# How come there are Schrödinger Picture operators with explicit time dependence?

In the Schrödinger picture, observables are said to be time independent (see Cohen, for example) operators. However, when deriving the Heisenberg Equation of Motion $$i\hbar\frac{d}{dt}A_H(t)=[A_H(t),H_H(t)]+i\hbar\Big(\frac{\partial}{\partial t}A_S(t)\Big)_H.$$ a term with an explicit time dependence of the operator in the Schrödinger picture appears. I looked at other related questions and some argued that in the S-picture, only operators that are related to observables are time-independent. Is this really the case? If so, is this equation a general description of dynamics of operators and reduces to $$i\hbar\frac{d}{dt}A_H(t)=[A_H(t),H_H(t)]$$ if $$A_S$$ is an observable? Furthermore, is the existence of (explicit) time dependence equivalent to time evolution?

• Which other related question? Nov 18, 2018 at 12:26
• @Qmechanic this one physics.stackexchange.com/q/351020 Nov 18, 2018 at 13:27

In the Schrodinger picture operators carry no time-dependence introduced by a unitary transformations (as opposed to the Interaction or Heisenberg picture). Operators in the Schrodinger picture can still have a time dependence if something is physically changing$$^*$$. An example of this is if we have a particle in a time dependent electric field. The Hamiltonian will have time dependence due to the field actually changing, not because of a unitary time evolution (if we treat the field as external to the system). The eigenvalues (possible measurement outcomes) in this case can have a time dependence.

So, in the Schrodinger picture, unitary transformations are what cause the state vector to change over time. Schrodinger operators do not experience "time evolution" in this way; if an operator does have explicit time dependence, the time-dependence must be due to something observable physically changing as described earlier. Therefore, the physical time-dependence in Schrodinger operators due to measurable observables is distinct from the time evolution introduced by unitary transformations in other pictures. Thus explicit time-dependence in the S-picture is not equivalent to time evolution.

$$^*$$ This seems to have gained some confusion here. I am not saying that unitary transformations do not have physical consequences. I am saying they do not represent physical changes themselves; they only represent changes in the probability of measuring the system to be in some state. State vectors and operators are not physical things, so unitary transformations that cause them to change are not direct physical changes. On the other hand, in my example fields are physical, directly measurable things. In the Schrodinger picture operators that depend on the field can have explicit time dependence, and so can the eigenvalues associated with those operators.

• The "time dependence" of states and operators "due to unitary transformations" is exactly the difference between each picture, right? And these unitary transformations are time evolution, right? What I can't seem to grasp is the difference between this time dependence and an explicit time dependence. Explicit time dependencies cannot be seen as time evolutions? What is the difference between them, physically speaking? Nov 18, 2018 at 3:40
• @JoãoPedroGomide are you asking what the physical difference is between the Shrodinger and Heisenberg pictures are? Nov 18, 2018 at 4:37
• @JoãoPedroGomide Or are you asking for a physical explanation of a unitary transformation/evolution? Nov 18, 2018 at 4:51
• This is very similar to the difference between explicit and implicit time dependence of the Lagrangian in classical mechanics. Here by implicit I mean the dependence through the time-dependence of coordinates and velocities on the path that solves Euler-Lagrange equations. Nov 18, 2018 at 12:28
• @AaronStevens I was asking for a "physical" explanation of a unitary transformation/evolution. But both your edit and Andrew Steane's comment made it really much more clear. Thank you! Nov 18, 2018 at 13:31

It is true that explicitly time-dependent observables (in the Schroedinger picture) may be due to time-changes in the environment. However, observables that explicitly depends on time pop out naturally also for isolated systems whose (finite dimensional) Lie group of symmetries $$G$$ includes time evolution $$U(t)=e^{-iH}$$ and the group is not the trivial product of time evolution and an independent subgroup. In this case, there are observables (selfadjoint operators) $$B$$ in the Lie algebra $$\mathfrak{g}$$ which do not commute with the (selfadjoint) Hamiltonian $$H$$. These observables give rise to other observables, still in the Lie algebra, which explicitly depend on time and are conserved quantities.

(All I am saying is rigorously valid in a common dense domain of all operators where they are essentially selfadjoint, for instance the Nelson or Garding domain. I will not indicate the domain explicitly.)

This situation is common in classical and relativistic quantum theory, even if the system is isolated. The typical observable of the Lie algebra of symmetries is the boost generator.

The boost generator $$B$$ (or some other observables constructed out of the boost and other generators of the Lie algebra of symmetries) cannot commute with the Hamiltonian just because the Hamiltonian depends on the used reference frame which is changed by the boost transformation.

On the other hand, since the group $$G$$ is finite dimensional and the action of the time evolution preserves the Lie algebra structure, it must be $$U(t)^\dagger B U(t)= \sum_{k=1}^n c_k(t) A_k$$ for some real coefficients $$c_k(t)$$, where the $$A_k$$ are a basis of observables of the Lie algebra.

If we define the observable that explicitly depends on time $$B(t) := \sum_{k=1}^n c_k(-t) A_k$$ we have the conservation law $$B(t)_H = U(t)^\dagger B(t) U(t)= B(0)\:. \tag{1}$$

This relation can be written into (the horrible differential version where a number of useless mathematical details should be fixed in comparison with the adamantine (1)) (I use $$\hbar=1$$)

$$\frac{d}{dt} B(t)_H = -i[B(t)_H, H] + \frac{\partial B_H(t)}{\partial t}=0$$

When $$B$$ is the boost generator, $$B(t)$$ is commonly called the boots at time $$t$$. For instance, in relativistic theory, $$B_a(t) = B_a -tP_a\quad \mbox{where a=x,y,z.}$$ Actually, the same identity holds also in the classical case, where now $$P_a$$ is the momentum and not the spatial components of the four-momentum.

Let's consider a typical derivation of the Fermi Golden Rule, which starts with a Hamiltonian like this: $$H = H_0 + V\cos(\omega t)$$ ...which has the explicit time dependence.

Yes, in some ways it is introduced by hands, we wouldn't have it if we considered an atom coupled to a quantized Em field, although even then it could slip in via boundary conditions or other parameters, characterizing coupling to environment, measurement device, etc.