1D scattering: Taylor expansion

[WWCY]

Hi all, I'm having a problem understanding a step in an arxiv paper (https://arxiv.org/pdf/0808.3566.pdf) and would like a bit of help.

In equation (29) the authors have
$$R = \frac{\sigma}{\sqrt{\pi}} \int dk \ e^{-(k - k_0)^2 \sigma^2} \ \Big( \frac{ k - \kappa}{ k+ \kappa} \Big)^2$$
where the ##k##-space peak is sharp about ##k_0##

They then state that "we can Taylor expand the complicating factors about ##k = k_0## to get a series of standard integrals", and wrote the result as
$$R \approx \Big( \frac{k_0 - \kappa _0}{k_0 + \kappa _0} \Big)^2 + \Big( \frac{2 k_0}{\kappa _0 ^3 } + \frac{8}{\kappa _0 ^2} \Big) \Big( \frac{k_0 - \kappa _0}{k_0 + \kappa _0} \Big)^2 \frac{1}{\sigma ^2}$$

How did they perform the taylor expansion and on which term? My guess is that they expanded something up to second order in ##k - k_0## before integrating but I can't figure out how they did it. Assistance is greatly appreciated!

 

[Charles Link]

In the ## (\frac{k-\kappa}{k+\kappa})^2 ## let's change ## \kappa ## to ##y ##.
Then ##R=R(y) ##, and we are actually finding ##R(y_o) ##.
I haven't solved it exactly yet, but we now have, with a little algebra and letting ## y=y_o ##,
##(\frac{k-y}{k+y})^2= (\frac{k-k_o+k_o-y_o}{k-k_o+k_o+y_o})^2 ##.
Now change ## k-k_o ## to ## x ## everywhere, including in the ## e^{-(k-k_o)^2 \sigma^2} ##.
Then we have the factor on the exponential is ## (\frac{x+A}{x+B})^2 ## where ## A=k_o-y_o ## and ## B=k_o+y_o ##.
Let ## f(x)=(\frac{x+A}{x+B})^2 ##, and ## f(x)=f(0)+f'(0)x+f''(0) \frac{x^2}{2} +... ## where ## x ## is small.
Then we have ## R(y_o)=\frac{\sigma}{\sqrt{\pi}} \int\limits_{-\infty}^{+\infty} e^{-x^2 \sigma^2} f(x) \, dx ##.
The ## x ## (middle) term integrates to ## 0 ## because ## x ## is odd.
## f(0)=(\frac{A}{B})^2 ## and the integral is normalized to unity.
It remains to compute ## f''(0) ## correctly, and evaluate ##I= \int\limits_{-\infty}^{+\infty} x^2 e^{-x^2 \sigma^2} \, dx ##.
It integrates by parts, and I think it gives ##I= \frac{\sqrt{\pi}}{2 \sigma^3}##.
I haven't gotten complete agreement yet with the textbook, but it's looking similar to what they have, and with a careful evaluation of ## f''(0) ## the results might agree.## \\ ## Edit: I get the ## \sigma^2 ## in the denominator like they do in the second term of the expression for ## R(\gamma_o) ##,(and that is the key term because ## \sigma ## is large in the numerator of the Gaussian, making for a narrow peak, (usually ## \sigma^2 ## is in the denominator of the exponential) , but so far I don't agree with the rest of what they have.

 

[Charles Link]

In general ## k_o \neq \gamma_o ## so I don't think the second term needs to have the second order form of ## (k_o-\gamma_o) ## . (I've substituted ## \gamma ## for ## \kappa ##). Perhaps I'm expanding it a different way, but I could not duplicate the form that they have.## \\ ## In addition, I do not get ## \gamma_o^2 ## or higher power in the denominator of the second term. Instead, I get a ## \gamma_o ## in the numerator. As ## \gamma_o \rightarrow 0 ##, I think their expression incorrectly diverges. The peak is at ## k=k_o ##, but I don't see any restriction on ## \gamma ## (= ## \kappa##).

 

[WWCY]

Hi @Charles Link , thanks for your assistance, I will try to work through and understand your work!

Edit: I have just realized that I have completely failed to mention that ##\kappa = \sqrt{k^2 + \frac{2mV_0}{\hbar ^2} }##

Sincerest apologies if I have wasted your efforts!

 

[Charles Link]

Hi @Charles Link , thanks for your assistance, I will try to work through and understand your work!

Edit: I have just realized that I have completely failed to mention that ##\kappa = \sqrt{k^2 + \frac{2mV_0}{\hbar ^2} }##

Sincerest apologies if I have wasted your efforts!

This last part is extra info and not needed to do this calculation. What it says is that in general, ## k \neq \kappa ##.

 

[WWCY]

This last part is extra info and not needed to do this calculation. What it says is that in general, ## k \neq \kappa ##.

In this case I'm not sure I understand the rationale behind ##(\frac{k-y}{k+y})^2= (\frac{k-k_o+k_o-y_o}{k-k_o+k_o+y_o})^2## and ##(\frac{x+A}{x+B})^2##, ##A=k_o-y_o, \ B=k_o+y_o##. Integrating over these functions already seems to assume that ##\kappa = y## is a constant, and not a function of ##k##.

Do you mind elaborating? Thanks.

 

[Charles Link]

Yes. That is correct: ## R=R(y) ##, and we are really interested in solving for ## R(y_o) ##. That makes ## \kappa=y_o ## is a constant, and the expansion in the first equation after "rationale behind" is just some algebra to simplify the function.
Since in the narrow Gaussian we have ## e^{-\sigma^2 (k-ko)^2 }##, we can do a change of variable and let ## x=k-k_o ##. That gives ## dk=dx ##. Meanwhile the limits of integration are still ## -\infty ## to ## +\infty ##.
[i.e. To a good approximation, since ## k ## in actuality does not become negative, but for mathematical simplicity we let ##k ## go to ## -\infty ##] ,

The algebra could be done simply by letting ## k=x+k_o ## instead of putting ## -k_o+k_o ## in both the numerator and denominator.
The narrow Gaussian peaks at ## x=0 ##, so the other term, (## (\frac{x+A}{x+B})^2 ##), requires a Taylor expansion about ## x=0 ##. You should be able to Taylor expand ##f(x)=(\frac{x+A}{x+B})^2 ## about ## x=0 ## , if your calculus is reasonably good.