Consider a general-form time-dependent Schrödinger equation:

$$i\partial_tv=\hat Hv,$$

where Hamiltonian $\hat H$ is an Hermitian matrix (finite-dimensional for simplicity), and $v(t)$ is a complex vector.

Now suppose we can easily solve this equation for any real symmetric $\hat H$, but not for general complex case. Is there any way to somehow transform $\hat H$ to such a form $\hat H'$ (which may have different dimensions), that the equation preserves its form, but $\hat H'$ is real symmetric?

I know the trick of $a+ib\to\begin{pmatrix}a&-b\\ b&a\end{pmatrix},$ but if I apply it to $\hat H$, I'll have to also replace $i$ on the left side of equation by $\begin{pmatrix}0&-1\\1&0\end{pmatrix}$, so the equation changes its form, which is undesirable.


We have

\begin{equation*} \hat{H}=\sum_{n}\lambda _{n}\left|u_{n}\right\rangle\left\langle u_{n}\right| \end{equation*} $\{u_{n}\}$ is in general a basis of complex eigenvectors. Let $\{w_{n}\}$ a second basis with real eigenvectors. Then there is a unitary operator $U$ such that \begin{equation*} u_{n}=Uw_{n} \end{equation*} Now \begin{equation*} \hat{H}=U\sum_{n}\lambda _{n}\left|w_{n}\right\rangle\left\langle w_{n}\right|U^{-1}=UH_{r}U^{-1} \end{equation*} and \begin{equation*} i\partial _{t}U^{-1}v=H_{r}U^{-1}v \end{equation*}

where $H_r$ is real.

Although this works in principle, finding $U$ may be difficult.


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