# Intrinsic time-dependence

1. Jun 28, 2014

### carllacan

There are many things in this sentence that I don't get
What does exactly mean intrinsic dependenc? I think I know the answer, but I'd like a formal definition, if possible.

Now, the sentence come up in the context of Schrödinger / Heisenberg picture. How is it possible for an Shcrödinger-picture operator to be time-dependent, be it explicitly or implicitly?

2. Jun 29, 2014

### Bill_K

Here's an example. Say an atom has the Hamiltonian H = H0. It's a function of whatever internal atomic variables you want to use, H(p, q) say.

Put the atom in a uniform external electric field and the Hamiltonian becomes H = H0 + Ez. Npw suppose the electric field is time varying, E = E0 cos ωt. Then the Hamiltonian H = H0 + E0z cos ωt is explicitly time-dependent.
It's now H(p, q, t).

3. Jun 29, 2014

### carllacan

And how does that differ from the time-dependence of a Heisenberg operator?

4. Jun 29, 2014

### Matterwave

The time-dependence of a Heisenberg operator has 2 parts in it. 1 part is the explicit time dependence, as Bill_K noted, and the other part comes from the time evolution due to the Hamiltonian. The Heisenberg picture is one in which the state vectors are not time dependent, they stay still, so all the time evolution, explicit and implicit, has to be encoded into the operators.

If in Bill_K's post, you had H(p,q)=H0 and no explicit time dependence, there are still operators which ARE time dependent if they don't commute with H.

5. Jun 29, 2014

### WannabeNewton

In the Schrodinger picture, $\frac{d}{dt}A_S = \frac{\partial}{\partial t}A_S$ which is what we mean when we say a Schrodinger operator is at most explicitly time dependent. The $\frac{d}{dt}$ is a total derivative and is to be contrasted with the partial derivative $\partial_t$.

To touch base with something you're probably already familiar with, consider the velocity field $\vec{v}(t,\vec{x})$ of some classical fluid. Then $\frac{d}{dt}\vec{v} = \frac{\partial}{\partial t}\vec{v} + (\frac{d \vec{x}}{dt}\cdot\vec{\nabla}) \vec{v} = \frac{\partial}{\partial t}\vec{v} + (\vec{v}\cdot\vec{\nabla}) \vec{v}$; what $\frac{d}{dt}\vec{v}$ does is follow the dynamical time evolution of $\vec{v}$ along the flow lines of $\vec{v}$ which means it considers not only the explicit time dependence of $\vec{v}$ but also the dynamics of $\vec{v}$ governed by the evolution of $\vec{x}$ under forces (Newton's 2nd law), whereas $\partial_t$ simply considers the explicit time dependence of $\vec{v}$. The operator $\frac{d}{dt}$ is also called the convective derivative in fluid mechanics.

In that same spirit, for a Heisenberg operator we have $\frac{d}{dt}A_H = \frac{\partial}{\partial t}A_H + \frac{i}{\hbar}[H,A_H]$ so $\frac{d}{dt}A_H$ tracks the dynamical evolution of $A_H$ whereas $\partial_t$ just evaluates the explicit time dependence of $A_H$. On the other hand a Schrodinger operator is non-dynamical so $\frac{d}{dt}$ only evaluates the explicit time dependence of the operator. Therefore when we say that the states $|\psi \rangle$ contain all the time-dependence in the Schrodinger picture what we really mean is, in the Schrodinger picture it is the states that are dynamical i.e. the operators are non-dynamical; in other words the states evolve dynamically under the Schrodinger equation $i \hbar \frac{d}{dt}|\psi \rangle = H |\psi \rangle$ whereas there are no equations of motion for the Schrodinger operators $A_S$ as the only quantity of relevance for them is $\frac{\partial}{\partial t}A_S$. In the Heisenberg picture it is the other way around.

6. Jun 29, 2014

### carllacan

OK, I think I get the idea.

Thank you very much, all of you.