# Mathematical problem related to spectral method for time evolution

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.
• Maybe I am barking up the wrong tree, but your answer would be ok for me if the rows of $P$ would be $|\phi_1\rangle,\dots,|\phi_N\rangle$, because of the matrix-vector product. Commented Nov 26, 2020 at 0:14
• @Johnpiton It seems as though you have a misunderstanding of how matrix multiplication works. If $v_1,\dots,v_N$ are the columns of a matrix $A$ and $x_1,\dots,x_N$ are the entries of a column vector $x$, then $$Ax = x_1 v_1 + \cdots + x_N v_N.$$ Commented Nov 26, 2020 at 1:46