# Schrödinger equation for the evolution operator transform under unitary transformation?

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?

The transformation for the evolution operator you wrote is wrong. The correct one is, $$$$\breve{U}(t_2,t_1)=T(t_2)U(t_2,t_1)T^\dagger(t_1)$$$$ note the explicit two-time dependence in case of the time-dependent Hamiltonian, $$$$i\frac{d}{dt_2}U(t_2,t_1)=H(t_2)U(t_2,t_1),\quad i\frac{d}{dt_1}U(t_2,t_1)=-U(t_2,t_1)H(t_1)$$$$ So in essence, you first transform the initial wavefunction to the old basis at the moment $$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.