I'm working through a book on Quantum Computing. The section is regarding the Time Evolution Postulate of quantum mechanics, and it has sort of thrown me a curveball. Given time-independent Schrödinger equation: $${d| {\psi(t)} \rangle \over dt} = -\frac{i}{\hbar} H (t) | {\psi (t)} \rangle $$
and supposed particular solution: $$ | {\psi (t_2)} \rangle = e ^ {-\frac{i}{\hbar} H (t_2-t_1)} | {\psi (t_1)} \rangle $$
Show that the above particular solution is indeed a solution to the time-independent Schrödinger equation.
My most fruitful attempt follows:
Noting, that $(t_2 - t_1)$ is not a parameter to H, but rather a duration scalar, and given the Hamiltonian, H, is taken to be constant (re-written K):
Taking the natural logarithm and limit of both sides:
$$ log( | {\psi (t_2)} \rangle ) - log( | {\psi (t_1)} \rangle ) = -\frac{i}{\hbar} K (t_2 - t_1)$$
some scalar division and the limit statement yields:
$$ \lim_{t_1 \rightarrow t_2} {{log( | {\psi (t_2)} \rangle ) - log( | {\psi (t_1)} \rangle )} \over {t_2 - t_1}} = -\frac{i}{\hbar} K $$
based on definition of derivative of vector-valued function is then:
$$ {d \space log( | {\psi (t)} \rangle ) \over {dt}} = -\frac{i}{\hbar} K $$
It would seem that my liberal application of basic algebra has given me a vector that is somehow equal to a matrix.
I'm stuck here, I'm not sure how to proceed, and I'm certain what I've written thus far is even valid, since I obviously haven't drilled down very far into vector calculus for formality. Nothing even remotely similar is present throughout the book (i.e., moving from the specific case we show here back up to the broader generality present in the Schrödinger equation), but nevertheless I want to be thorough because it is a subject I'm very interested in. Obviously the author is trying to help me demonstrate to myself that the quantum information case we are interested in is indeed a subset of the more general claims Quantum Mechanics ultimately makes (i.e., discontinuous time steps, and finite-dimensional state vectors). Help!
Given $A\in\mathbb C^{n\times n}$ and $x\in\mathbb C^n$, $e^{A}x\in\mathbb C^n$ is well-defined but you tried to apply the multiplication identity $\log(ab)=\log(a)+\log(b)$, $a,b\geq 0$ which does not work here for multiple reasons. The most obvious probably is that the sizes do not match up: $$ \underbrace{\log(e^Ax)}_{\in\mathbb C^n}=\log(e^A)+\log(x)=\underbrace{A}_{\in\mathbb C^{n\times n}}+\underbrace{\log(x)}_{\in\mathbb C^n}\quad $$ Now if the Hamiltonian $H(t)=H$ is constant then the solution of your ordinary differential equation is given by $|\psi(t)\rangle=e^{-\frac{i}{\hbar}Ht}|\psi(0)\rangle$ or, more generally, $|\psi(t)\rangle=e^{-\frac{i}{\hbar} H(t-t_0)}|\psi(t_0)\rangle$ for any initial time $t_0\geq 0$. This is readily verified by differentiating: $$ \begin{align} \frac{d}{dt}|\psi(t)\rangle&=\frac{d}{dt}e^{-\frac{i}{\hbar} H(t-t_0)}|\psi(t_0)\rangle=\Big(\frac{d}{dt}e^{-\frac{i}{\hbar} Ht}\Big)\overbrace{e^{\frac{i}{\hbar} Ht_0}|\psi(t_0)\rangle}^{\text{independent of }t}\\ &=(-\frac{i}{\hbar} H)e^{-\frac{i}{\hbar} H(t-t_0)}|\psi(t_0)\rangle=-\frac{i}{\hbar} H|\psi(t)\rangle \end{align} $$ Indeed because the matrix $H$ in $t\mapsto e^{tH}$ acts like a constant the time-derivative is the same as if $H$ was a scalar.
Side note: if $H=H(t)$ depends on time (in a differentiable manner) then the solution, formally, is given by $$ |\psi(t)\rangle=e^{-\frac{i}{\hbar}\int_{t_0}^t H(t')\,dt'}|\psi(t_0)\rangle\,. $$ Actually for this expression to be correct one needs a time-ordering operator applied to the exponential, but that is way beyond what you asked about here.