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.

  • 1
    $\begingroup$ 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

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.