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How do I / can I actually prove the relationship

$[a,b]=0 \Rightarrow [f(a),g(b)]=0$ for all functions $f,g$.

I'm asking because the following sentence in the solution to my quantum mechanics homework irritates me:

For $i \neq j$ , the $\hat{n}_i$ commute with one another, and therefore functions of the $\hat{n}_i$ always commute with one another.

Where $\hat{n}_i = \hat{a}_i^\dagger \hat{a}_i $ with the Bose-Operators $\hat{a}_i^\dagger ,\hat{a}_i $. It is not my task to prove that relation, but the relation itself was required for being able to solve the exercise.

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    $\begingroup$ The usual physicist's proof of this proceeds straightforwardly by Taylor expanding $f,g$. $\endgroup$
    – ACuriousMind ♦
    Commented Jul 16, 2015 at 22:00
  • $\begingroup$ But $f$ and $g$ need not be analytic, so the physicist would be left with a terrible headache :P $\endgroup$
    – Phoenix87
    Commented Jul 16, 2015 at 22:05
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    $\begingroup$ First one needs to define what $f(a)$ means for operators. As AcuriousMind pointed out, Taylor expansion is the usual way to go and that probably covers all the functions of operators you'll ever encounter. But there are non-analytical ones, for example $f(x)=e^{-1/x}$. A more general definition is to go to a basis where $A$ is diagonalized (if $A$ is a observable then this should always be possible), and in the eigen-basis of $A$, $f(A)$ acts diagonally and is perfectly defined. With this one can write down $f(A)$ in any other basis. $\endgroup$
    – Meng Cheng
    Commented Jul 16, 2015 at 23:45
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    $\begingroup$ (continuing) Using this definition, since $a,b$ commute, they can be simultaneously diagonalized and $[f(A), f(B)]=0$ follow trivially from the fact that diagonal matrices always commute with each other. Although we prove it in a particular basis, this is a basis independent statement and we are done. $\endgroup$
    – Meng Cheng
    Commented Jul 16, 2015 at 23:45
  • $\begingroup$ @MengCheng Sadly I can't mark your answer as the accepted one, but thank you for explaining in an easily understandable way. $\endgroup$
    – ari
    Commented Jul 17, 2015 at 12:14

1 Answer 1

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For normal elements in a C*-algebra you can do continuous functional calculus, that is, if $a$ is a normal operator, then $f(a)$ is well-defined for any $f\in C(\sigma(a))$. Since $\sigma(a)$ is always compact you can use Stone-Weierstrass to write $f$ as a uniform limit of polynomials in one complex variable and its complex conjugate. Hence you can verify what you need on polynomials. If $a$ and $b$ commute, then $a^2$ and $b^2$ commute and so on. Hence $f(a)$ and $g(b)$ commute for any $f\in C(\sigma(a))$ and $g\in C(\sigma(b))$. For von Neumann algebras one can push this argument to Borel functions.

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  • $\begingroup$ This is a very nice answer, but I would change a couple of things to make it clearer. First, I think there may be a typo in "Hence you can ....", because it presently reads as though the verification for polynomials is deduced from your Stone-Weierstrass argument: I know you're not saying that but I think it could imply that. You're using SW to extend a result valid for polynomials to $C(\sigma(a))$. So why not write the polynomials bit first (especially since the OP is likely to have seen that argument in a textbook) and then pass to the uniform limit using SW. $\endgroup$
    – Selene Routley
    Commented Jul 16, 2015 at 23:29
  • $\begingroup$ I also think you need to say somewhere that $a$ is assumed to be bounded, therefore $\sigma(a)$ is compact. It's an assumption the OP hasn't stated and also "$\sigma(a)$ is always compact" is otherwise likely to come out of left field for the OP. $\endgroup$
    – Selene Routley
    Commented Jul 17, 2015 at 0:46
  • $\begingroup$ Also, to complement, the answer as it is does not apply to the number operator of the OP, that is self-adjoint but unbounded. For self-adjoint commuting operators (and that means not only $[a,b]=0$ on a suitable dense domain, but that all the spectral projections commute) the proof is given immediately by the spectral theorem in its functional calculus form for all the measurable functions (with respect to the common spectral family). $\endgroup$
    – yuggib
    Commented Jul 17, 2015 at 6:04
  • $\begingroup$ @WetSavannaAnimalakaRodVance Thanks for your comments. Indeed you're right that one first checks the property on polynomials and then by SW one can extend it to continuous functions. By my assumption that the operators are from a C*-algebra it follows that they are bounded, so that condition is already stated in the answer. $\endgroup$
    – Phoenix87
    Commented Jul 17, 2015 at 11:28
  • $\begingroup$ Ah, continuity of the C* operators gives you this right?: got it! Thanks. $\endgroup$
    – Selene Routley
    Commented Jul 17, 2015 at 11:51

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