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Operators and commutators

  1. May 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ## \hat{A} = x ## and ## \hat{B} = \dfrac{\partial}{\partial x} ## be operators
    Let ## \hat{C} ## be defined ## \hat{C} = c ## with c some complex number.

    A commutator of two operators ## \hat{A} ## and ## \hat{B} ## is written ## [ \hat{A}, \hat{B} ] ## and is defined ## [ \hat{A}, \hat{B} ] = \hat{A} \hat{B} - \hat{B} \hat{A}##
    A common way to evaluate commutators is to apply them to a general test function.

    Evaluate ## [ \hat{A}, \hat{B} ] ##
    Evaluate ## [ \hat{C}, \hat{B} ] ##

    2. Relevant equations

    The definition of a commutator

    3. The attempt at a solution

    I'm going to try f(x) as my tet function:

    ## [ \hat{A}, \hat{B} ] = x \dfrac{\partial}{\partial x} f(x) - \dfrac{\partial}{\partial x} x f(x) ##

    ## [ \hat{A}, \hat{B} ] = x f'(x) - x f'(x) - f(x) ##

    ## [ \hat{A}, \hat{B} ] = -f(x) ##

    And so removing the test function:

    ## [ \hat{A}, \hat{B} ] = -1 ##

    And for the second question:

    Evaluate ## [ \hat{C}, \hat{B} ] = c \dfrac{\partial}{\partial x} f(x) - \dfrac{\partial}{\partial x} c f(x)##

    Evaluate ## [ \hat{C}, \hat{B} ] = c f(x) - c f(x) = 0##

    Is that right?
     
  2. jcsd
  3. May 25, 2013 #2

    CompuChip

    User Avatar
    Science Advisor
    Homework Helper

    That's right.
    You will see these things a lot in Quantum Mechanics, where the momentum operator
    [tex]\hat p \propto \frac{\partial}{\partial x}[/tex]
    (and the proportionality factor involves the imaginary unit i and the physical constant [itex]\hbar[/itex]) so that the position operator and the momentum operator can no longer be interchanged as in classical mechanics:
    [tex][\hat x, \hat p] \neq 0[/tex]

    Constants (complex numbers) commute with pretty much anything, which you have now shown for the special case where anything = x.
     
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