Delta\varepsilon$$). Given $$H=\sum_\varepsilon\varepsilon\Pi(\varepsilon)$$, I should prove that the $$A_\alpha(\Delta\varepsilon)$$ are eigenvectors of the superoperator $$[H,\cdot]$$.

My own calculations don't quite give the correct result... could anyone give a quick proof?

• What is $\Delta\varepsilon$? How do you define subtraction of subspaces? Are subspaces assumed to be $(n-1)$-dimensional and hence identified by a vector? Sep 10, 2021 at 18:54
• @AdamZalcman Edited for clarity. I'm working with finite-dimensional spaces; as I stated $\Delta\varepsilon:=\varepsilon-\varepsilon'$ is the gap between the two eigenvalues $\varepsilon$ and $\varepsilon'$. Sep 10, 2021 at 19:31
• out of curiosity, where did this encounter this problem/statement?
– glS
Sep 11, 2021 at 14:13
• @glS It is useful to show that the Lamb Shift hamiltonian commutes with the original local hamiltonian in the derivation of Lindblad's equation. Sep 13, 2021 at 11:21

I'll assume the subspaces are orthogonal, i.e. $$\Pi(\epsilon)\Pi(\eta)=\delta_{\epsilon\eta} \Pi(\epsilon)$$.
You are asking what is $$[H,A(\Delta\epsilon)]$$, with $$H\equiv \sum_\epsilon \epsilon \Pi(\epsilon), \qquad A(\Delta\epsilon)\equiv \sum_\eta \Pi(\eta+\Delta\epsilon)A\Pi(\eta),$$ for some arbitrary Hermitian matrix $$A$$.
Observe that $$[H,A(\Delta\epsilon)] = \sum_{\epsilon\eta} \epsilon [ \Pi(\epsilon), \Pi(\eta+\Delta\epsilon)A\Pi(\eta) ] \\ = \sum_\eta \left((\eta+\Delta\epsilon) - \eta \right) \Pi(\eta+\Delta\epsilon)A\Pi(\eta) = \Delta\epsilon \sum_\eta\Pi(\eta+\Delta\epsilon)A\Pi(\eta) \\ \equiv \Delta\epsilon A(\Delta\epsilon),$$ which proves the statement: $$A(\Delta\epsilon)$$ is an eigenvector for $$\operatorname{ad}(H)$$ (that is, for the operator $$X\mapsto [H,X]$$), with eigenvalue $$\Delta\epsilon$$.
Note that the result isn't surprising: generally speaking, the eigenvalues of $$\operatorname{ad}(H)$$ are differences of eigenvalues of $$H$$, which is a result used e.g. when studying Lie algebras, see e.g. this post on math.SE.
• Thank you, writing $\eta+\Delta\epsilon$ instead of keeping that condition on the sum makes it easy. Sep 13, 2021 at 11:19
• It looks like this proof doesn't need/use hermitianity of $A$ anywhere, does it? So the statement should be valid for any (bounded?) linear operator on the Hilbert space? May 24 at 19:31