# Unitary Transformations in Quantum Mechanics - Changes Meaning of Operators?

In my quantum-optics lecture, my Professor wanted to derive some quantities regarding the time evolution of an electromagnetic field, coupled to an electron. He started with the usual Hamiltonian in the Schroedinger Picture: $$(\hat{\vec{p}}-\frac{e}{c} \hat{\vec{A}}(0))^2\frac{1}{2m} + V(\hat{\vec{x}}) + \sum_{\vec{k}, \lambda} \hat{a}_{\vec{k}, \lambda}^\dagger \hat{a}_{\vec{k}, \lambda}$$ All the operators are being time-independent schroedinger-operators there. In order to simplify calculations, he then applied two Unitary transformations, one being $\hat{U} = e^{\frac{e}{\hbar}\hat{i\vec{x}}\hat{\vec{E}}}$, the other one being $\hat{T}(t) = D(-\{\alpha_{\vec{k}}e^{-i \omega_{\vec{k}}t} \})$ with D being the displacement operator for multiple modes. Applying those transformations, the time depence of states changes to: $$i \hbar \frac{d}{dt} | \Psi(t) \rangle = \frac{1}{2m}\hat{p}^2 + V(\hat{\vec{x}}) - e \hat{\vec{x}}(\hat{\vec{E}}(0) + \vec{E}_\mathrm{classical}(0, t) ) | \Psi(t) \rangle$$

Of course I can solve the time evolution of a given state with that, but what bothers me is (and what my question is):

what does this state mean? Since I did 2 transformations, and I want to know for example the position of the electron, the operator for that is no longer $\hat{\vec{x}}$, but instead $\hat{U}^{-1} \hat{T}^{-1} \hat{\vec{x}} \hat{U}^{-1} \hat{T}^{-1}$ Is it true to say something like that? Furthermore, what meaning to eigenstates of $\hat{\vec{x}}$ have, since it is no longer $\hat{\vec{x}}$, measuring position, but instead $\hat{U}^{-1} \hat{T}^{-1} \hat{\vec{x}} \hat{U}^{-1} \hat{T}^{-1}$?