# Representation of linear operator using series ?

1. May 19, 2013

### 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?

2. May 19, 2013

### micromass

Staff Emeritus
3. May 19, 2013

### MisterX

It's right.

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: