Why do commuting operators imply that A=A(a) will commute with b?

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The discussion focuses on the commutation of operators in quantum mechanics, specifically addressing why the operator A=A(a) commutes with b when [a,b] = 0, indicating that a and b are commuting functions. The user suggests using Taylor expansion of A(a) to demonstrate this relationship, leading to the conclusion that the series expansion results in zero commutation. This highlights the importance of understanding operator algebra in quantum mechanics.

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Nikitin
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Hi. Say a, A(a) and b are well behaving functions. Then say [a,b] = 0, i.e. a and b commute.

Why will this automatically mean that A=A(a) will commute with b? Can somebody give me an intuitive explanation, or link me to some proof?
 
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I think one way is taylor expanding A(a). Then [itex][A(a),b]=c_0 [1,b]+c_1 [a,b]+c_2[a^2,b]+c_3[a^3,b]+...=0[/itex].
 
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