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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?

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

Yes. The outcome of an integral of the form $$\int_{\mathbb{R}} f(x,t) \, dx \tag{1}$$ 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: $$\frac{\text{d}}{\text{d}t} \int_{\mathbb{R}} f(x,t) \, dx = \int \frac{\partial f}{\partial t}(x,t) \, dx$$ as dictated by Leibniz Integral Theorem (do note that I've assumed some weak assumptions on the behaviour of $f$, but its not of incredible interest here). A trivial application of this in $$\langle x \rangle := \int_{\mathbb{R}} x |{\Psi(x,t)}|^2 \, dx$$ yields the desirable result.

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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.

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    $\begingroup$ I suppose strictly speaking in the Schrodinger picture, observables have no implicit dependence on time, but they could have explicit time dependence. The OP might find this question and its answers helpful. $\endgroup$ – Philip Nov 21 '20 at 8:53

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