0

The book "Introduction to Quantum Mechanics (Second edition)" by Griffiths says the following on pg 103:

$$|A|^2 \int\limits_{-\infty}^\infty e^{i(p-p')x/h}=|A|^2 2\pi h\delta(p-p')$$

Here $\delta$ is the delta function. How is this true? If $p\neq p'$, then the above formula says that the integral $\int\limits_{-\infty}^\infty e^{i(p-p')x/h}=0$. However, I think that the integral is not well-defined, as opposed to $0$.

Ryan Hendricks
  • 339
  • 3
    You are right it's not a well defined Riemann integral. However meant as a Fourier transform of a constant function it does makes sense. See e.g. https://math.stackexchange.com/questions/1595771/how-can-i-find-the-fourier-transform-of-constant-value-like-1 ; https://math.stackexchange.com/questions/2268921/what-is-the-fourier-transform-of-ft-1-or-simply-a-constant ; https://math.stackexchange.com/questions/559194/how-to-prove-that-inverse-fourier-transform-of-1-is-delta-funstion ; https://math.stackexchange.com/q/558077/147873 – Winther Jun 13 '19 at 21:32

1 Answers1

2

The meaning of this equality is that $$ \lim_{M\rightarrow\infty} \int_{-M}^M e^{i(p-p')x/\hbar} dx = 2\pi \hbar \delta(p-p') $$ where the limit was taken in the sense of distributions, that is for any function (that is sufficiently smooth and integrable) we have $$ \lim_{M\rightarrow\infty} \int_{-\infty}^{\infty} \Big(\int_{-M}^M e^{i(p-p')x/\hbar} dx \Big) f(p') dp' = \int_{-\infty}^\infty \Big(2\pi \hbar \delta(p-p') \Big) f(p') dp' = 2\pi \hbar f(p) $$ (in this equality $\lim$ is a normal limit in $\mathbb C$). You can check that $$ \int_{-M}^M e^{i(p-p')x/\hbar} dx = \frac{2 \sin\big((p-p')M/\hbar\big)}{(p-p')/\hbar}$$ and \begin{align} & \lim_{M\rightarrow\infty} \int_{-\infty}^{\infty} \frac{2 \sin\big((p-p')M/\hbar\big)}{(p-p')/\hbar} f(p') dp' =^{z:= (p-p')M/\hbar} \\ &= \lim_{M\rightarrow\infty} \hbar \int_{-\infty}^\infty \frac{2\sin z}{z} f(p - \frac{z\hbar}{M}) dz =^{\text{under some assumptions about }f} \\ &= \hbar \int_{-\infty}^\infty \frac{2\sin z}{z} f(p) dz = \\ &= 2\pi\hbar f(p)\end{align}

Adam Latosiński
  • 10,723