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A If [A,B]=0, are they both functions of some other operator?

  1. Jul 20, 2017 #1
    In other words, if we are told that A and B commute, then does that mean that there exists some other operator X such that A and B can both be written as power series of X? My instinct is yes but I haven't been able to prove it.
     
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  3. Jul 20, 2017 #2

    fresh_42

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    If you have ##[.,.]## defined as a multiplication with certain properties, it simply says ##[A,B]=0##, nothing more than this. However, we often have an additional multiplication ##A \cdot B## and define ##[A,B]=A\cdot B - B\cdot A##. This is often the case, but not automatically.

    What ##A,B## are or where they're from depends on your example. They can be functions, matrices or whatever. Usually they are vectors, which is no contradiction as function and matrices can be vectors, too. Obviously we need a zero to make sense of the equation, and this naturally leads to vector spaces, resp. algebras.

    Your question is settled somewhere in nowhere. I suspect that you confuse some concepts which cannot be condensed the way you suggest, but this is guesswork. So let me give an example: ##A,B \in \mathbb{R}## and ##A\cdot B## is the usual multiplication of real numbers. Then ##[A,B]=A\cdot B - B\cdot A = 0## for all choices of ##A,B##. Where do you see ##X## in here?
     
  4. Jul 27, 2017 #3

    tom.stoer

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    Sometimes there's a tendency overcomplicate questions ...

    My interpretation is that pantheid is asking if for two operators A,B (defined on a Hilbert space) satisfying [A,B] = AB - BA = 0 one can prove that there must exist some other operator X and two operator functions f,g with A = f(X) and B = g(X).
     
  5. Jul 27, 2017 #4

    Orodruin

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    The obvious answer is then yes unless you add additional requirements on the function. You just let ##f(X)## and ##g(X)## be constant functions equal to ##A## and ##B##, respectively.

    Now, if you for example want the functions to be defined through power series in ##X## with scalar coefficients, that is a completely different issue and it is trivial to find counter examples.
     
  6. Jul 27, 2017 #5

    andrewkirk

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    I got the impression that the power series the OP had in mind had scalar coefficients. In that case those constant functions would not be power series in the same operator unless one of A, B is a scalar multiple of the other.

    EDIT: I just saw the second part of your post. For some reason I didn't register it at first, so the foregoing is probably not applicable, and it is the following that puzzles me:.

    Can you offer a counter-example, because I couldn't think of one (assuming the operators are linear)? [so I hope it's not too trivial! :nb)]
     
  7. Jul 27, 2017 #6

    fresh_42

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    How about smooth functions in two variables as vector space and ##A=\frac{d}{dx}## and ##B=\frac{d}{dy}##. In this case I cannot see an ##X##.

    But if we only take, what actually has been provided by the OP, who didn't mention any vector spaces at all, one can at least in the finite dimensional, and probably the countable cases say
    $$
    \mathbb{F}^n \cong \{A\,\vert \, [A,B]=0\} \cong \mathbb{F}[X]/(X^{n})
    $$
     
    Last edited: Jul 27, 2017
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