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Schrodinger's picture is that operators are time-independent. But time evolution operator $U(t)$ is time-dependent. Why is that?

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Other people have given great answers, I just wanted to add my 2 cents in case that combined with the other answers helps you out.

In the Schrodinger picture, the claim is not that all operators are time-independent. Rather, the claim is that: states evolve with time, Observable operators are time-independent. So my state in the Schrodinger picture $|\psi_S\rangle$ will depend on time: $|\psi_S\rangle = |\psi_S(\vec{r}, t)\rangle$ and my Observable operators will not depend on time: $S_x \neq S_x(t)$.

However, operators that correspond to non-observables can depend on time. $U(t)$ is not an operator corresponding to an observable (this is obvious since $U(t)$ is not Hermitian).

In the Heisenberg picture, it is states that don't evolve with time and Observable operators that do.

Hope that helps!

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The time evolution operator $U(t)$ is not really an operator observable in the Schrödinger nor the Heisenberg pictures per se, but rather an intertwining operator between the two pictures.

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As an operator, the time evolution operator reflects the time evolution of a state ket. Suppose you have a state ket given by $\vert\alpha,t_0\rangle$ at $t=t_0$. The time evolution operator tells you the evolution of the state ket at a later time $t$, $\vert\alpha,t_0;t\rangle$.

For the simple case in which the Hamiltonian of the system is time-independent, the time-evolution operator has the form

$$U(t,t_0)=exp\left[\frac{-iH(t-t_0)}{\hbar}\right]$$

which follows from the Schrodinger equation for the time-evolution operator:

$$i\hbar\frac{\partial}{\partial t}U(t,t_0)=HU(t,t_0)$$

As you can see, the time evolution operator does not depend on the absolute values of either $t$ or $t_0$, instead it depends only on the difference between the two. So, the time evolution operator has no explicit dependence on time.

That is, given the time interval in which the evolution happens, the matrix elements of the operator with respect to base kets of the Hamiltonian are not time-varying. You may be confused with the term $t-t_0$ in the operator. This is necessary there because it tells about the time evolution that happens to the state ket over that time interval, provided you know the initial state of the system. But, within that particular time interval, the operator form will not change.

An analogy can be made with the case of stationary kets. As the name indicates, they do not change with respect to time. However, if you analyse the wavefunction,

$$\psi(x,t)=\psi(x)e^{-\frac{iHt}{\hbar}}$$

there is a time-dependent part. So does this mean that the stationary state is evolving in time? It's a somewhat yes. Although the states themselves evolve according to time, any measurable quantity or the expectation values of observable are still time-independent. How is this possible? The time-dependent part can only give rise to a change in he phase factor- that's all.

In short, for a given time interval of evolution, the wavefunction, in general, changes its form, but the operator will not. This is exactly the Schrodinger picture.

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  • $\begingroup$ what you mean is, depending on time interval is not the same thing with the time-dependent? So, can I think of the time interval as another variable not time? $\endgroup$ – Orient Aug 9 '17 at 6:07
  • $\begingroup$ No, for the time interval for which you define the time evolution operator, it remains the same. The operator suffers no change at all. $\endgroup$ – UKH Aug 9 '17 at 7:15
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Conceptually, it's usually better to think of the time-evolution operator $U(t_f, t_i)$ as depending on two times: an initial and a final one. When we write it in the form $U(t)$ as depending on a single time, we are either (a) letting $t := t_f$ and implicitly setting $t_i$ to some understood reference time such as $t_i = 0$, and/or (b) if the system is time translationally invariant, then $U$ only depends on the difference $t_f - t_i$, and we are denoting this difference by $t$. (Case (b) can be thought of as a special case of case (a).) While we can often get away with doing this in many useful situations, strictly speaking it only applies in special cases.

For trying to understand general properties of the time-evolution operators, it's better to consider the general case with two times and think of $U(t_f, t_i)$. When we do so, we see that the time evolution operator doesn't actually have any specific value at any given instant in time - instead, it inherently describes the relationship between two different times. It's more like a displacement vector than a position vector. (Even in the time-translationally invariant case, the one-argument version $U(t_f - t_i)$ is technically defined over a one-dimensional affine space, rather than over a one-dimensional vector space as you might expect.)

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