Is there a unitary transformation such that the Hamiltonian in the time-dependent Schrödinger equation becomes real symmetric?

The time-dependent Schroödinger equation is given as (with $$\hbar=1$$): $$i\dfrac{d}{dt}\psi(t)=H(t)\psi(t)\ ,$$ where $$\psi$$ is some normalized column vector and $$H(t)$$ is a Hermitian matrix with time-dependent elements.

Let $$\Psi(t)=U(t)\psi(t)$$, where $$U(t)$$ is unitary. It can be shown that the time-dependent Schrödinger equation in terms of $$\Psi$$ can be written as: $$i\dfrac{d}{dt}\Psi(t)=\left[UHU^{-1}-iU\dot{\left(U^{-1}\right)}\right]\Psi(t)\ ,$$ where the overdot indicates element-wise time derivative. Is it possible to find a $$U$$ such that this new Hamiltonian $$UHU^{-1}-iU\dot{\left(U^{-1}\right)}$$ is real symmetric?

A simple solution can be found when $$H$$ is 2 x 2, by assuming that $$U$$ is diagonal. But, this method fails for higher dimensional cases. Can it be done under some special conditions? Can it be done if $$U$$ is invertible, but not necessarily unitary?

• Choose $U$ wih each column is the energy eigen-states of $H$. Essentially, diagonalize the Hamiltonian. – K_inverse Oct 2 '18 at 7:57
• @K_inverse that only works if $U$ does not depend on time, but here the additional $-iU\dot{\left(U^{-1}\right)}$ term messes that up. – Feel My Black Hole Oct 2 '18 at 8:51
• I would expect it to work, since unitary $U$ has $n^2$ degrees of freedom, and making $UHU^\dagger + i \dot U U^\dagger$ real is only $n^2 / 2 - n / 2$ conditions. But I don't have a proof. – Noiralef Oct 2 '18 at 17:22
• Or rather, $n^2 - 1$ because a global phase doesn't do anything. Note that diagonal unitary $U$ has only $n-1$ d.o.f., so it should not work in any $n>2$. – Noiralef Oct 2 '18 at 17:27
• @Noiralef I thought that d.o.f. argument only works if the unknown variables are in a linear system. If the elements of $U$ are not bijective functions, then that argument can easily fail. – Feel My Black Hole Oct 3 '18 at 5:16

Yes, this can be done in general in a kind of trivial way. Let $$U$$ be $$V^\dagger$$, where $$V$$ is the time-evolution operator. The Schrodinger equation gives that $$i \frac{dV}{dt} = H V$$, so $$U H U^\dagger - i U \frac{dU^\dagger}{dt} = V^\dagger H V - i V^\dagger \frac{dV}{dt} = V^\dagger H V - V^\dagger H V = 0,$$ and the zero matrix is certainly real symmetric. This works regardless of whether the Hamiltonian is time-dependent.