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At most explicit time-dependent operator

  1. Jun 28, 2014 #1
    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"

    Thank you for your time.

    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. jcsd
  3. Jun 28, 2014 #2

    Bill_K

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    A bit more context would help.
     
  4. Jun 28, 2014 #3
    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?
     
  5. Jun 28, 2014 #4

    Bill_K

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    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.
     
  6. Jun 28, 2014 #5
    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?
     
  7. Jun 28, 2014 #6

    Bill_K

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    Usually, but not always.

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

     
  8. Jun 28, 2014 #7
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