Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equality of operators

  1. Dec 22, 2012 #1

    ShayanJ

    User Avatar
    Gold Member

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

    Thanks
     
  2. jcsd
  3. Dec 22, 2012 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

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

    [tex]T=S[/tex]

    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

    ShayanJ

    User Avatar
    Gold Member

    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?
    Thanks
     
  5. Dec 22, 2012 #4

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    @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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    I see. But in that case, [A,B] is only densely defined. So

    [tex][A,B]\psi[/tex]

    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

    ShayanJ

    User Avatar
    Gold Member

    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.

    P.S.
    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equality of operators
  1. Equality of spans (Replies: 5)

  2. Is this equal to that? (Replies: 1)

Loading...