# What is the physical states in Heisenberg picture?

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?

• You may find this post useful physics.stackexchange.com/questions/173219/… – user929304 May 18 '16 at 12:13
• If you downvote this question, please at least tell me why. – Shing May 18 '16 at 14:59
• 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. – alanf Jun 13 at 12:27

## 2 Answers

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.

• Thanks for answering, I am quite satisfied by your answer, but I do not understand how operators themselves can't be physical states? – Shing May 19 '16 at 8:40
• @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) – OON May 19 '16 at 17:14
• @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. – OON May 19 '16 at 17:26
• I see it now... indeed, we can't take measurement as the physical states of the particles. Thanks for the elaboration :) – Shing May 20 '16 at 9:20

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.