# The Born series expansion

1. Jan 30, 2012

### PineApple2

Hello. I read about the born series in scattering,
$$|\psi> = (1+G_0V+\ldots)|\psi_0>$$
Now when I want to move to spatial representation, the textbook asserts I should get
$$\psi(\vec{r})=\psi_0(\vec{r}) + \int dV' G_0(\vec{r},\vec{r'}) V(\vec{r'})\psi_0(\vec{r'})+\ldots$$
by operating with $<\vec{r}|$ from the left. However I don't know how to get the 2nd term (the integral). I tried to insert a complete basis like this:
$$<\vec{r}|G_0V|\psi_0> = \int d^3r'<\vec{r}|G_0|\vec{r'}><\vec{r'}|V|\psi_0>$$
however I don't know how to get $V(\vec{r'})$ from the second bracketed term. Any help?

By the way: is there a "nicer" way to write 'bra' and 'ket' in this forum?

Last edited: Jan 30, 2012
2. Jan 30, 2012

### vanhees71

Introduce another unit operator in terms of the completeness relation for the position-eigenbasis. Then you use

$$\langle \vec{x}' |V(\hat{\vec{x}}\vec{x}'' \rangle = V(\vec{x}'') \langle \vec{x}'| \vec{x}'' \rangle=V(\vec{x}') \delta^{(3)}(\vec{x}'-\vec{x}'').$$

Then one of the integrals from the completeness relations can be used to get rid of the $\delta$ distribution, and you arrive at Born's series in the position representation as given by your textbook.

3. Jan 30, 2012

### PineApple2

I see. And then $V(\vec{x'})$ can be taken out as $|\vec{x'}>$ are its eigenstates and it is taken out as a scalar.
Thanks!