Question

Suppose A^ and B^ are commuting operators on some Hilbert space, [A^,B^]=0, chosen suitably to ensure operators 1/A^, 1/B^ and A^/B^

are suitably well defined and bounded (for the purposes of convergence - whatever form that may take?). Does the following Newtonian generalisation of the binomial theorem to negative integers hold, and under what conditions might this be valid:

(A^+B^)−1=1A^(1^+B^A^)−1=1A^⎛⎝1^−B^A^+(B^A^)2−(B^A^)3+⋯⎞⎠

There are many confusion tucked into this query: Is the operator 1/A^
typically the inverse of the operator A^
under composition? Is the first equality in the above a valid thing to do, and under what conditions is this possible? Is such a series expansion valid given that A^ and B^ commute? What other conditions may be necessary?

Answer 1

Valid binomial expansions for the inverse of commuting operators require operators to be bounded and their inverses well-defined, with convergence of the series. Such expansions are a generalization of the classical binomial theorem.

Step-by-step explanation:

The question concerns the validity of a binomial expansion for the inverse of a sum of commuting operators A^ and B^ in a Hilbert space, where [A^,B^]=0. For such a series expansion to be valid, the operators must be bounded and their inverses must exist such that operators like 1/A^ and A^/B^ are well-defined. The first equality involving binomial expansion to negative integers is indeed a generalization of the binomial theorem and can be expressed as (A^+B^)−1 = 1/A^ (1 + B^/A^)−1 = 1/A^ (1 - B^/A^ + (B^/A^)2 - (B^/A^)3 + ...). The series expansion can be considered a formal power series, assuming further that A^ and B^ are such that A^+B^ is invertible and this series converges. Otherwise, the result could be invalid or undefined.