The Schrödinger equation for states is known as $\frac{d}{dt} \psi(t)= – iH(t)\psi(t) $. The solution can be expressed via the time evolution operator $U(t)$ so that $\psi(t)=U(t)\psi(0)$ where now also $U(t)$ satisfies the Schrödinger equation \begin{align} \frac{d}{dt} U(t) = -i H(t) U(t)\end{align}
Applying a time-dependent unitary basis transformation $T(t)$, the Hamiltonian can be shown to transform as \begin{align}
\breve{H} = THT^\dagger + i \dot{T}T^\dagger
\end{align}
yielding a transformed Schrödinger equation \begin{align} \frac{d}{dt} \breve{\psi} = -i\breve{H}(t)\breve{\psi}(t) \end{align} where $\breve{\psi}(t)=T\psi$
Question: Does the time evolution operator solving the transformed Schrödinger equation also satisfy the transformed Schrödinger equation? I.e. $\frac{d}{dt}\breve{U}=-i\breve{H}\breve{U}$?
Background: I am actually interested in introducing a time-dependent basis transformation of the operator basis that $U$ may be written in. Naively, such a basis transform would transform $U$ as $U\mapsto \breve{U} = T U T^\dagger$. However the equation of motion for this $\breve{U}$ takes the form
\begin{align*}
\frac{d}{dt} \breve{U} =& \dot{T}UT^{\dagger}+T\dot{U}T^{\dagger}+TU\dot{T^{\dagger}}\\
= & \dot{T}T^{\dagger}TUT^{\dagger}-iTHUT^{\dagger}+TUT^{\dagger}T\dot{T^{\dagger}}\\
= & \dot{T}T^{\dagger}\breve{U}-iTHUT^{\dagger}+\breve{U}T\dot{T^{\dagger}}\\
= & \dot{T}T^{\dagger}\breve{U}-iTHT^{\dagger}TUT^{\dagger}+\breve{U}T\dot{T^{\dagger}}\\
= & \dot{T}T^{\dagger}\breve{U}-iTHT^{\dagger}\breve{U}+\breve{U}T\dot{T^{\dagger}}\\
= & \dot{T}T^{\dagger}\breve{U}-iTHT^{\dagger}\breve{U}+\breve{U}T\dot{T^{\dagger}}
\end{align*}
and upon replacing $THT^{\dagger}=\breve{H}-i\dot{T}T^{\dagger}$ we find \begin{align*}
\frac{d}{dt}\breve{U}= & \dot{T}T^{\dagger}\breve{U}-i\left(\breve{H}-i\dot{T}T^{\dagger}\right)\breve{U}+\breve{U}T\dot{T^{\dagger}}\\
= & \dot{T}T^{\dagger}\breve{U}-i\breve{H}\breve{U}-\dot{T}T^{\dagger}\breve{U}+\breve{U}T\dot{T^{\dagger}}\\
= & -i\breve{H}\breve{U}+\breve{U}T\dot{T^{\dagger}}
\end{align*}
The final equation of motion for $\breve{U}$ almost looks like the Schrödinger equation with transformed Hamiltonian $\breve{H}$ if it were not for the addition term featuring $T \dot{T^\dagger}$ operated from the right on $\breve{U}$. How does one handle this equation? I am used to factoring out $U$ or $\psi$ in any equation of motion. This does not seem to be possible here.
I am wondering: is there any literature/ mathematical theories treating such a time-dependent transformation?
Best Answer
The transformation for the evolution operator you wrote is wrong. The correct one is, \begin{equation} \breve{U}(t_2,t_1)=T(t_2)U(t_2,t_1)T^\dagger(t_1) \end{equation} note the explicit two-time dependence in case of the time-dependent Hamiltonian, \begin{equation} i\frac{d}{dt_2}U(t_2,t_1)=H(t_2)U(t_2,t_1),\quad i\frac{d}{dt_1}U(t_2,t_1)=-U(t_2,t_1)H(t_1) \end{equation} So in essence, you first transform the initial wavefunction to the old basis at the moment $t_1$, then evolve it to $t_2$ and then transform it to the new basis with a different operator $T(t_2)$. When you differentiate $\breve{U}(t_2,t_1)$ by $t_2$, $T^\dagger(t_1)$ doesn't depend on $t_2$ and therefore does not contribute extra term.