Assume that the Hamiltonian
has a complete
set of eigenfunctions
with eigenvalues
:

For the Hermitian operator
we define the
repeated commutator
iteratively by:
![{\displaystyle {\begin{aligned}{\hat {C}}^{(0)}&\equiv {\hat {A}}\\{\hat {C}}^{(1)}&\equiv [{\hat {H}},{\hat {A}}]={\hat {H}}{\hat {A}}-{\hat {A}}{\hat {H}}\\{\hat {C}}^{(k)}&\equiv [{\hat {H}},{\hat {C}}^{(k-1)}],\ \ \ k=1,2,\ldots \end{aligned}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/2e38c4e3192a59e4492e20010700727a6e2b1f18)
The operator
is Hermitian since
is defined to be Hermitian. The operator
is
anti-Hermitian:

By induction one finds:

and also

For a Hermitian operator we have

Using this relation we derive:
![{\displaystyle {\begin{aligned}\langle m|[{\hat {A}},{\hat {C}}^{(k)}]|m\rangle &=\langle m|{\hat {A}}{\hat {C}}^{(k)}|m\rangle -\langle m|{\hat {C}}^{(k)}{\hat {A}}|m\rangle \\&=\sum _{n}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {C}}^{(k)}|m\rangle -\langle m|{\hat {C}}^{(k)}|n\rangle \langle n|{\hat {A}}|m\rangle \\&=\sum _{n}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle (E_{n}-E_{m})^{k}-(E_{m}-E_{n})^{k}\langle m|{\hat {A}}|n\rangle \langle n|{\hat {A}}|m\rangle \\&=\sum _{n}(1-(-1)^{k})(E_{n}-E_{m})^{k}|\langle m|{\hat {A}}|n\rangle |^{2}.\end{aligned}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/b0856da3cfd137892a729122ae07b0d76030fd99)
The result can be written as
![{\displaystyle \langle m|[{\hat {A}},{\hat {C}}^{(k)}]|m\rangle ={\begin{cases}0,&{\mbox{if }}k{\mbox{ is even}}\\2\sum _{n}(E_{n}-E_{m})^{k}|\langle m|{\hat {A}}|n\rangle |^{2},&{\mbox{if }}k{\mbox{ is odd}}.\end{cases}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/9170c4d8b9085011605cca49e3810a2e64e4a525)
For
this gives:
![{\displaystyle \langle m|[{\hat {A}},[{\hat {H}},{\hat {A}}]]|m\rangle =2\sum _{n}(E_{n}-E_{m})|\langle m|{\hat {A}}|n\rangle |^{2}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/36320be103235cab1fcbb667f753d87d88f9507c)