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Fourier transform of a gaussian

skatenerd

Active member
Oct 3, 2012
114
I'm given a Gaussian function to apply a Fourier transform to.
$$f(x)=\frac{1}{\sqrt{a\sqrt{\pi}}}e^{ik_ox}e^{-\frac{x^2}{2a^2}}$$
Not the most appetizing integral...
$$g(k)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{a\sqrt{\pi}}}\int_{-\infty}^{\infty}e^{ik_ox}e^{-\frac{x^2}{2a^2}}e^{-ikx}dx$$
$$=\frac{1}{\sqrt{2\pi{a}\sqrt{\pi}}}\int_{-\infty}^{\infty}exp[-\frac{x^2}{2a^2}+i(k_o-k)x]dx$$
I don't really want to write out the whole process, but I multiplied this expression by
$$\frac{e^{A^2(k_o-k)^2}}{e^{A^2(k_o-k)^2}}$$
and then I found \(A\) in order to complete the square in the exponent inside the integrand,
$$A=-\frac{i\sqrt{2}a}{2}$$
So I now have the integral
$$g(k)=\frac{1}{\sqrt{2\pi{a}\sqrt{\pi}}}e^{-\frac{a^2}{2}(k_o-k)^2}\int_{-\infty}^{\infty}exp[-(\frac{x}{\sqrt{2}a}-\frac{i\sqrt{2}a}{2}(k_o-k))^2]dx$$
But this is where I get stuck. I don't really understand how to integrate this expression. I learned from my lecture that this whole completing the square process is how we simplify the integrand, but I am iffy on the process after this point. Any hints in the right direction would be nice. (Also there is definitely a chance that this integrand isn't quite correct. There is a lot of room for error to arrive at this point, though I did check my work multiple times)
 
Last edited:

skatenerd

Active member
Oct 3, 2012
114
Re: fourier transform of a gaussian

Also, it may help to know that I am supposed to verify that the result is
$$g(k)=\sqrt{\frac{a}{\sqrt{\pi}}}e^{-\frac{(k_o-k)^2a^2}{2}}$$
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: fourier transform of a gaussian

I'm given a Gaussian function to apply a Fourier transform to.
$$f(x)=\frac{1}{\sqrt{a\sqrt{\pi}}}e^{ik_ox}e^{-\frac{x^2}{2a^2}}$$
Not the most appetizing integral...
$$g(k)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{a\sqrt{\pi}}}\int_{-\infty}^{\infty}e^{ik_ox}e^{-\frac{x^2}{2a^2}}e^{-ikx}dx$$
$$=\frac{1}{\sqrt{2\pi{a}\sqrt{\pi}}}\int_{-\infty}^{\infty}exp[-\frac{x^2}{2a^2}+i(k_o-k)x]dx$$
I don't really want to write out the whole process, but I multiplied this expression by
$$\frac{e^{A^2(k_o-k)^2}}{e^{A^2(k_o-k)^2}}$$
and then I found \(A\) in order to complete the square in the exponent inside the integrand,
$$A=-\frac{i\sqrt{2}a}{2}$$
So I now have the integral
$$g(k)=\frac{1}{\sqrt{2\pi{a}\sqrt{\pi}}}\int_{-\infty}^{\infty}exp[-(\frac{x}{\sqrt{2}a}-\frac{i\sqrt{2}a}{2}(k_o-k))^2]dx$$
But this is where I get stuck. I don't really understand how to integrate this expression. I learned from my lecture that this whole completing the square process is how we simplify the integrand, but I am iffy on the process after this point. Any hints in the right direction would be nice. (Also there is definitely a chance that this integrand isn't quite correct. There is a lot of room for error to arrive at this point, though I did check my work multiple times)
I would probably set it up this way.

You're given that $f(x) = \dfrac{1}{\sqrt{a\sqrt{\pi}}} \exp(ik_ox) \exp\left(-\dfrac{x^2}{2a^2}\right) = \dfrac{1}{\sqrt{a\sqrt{\pi}}} \exp\left(-\dfrac{x^2}{2a^2} + ik_o x\right)$.

The integral formula you use to compute $g(k)$ is the one that restores the symmetry of the Fourier transform and it's inverse. So, using that formula, we have that

\[\begin{aligned}g(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f(x) \exp(-ikx)\,dx \\ &= \frac{1}{\sqrt{2\pi a\sqrt{\pi}}} \int_{-\infty}^{\infty} \exp\left(-\dfrac{x^2}{2a^2} -i(k-k_o)x\right)\,dx \\ &= \frac{1}{\sqrt{2\pi a\sqrt{\pi}}} \int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{2a^2}\right) \exp\left(-i(k-k_o)x\right) \\ &= \frac{1}{\sqrt{2\pi a\sqrt{\pi}}}\left[ \int_{-\infty}^{\infty}\exp\left(-\frac{x^2}{2a^2}\right) \cos\left((k-k_o)x\right)\,dx - \underbrace{i\int_{-\infty}^{\infty} \exp\left(-\frac{x^2}{2a^2}\right) \sin\left( (k-k_o)x\right)\,dx}_{ =0\text{ since integrand is odd.}}\right] \\ &= \frac{1}{\sqrt{2\pi a\sqrt{\pi}}} \int_{-\infty}^{\infty} \exp\left(-\frac{x^2}{2a^2}\right) \cos\left((k-k_o)x\right)\,dx \end{aligned}\]

Now, I'm going to cheat a little bit (because I don't want to evaluate that integral; if someone wants to fill in the details of that integration, I'd be intrigued to see how it would be done). Anyways, due to Abramowitz and Stegun, we know that

\[\int_{-\infty}^{\infty} e^{-bx^2}\cos(2\pi y x)\,dx = \sqrt{\frac{\pi}{b}} \exp\left(-\frac{\pi^2 y^2}{b}\right).\]

To apply this formula, take $b=\dfrac{1}{2a^2}$ and $y=\dfrac{k-k_o}{2\pi}$.

Thus,

\[\begin{aligned}\frac{1}{\sqrt{2\pi a\sqrt{\pi}}} \int_{-\infty}^{\infty} \exp\left(-\frac{x^2}{2a^2}\right) \cos\left((k-k_o)x\right)\,dx &= \frac{1}{\sqrt{2\pi a\sqrt{\pi}}} \cdot \sqrt{2\pi a^2}\exp\left(-\frac{2a^2\pi^2 (k-k_o)^2}{4\pi^2}\right) \\ &= \sqrt{\frac{a}{\sqrt{\pi}}} \exp\left(-\frac{a^2(k-k_o)^2}{2}\right)\end{aligned}\]

and therefore

\[g(k) = \sqrt{\frac{a}{\sqrt{\pi}}} \exp\left(-\frac{a^2(k-k_o)^2}{2}\right).\]

(Whew)

I'm actually kind of glad to see that this matches up with the answer you're supposed to get. (Nod)

I hope this makes sense!
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,856
Re: fourier transform of a gaussian

So I now have the integral
$$g(k)=\frac{1}{\sqrt{2\pi{a}\sqrt{\pi}}}\int_{-\infty}^{\infty}exp[-(\frac{x}{\sqrt{2}a}-\frac{i\sqrt{2}a}{2}(k_o-k))^2]dx$$
But this is where I get stuck. I don't really understand how to integrate this expression. I learned from my lecture that this whole completing the square process is how we simplify the integrand, but I am iffy on the process after this point. Any hints in the right direction would be nice. (Also there is definitely a chance that this integrand isn't quite correct. There is a lot of room for error to arrive at this point, though I did check my work multiple times)
I think you're supposed to use that:
$$\int_{-\infty}^{\infty} e^{-u^2} du = \sqrt{\pi}$$

To get to the point that you can, use the substitution:
$$u = \frac{x}{\sqrt{2}a}-\frac{i\sqrt{2}a}{2}(k_o-k)$$
 

skatenerd

Active member
Oct 3, 2012
114
@Chris L T521:
That's an interesting and impressive solution you got there...I appreciate you going through all that. However I think since I already went through all the trouble of the method with completing the square inside the exponent and finding the constant I needed and all that I will go with I Like Serena's method.
Pretty much that substitution was all I needed to get to the answer. It was smooth sailing from that point. Thanks for all the help you guys! Such a cool problem in the end.