# Associativity of operators in quantum mechanics

1. Nov 24, 2012

### p6.626x1034js

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

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

3. 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?

2. Nov 24, 2012

### dextercioby

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 $\left\langle \phi, \frac{dA(t)}{dt}\phi\right\rangle$