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I don't understand something about the Heisenberg and interaction picture, in my notes the time evolution of operators for the Heisenberg and interaction picture is derived, by inserting them into the Schrödinger equation...

My question is: Doesn't the SE only give the time evolution of states and not operators?

In my notes the time-dependent state $U .\psi (x,t=0)$ is inserted into the SE, and the time evolution of the $U$ operator is derived ($i \hbar \frac{ \partial U}{\partial t}=H \space U$), the following argument is given at the end:

"Because this ($i \hbar \frac{\partial \space U \psi}{\partial t} = H\space U\psi$) holds for any wavefunction, the equation above must hold also for the operators themselves."

$\uparrow$ I don't understand this.

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It is a change of view. In the Schrödinger picture your operator is not time dependet (or only explicitly time dependent) but your states evolve with time. On the other side in the Heisenberg picture your states are fixed but the operators depend on time. There you use replace the Schödinger equation with the Heienberg equation

$$\frac{d}{dt} A = \frac{i}{\hbar} [H,A] +\partial_t A $$

which describes the same physics.

You can think of it by an classical analogue:

From David J. Toms "The Schwingers action principle and effective action":

There is an analogy between the two pictures in quantum mechanics and the treatment of classical mechanics in a rotating frame which might make things a bit clearer. A vector in a rotating frame can be considered from two points of view. We may either consider the vector to be rotating with respect to a set of fixed axes (analogous to the Schr¨odinger picture with the state vector evolving in time but with the base kets fixed) or consider the vector to be fixed but take the coordinate axes to be rotating (analogous to the Heisenberg picture with the state vector fixed). In order that these two different views describe exactly the same physics, in the second case the axes must rotate in the opposite direction to that of the vector in the first case.

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  • $\begingroup$ I get that... I don't understand how the heisenberg equation is derived... $\endgroup$ – Luka8281 Mar 3 '17 at 10:09
  • $\begingroup$ Look at: en.wikipedia.org/wiki/…. But to get this time evoultuion operator you follow he way you described. Since it should not depend on the choice of $\psi (t_0)$ it gives an equation for $U$. In other words the time evoultion should be a basic principle depending on the system, i.e. $H$, not on a specific wave function. $\endgroup$ – Alpha001 Mar 3 '17 at 10:26

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