# Interaction picture: why the Hamiltonian describing the dynamic doesn't change with the same law as other observables?

First: what happens in a general change of picture?

If I have the following equation:

$$A | x \rangle = | y \rangle .$$

To do a change of picture is to apply a unitary $$U$$ on all vectors of the Hilbert space.

For any vector in my first picture $$|e_i\rangle$$ there will be a corresponding vector in my new picture : $$|e_i'\rangle = U |e_i \rangle$$.

In the equation I wrote above, if I apply $$U$$ on my equation, I have:

$$U A | x \rangle = U | y \rangle$$

$$U A U^{\dagger} U |x \rangle = U | y \rangle$$

$$A' |x'\rangle = |y' \rangle .$$

I found the law of transformation of observables under change of picture:

$$A \rightarrow U A U^{\dagger}.$$

Thus, in summary, changing of picture is basically doing an active transformation on all the Hilbert space. Then we work in an isomorphic Hilbert space and we do all the physics there.

However, the thing that really confuses me is the interaction picture.

I consider: $$H = H_0 + H_{int}$$. I will go to interaction picture with respect to $$H_0$$.

This means that $$U=e^{+\frac{i}{\hbar} H_0 t}$$. Allright.

So all the observables I consider will change like this:

$$A \rightarrow A'=e^{+\frac{i}{\hbar} H_0 t} A e^{-\frac{i}{\hbar} H_0 t}.$$

The hamiltonian is a just a special case of "any observable", thus it should change with the same law:

$$H \rightarrow H'=e^{+\frac{i}{\hbar} H_0 t} H e^{-\frac{i}{\hbar} H_0 t}=H_0 + H^I_{int}$$

Where I wrote: $$H^I_{int} = e^{+\frac{i}{\hbar} H_0 t} H_{int} e^{-\frac{i}{\hbar} H_0 t}.$$

However, in interaction picture, the Schrodinger equation is:

$$i \hbar \frac{\partial |\psi \rangle}{\partial t} = H^I_{int} |\psi \rangle.$$

I know how we find this new Schrödinger equation, I am just confused with the two contradictory things that:

• The interaction picture hamiltonian should be $$e^{+\frac{i}{\hbar} H_0 t} H e^{-\frac{i}{\hbar} H_0 t}.$$

• When I compute the new Schrödinger equation that interaction picture vectors follow, the Hamiltonian appearing is not this one.

What does that mean physically?

I would say that the "true" Hamiltonian is $$e^{+\frac{i}{\hbar} H_0 t} H e^{-\frac{i}{\hbar} H_0 t}$$ in the sense that it is the one I have to consider if I wonder what the energy of my system is because the eigenvalues are "conserved" under the change of picture.

Indeed, I have:

$$H|\psi \rangle = E |\psi \rangle \rightarrow H' |\psi'\rangle = E |\psi'\rangle.$$

However, the dynamic is not given by this new Hamiltonian. And I find it physically weird as it is the corresponding observable.

Is there a way to understand this "paradox" if I can call it this way...

• The Hamiltonian is not just an observable, it is also the generator of time evolution. So it's hardly surprising that the generator transforms differently under a time-dependent unitary transformation. – Mark Mitchison Feb 19 at 20:47
• @Mark Mitchinson My question is : you have two quantities that we call call hamiltonians in interaction picture. The one describing the evolution H_I and the observable I called H' in my question. H' gives the energy of the system. There is something physical to understand as usually those quantities coincide. Here they dont. How to interpret it physically as you would expect the energy observable being the same as the time translation operator. – StarBucK Feb 19 at 23:14
• What you said is that if you do the calculation it is obvious that the quantity describing the dynamic will not be H'. And I agree with that. However I would like to know if there are some physical interpretation about the fact the time translation operator and the energy observable do not coincide. – StarBucK Feb 19 at 23:18