# Getting the time evolution relationship in QM

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.

• What is $P$, what is $\vec{c}$ and why exactly do you think you need to commute $P^{-1}$ past $D$? Commented Nov 19, 2020 at 11:25
• @BySymmetry I assume $P$ is the similarity matrix (OP wants to diagonalise $U$ for some reason) Commented Nov 19, 2020 at 11:33
• @NiharKarve How to obtain 1 without doing some diagonalization? It was a try otherwise I do not know how to start, sincerely. Commented Nov 19, 2020 at 11:37

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}