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Homework Help: Associativity of operators in quantum mechanics

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data
    What is the correct interpretation of
    [tex] < \frac{\partial {A}}{\partial t} >[/tex], where A is an operator?

    2. Relevant equations
    for a wave function [tex]\phi[/tex] and operator A,
    [tex]<A> = \int_{V}\phi^{*}(A\phi)dV[/tex]

    3. The attempt at a solution
    I thought it could mean
    [tex] < \frac{\partial {A}}{\partial t} > = \int_{V}\phi^{*}\frac{\partial}{\partial t}(A\phi)dV[/tex]
    but then again it might mean
    [tex] < \frac{\partial {A}}{\partial t} > = \int_{V}\phi^{*}(\frac{\partial A}{\partial t})(\phi)dV[/tex].

    I read an article saying that operators are associative. But, when I think about the operators [tex]t[/tex] and [tex]\frac{\partial}{\partial t}[/tex], then,

    [tex]\frac{\partial}{\partial t}\left(t\phi\right) = t\frac{\partial\phi}{\partial t} + \phi \neq \left(\frac{\partial t}{\partial t}\right)\phi = \phi[/tex]

    any thoughts?
  2. jcsd
  3. Nov 24, 2012 #2


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    Science Advisor
    Homework Helper

    The d/dt is not an operator in the normal sense. A(t) is interpreted as a set of operators which depend on a parameter, t. The derivative of this A(t) wrt the parameter is defined by means of a limiting procedure always in the presence of vectors in the domain of all A(t).

    [tex] \frac{d A(t)}{dt} \psi = \lim_{t\rightarrow 0} \frac{A(t)\psi - A(0)\psi}{t} [/tex]

    The expectation value is then simply [itex] \left\langle \phi, \frac{dA(t)}{dt}\phi\right\rangle [/itex]
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