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Evaluating integral born approximation

  1. Dec 29, 2004 #1
    I'm trying to evaluate the following integral to calculate the scattering cross section for a spherically symmetrical potential [tex]e^{\frac{-r^2}{a^2}}[/tex]?

    [tex]f(\theta)=\int r e^{\frac{-r^2}{a^2}} sin(kr) dr[/tex] where a is a constant.

    What is the easiest way to evaluate this? I was able to get the answer by doing the integral using Mathematica but I don't know how to do this by hand.

    Also, is the approximation better for low energies or high energies??
    Last edited: Dec 30, 2004
  2. jcsd
  3. Dec 30, 2004 #2


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    IIRC,the scattering amplitude and the differential cross section are given by:
    [tex] \frac{d\sigma}{d\Omega}=|f(\vartheta)|^{2} [/tex] (1)

    For your potential:[itex] V(r)=V_{0}\exp(\frac{-r^{2}}{a^{2}}) [/tex]
    ,the scattering amplitude is
    [tex] f(\vartheta)=-\frac{2mV_{0}}{\hbar^{2}}\int_{0}^{\infty} r^{2} [\exp(\frac{-r^{2}}{a^{2}})] \frac{\sin(Kr)}{Kr} dr [/tex] (2)
    [tex] K=2k\sin\frac{\vartheta}{2} [/tex]

    I suggest u plug the integral from (2) in 'Mathematica' and use the result the software gives to find (1).

    Your formula was wrong then.

  4. Dec 30, 2004 #3


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    For small energies scattering ('k' very small=>K very small),the result is very easy to get:
    [tex]\frac{\sin Kr}{Kr} \sim 1 [/tex]
    ,and your integral is immediate:
    [tex]f(\vartheta)=-\frac{2mV_{0}}{\hbar^{2}}\int_{0}^{\infty} r^{2}\exp[\frac{-r^{2}}{a^{2}}] dr =...=-\frac{mV_{0}a^{3}\sqrt{2\pi}}{\hbar^{2}} [/tex]
    ,where i made use of part integration and the value for the Poisson integral.

    So the differential scattering cross section is:
    [tex] \frac{d\sigma}{d\Omega} =[-\frac{mV_{0}a^{3}\sqrt{2\pi}}{\hbar^{2}}]^{2} =\frac{2\pi m^{2}V_{0}^{2} a^{6}}{\hbar^{4}}[/tex](1)

    Integrating (1),u get the integral scattering cross section:
    [tex]\sigma =\frac{8\pi^{2} m^{2}V_{0}^{2} a^{6}}{\hbar^{4}}[/tex] (2)
    ,which should be the answer given by the Born approximation for low,very low energies.

  5. Dec 30, 2004 #4
    I made a mistake in my equation. The correct one is this:

    [tex]f(\theta)=\int r e^{\frac{-r^2}{a^2}} sin(kr) dr[/tex] where a is a constant.

    This equation gives the scattering amplitude for spherical symmetry.
    If I write this as:
    [tex]f(\theta)=\int r^2 e^{\frac{-r^2}{a^2}} \frac{sin(kr)}{r} dr[/tex]

    [tex]\frac{sin(kr)}{r}[/tex] is approximately 1

    [tex]f(\theta)=\int r^2 e^{\frac{-r^2}{a^2}} dr[/tex]

    However, this gives me an answer that is not dependent on k. Low-energy scattering wasn't specified.

    This is how I did the problem:
    The potential is [tex]V(r)=V e^{\frac{-r^2}{a^2}}[/tex]

    [tex]f(\theta) = -\frac{2m}{\hbar^{2} \kappa} \int_{0}^{\infty} r V e^{\frac{-r^2}{a^2}} sin(kr) dr

    = \frac{-a^3 V e^{-\frac{1}{4} a^{2} \kappa^2} m \sqrt{\pi} v}{2 \hbar^2} [/tex]

    Since [tex]\kappa = 2 k {sin(\theta/2)}[/tex]

    [tex] \frac{d\sigma}{d\Omega} = |f(\theta)|^2 =\frac{a^6 m^2 V^2 \pi}{4 \hbar^4} e^{-\frac{a^2}{2} (2 k sin(\theta/2))^2}[/tex]

    where [tex] E = \frac{\hbar^2 k^2}{2m} [/tex]

    I get the correct answer, but my question is how do you evaluate the integral by hand for [tex]f(\theta)[/tex]. I did it on Mathematica but this is supposed to be a sample exam question where we're supposed to do it by hand... the solution says one way to do it is to transform into cartesian coordinates, but I'm confused about how to do this.


    Last edited: Dec 30, 2004
  6. Jan 2, 2005 #5


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    Okay,let's do it by hand.
    U wanna compute this function
    [tex] f(\theta)=-\frac{2mV}{\hbar^{2}\kappa}\int_{0}^{+\infty} re^{-\frac{r^{2}}{a^{2}}} \sin(\kappa r) dr [/tex] (1)

    Let's concentrate upon the integral.Call it 'I'.We can write
    [tex]f(\theta)=-\frac{2mV}{\hbar^{2}\kappa} I [/tex] (2)
    ,where I is given by
    [tex] I=:\int_{0}^{+\infty} re^{-\frac{r^{2}}{a^{2}}} \sin(\kappa r) dr [/tex] (3)

    Using part integration,we find:
    [tex] I=\int_{0}^{+\infty} \sin(\kappa r)(-\frac{a^{2}}{2}) d(e^{-\frac{r^{2}}{a^{2}}}) =-\frac{a^{2}}{2}\{[\sin (\kappa r)e^{-\frac{r^{2}}{a^{2}}}]|_{0}^{+\infty} -k\int_{0}^{+\infty} \cos(\kappa r)e^{-\frac{r^{2}}{a^{2}}} dr \} =\frac{ka^{2}}{2}\int_{0}^{+\infty} \cos(\kappa r)e^{-\frac{r^{2}}{a^{2}}} dr [/tex] (4)

    Call the last integral of (4) by J.Therefore:
    [tex] I=\frac{ka^{2}}{2} J [/tex] (5)
    We need to find J.
    We use Euler's formula to get
    [tex]J=Re(\int_{0}^{+\infty} e^{i\kappa r-\frac{r^{2}}{a^{2}}} dr) [/tex] (6)
    ,which can be written as follows
    [tex]J=Re(\int_{0}^{+\infty} e^{-[(\frac{r}{a})^{2}-2\frac{r}{a}\frac{i\kappa a}{2}+(\frac{i\kappa a}{2})^{2}]} e^{+(\frac{i\kappa a}{2})^{2}} dr =a e^{-\frac{\kappa^{2}a^{2}}{4}}Re[\int_{0}^{+\infty} e^{-(\frac{r}{a}-\frac{i\kappa a}{2})^{2}} d(\frac{r}{a}-\frac{i\kappa a}{2})] [/tex](7)

    In the last integral make the obvious substitution [itex] (\frac{r}{a}-\frac{i\kappa a}{2})\rightarrow z [/itex] (8)
    ,and u'll be getting
    [tex]J=a e^{-\frac{\kappa^{2}a^{2}}{4}}Re[\int_{0}^{+\infty} e^{-z^{2}} dz] (8)=\frac{a\sqrt{\pi}}{2}e^{-\frac{\kappa^{2}a^{2}}{4}} [/tex](9)

    Go with (9) in (5) and get I:
    [tex]I=\frac{\kappa a^{3}\sqrt{\pi}}{4} e^{-\frac{\kappa^{2}a^{2}}{4}} [/tex] (10)

    Go with I given by (10) into (2) to find the amplitude of scattering
    [tex]f(\theta)=-\frac{mVa^{3}\sqrt{\pi}}{2\hbar^{2}}e^{-\frac{\kappa^{2}a^{2}}{4}} [/tex] (11)

    ,which is just the result 'Mathematica' gives u.So the software is correct.

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