I 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.