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I'm looking at Fourier Transforms in a Quantum Physics sense, and it's useful to associate the Fourier Series with the Dirac Delta. The book I'm using follows this argument (Shankar, Quantum Mechanics):

The Dirac Delta has the following property:

$\int \delta(x-x^{\prime})f(x^{\prime}) \,\,\mathrm{d}x^{\prime} = f(x)$

We can represent a function by it's transform in the frequency domain:

$\hat{f}(k) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x)\,\, \mathrm dx$

and can also perform an inverse transform:

$f(x^{\prime})=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \hat{f}(k)\,\,\mathrm{d} k$

Substituting the first transform in the second:

$f(x^{\prime})=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x) \,\,\mathrm{d} x \right) \,\,\mathrm dk$

Now, in the discussion that I'm reading, this is rearranged as:

$f(x^{\prime})=\int_{-\infty}^{\infty} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathrm{d} k\,\, e^{jk(x^{\prime}-x)}\right)f(x) \,\,\mathrm dx$

Which, by comparison with the first function allows us to associate the parenthesized part of this equation with the dirac delta.

My Question:

Why can we remove the integrand from the inside of the dk integral, since it clearly depends on k, and $x-x^{\prime}$ is not necessarily zero?

kypalmer
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    Are they totally consistent with standard $\int (\mbox{integrand})\ dx$ through the rest of the book? I have seen physicists write $\int dx\ (\mbox{integrand})$ before. If that's the case, then they just changed the order of integration. – Neal Jan 23 '15 at 01:41
  • Should you have i in the exponential instead of j? Also, if you want to put a hat symbol on top of the f, to denote the Fourier transform of f, use \hat{f} : $\hat{f}$ – Clyde Jan 23 '15 at 01:42
  • I also suspect exactly what Neal has implied. Since this is a math forum, FYI, Fubini's theorem, http://en.wikipedia.org/wiki/Fubini%27s_theorem, allows you to switch the order of integration – Clyde Jan 23 '15 at 01:44
  • That should be $dx'$ in the first integral. – Thomas Andrews Jan 23 '15 at 01:49
  • There's a lot of notation error in the above. It is not true that $f(k)=\dots$ in the second integral, either, but rather $\hat{f}(k)=\dots$, and the next integral also should be $\hat{f}(k)$ inside the integral. – Thomas Andrews Jan 23 '15 at 01:52
  • Neal, this may be the case. The notation in the book, however, is exactly as it appears in my last equation, with the transposition of the inner integrand, while not on the outer. – kypalmer Jan 23 '15 at 02:07
  • What @Neal said is correct, this is common notation in many physics texts. One can see this as $e^{ik(x-x')}$ does not make sense outside of an integeral (since $k$ does not have a meaning outside of the integral) – Winther Jan 23 '15 at 02:20
  • Fubini's theorem does not justify these shenanigans. It tells us that $\int \int g(x,k) dx dk = \int \int g(x,k) dk dx$ provided that $\int \int |g(x,k)| dx dk < \infty$. But that is not the case here. Putting $g(x,k) = e^{ikx'}e^{-ikx}f(x)$, we have $|g(x,k)| = |f(x)|$, so the iterated integral of $|g|$ won't be finite unless $f$ is zero almost everywhere. –  Jan 23 '15 at 02:26
  • So @Bungo, as long as $f(x)$ is Lebesgue integrable, we should have no problem, right? What about if $x \ne x^{\prime}$? – kypalmer Jan 23 '15 at 17:16
  • @kypalmer: the integral $\int_{-\infty}^{\infty}e^{ik(x-x')}dk$ does not converge, regardless of whether $x=x'$ or $x \neq x'$. In the first case, the integrand is simply $1$, and so the integral is $\infty$. In the second case, the integrand is $\cos(k(x-x'))+i\sin(k(x-x'))$ so both the real and imaginary parts oscillate periodically and therefore the integral from $-\infty$ to $\infty$ does not converge. None of this changes if you assume Lebesgue vs. Riemann integration. –  Jan 23 '15 at 18:01
  • @Bungo If we require that $f(x)$ converge, should not the product with a sinusoid converge? – kypalmer Jan 23 '15 at 23:59
  • @kypalmer: $\int |f(x)| dx$ converges if $f$ is integrable, but if the value isn't zero then the double integral $\int \left(\int |f(x)| dx\right) dk$ does not converge. –  Jan 24 '15 at 02:53

1 Answers1

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The steps missing:

$$\begin{align}f(x^{\prime})&=\int_{-\infty}^{\infty}e^{jkx^{\prime}} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-jkx}f(x) \,\,\mathrm{d} x \right) \,\,\mathrm dk\tag{0}\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dx\,dk\tag{1}\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dk\,dx\tag{2}\\ &=\int_{-\infty}^{\infty} \left(\frac{1}{2\pi} \int_{-\infty}^{\infty}\mathrm{d} k\,\, e^{jk(x^{\prime}-x)}\right)f(x) \,\,\mathrm dx\tag{3} \end{align}$$

The original proof jumped from (0) to (3).

(1) is bringing the $e^{jkx'}$into the integral, which you can do by distributive law.

(2) is a switching in the order of integration.

(3) is pulling out the $f(x)$ and $\frac{1}{2\pi}$ out, because it is a constant in the inner integral:

$$\int_{-\infty}^{\infty}\frac{1}{2\pi}e^{jk(x'-x)}f(x)\,dk$$

None of this is valid, mathematically, but it all can be made rigorous by doing the work in "distribution theory." Most physicists don't give a damn about that part, though, because they are mostly dealing with wave functions that are "close enough to" $\delta(x)$, not actually $\delta(x)$.

Thomas Andrews
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  • In Eq. 3, you have the integrand transposed with the differential, is this just a matter of style as Neal suggested in the comments of my question? – kypalmer Jan 23 '15 at 02:20
  • Yeah, that's irrelevant. I just copied to the expression directly from your proof, so I didn't even notice, but that order is irrelevant. Was just watching MIT QM lectures the other day where the prof kept writing $\int dx f(x)$. Makes the variable of integration clearer. – Thomas Andrews Jan 23 '15 at 02:22