I have a mathematical problem related to spectral method for time evolution in QM.
Considering the standard evolution for a generic quantum state $\psi(t) \in \mathbb{C}^N $ which is not express in a basis where U is diagonal, setting $\hbar=1$ we have: $$| \psi(t) \rangle = U |\psi(0) \rangle \hspace{5em} \text{where} \hspace{2em} U^\dagger U = UU^\dagger = 1_N \hspace{1em} \text{and} \hspace{1em} U=\exp[-iH(t-t_0)] $$ it is known that in the basis of the eigenvectors it can be written as: $$| \psi(t) \rangle =\sum_{n}c_n \exp[-iE_n(t-t_0)] | \phi_n \rangle \hspace{5em} (1)$$ I'm stuck in derive this simply expression, supposing to store the eigenvectors in the columns of P: $$| \psi(t) \rangle = U |\psi(0) \rangle = P D P^{\dagger} |\psi(0) \rangle $$ because of $ P^{\dagger} |\psi(0) \rangle=\vec{c} \hspace{2em} \text{where} \hspace{2em} c_n=\langle \phi_n|\psi(0) \rangle $ $$| \psi(t) \rangle =P D \vec{c} \neq \vec{c} D P $$ The commutation relation beacuse of the elements on the diagonal are different does not seem to me to be valid. I'm stuck
So indeed, we have $$ | \psi(t) \rangle = PD \vec c. $$ Note that the diagonal entries of $D$ are $\exp(-iE_n(t-t_0))$. So, $D\vec c$ is the vector $$ D\vec c = \pmatrix{c_1\exp(-iE_1 (t-t_0))\\ \vdots \\ c_N \exp(-iE_N(t-t_0))}. $$ Since the columns of $P$ are $|\phi_1\rangle,\dots,|\phi_N\rangle$, it follows that $$ PD\vec c = \sum_{n=1}^N c_n \exp(-iE_n(t-t_0)) \cdot |\phi_n\rangle, $$ which is what we wanted.