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The physics states in Quantum mechanics is represented by vectors in Hilbert space, however in Heisenberg's picture, the equation of motion

$$ \frac{d}{dt}A_H(t) = \frac{i}{\hbar}[H,A_H(t)]+\frac{\partial}{\partial t}A_H(t) $$

only deals with operators' time evolution.

I am confused that what is the physical state in Heisenberg picture?

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  • $\begingroup$ You may find this post useful physics.stackexchange.com/questions/173219/… $\endgroup$
    – user929304
    Commented May 18, 2016 at 12:13
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    $\begingroup$ If you downvote this question, please at least tell me why. $\endgroup$
    – Shing
    Commented May 18, 2016 at 14:59
  • $\begingroup$ This paper describes the way reality works in terms of the Heisenberg picture arxiv.org/abs/quant-ph/0104033. The PDF may not render properly in Chrome, but it will work in a pdf viewer. $\endgroup$
    – alanf
    Commented Jun 13, 2019 at 12:27

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If you ask about what plays the role of the state of the system at some given time than the answer is: nothing. You talk only about the initial state and what you get in the measurements (expectation values or probabilities of outcomes for observables). The Heisenberg picture is very "Copenhagen" in its spirit and abstracts itself from what's happening with the system itself.

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  • $\begingroup$ Thanks for answering, I am quite satisfied by your answer, but I do not understand how operators themselves can't be physical states? $\endgroup$
    – Shing
    Commented May 19, 2016 at 8:40
  • $\begingroup$ @Shing For starters, I can use the same operators no matter what was the initial state of the system (and thus the corresponding Schrodinger state at that time) $\endgroup$
    – OON
    Commented May 19, 2016 at 17:14
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    $\begingroup$ @Shing I think there's very simple idea that helps very much to get various "strange" stuff about quantum theory (like e.g. about identical particles). Classically we used to think about observables as the properties of the system. But when we do quantum theory the observables are the measurements we do. $\endgroup$
    – OON
    Commented May 19, 2016 at 17:26
  • $\begingroup$ I see it now... indeed, we can't take measurement as the physical states of the particles. Thanks for the elaboration :) $\endgroup$
    – Shing
    Commented May 20, 2016 at 9:20
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The physical states in the Heisenberg picture are frozen in time, and can be made to coincide with the Schrodinger-picture state at any given time $t_0$. In other words, $$|\psi(t_0)\rangle^S=|\psi\rangle^H$$ which doesn't evolve with time.

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