# How does the expectation value of a partial derivative of a random variable make mathematical sense in QM?

In probability theory, I'm familiar with the definition of the expectation value of a random variable $$X \colon \Omega \rightarrow \mathbb{R}$$ being:

$$\langle X\rangle= \int_{-\infty}^{\infty} x f_X(x) dx$$

where $$f_X(x)$$ is the probability density function of $$X$$. I'm currently learning QM and I keep seeing quantities in the format: $$\langle\frac{\partial X}{\partial t}\rangle$$. (Such as in Ehrenfest's theorem, where the RHS includes $$\langle V'(x)\rangle$$.) I'm not sure how taking the derivative of a random variable mathematically makes sense. The domain of a random variable is the sample space, so what does its derivative even mean?

• If $dx/dt$ is velocity and you can measure velocity, then you can have a sample space over $dx/dt$, right? In other words, $dx/dt$ is itself a random variable. Feb 24, 2019 at 23:48

The symbols like $$A$$ or $$X$$ in the Ehrenfest theorem are (usually) not random variables, they are supposed to be linear operators that take one vector to another. Such operators may be functions of time, or not. In either case, the time derivative is defined in the natural way, as a limit of operators
$$\frac{A(t+\Delta t) - A(t)}{\Delta t}$$ as $$\Delta t$$ goes to zero.
$$H=-\frac{\hbar^2}{2m}\frac {\partial^2}{\partial x^2} - qE(t)x$$ where $$E(t)$$ is some function of time (for example, a harmonical oscillation).
The derivative $$\partial H/\partial t$$ is, by the above definition, the operator
$$\frac{\partial H}{\partial t} = -qE'(t)x.$$