Delta function – Fourier transform

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.26.

The Dirac delta function is defined by the two conditions

\displaystyle   \delta(x) \displaystyle  = \displaystyle  0\;\mathrm{if}\;x\ne0\ \ \ \ \ (1)
\displaystyle  \int_{-\infty}^{\infty}\delta(x)dx \displaystyle  = \displaystyle  1 \ \ \ \ \ (2)

Using Plancherel’s theorem, we can find the Fourier transform {\Delta(k)} of {\delta(x)}:

\displaystyle   \Delta(k) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\delta(x)e^{-ikx}dx\ \ \ \ \ (3)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}} \ \ \ \ \ (4)

The inverse of this relation is

\displaystyle   \delta(x) \displaystyle  = \displaystyle  \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\Delta(k)e^{ikx}dk\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2\pi}\int_{-\infty}^{+\infty}e^{ikx}dk \ \ \ \ \ (6)

This formula is clearly nonsense, since we’re integrating an oscillating function over an infinite range, so it doesn’t converge. Griffiths states that the formula “can be extremely useful, if handled with care.” I’m not entirely convinced by this; a dodgy formula is still a dodgy formula, no matter how carefully you handle it. Nevertheless, it’s used in a lot of quantum mechanics and the results do seem to be verified by experiment, so I guess we’ll have to accept it.

[A derivation of {\delta\left(x\right)} that makes this result look a bit more believeable is given here.]

11 thoughts on “Delta function – Fourier transform

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  3. John Boersma

    Also, it seems like the claim “The source of the problem is that the delta function doesn’t meet the requirement (square-integrabilty) for Plancherel’s theorem” is not correct. Shouldn’t delta^2 just integrate to one? Probably it’s safer to say that Plancherel’s theorem applies to true functions, and for generalized functions we just have to analyze the specifics, unless we really want to learn the theory for generalized functions. Make sense?

  4. John Boersma

    On further thought – I see that the the delta function is indeed not square-integrable. Limit of h^2 (hight) times 1/h (width) as h goes to infinity is infinite. Please disregard! Thanks for the excellent commentary on Griffiths on your site – very helpful!

  5. Anonymous

    “This formula is clearly nonsense, since we’re integrating an oscillating function over an infinite range, so it doesn’t converge. ”
    My idea is that if you put e^(ikx) = cos(kx)+isin(kx) (Euler’s formula),
    then the integration from infinity to -infinity simply means integrating one period of cos(kx)+isin(kx), that is integrate from pi -> -pi. This way the integral gives you 2pi only if k=0, 2pi/2pi = 1, and it is thus the same as the delta function.

    1. growescience

      I have tried to justify the formula using that very argument, but mathematically it still doesn’t work. A finite limit for the integral {lim_{kapparightarrowinfty}int_{-kappa}^{kappa}e^{ikx}dk} would require the integrand to tend to zero (uniquely) as {krightarrowpminfty}, which {e^{ikx}} just doesn’t do. The decision to cut the integral off after some number of complete cycles of the exponential is arbitrary; we could just as easily stop it at any other point, in which case the integral wouldn’t be zero. However, a delta function (in physics anyway) is never a final answer for an observable state; it’s always an intermediary in some calculation that involves an integral over the delta function, which does give a proper answer, so I suppose in that sense it’s useful.

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