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The Born series expansion

  1. Jan 30, 2012 #1
    Hello. I read about the born series in scattering,
    [tex]
    |\psi> = (1+G_0V+\ldots)|\psi_0>
    [/tex]
    Now when I want to move to spatial representation, the textbook asserts I should get
    [tex]
    \psi(\vec{r})=\psi_0(\vec{r}) + \int dV' G_0(\vec{r},\vec{r'}) V(\vec{r'})\psi_0(\vec{r'})+\ldots
    [/tex]
    by operating with [itex]<\vec{r}|[/itex] 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:
    [tex]
    <\vec{r}|G_0V|\psi_0> = \int d^3r'<\vec{r}|G_0|\vec{r'}><\vec{r'}|V|\psi_0>
    [/tex]
    however I don't know how to get [itex]V(\vec{r'})[/itex] 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. jcsd
  3. Jan 30, 2012 #2

    vanhees71

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    2016 Award

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

    [tex]\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}'').[/tex]

    Then one of the integrals from the completeness relations can be used to get rid of the [itex]\delta[/itex] distribution, and you arrive at Born's series in the position representation as given by your textbook.
     
  4. Jan 30, 2012 #3
    I see. And then [itex]V(\vec{x'})[/itex] can be taken out as [itex]|\vec{x'}>[/itex] are its eigenstates and it is taken out as a scalar.
    Thanks!
     
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