I want to expand a Gaussian wavepacket in terms of sines

  • #1
saadhusayn
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TL;DR Summary
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?
 
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  • #2
It looks like the paper sets up the problem so that the eigenfunctions are orthogonal on the domain 0 to 1, so start there. I'd also try to do a change variable in the integral in such a way to see that the upper limit will go to infinity while the lower limit stays zero.
 
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  • #3
Yes, you're right about the upper limit being ##1##. Also, the fact that ##\sigma << 1## implies that we might as well take the upper limit to ##+\infty##. But other than a factor in front, I don't see how that changes my answer.
 

FAQ: I want to expand a Gaussian wavepacket in terms of sines

What is a Gaussian wavepacket?

A Gaussian wavepacket is a quantum mechanical wave function that describes a particle localized in both position and momentum space. It has the shape of a Gaussian function, which means it is characterized by a bell-shaped curve. This type of wavepacket is often used because it provides a good approximation for the behavior of particles in quantum mechanics.

Why would I want to expand a Gaussian wavepacket in terms of sines?

Expanding a Gaussian wavepacket in terms of sines (or sine functions) is useful because sine functions form a complete orthogonal basis set. This means any function, including a Gaussian wavepacket, can be represented as a sum of sine functions. Such expansions are particularly useful in solving differential equations and in analyzing wave behavior in bounded systems like potential wells.

How do I mathematically expand a Gaussian wavepacket using sine functions?

To mathematically expand a Gaussian wavepacket using sine functions, you would typically use a Fourier sine series. The Fourier coefficients are calculated by integrating the product of the Gaussian wavepacket and the sine functions over the domain of interest. The general form of the expansion is given by \( f(x) = \sum_{n=1}^{\infty} b_n \sin(n \pi x / L) \), where \( b_n \) are the Fourier coefficients determined by \( b_n = \frac{2}{L} \int_0^L f(x) \sin(n \pi x / L) dx \).

What are the advantages of using a sine expansion for a Gaussian wavepacket?

Using a sine expansion for a Gaussian wavepacket offers several advantages: it simplifies solving boundary value problems, allows for easier numerical computations, and provides a clear physical interpretation of the wavepacket in terms of its constituent wave components. This approach can also make it easier to apply boundary conditions and analyze the behavior of the wavepacket in confined systems.

Are there any limitations or challenges in expanding a Gaussian wavepacket in terms of sines?

One limitation of expanding a Gaussian wavepacket in terms of sines is that the sine functions are not localized in space, which can make the expansion less intuitive for representing localized wavepackets. Additionally, the convergence of the series can be slow, especially if the wavepacket has sharp features. Calculating the Fourier coefficients accurately can also be challenging, particularly for more complex or multi-dimensional wavepackets.

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