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?

  • $\begingroup$ 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? $\endgroup$ Sep 10, 2021 at 18:54
  • $\begingroup$ @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 $\varepsilon'$. $\endgroup$ Sep 10, 2021 at 19:31
  • $\begingroup$ out of curiosity, where did this encounter this problem/statement? $\endgroup$
    – glS
    Sep 11, 2021 at 14:13
  • $\begingroup$ @glS It is useful to show that the Lamb Shift hamiltonian commutes with the original local hamiltonian in the derivation of Lindblad's equation. $\endgroup$ Sep 13, 2021 at 11:21

1 Answer 1


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.

  • $\begingroup$ Thank you, writing $\eta+\Delta\epsilon$ instead of keeping that condition on the sum makes it easy. $\endgroup$ Sep 13, 2021 at 11:19
  • $\begingroup$ 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? $\endgroup$ May 24 at 19:31

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