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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Neal Jan 23 '15 at 1:41
  • $\begingroup$ 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} : $\hat{f}$ $\endgroup$ – Nick Jan 23 '15 at 1:42
  • $\begingroup$ I also suspect exactly what Neal has implied. Since this is a math forum, FYI, Fubini's theorem, en.wikipedia.org/wiki/Fubini%27s_theorem, allows you to switch the order of integration $\endgroup$ – Nick Jan 23 '15 at 1:44
  • $\begingroup$ That should be $dx'$ in the first integral. $\endgroup$ – Thomas Andrews Jan 23 '15 at 1:49
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Jan 23 '15 at 1:52

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:


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 $\delta(x)$.

  • $\begingroup$ 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? $\endgroup$ – kypalmer Jan 23 '15 at 2:20
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews Jan 23 '15 at 2:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.