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I don't understand the difference between the Schrödinger picture and the Heisenberg picture in quantum mechanics. Here's some of my doubts:

  1. If in the Heisenberg picture state vectors are constant in time and in the Schrödinger picture operators are constant in time, in which picture am I if I study a system where the potential is time dependent, such as the interaction between an Hydrogen atom and radiation? In this particular example, the Schrödinger equation is $$i\hbar \dfrac{\partial}{\partial t}\Psi(\vec{r},t)=\left( -\dfrac{\hbar^2}{2m}\nabla^2 -\dfrac{e^2}{4\pi\varepsilon_0r} -\dfrac{i\hbar e}{m}\vec{A}(\vec{r},t)\cdot \vec{\nabla} +\dfrac{e^2}{2m}A^2(\vec{r},t) \right)\Psi(\vec{r},t) $$ where $\vec{A}(\vec{r},t)$ is the vector potential. Here both the wave function and the operators which contain the vector potential are time dependent.

  2. According to this wikipedia page https://en.wikipedia.org/wiki/Heisenberg_picture an operator in the Heisenberg picture satisfies the equation $$\frac{d}{dt}A_\text{H}(t)=\frac{i}{\hbar}[H_\text{H},A_\text{H}(t)]+\left( \frac{\partial A_\text{S}}{\partial t} \right)_H,$$ where the $H$ and the $S$ indicate the operator in the Schrödinger or in the Heisenberg picture. In this equation appears the time derivative of an operator in the Schrödinger picture, which I don't understand since operators in the Schrödinger picture are time independent.

Qmechanic
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Rhino
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    Your equation is in the Schrödinger picture with an explicitly time dependent Hamiltonian. The $A$ in the second equation is just a place holder for any operator and is not necessarily the same as the one in the first equation. An analogy is the following. Imagine you want to take pictures of an object in 360 degree. You could rotate the object(state of the system) while holding your camera(<- operator) fixed and take the snapshots this would correspond to the Schrödinger picture. In the Heißenberg picture you would hold the object fixed while letting the camera rotate around the object ... – Hans Wurst Sep 13 '21 at 17:44
  • ... This is of course a simplified analogy but I hope it gets the idea across. The rotation corresponds to time evolution, which is facilitated by the time evolution operator. – Hans Wurst Sep 13 '21 at 17:46
  • Possible duplicates: https://physics.stackexchange.com/q/11264/2451 and links therein. – Qmechanic Sep 13 '21 at 18:05

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The operator equation here has close parallels in Hamilton equations of motion in classical mechanics (up to replacement of the commutator by a Poisson bracket with appropriate coefficient): $$ \frac{d}{dt}f(p,q,t)=\frac{\partial}{\partial t}f + \{f,H\}, $$ which also distinguish the explicit time dependence of the operators from their inherent time dependence due to the Hamiltonian evolution.

Thus, regaring the point 1. - we have to distinguish the explicit time dependence that appears, e.g., in operator $\mathbf{A}(\mathbf{r},t)$ even in the Schrödinger picture, and the implicit time dependence that will arize only in the Heisenberg picture.

Regarding the second point - I believe that the subscript $S$ in the term with the partial derivative is an error. In fact the equation is a closed equation for the Heisenberg operators.

I would suggest working first through simple examples, such as free particle: $H=\frac{p^2}{2m}$ and harmonic oscillator $H=\frac{p^2}{2m}+\frac{m\omega^2x^2}{2}$,w here the equations for the operators $x,p$ can be explicitly written and solved in the Heisenberg picture.

Roger V.
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  • It's clear to me the difference between implicit and explicit time dependence for a function in classical mechanics, but what is the difference between implicit and explicit time dependence for an operator? In what sense can an operator have implicit time dependence? – Rhino Sep 13 '21 at 15:41
  • @Rhino it is the same as in classical mechanics. If you take a Hamiltonian without explicit time dependence, then in Schrödinger picture operators are time-independent - all the time dependence is in the wave function. In Heisenberg picture, the wave function is time-independent, but the operators obey the same equations as if the Hamiltonain was a classical one (if we speak about position and momentum). – Roger V. Sep 13 '21 at 15:44