Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Must the Fourier transform exist for Dirac delta functions?

  1. Jul 28, 2012 #1
    I originally asked this in the Calculus & Analysis forum. But perhaps this is better suited as a question in Abstract algebra.

    For the set of all Dirac delta functions that have a difference for an argument, we have the property that:

    [tex]\int_{ - \infty }^\infty {{\rm{\delta (x - }}{{\rm{x}}_1}){\rm{\delta (}}{{\rm{x}}_1}{\rm{ - }}{{\rm{x}}_0})d{x_1}} = {\rm{\delta (x - }}{{\rm{x}}_0})[/tex]
    But the above equation can be seen as a convoltuion integral which can be solved with the Fourier transforms method. All three delta functions do exist in this set. And I'm told that the Fourier transform maps this set of Dirac deltas into this same set of Dirac deltas that have a difference as input. So does that gaurantee the existence of the Fourier transform for this set of Dirac deltas? I'm trying to determine if the FT for this set of Dirac delta functions is a necessity or optional added feature that could be used. And in my reading I'm told that the Fourier transform is an automorphism which transforms the deltas into themselves. Does that gaurantee the existence of the FT? What if you add the requirement that the convolution exist as seen in the above equation? If the convolutions exist and the FTs also exist in the same set of Dirac delta functions, does the above equation imply that the FT must also exist? Maybe this is the same as asking whether all automorphisms necessarily exist? Or maybe the FT is an equivalent automorphism to the convolution automorphism for this set of Dirac delta functions? Thanks.
     
    Last edited: Jul 28, 2012
  2. jcsd
  3. Jul 29, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, that equation doesn't "guarentee" the existance of a Fourier transform for the delta function. It is the fundamental formula for the "Fourier transform" shows that it exist:
    [tex]\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-st}\delta(t- a)dt= \frac{e^{-as}}{\sqrt{2\pi}}[/tex]
     
    Last edited: Jul 30, 2012
  4. Jul 30, 2012 #3
    Yes, that would gaurantee the existence of the Fourier Transfrom of the dirac delta with an argument of a difference... if one should decide to use the FT for such. That's a start. Thank you.

    But what I'd really like to see is some sort of necessary requirement of the use of the FT for this set of dirac deltas that have a difference as input. For I'm trying to justify the use of complex numbers in quantum mechanics, and I'm thinking that the use of the FT will introduce those complex numbers.

    What I start with is the self-convolution integral in my first post. It's obvious that this convolution integral exists. And it's also true that for any dirac delta in the set, the FT also exists. And it is also true that the FT can be used to solve the convolution integral. So how close is all this to proving that the FT must be implied by the convolution integral for this set?

    What's the logic I'm looking for? The convolution exists and is solved/derived/proved by the FT which also exists. So... if the FT exists and proves the convolution. Does that mean the convolution proves the FT? If I assert the convolution exists, do I also assert in the process that the FT must also exist since it solves the convolution? Or something like that.
     
    Last edited: Jul 30, 2012
  5. Jul 31, 2012 #4
    I think this is the same as asking: Can the convolution exist without the Fourier Transform?
     
  6. Jul 31, 2012 #5

    rbj

    User Avatar

    i think the equation above is just an obfuscation of the essential sampling property of the dirac delta.

    [tex]\int_{-\infty}^{\infty}f(x) \delta(x-x_0) \ dx = f(x_0) [/tex]

    if your equation can be expressed as the above, it's true, if it can't it's not. it's sorta nasty sampling the dirac delta itself, but keeping notational consistency...

    [tex]\int_{-\infty}^{\infty}\delta(x) \delta(x - x_0) \ dx = \delta(x_0) [/tex]


    [tex]\int_{-\infty}^{\infty}\delta(x - x_1) \delta(x - x_0) \ dx = \delta(x_0 - x_1) [/tex]

    and it's also a convolution. the sampling property is essentially the same integral with the impulse response an impulse (what us electrical engineers call a "wire").

    is the question, can you get to the convolution property, from the time-domain properties of Linear and Time-Invariant systems (i dunno what they call it in Linear Algebra) without using the Fourier Transform, the answer is yes. but you have to understand the Dirac delta "function" and use its properties to advantage.


    essentially convolution as an operation exists without Fourier Transform in its definition and the Fourier Transform exists without reference to the concept of convolution. but you could say that about the operation of multiplication and the transform we call "logarithm". they exist separately, but there are salient and handy theorems regarding their relationship to each other.
     
    Last edited: Jul 31, 2012
  7. Aug 4, 2012 #6
    If we start with the self-convolution integral:

    1)...... [itex]\delta (x - {x_0}) = \int_{ - \infty }^{ + \infty } {\delta (x - {x_1})\delta ({x_1} - {x_0})} \,d{x_1}[/itex]

    And we let the Dirac delta be represented by the guassian form:

    2)...... [itex]\delta (x - {x_1}) = \mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{{\textstyle{1 \over 2}}}}}}\,{e^{\frac{{ - {{(x - {x_1})}^2}}}{{{\Delta ^2}}}}}[/itex]

    Then if we put 2) into equation 1), we get

    3)..... [itex]\int_{ - \infty }^{ + \infty } {\mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{{\textstyle{1 \over 2}}}}}}\,{e^{\frac{{ - {{(x - {x_1})}^2}}}{{{\Delta ^2}}}}}\delta ({x_1} - {x_0})} \,d{x_1}[/itex]

    But by the sifting property of the Dirac delta, 3) becomes:

    4)..... [itex]\mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{{\textstyle{1 \over 2}}}}}}\,{e^{\frac{{ - {{(x - {x_0})}^2}}}{{{\Delta ^2}}}}} = \,\,\delta (x - {x_0})[/itex]

    So gaussian Dirac deltas are self-convolutions of each other.

    But we also have:

    5)..... [itex]{e^{\frac{{ - {{(x - {x_1})}^2}}}{{{\Delta ^2}}}}} = \,\,\,{e^{\frac{{ - ({x_1} - x)}}{{{\Delta ^2}}}({x_1} - x)}}\,\, = \,\,{e^{ - ip({x_1} - x)}}\, = \,\,{e^{ - ipz}}[/itex]

    where [itex]p = \frac{{ - i({x_1} - x)}}{{{\Delta ^2}}}[/itex] and [itex]z = {x_1} - x[/itex], so that [itex]{x_1} = z + x[/itex] and [itex]d{x_1} = dz[/itex]

    And when 5) is put into 4), then equation 3) becomes:

    6)..... [itex]\int_{ - \infty }^{ + \infty } {\delta (x - {x_1})\delta ({x_1} - {x_0})} \,d{x_1}\,\, = \,\,\mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{{\textstyle{1 \over 2}}}}}}\int_{ - \infty }^{ + \infty } {{e^{ - ipz}}\delta (z + x - {x_0})} \,d{z}\, = \,\mathop {\lim }\limits_{\Delta \to 0} \frac{1}{{{{(\pi {\Delta ^2})}^{{\textstyle{1 \over 2}}}}}}\Im \{ \delta (z + x - {x_0})\} [/itex]

    which is [itex]\delta (x - {x_0})[/itex] by equation 1) above.

    I think this shows that Fourier Transform of a Dirac delta is another Dirac delta. Is this what is meant when I read that the FT is an automorphism for a tempered distribution like the Dirac deltas?
     
    Last edited: Aug 4, 2012
  8. Aug 5, 2012 #7

    rbj

    User Avatar

    3) is not an equation.

    can you make this appear more clear by not using subscripted variables for dummy variables? try using something else for [itex]x_1[/itex].

    also,

    [tex] \sqrt{x} = x^{\frac{1}{2}}[/tex]


    is how to write square root in LaTeX.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Must the Fourier transform exist for Dirac delta functions?
Loading...