In statistics and probability I came across something I don't quite believe or understand. How come the derivative of the expectation value is the expectation value of the derivative?

Question

empty · Answer

Specifically what I'm saying is I don't think these two things are equal:

$$\frac{d \langle x \rangle}{dt} \ne \langle \frac{dx}{dt} \rangle $$

empty · Answer

Where I'm saying for any arbitrary thing, the expectation value can be calculated as: 

$$\langle a \rangle = \int a \rho(x) dx $$

anonymous · Answer

the general idea has to do with the fact that both are linear operators

empty · Answer

Hmmm I realized I have made a mistake in what I am saying. What I am saying is subtle so I'm having trouble articulating it, in that I do believe:

$$\frac{d \langle x \rangle}{dt} = \langle \frac{dx}{dt} \rangle $$

However I don't believe that these two expressions that result within the integral are necessarily identical. Although now I realize my argument is sort of going back to where I originally found it which is not entirely within the realm of mathematics.

$$\frac{d}{dt} \int_{-\infty}^\infty x \Psi^*(x,t) \Psi(x,t) dx =  \int_{-\infty}^\infty x \Psi^*_t(x,t) \Psi_t(x,t) dx$$

I don't exactly believe the fact that we are essentially dropping out the term $$\frac{dx}{dt}=0$$ in differentiation under the integral sign since it seems to me that position could be thought of as depending on time... Although I understand there's the ensemble interpretation, I just don't feel convinced that:

"The derivative of the expectation value of position is the expectation value of velocity"

These seem to me to be two different things, the first being the velocity of the average particle and the second is the average velocity. So while these integrals for expectation values align in the same way that:

$$\langle v \rangle = \frac{1+2+3}{3}=2$$
$$\frac{d}{dt} \langle x \rangle = \frac{2+2+2}{3}=2$$

I guess what I'm saying is I don't think the momentum operator produces the momentum, even though it ends up in the integral evaluating to the same expected value.

empty · Answer

Ok I will rephrase the last bit since I think that really comes to the core of my question.

anonymous · Answer

are you learning basic QM?

empty · Answer

Yep

anonymous · Answer

ehrenfest theorem? https://en.wikipedia.org/wiki/Ehrenfest_theorem

anonymous · Answer

i'd be careful about trying to dive too deep into the math in QM since it quickly becomes very advanced but

anonymous · Answer

the Hamiltonian gives time evolution so the commutator $[A,H]$ means that $A$ is conserved i think

empty · Answer

Nothing to be careful about, just gotta dive in and wreck stuff up and get my hands dirty. :P

Specifically I'm just trying to understand how in this alleged derivation of the momentum operator ends up having this first term equal to zero. 

$$\frac{d}{dt} \int\limits_{-\infty}^\infty x \psi^* \psi dx = \int\limits_{-\infty}^\infty \frac{\partial x}{\partial t} \psi^* \psi dx +\int\limits_{-\infty}^\infty x \psi^*_t \psi dx + \int\limits_{-\infty}^\infty x \psi^* \psi_t dx$$

So here they say that this partial derivative of x is zero,

$$\frac{\partial x}{\partial t}=0$$ so therefore 
$$\int_{-\infty}^\infty \frac{\partial x}{\partial t} \psi^* \psi dx = 0 $$

empty · Answer

I've taken quantum mechanics in the past, however this time I'm going through Griffith's Introduction to Quantum Mechanics at my own pace doing every example in a very linear fashion and taking time to appreciate every question and comment. I've done a fair bit of spectroscopy in the past and in a lot of senses this book is mostly review.

As far as it goes, I have a handful of differential geometry from studying clifford algebra and tensor calculus, so I'm at this point I don't care about definitions unless they are intuitive and meaningful to me in some way, since this is at my pace on my own terms. 

At the end of the day however, the Schrodinger equation seems to remain something unmotivated, from wikipedia they have a nice quote by Feynman, "Where did we get that (equation) from? Nowhere. It is not possible to derive it from anything you know. It came out of the mind of Schrödinger."

And so this is the wall I am looking at and through trying to break down will show me the limitations of it so that I can understand it better in the process.

anonymous · Answer

consider in 1 dimension$$\langle x\rangle=\int_{\mathbb{R}} dx\,\psi^* x \psi\\frac{d}{dt}\langle x\rangle=\frac{d}{dt}\int_{\mathbb{R}}dx\,\psi^* x\psi=\int_\mathbb{R} dx\frac{\partial}{\partial t}\left[\psi^* x\psi\right]\\qquad\quad=\int_\mathbb{R}dx\,\frac{\partial \psi^*}{\partial t}x\psi+\int_\mathbb{R}dx\,\psi^*\frac{\partial x}{\partial t}\psi+\int_\mathbb{R}dx\,\psi^* x\frac{\partial\psi}{\partial t}$$

well, here $x$ is merely the variable of integration, so it has no possible dependency on $t$, and the 'edge' terms from differentiating under the integral disappear because of the fact $\psi\in L^2(\mathbb{R})$ and is thus square integrable (and so cannot have even thick tails)

anonymous · Answer

but you're probably going to need to learn more functional analysis and familiarize yourself with the theory of symplectic manifolds to try to 'get' a lot of the mathematical formalism in QM; I would not attempt to learn QM without a solid basis in analytical treatments of classical mechanics (Lagrangian, Hamiltonian formalisms)