Associativity of operators in quantum mechanics

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Homework Statement


What is the correct interpretation of
< \frac{\partial {A}}{\partial t} >, where A is an operator?

Homework Equations


for a wave function \phi and operator A,
<A> = \int_{V}\phi^{*}(A\phi)dV

The Attempt at a Solution


I thought it could mean
< \frac{\partial {A}}{\partial t} > = \int_{V}\phi^{*}\frac{\partial}{\partial t}(A\phi)dV
but then again it might mean
< \frac{\partial {A}}{\partial t} > = \int_{V}\phi^{*}(\frac{\partial A}{\partial t})(\phi)dV.

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

\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

any thoughts?
 
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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).

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

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