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Equality of operators

  1. Dec 22, 2012 #1


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    Imagine we have two operators A and B on a complex hilbert space H such that:
    [A,B] \psi = (AB-BA) \psi=c \psi \ \ \ \ \psi \epsilon H \mbox{ and } c \epsilon C
    Then can we say that [A,B] is the same as cI when I is the identity operator?Why?

  2. jcsd
  3. Dec 22, 2012 #2
    Let [itex]T,S:H\rightarrow H[/itex] (if you're working with unbounded operators then things change). By definition, we have


    if and only if

    [tex]T\psi = S\psi[/tex]

    for all [itex]\psi\in H[/itex].

    So yes, in that case we can say [itex][A,B]=cI[/itex].
  4. Dec 22, 2012 #3


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    So if we have unbounded operators,we should deal with the domains too?
    Let D(I)=H and let one of A and B be unbounded.
    What you tell now?
  5. Dec 22, 2012 #4


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    @micromass: Shyan's questions are motivated by this thread, where there's some discussion about what's wrong with calculations like this one:
    I don't understand the issues well enough (or just haven't thought it through enough) to write down a complete explanation of everything that's going on here, but it's clear that it has something to do with the domains of the operators.
  6. Dec 22, 2012 #5
    I see. But in that case, [A,B] is only densely defined. So


    doesn't even make sense for all [itex]\psi \in H[/itex].

    And since [itex]I[/itex] is everywhere defined, we can never have [itex][A,B]=cI[/itex]. We can only have [itex][A,B]\subset cI[/itex].
  7. Dec 22, 2012 #6


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    Yes,But we can define an identity operator with the same domain as [A,B] and do the calculations.Or we can choose a vector from [A,B]'s domain.So what you say doesn't solve the problem.

    Ok guys,looks like here's going to be the same discussion as the topic related to that paradox,so I guess you guys may want to lock this one.
    Last edited: Dec 22, 2012
  8. Dec 22, 2012 #7
    What problem??
    It's probably not a good idea to have two threads on the same topic. So I'm going to lock this thread and ask that we discuss this further in the thread that Fredrik linked.
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