# Exponential of operator

1. Jul 31, 2012

### KFC

Hi there,
I learn from the text that the exponential of an operator could be expanded with a series such that

$e^{\hat{A}} = \sum_{n=0}^{\infty} \frac{\hat{A}^n}{n!}$
So if the eigenvalue of the operator $\hat{A}$ is given as $a_i$
$e^{\hat{A}}|\psi\rangle$

will be a matrix with diagonal elements given as $$\exp(a_i)$$, is that right?

So I am wondering what happen if $\hat{A}$ is now written as a power form, i.e. $\hat{A}^n$, can we conclude that
So if the eigenvalue of the operator $\hat{A}$ is given as $a_i$

$e^{\hat{A}^n}|\psi\rangle$

gives a matrix with diagonal elements as $\exp(a_i^n)$ ?

Last edited: Jul 31, 2012
2. Jul 31, 2012

### Staff: Mentor

That's correct - for operators that can be diagonalised.

Its basically how functions of diagonalisable operators are defined - when diagonalised its a function of the eigenvalues.

Non diagonalisable operators are however a bit trickier - there are a few definitions I have seen and even a paper showing they are basically all equivalent - but the point is they are a more difficult proposition.

Thanks
Bill

3. Jul 31, 2012

### lugita15

If $\hat{A}|\psi\rangle = a_i|\psi\rangle$, then $e^{\hat{A}}|\psi\rangle= \exp(a_i)|\psi\rangle$ and $e^{\hat{A}^n}|\psi\rangle = \exp(a_i^n)|\psi\rangle$. You should be able to prove this pretty easily using the Taylor series definition you mentioned.

4. Aug 1, 2012

### tom.stoer

afaik the proof of

$$f(A) = \sum_n f_n A^n$$

where fn are the Taylor coefficients is rather involved mathematically. What you need is the spectral theorem for self-adjoint operators. For us physicists it reads

$$f(A) = f(A) \sum_n |a\rangle\langle a| = \sum_n f(a) |a\rangle\langle a|$$

where the states |a> are the eigenstates of A and form a complete set.