Why does $\frac{\partial x}{\partial t} = 0$ here?

Question

Perhaps I am forgetting something ‘simple’ from Multivariable calculus, but could someone point to a reference or provide an explanation as to why $\frac{\partial x}{\partial t} = 0$ in the following manipulation:

(Taken from chapter 1 of Griffiths’ Introduction to Quantum Mechanics 3rd Ed., we are trying to find the time average of x)

$$\frac{d\langle x\rangle}{dt} = \frac{d}{dt} \int_{-\infty}^{\infty} x|\Psi|^2 dx $$ $$ = \int_{-\infty}^{\infty}\frac{\partial}{\partial t} (x|\Psi|^2)dx $$

$$ = \int_{-\infty}^{\infty}\frac{\partial x}{\partial t}|\Psi|^2 + \frac{\partial |\Psi|^2 }{\partial t}xdx $$

Here, Griffiths tosses away the first term, implying that $\frac{\partial x}{\partial t} = 0$. And thus,

$$ = \int_{-\infty}^{\infty}\frac{\partial |\Psi|^2 }{\partial t}xdx $$

For context: I am an undergraduate, taking quantum for the first time, with this book. I have not gotten past chapter 1 yet.

Answer 1

The book is working in the Schroedinger picture in which the states $|\Psi\rangle$ and their wavefunctions are time dependendent, so $$ \Psi(x,t)\equiv \langle x|\Psi(t)\rangle = \langle x|e^{iHt}|\Psi\rangle $$ depends on time $t$, but the operators such as $\hat x$, of which the variable $x$ in your integral is the eigenvalue, are time independent, so $dx/dt=0$.

There is another way of doing things called the Heisenberg picture in which we make the states time independent, but put the time dependence into the operators $$ \hat x(t) = e^{iHt}x(0) e^{-iHt}. $$ It gives the same answers, but it is less used in introductory books.

Answer 2

From the useful comments below the main post, I will write the answer to my own question.

In this context, $x$ and $t$ are independent variables. In general, if there is no relation between the variables of a function, be it that one of them is a function of the other, or they are both a function of some other parameter (like here https://math.stackexchange.com/questions/2425481/partial-derivative-of-independent-variable#2425508), this derivative is 0.

The better question is why $x$ and $t$ are independent. To see this, we have to consider what $x$ is; Griffiths doesn’t do such a good job explaining that. $x$ is not the position of some particle. $x$ is just the distance from an origin here. $|\Psi|^2$ is the probability we measure our particle at a point in space, $x$.

Answer 3

As @danielc said, x and t are independent. X, y, z, t are independent coordinates. Griffiths should not even have brought it up.