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?

I have confirmed that it is possible for the 3x3 case by doing a brute-force computation using the parametric form for a 3x3 special unitary matrix.

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    $\begingroup$ Choose $U$ wih each column is the energy eigen-states of $H$. Essentially, diagonalize the Hamiltonian. $\endgroup$
    – K_inverse
    Oct 2, 2018 at 7:57
  • $\begingroup$ @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. $\endgroup$ Oct 2, 2018 at 8:51
  • $\begingroup$ 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. $\endgroup$
    – Noiralef
    Oct 2, 2018 at 17:22
  • $\begingroup$ 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 $n>2$. $\endgroup$
    – Noiralef
    Oct 2, 2018 at 17:27
  • $\begingroup$ @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. $\endgroup$ Oct 3, 2018 at 5:16

1 Answer 1


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.


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