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Considering the standard evolution for a generic quantum state $\psi(t)$, setting $\hbar=1$ we have: $$| \psi(t) \rangle = U |\psi(0) \rangle \hspace{5em} \text{where}\hspace{1em} U=\exp[-iH(t-t_0)] $$ 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: $$| \psi(t) \rangle = U |\psi(0) \rangle = P^{-1}D P |\psi(0) \rangle $$ because of $c_n=\langle \phi_n|\psi(0) \rangle $ $$| \psi(t) \rangle =P^{-1}D \vec{c} \neq D P^{-1} \vec{c}$$ where $\vec{c}=\{c_n\}$ and P is the matrix with on the columns the eigenstates of the hamiltonian H. The commutation relation does not seem to me to be valid.

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    $\begingroup$ What is $P$, what is $\vec{c}$ and why exactly do you think you need to commute $P^{-1}$ past $D$? $\endgroup$ Nov 19 '20 at 11:25
  • $\begingroup$ @BySymmetry I assume $P$ is the similarity matrix (OP wants to diagonalise $U$ for some reason) $\endgroup$ Nov 19 '20 at 11:33
  • $\begingroup$ @NiharKarve How to obtain 1 without doing some diagonalization? It was a try otherwise I do not know how to start, sincerely. $\endgroup$
    – Johnpiton
    Nov 19 '20 at 11:37
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The state $|\psi(t)\rangle$ and $U$ are not expressed in any particular basis, so the notion of diagnoalising does not really make sense. However not being stuck in any particular basis is a good thing! It means we can pick any basis that is convenient for our problem.

In particular we can choose to work in a basis such that $H$ (and so $U$) is already diagonal. Expressing $|\psi(0)\rangle$ in terms of energy eigenstates we obtain $$ |\psi(0)\rangle = \sum_n c_n |\phi_n\rangle\;. $$ Now all we do is simply apply $U$ and use linearity \begin{align} |\psi(t)\rangle =& U|\psi(0)\rangle\\ =& \sum_n c_n H |\phi_n\rangle\\ =& \sum_n c_n e^{-\imath E_n(t-t_0)} |\phi_n\rangle\;. \end{align}

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