Why Does the Domain Transform to Keep the Time Dependent Operator Self Adjoint?

[tommy01]

Hi together ...

I encountered the following statement:
Operator A is self adjoint on D(A) then $$A(t) \equiv \exp(iHt) A \exp(-iHt)$$ is self adjoint on $$D(A(t)) \equiv \exp(-iHt) D(A)$$.

H is self adjoint, so that exp(...) is a unitary transformation. But why does the domain transform this way to keep the time dependent operator self adjoint? I don't get an expression like $$(\Psi,A(t)\Phi)=(A(t)\Psi,\Phi) ~~~ \Psi, \Phi \in D(A(t))$$ if i use the definitions.

greetings.

 

[dextercioby]

Of course you don't get the last expression, because the "t" dependence on the vectors $\Psi$ and $\Phi$ is missing. With this <t> put into place, your expression should be

$$\left\langle U(t)\Psi, U(t)AU^{\dagger}(t) U(t)\Phi\right\rangle$$ which shows the needed invariance and s-a of the <evolved> operator.

The domain of the A(t) should be U(t)D(A), of course. The range of A(t) should be U(t) Range(A).

EDIT: Ammended by the post #4 of this thread below.

 

[tommy01]

Thanks, but in the statement i quoted the domain of A(t) isn't U(t)D(A) but $$U(t)^+ D(A)$$ and then we have $$(U(t)^+ \Psi, U(t)AU(t)^+ U(t)^+ \Phi)$$. Is this a typo?

 

[dextercioby]

The issue in post #1 is addressed by the Lemma 4.3, page 225 of E. Prugovecki's "Quantum Mechanics in Hilbert Space".

 

[tommy01]

Thanks a lot for the hint. So it was an error in the book. For the record: B. Thaller - The Dirac Equation. Section 1.2.2.

Greetings.
Tommy

 

[dextercioby]

Please, see the screenshot attached. It's a small error by Thaller.