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I am working through a paper (Phys Rev A 49 4, 1993) and have been stuck on one step for some time now. The authors make an approximation that I can not reproduce - specifically they are expanding the following function in terms of the inverse of its argument:

$$n(\xi)=\left(1 + \left[\xi^2+2\frac{\omega_0}{\omega_p}\xi + i\frac{\gamma}{\omega_p}\xi + i\frac{\gamma\omega_0}{\omega_p^2}\right]^{-1} \right)^{1/2} \approx 1 + \frac{1}{4\xi}\left(\frac{\omega_p}{\omega_0}\right)-i\frac{1}{8\xi^2}\left(\frac{\gamma}{\omega_p}\right)\left(\frac{\omega_p}{\omega_0}\right)$$

Here $\vert\xi\vert$ is a dimensionless frequency $(\omega-\omega_0)/\omega_p$ and is of order unity.

For the region of interest, $\gamma\ll\omega_p\ll\omega_0$. Thus the parameters $\frac{\gamma}{\omega_p}$ and $\frac{\omega_p}{\omega_0}$ are small. The idea is a series expansion in terms of these small parameters about zero, but can not get an expression of this form.

I realize this might be trivial and I'm just missing something but if anyone has any pointers I would be very grateful to hear them.

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  • $\begingroup$ For whatever reason, this is the first order expansion in $x\equiv \omega_p/\omega_0$. Is this evident to you? $\endgroup$ Apr 2, 2021 at 16:15

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$$n(\xi)=\left(1 + \left[\xi^2+2\frac{\omega_0}{\omega_p}\xi + i\frac{\gamma}{\omega_p}\xi + i\frac{\gamma\omega_0}{\omega_p^2}\right]^{-1} \right)^{1/2}\\= \left(1 + \frac{\omega_p/\omega_0}{2 \xi + (\xi^2+i\frac{\gamma}{\omega_p}\xi)(\omega_p/\omega_0) + i\frac{\gamma}{\omega_p} }\right)^{1/2} \\ \approx \left(1 + \frac{\omega_p/\omega_0}{2 \xi + i\frac{\gamma}{\omega_p} }\right)^{1/2} \approx \left(1 + \frac{(\omega_p/\omega_0)\left(1- i\frac{\gamma}{2\xi\omega_p}\right )}{2 \xi }\right)^{1/2} \\ \approx 1 + \frac{1}{4\xi}\left(\frac{\omega_p}{\omega_0}\right)-i\frac{1}{8\xi^2}\left(\frac{\gamma}{\omega_p}\right)\left(\frac{\omega_p}{\omega_0}\right).$$

From the 2nd to the 3rd line, we have eliminated terms of $O((\omega_p/\omega_0)^2)$, anticipating the expansion to the last line, and, next, eliminated subdominant terms of $O((\gamma/\omega_p)^2)$. The last line, then, is self-explanatory to $O((\omega_p/\omega_0)^2)$. The important step is in the first line, where no large quantities appear among the terms of the sum in the denominator, so we can toss higher-order small terms.

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  • $\begingroup$ This is perfect. Indeed I was failing to see the initial step that allows for line 2, which in turn makes the rest of the steps more straightforward. Thank you for your concise answer! $\endgroup$ Apr 2, 2021 at 19:35

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