- #1
saadhusayn
- 17
- 1
- TL;DR
- I have a Gaussian wavepacket $$ \phi(x) = \frac{1}{(2\pi\sigma^2)^{\frac{1}{4}}}\exp \Big(-\frac{(x-x_{0})^{2}}{4\sigma^{2}} + ik_{0}(x-x_{0})\Big) $$
I want to find the Fourier sine coefficients when expanding this in terms of $$\psi_{n}(x) = \sqrt{2} \sin(n \pi x)$$.
From this paper, I am trying to compute the coefficients in the expansion of the Gaussian wavepacket
$$\phi(x) = \frac{1}{(2\pi\sigma^2)^{\frac{1}{4}}}\exp \Big(-\frac{(x-x_{0})^{2}}{4\sigma^{2}} + ik_{0}(x-x_{0})\Big) $$ where $$\sigma << 1$$and $$k_{0} >> \frac{1}{\sigma}$$
in terms of the functions $$ \psi_{n}(x) = \sqrt{2}\sin(n\pi x).$$
My attempt is as follows:
$$\phi(x) = \sum_{m}\alpha_{m}\psi_{m}(x) $$
From the orthogonality of sines, we have $$ \int_{-1}^{1}\psi_{m}(x) \psi_{n}(x) dx = 2 \delta_{mn}.$$ This gives us
$$ \alpha_{n} = \frac{\sqrt{2}}{2}\frac{1}{(2\pi\sigma^2)^{\frac{1}{4}}} \int_{-1}^{1} dx \sin(n \pi x) \exp \Big(-\frac{(x-x_{0})^{2}}{4\sigma^{2}} + ik_{0}(x-x_{0})\Big).$$
My next step is to write the sine in terms of exponential functions and to complete the square on both terms to get:
$$ \alpha_{n} = \frac{1}{2i}\frac{1}{(8\pi \sigma^{2})^{\frac{1}{4}}}\int_{-1}^{+1}dx\Bigg(\exp\Big[-\frac{1}{4\sigma^{2}}(x-a_{+})^{2} -\sigma^2(k_{0}+\pi n)^2 +2in \pi x_{0}\Big] - \exp\Big[-\frac{1}{4\sigma^{2}}(x-a_{-})^{2} -\sigma^2(k_{0}-\pi n)^2 -2in \pi x_{0}\Big]\Bigg)$$
Here, $$a_{\pm} = 2\sigma^{2}\Big(\pm i n \pi + \frac{x_{0}}{2 \sigma^{2}} +ik_{0}\Big)$$
The next step is to use the fact that $$\sigma << 1$$ to extend the limits to $$\pm \infty$$ and treat the integral as approximately Gaussian. I get the following approximate result:
$$\alpha_{n} \approx -i(2 \pi \sigma^{2})^{\frac{1}{4}}\Bigg(\exp\Big[-\sigma^2(k_{0}+\pi n)^2 +2in \pi x_{0}\Big] - \exp\Big[-\sigma^2(k_{0}-\pi n)^2 -2in \pi x_{0}\Big]\Bigg)$$
But the correct answer according to the paper is (page 15, equation 5.9)
$$\alpha_{n} \approx i(2 \pi \sigma^{2})^{\frac{1}{4}} \exp\Big[-(k_{0}-\pi n)^{2} \sigma^{2} + i (k_{0} - \pi n) x_{0}\Big]$$
Would this problem be better solved using the stationary phase method or using the error function? Or do I have the right idea here?
$$\phi(x) = \frac{1}{(2\pi\sigma^2)^{\frac{1}{4}}}\exp \Big(-\frac{(x-x_{0})^{2}}{4\sigma^{2}} + ik_{0}(x-x_{0})\Big) $$ where $$\sigma << 1$$and $$k_{0} >> \frac{1}{\sigma}$$
in terms of the functions $$ \psi_{n}(x) = \sqrt{2}\sin(n\pi x).$$
My attempt is as follows:
$$\phi(x) = \sum_{m}\alpha_{m}\psi_{m}(x) $$
From the orthogonality of sines, we have $$ \int_{-1}^{1}\psi_{m}(x) \psi_{n}(x) dx = 2 \delta_{mn}.$$ This gives us
$$ \alpha_{n} = \frac{\sqrt{2}}{2}\frac{1}{(2\pi\sigma^2)^{\frac{1}{4}}} \int_{-1}^{1} dx \sin(n \pi x) \exp \Big(-\frac{(x-x_{0})^{2}}{4\sigma^{2}} + ik_{0}(x-x_{0})\Big).$$
My next step is to write the sine in terms of exponential functions and to complete the square on both terms to get:
$$ \alpha_{n} = \frac{1}{2i}\frac{1}{(8\pi \sigma^{2})^{\frac{1}{4}}}\int_{-1}^{+1}dx\Bigg(\exp\Big[-\frac{1}{4\sigma^{2}}(x-a_{+})^{2} -\sigma^2(k_{0}+\pi n)^2 +2in \pi x_{0}\Big] - \exp\Big[-\frac{1}{4\sigma^{2}}(x-a_{-})^{2} -\sigma^2(k_{0}-\pi n)^2 -2in \pi x_{0}\Big]\Bigg)$$
Here, $$a_{\pm} = 2\sigma^{2}\Big(\pm i n \pi + \frac{x_{0}}{2 \sigma^{2}} +ik_{0}\Big)$$
The next step is to use the fact that $$\sigma << 1$$ to extend the limits to $$\pm \infty$$ and treat the integral as approximately Gaussian. I get the following approximate result:
$$\alpha_{n} \approx -i(2 \pi \sigma^{2})^{\frac{1}{4}}\Bigg(\exp\Big[-\sigma^2(k_{0}+\pi n)^2 +2in \pi x_{0}\Big] - \exp\Big[-\sigma^2(k_{0}-\pi n)^2 -2in \pi x_{0}\Big]\Bigg)$$
But the correct answer according to the paper is (page 15, equation 5.9)
$$\alpha_{n} \approx i(2 \pi \sigma^{2})^{\frac{1}{4}} \exp\Big[-(k_{0}-\pi n)^{2} \sigma^{2} + i (k_{0} - \pi n) x_{0}\Big]$$
Would this problem be better solved using the stationary phase method or using the error function? Or do I have the right idea here?