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Homework Help: Explicit Fourier transform of sin(x)/x.

  1. Feb 5, 2010 #1

    collinsmark

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    1. The problem statement, all variables and given/known data

    I'm reviewing my Fourier transforms (useful in quantum mechanics, in this case 1-dimensional representation), and I'm having a heck of time *explicitly* solving the Fourier transform of

    [tex]\psi(x) = sinc(x)[/tex]

    [tex]\phi(p) = F \left\{ \psi(x) \right\} = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{sin(x)}{x} e^{\frac{-ipx}{\hbar}} dx [/tex]

    Sure, I know quite well that the answer is a scaled box function. So I could start with

    [tex]\phi(p) = \left\{ \stackrel{\sqrt{ \frac{\pi}{2 \hbar}}; - \hbar < p < \hbar}{0; otherwise} [/tex]

    and then just take the inverse Fourier transform of that, giving me my original sinc(x) function. Starting with a box box function and ending with a sinc function is pretty easy. So that sort of proves that the forward Fourier transform is the above box function.

    But that's not what I want. Sure, I already know the answer, but I'd like to explicitly derive the box function (starting with the sinc function), without doing the round-a-bout inverse Fourier transform thing.

    3. The attempt at a solution

    [tex]\phi(p) = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{sin(x)}{x} e^{\frac{-ipx}{\hbar}} dx [/tex]

    [tex] = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{( e^{ix} - e^{-ix} )}{2 i x}} e^{\frac{-ipx}{\hbar}} dx [/tex]

    [tex] = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{( e^{ix - \frac{ipx}{ \hbar}} - e^{-ix- \frac{ipx} {\hbar}} )}{2 i x}} dx [/tex]

    [tex] = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{( e^{ix(1-p / \hbar)} - e^{-ix(1+p/ \hbar)} )}{2 i x}} dx [/tex]

    [tex] = \frac{1}{\sqrt{2 \pi \hbar}} \left( \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {2 i x} dx + \int^{\infty} _{- \infty} \frac{ - e^{-ix(1+p/ \hbar)}}{2 i x } dx \right) dx [/tex]

    [tex] = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {2 i x} dx + \frac{1}{\sqrt{2 \pi \hbar}} \int^{\infty} _{- \infty} \frac{ - e^{-ix(1+p/ \hbar)}}{2 i x } dx [/tex]

    [tex] = \frac{1 - p / \hbar}{2 \sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {i x (1-p / \hbar)} dx + \frac{1 + p/ \hbar}{2 \sqrt{2 \pi \hbar}} \int^{\infty} _{- \infty} \frac{ - e^{-ix(1+p/ \hbar)}}{i x (1+p\hbar)} dx [/tex]

    Substituting [tex] U = x(1-p/\hbar) [/tex], with [tex] dU = (1-p/\hbar)dx [/tex]

    and [tex]V = x(1+p/\hbar) [/tex], with [tex] dV = (1+p/\hbar)dx [/tex]

    [tex] = \frac{1}{2 \sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{iU}} {i U} dU + \frac{1}{2 \sqrt{2 \pi \hbar}} \int^{\infty} _{- \infty} \frac{ - e^{-iV}}{i V} dV [/tex]

    Then I start to have trouble. I can't figure out a way to evaluate the above integrals. I could try to use a "tempering" function, such as multiplying the whole thing by [tex] e^{-ax^{2}} [/tex], complete the square, evaluate the integral, and then later take the limit as a -> 0. But that doesn't really help me evaluate the integral, because there is still that x in the denominator (or U or V, after the substitution). I've tried a number of other tempering functions, but I haven't had any luck.

    I was speculating that maybe, just maybe, there is a tempering function that after multiplying it by the integrand, allows me to evaluate the integral, and then leaves a sigmoid function (or similar) for each integral (total of 2); which after taking a limit, leaves the resulting desired box function. But if there is such a tempering function, I haven't found it yet.

    I'm kinda feeling foolish. The sinc function is one of the fundamental waveforms of the Fourier transform. I had no idea it would be so tricky. Can anybody point me in the general direction on how to derive the Fourier transform of a sinc function? Even a quick blurb on the general jist of the procedure would be appreciative (a link to where this has been solved before would be ideal, of course). Or if you happen to know that no closed form solution has been discovered yet, please let me know that too. And any advise doesn't need to be in the quantum mechanics versions either. If you'd prefer to interpret what I've done so far in non-quantum mechanics terminology, just set [tex] \hbar [/tex] = 1 in all the above equations, and then everything reduces to the run-of-the-mill Fourier transform. At this point, I'm looking for almost anything.

    (please don't make the effort of replying with the approach involving the inverse Fourier transform of a box function. -- I already know that much.)

    Thanks,
    -Mark
     
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2

    vela

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    How about a contour integration in the complex plane?
     
  4. Feb 6, 2010 #3

    collinsmark

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    Ah, contour integration. After googling "contour integration" I gather that it's the approach I should probably take. Unfortunately, I know nothing about contour integration except what I've learned in the last half hour, which isn't much. Before you replied I've never even heard of contour integration. It looks complicated. :uhh:

    I'm presently not a formal student. I'm just teaching myself much of this stuff for fun. 'Looks like I've got a lot of fun ahead of me with this contour stuff. :redface: Or should I say lots of phun.
     
  5. Feb 6, 2010 #4

    collinsmark

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    Alright. So I've studied contour integrals over an hour, and here's what I've come up with.

    But first, let's first break up the integrals so we can work with them one at a time.
    define

    [tex]
    \phi _{1} = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {x} dx, [/tex]

    [tex]
    \phi _{2} = \frac{1}{2 i\sqrt{2 \pi \hbar}} \int^{\infty} _{- \infty} \frac{ - e^{-ix(1+p/ \hbar)}}{x } dx,
    [/tex]

    such that

    [tex] \phi = \phi _{1} + \phi _{2} [/tex].

    Now the contour integral stuff.
    I'll use the standard semicircle in the upper half of the complex plane. The integrand has a single pole at zero.

    Since
    [tex] \oint _{C} f(z) dz = \frac {Lim}{a \rightarrow \infty} \int _{-a} ^{a} f(z) dz + \frac {Lim}{\left| z \right| \rightarrow \infty} \int _{ARC} f(z) dz, [/tex]

    thus,

    [tex] \frac {Lim}{a \rightarrow \infty} \int _{-a} ^{a} f(z) dz = \oint _{C} f(z) dz - \frac {Lim}{\left| z \right| \rightarrow \infty} \int _{ARC} f(z) dz. [/tex]

    Just to be clear, the left side of the equation directly above is what we are looking for. It's just that that integral is a real bugger. But hopefully we can evaluate it indirectly by evaluating the integrals on the right side of the equation (which I gather is the whole point of this contour integral stuff).

    First, let's start with the [tex] \oint _{C} f(z) dz [/tex] integral.

    I'll use the unit circle as the path. It certainly contains the pole at zero. It can be characterized by [tex] e^{i \theta} [/tex]. So define

    [tex] x = e^{i \theta} [/tex]

    and thus

    [tex] dx = i e^{i\theta} d \theta[/tex].

    Recall,

    [tex]
    \phi _{1} = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {x} dx. [/tex]

    This gives us,

    [tex] \oint _{C} f(z) dz = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{2 \pi} _{0} \frac{ e^{ie^{i \theta}(1-p/ \hbar)}} {e^{i \theta} } i e^{i \theta} d\theta, [/tex]



    (Holy moly! [tex] e^{ie^{i \theta}} [/tex].... !?! :bugeye:)

    [tex] = \frac{1}{2 \sqrt{2 \pi \hbar}} \int ^{2 \pi} _{0} e^{ie^{i \theta}(1-p/ \hbar)}}} d\theta, [/tex]

    [tex]
    = \sqrt{\frac {\pi}{2 \hbar}} [/tex]

    So,

    [tex] \oint _{C} f(z) dz = \sqrt{\frac {\pi}{2 \hbar}}. [/tex]

    Next is to discuss the

    [tex] \frac {Lim}{\left| z \right| \rightarrow \infty} \int _{ARC} f(z) dz. [/tex]

    part.

    Recall,

    [tex]
    \phi _{1} = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {x} dx. [/tex]

    The magnitude of the numerator is always 1. It's a complex spiral, and just twirls around, never getting larger than 1. But the denominator gets large as |x| gets large. Even if we add the [tex] - i \epsilon [/tex], it makes no significant difference. So it's pretty obvious that the ARC integral goes to zero as |x| goes to infinity. So now, we can say,

    [tex] \phi _{1} = \sqrt{\frac {\pi}{2 \hbar}}. [/tex]

    Now for [tex] \phi _{2} [/tex]. Rather than repeat everything (this post is getting long as it is), I'm going to go out on a limb and suspect that it's another constant.

    And that would make

    [tex] \phi (p) = [/tex] some constant.

    But that's not a box function.

    Maybe I'm just too new to the contour integrals. I must admit, I'm not very familiar with them, and all I know of them I just learned tonight on the Internet.

    Any ideas where I might want to go from here?

    Thanks,
    -Mark
     
    Last edited: Feb 6, 2010
  6. Feb 6, 2010 #5

    collinsmark

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    I may have found a mistake in my logic, maybe. Before, I said that the ARC integral can be ignored. I may have spoken too soon.

    In my naivety, I was treating 'x' as a real number all the time, when determining whether to ignore the ARC integral. But x needs to be treated as a complex number, not just a real number!

    Looks like I need to investigate evaluating the ARC integral as |x| goes to infinity.

    -Mark
     
  7. Feb 6, 2010 #6

    collinsmark

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    Okay, I think I may have got it. Let me backtrack a little, and forget what I said earlier about ignoring the ARC integral. It *cannot* be ignored.

    So, what we're trying to solve is:

    [tex] \frac {Lim}{a \rightarrow \infty} \int _{-a} ^{a} f(z) dz = \oint _{C} f(z) dz - \frac {Lim}{\left| z \right| \rightarrow \infty} \int _{ARC} f(z) dz. [/tex].

    We've already determined the contour integral (first integral on the right side) to be

    [tex] \oint _{C} f(z) dz = \sqrt{\frac {\pi}{2 \hbar}} [/tex]

    So we have one more integral to solve (then the desired integral to the left is simply the sum of the two we solved here).

    The ARC integral is similar to the contour integral, except instead of setting [tex] x = e^{i \theta} [/tex], we set [tex] x = \epsilon e^{i \theta} [/tex]; and we only integrate half way around the circle, from 0 to [tex] \pi [/tex]. We'll eventual take the limit as [tex] \epsilon \rightarrow \infty [/tex].

    Recall,

    [tex] \phi _{1} = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{ e^{ix(1-p/ \hbar)}} {x} dx. [/tex]

    We set

    [tex] x = \epsilon e^{i \theta} [/tex]

    and thus [tex] dx = i \epsilon e^{i \theta} d\theta. [/tex]

    [tex] \int _{ARC} f(z)dz = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{\pi} _{0} \frac{ e^{i \epsilon e^{i \theta}(1-p/ \hbar)}} {\epsilon e^{i \theta} } i \epsilon e^{i \theta} d\theta,
    [/tex]

    [tex] = \frac{1}{2 \sqrt{2 \pi \hbar}} \int ^{\pi} _{0} e^{i \epsilon e^{i \theta}(1-p/ \hbar)} d\theta,
    [/tex]

    [tex] = \frac{1}{2 \sqrt{2 \pi \hbar}} \left( \pi - 2 Si \left\{ \epsilon (1 - \frac{p}{\hbar}) \right\} \right) [/tex]

    Where Si is the Sine integral. The sine integral. Here you can see it's properties:
    http://mathworld.wolfram.com/SineIntegral.html
    And it's pretty much just what we want.

    The above function (the ARC integral result), is near zero for very large negative values of p. The function is equal to the contour integral result for very large positive values of p. And that's the result of the Si function. When [tex] (1 - p/\hbar) = 0 [/tex], meaning that [tex] p = \hbar [/tex], the function is at its midpoint. As p gets larger the function appraches maximum. When p gets smaller, the function approaches 0.

    But setting [tex] \epsilon \rightarrow \infty [/tex] speeds things up a lot. as a matter of fact, the function will jump from one extreme to the other whenever p crosses [tex] \hbar [/tex].

    So,

    [tex] \frac{Lim}{\epsilon \rightarrow \infty} \frac{1}{2 \sqrt{2 \pi \hbar}} \left( \pi - 2 Si \left\{ \epsilon (1 - \frac{p}{\hbar}) \right\} \right) = \sqrt{\frac{\pi}{2 \hbar}} H(p - \hbar), [/tex]

    where H() is the heavyside step function.

    So, then

    [tex] \phi _{1} = \oint _{C} f(z) dz + \int _{ARC} f(z)dz [/tex]

    [tex] = \sqrt{\frac{\pi}{2 \hbar}} - \sqrt{\frac{\pi}{2 \hbar}} H(p - \hbar). [/tex]

    What that means is that when p is greater than [tex] \hbar [/tex], [tex] \phi _1 [/tex] is zero. When p is less than that, [tex] \phi _1 [/tex] is a constant positive value.

    We could go through this again for [tex] \phi _2 [/tex], but due to symmetry, I speculate the results come out to be

    [tex] \phi _{2} = \sqrt{\frac{\pi}{2 \hbar}} - \sqrt{\frac{\pi}{2 \hbar}} H \left\{ -(p + \hbar)\right\}. [/tex]

    Noting that [tex] \phi = \phi _{1} + \phi _{2} [/tex]

    the final result is

    [tex]
    F \left\{ \psi(x) \right\} = \phi(p) = \frac{1}{\sqrt{2 \pi \hbar}} \int ^{\infty} _{- \infty} \frac{sin(x)}{x} e^{\frac{-ipx}{\hbar}} dx [/tex]

    [tex]
    = F \left\{ sinc(x) \right\} = 2\sqrt{\frac{\pi}{2 \hbar}} - \sqrt{\frac{\pi}{2 \hbar}} H(p - \hbar) - \sqrt{\frac{\pi}{2 \hbar}} H \left\{ -(p + \hbar)\right\}
    [/tex]

    Which is the same thing as saying,

    [tex]
    F \left\{ sinc(x) \right\} = \left\{ \stackrel{\sqrt{\frac{\pi}{2 \hbar}}; -\hbar < p <\hbar}{0; otherwise}
    [/tex]

    Which is the desired box function! Yeah. Oh, sweet fancy Moses. I am so glad I got that done.

    Thanks for everybody's help, particularly vela!

    Mark Collins
    Shady Crypt Observatory
    www.shadycrypt.com
     
    Last edited: Feb 6, 2010
  8. Feb 6, 2010 #7

    Redbelly98

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    I am impressed. http://www.websmileys.com/sm/happy/783.gif
     
  9. Feb 7, 2010 #8

    ideasrule

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    Wow, that was certainly an impressively long proof! Just a small note:

    This integral can be evaluated using the Cauchy's integral formula.
    [tex] \oint _{C} f(z) dz [/tex]
    [tex]=\oint _{C} \frac{F(z)dz}{z-0} [/tex]

    a=0 and F(z)=e^(i*theta*constant), so the integral is equal to 2*pi*i.
     
  10. Feb 8, 2010 #9

    collinsmark

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    Yes, you're right. Cauchy's integral formula would have saved me a small step, but the results are still the same (my original function involves an additional constant which is pulled out from under the integral). That said, your comment is valid and is is still useful for checking my answer for the contour integral (which I originally did without Cauchy's integral formula). Using it is something I probably should have done in the first place, as it is more clear in the end.

    [tex]
    \oint _{C} f(z) dz = \frac{1}{2 i \sqrt{2 \pi \hbar}} \oint _{C} \frac{ e^{ix(1-p/ \hbar)}} {x} dx.
    [/tex]

    Then using Cauchy's integral formula,

    [tex]
    = \frac{1}{2 i \sqrt{2 \pi \hbar}} \left[ (2 \pi i )e^{ix(1-p/ \hbar)} \right] \right|\right| _{x=0},
    [/tex]

    [tex]
    = \frac{1}{2 i \sqrt{2 \pi \hbar}} (2 \pi i )
    [/tex]

    [tex]
    = \sqrt{\frac {\pi}{2 \hbar}}
    [/tex]

    So yeah, everything still checks out. Thanks!

    I did notice one small error I made though, in post # 6, about 2/3 of the way down. It's more of a typo really. The math that follows doesn't follow the typo. But just to be clear in case anybody refers to this thread in the future, I stated that:

    [tex]
    \phi _{1} = \oint _{C} f(z) dz + \int _{ARC} f(z)dz
    [/tex]

    which has a plus sign instead of the proper minus sign. It should be:

    [tex]
    \phi _{1} = \oint _{C} f(z) dz - \int _{ARC} f(z)dz
    [/tex]

    It doesn't affect the rest of the post though.

    Thanks, again.

    -Mark
     
    Last edited: Feb 8, 2010
  11. Feb 9, 2010 #10

    collinsmark

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    Oops. I should make one more correction. (Serves me right, I suppose, for originally doing this at 6 AM, Saturday morning with no sleep).

    Previously I speculated about [tex] \phi _{2} [/tex] instead of actually deriving it. I speculated that

    [tex]
    \phi _{2} = \sqrt{\frac{\pi}{2 \hbar}} - \sqrt{\frac{\pi}{2 \hbar}} H \left\{ -(p + \hbar)\right\},
    [/tex]

    without providing any proof whatsoever. It turns out that my speculation was incorrect, regarding some of the +/- signs. (i.e. the equation above is incorrect.)

    I'll work it out in better detail here.

    Recall,

    [tex]
    \phi _{2} = \frac{1}{2 i\sqrt{2 \pi \hbar}} \int^{\infty} _{- \infty} \frac{ - e^{-ix(1+p/ \hbar)}}{x } dx.
    [/tex]

    The contour integral, using Cauchy's integral formula, is then

    [tex]
    \oint _{C} f(z)dz = \frac{1}{2 i \sqrt{2 \pi \hbar}} \left[ (2 \pi i ) \left( -e^{-ix(1+p/ \hbar)} \right) \right] \right|\right| _{x=0},
    [/tex]

    [tex]
    = -\sqrt{ \frac{\pi}{2 \hbar}}. [/tex]

    The ARC integral is

    [tex]
    \int _{ARC} f(z)dz = \frac{1}{2 i \sqrt{2 \pi \hbar}} \int ^{\pi} _{0} \frac{ -e^{-i \epsilon e^{i \theta}(1+p/ \hbar)}} {\epsilon e^{i \theta} } i \epsilon e^{i \theta} d\theta,
    [/tex]

    [tex]
    = \frac{-1}{2 \sqrt{2 \pi \hbar}} \int ^{\pi} _{0} e^{-i \epsilon e^{i \theta}(1+p/ \hbar)} d\theta,
    [/tex]

    [tex]
    = \frac{-1}{2 \sqrt{2 \pi \hbar}} \left( \pi + 2 Si \left[ \frac {\epsilon}{\hbar} \left(\hbar + p \right) \right] \right), [/tex]

    and after taking the limit as [tex] \epsilon \rightarrow \infty [/tex],

    [tex]
    \int _{ARC} f(z)dz = -\sqrt{ \frac{\pi}{2 \hbar}} H(\hbar +p). [/tex]

    Since

    [tex] \phi _{2} = \oint _{C} f(z)dz - \int _{ARC} f(z)dz, [/tex]

    then

    [tex] \phi _{2} = -\sqrt{ \frac{\pi}{2 \hbar}} + \sqrt{ \frac{\pi}{2 \hbar}} H(\hbar +p). [/tex]

    Putting [tex] \phi = \phi _{1} + \phi _{2} [/tex] together,

    [tex]
    F \left\{ sinc(x) \right\} = - \sqrt{\frac{\pi}{2 \hbar}} H(p - \hbar) + \sqrt{\frac{\pi}{2 \hbar}} H (p + \hbar)
    [/tex]

    which is really what we want. Below [tex] -\hbar [/tex], both terms are zero. Above [tex] \hbar [/tex], each term is non-zero, but they cancel each other (they add up to zero). Only in between [tex] -\hbar [/tex] and [tex] \hbar [/tex] is the resulting box function, since only the second term is non-zero. So the above equation is the same thing as saying:

    [tex]
    F \left\{ sinc(x) \right\} = \left\{
    \begin{tabular}{l}
    \sqrt{\frac{\pi}{2 \hbar}}; -\hbar < p <\hbar \\
    \left( \frac{1}{2} \right) \sqrt{\frac{\pi}{2 \hbar}}; p= \pm \hbar \\
    0; otherwise \\
    \end{tabular}
    [/tex]

    -Mark
    www.shadycrypt.com
     
  12. Nov 20, 2011 #11
    In your first line:
    [tex] \oint _{C} f(z) dz = \lim_{a \rightarrow \infty} \int _{-a} ^{a} f(z) dz + \lim_{\left| z \right| \rightarrow \infty} \int _{ARC} f(z) dz, [/tex]
    you defined the contour to be an arc in the upper half plane. But that contour does not enclose the pole at 0, but instead the pole at 0 lies on the boundary. Then you go on to change the contour to the unit circle, which seems to negate the equality because now in the original integral [tex] \oint _{C} f(z) dz[/tex] C is no longer the arc in the upper half plane.

    I'm doing a project for a complex analysis class which requires me to explicitly find the Fourier transform of [tex] \frac{sinx}{x} [/tex] and I tried something similar to this but I can't justify using the arc as the path because it does not enclose the pole.

    Let me know if you have any ideas.
    Mike
     
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