Representation of linear operator using series ?

[V0ODO0CH1LD]

representation of linear operator using "series"?

I was looking into the progression of quantum states with respect to time. From what I understood the progression of a state $\left|\psi(t)\right>$ is given by:
$$ \left|\psi(t)\right> = U(t)\left|\psi(0)\right> $$
I'm not sure if that's right. But it's okay if I haven't got that yet.. That was just to give some context to my actual question.

What is this representation of the linear operator $U(t)$ at $t = \epsilon$, where $\epsilon$ is an infinitesimal?
$$ U(\epsilon) = I - i\epsilon H $$
Where $i$ is the imaginary unit, $I$ is the identity matrix and I think $H$ is the hamiltonian.
It also apparently has more terms of order $\epsilon^2$ and so on. What "series" is this? Is it some first order approximation of $U(t)$? What should I look into to understand where those terms are coming from?

 

[micromass]

This seems relevant: http://en.wikipedia.org/wiki/Stone's_theorem_on_one-parameter_unitary_groups

 

[MisterX]

I was looking into the progression of quantum states with respect to time. From what I understood the progression of a state $\left|\psi(t)\right>$ is given by:
$$ \left|\psi(t)\right> = U(t)\left|\psi(0)\right> $$
I'm not sure if that's right.

It's right.

What is this representation of the linear operator $U(t)$ at $t = \epsilon$, where $\epsilon$ is an infinitesimal?
$$ U(\epsilon) = I - i\epsilon H $$
Where $i$ is the imaginary unit, $I$ is the identity matrix and I think $H$ is the hamiltonian.
It also apparently has more terms of order $\epsilon^2$ and so on. What "series" is this? Is it some first order approximation of $U(t)$? What should I look into to understand where those terms are coming from?

Those are the first two terms of the Taylor series expansion for $e^{-i\mathbf{H}\epsilon}$. It just so happens that $U(t) = e^{-i\mathbf{H}t/\hbar}$. This can be related to the idea that $E = \hbar \omega \rightarrow \omega = E/\hbar$. A state $\mid n\rangle$ with definite energy $E_n$ has the following time evolution.
$\mid n(t)\rangle = e^{-i\omega t}\mid n(0)\rangle = e^{-iE_n t/\hbar}\mid n(0)\rangle$

A function of an operator like $e^{-i\mathbf{H}t/\hbar}$ has the same eigenvectors as the operator, but the eigenvalues are replaced with the corresponding function of the eigenvalues.

Here are some relevant video lectures:
https://www.youtube.com/watch?v=cVbB6wFNqYc&list=PL9LRV0x7N1NCpd-ZerPxiTzm97y9VGPLp
https://www.youtube.com/watch?v=31XrxGMRwtw&list=PL9LRV0x7N1NCpd-ZerPxiTzm97y9VGPLp