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Consider two self-adjoint operators A and B with commutator [A,B]=C such that [A,C]=0.
Now I consider an operator which is a function of A and is defined by the series ## F(A)=\sum_n a_n A^n ## and try to calculate its commutator with B:
## [F(A),B]=[\sum_n a_n A^n,B]= \\ \sum_n a_n [A^n,B]=\sum_n a_n \left[ A^{n-1}[A,B]+A^{n-2}[A,B]A+\dots+A[A,B]A^{n-2}+[A,B]A^{n-1} \right]=\\ \sum_n a_n \left[ A^{n-1}C+A^{n-2}CA+\dots+ACA^{n-2}+CA^{n-1} \right]=\sum_n n a_n A^{n-1} C##
Now if I formally write ## \sum_n n a_n A^{n-1}=\frac{dF}{dA} ##, I can have ##[F(A),B]=\frac{dF}{dA} C ##.
But what happens if A is unbounded?
Thanks
Now I consider an operator which is a function of A and is defined by the series ## F(A)=\sum_n a_n A^n ## and try to calculate its commutator with B:
## [F(A),B]=[\sum_n a_n A^n,B]= \\ \sum_n a_n [A^n,B]=\sum_n a_n \left[ A^{n-1}[A,B]+A^{n-2}[A,B]A+\dots+A[A,B]A^{n-2}+[A,B]A^{n-1} \right]=\\ \sum_n a_n \left[ A^{n-1}C+A^{n-2}CA+\dots+ACA^{n-2}+CA^{n-1} \right]=\sum_n n a_n A^{n-1} C##
Now if I formally write ## \sum_n n a_n A^{n-1}=\frac{dF}{dA} ##, I can have ##[F(A),B]=\frac{dF}{dA} C ##.
But what happens if A is unbounded?
Thanks