# At most explicit time-dependent operator

1. Jun 28, 2014

### carllacan

Hi.

I have a little language problem. I'm studying in Germany, and my German is... nicht sehr gut, so I sometimes have problems understanding the exercises. The one I'm having issues right now has a part which says einen höchstens explizit zeitabhängigen Ope-rator im Schrödingerbild.

My translation for this is "An at most explicitly time-dependent operator in Schrödinger picture." This doesn't make any sense: how can a Schrödinger-picture operator be time dependent? And what does it mean by "at most explicitly"

PD: if you happen to speak German I'd would appreciate if you checked my translation.

PPD: I know I could simply ask the teacher, but... I should have made this exercise way earlier, and I'm kind of ashamed to openly tell him that I am doing it now, so I'd like to avoid it if possible.

Last edited: Jun 28, 2014
2. Jun 28, 2014

### Bill_K

A bit more context would help.

3. Jun 28, 2014

### carllacan

There is not much. I am given the Hamiltonian for an Harmonic oscillator and I am told to find the time-dependence of $O_H= U^{\dagger}(t, t_0)O_SU(t, t_0)$, where $O_S$ is the mentioned "at most explicitly time-dependent operator in Schrödinger picture.", and U is the time-development operator.

Maybe I am to just state the Heisenberg equation of motion for O?

4. Jun 28, 2014

### Bill_K

The Heisenberg equation of motion is $\frac{dO_H}{dt} = \frac{i}{\hbar}[H_H, O_H] + \frac{\partial O_S}{\partial t}$. I think he's saying, keep the $\frac{\partial O_S}{\partial t}$ term.

5. Jun 28, 2014

### carllacan

I've never seen it in that form. I don't understand it, isn't the point of the Schrödinger picture that operators are time-independent?

6. Jun 28, 2014

### Bill_K

Usually, but not always.

https://en.wikipedia.org/wiki/Heisenberg_picture#Mathematical_details

7. Jun 28, 2014