Time dependence of operators

Question

In Griffiths's Introduction to Quantum Mechanics, while studying the time evolution of the expectation value of position, the author wrote: $$\langle x\rangle=\int_{-\infty}^{+\infty}x|\Psi(x,t)|^2\,dx.$$

So $$\frac{d\langle x\rangle}{dt}=\int x\frac{\partial}{\partial t}|\Psi(x,t)|^2\,dx.$$

Did he just assume that $x$ has no time dependence? And why?

Answer 1

Did he just assumed that x has no time dependence? And why?

Yes. The outcome of an integral of the form $$\int_{\mathbb{R}} f(x,t) \, \text{d}x \tag{1}$$ is a function of time $t$; that is, a function of one real variable (or, loosely speaking, the integral will evaluate to a quantity that will not depend on $x$, only on $t$). Thus, upon differentiating $(1)$, one would get: $$\frac{\text{d}}{\text{d}t} \int_{\mathbb{R}} f(x,t) \, \text{d}x = \int \frac{\partial f}{\partial t}(x,t) \, \text{d}x$$ as dictated by Leibniz Integral Theorem (do note that I've assumed some weak assumptions on the behaviour of $f$, but it is not of incredible interest here). A trivial application of this in $$\langle x \rangle := \int_{\mathbb{R}} x \, |{\Psi(x,t)}|^2 \, \text{d}x$$ yields the desirable result.

Answer 2

The are two formulations of quantum mechanics :

• Schrödinger representation. The time evolution is encoded in the state vector, wavefunction - $\Psi(x,t)$, and the observables(operators) are constant in time
• Heisenberg representation. Now the operators evolve in time, and the state vectors are time-independent, kept fixed.

In the case of interacting theories there is a hybrid Interaction representation. Here the operators evolve with the non-interacting Hamiltonian $H_0$, and the states evolve via the interaction part $H_I$.

So in your case the author uses the Schrödinger representation.