Heisenberg's Uncertainty Principle — Question 4

Question 4

A perfectly sharp aperture has finite position width but divergent momentum variance

Immediately after an ideal slit, $\psi=1/\sqrt a$ for $|x|<a/2$ and zero elsewhere. Its momentum density is $|\widetilde\psi(p)|^2=[a/(2\pi\hbar)]\operatorname{sinc}^2(pa/2\hbar)$, where $\operatorname{sinc}z=\sin z/z$.

1. Find $\sigma_x$ and the first nonzero momentum at which the diffraction density vanishes.

2. Determine whether that first-zero scale is a finite standard deviation $\sigma_p$.

Question 4 - Solution

A perfectly sharp aperture has finite position width but divergent momentum variance

(a) Method and calculation

$$\langle x\rangle=0,\quad\sigma_x^2=\frac1a\int_{-a/2}^{a/2}x^2dx=\frac{a^2}{12},\quad\boxed{\sigma_x=a/\sqrt{12}}.$$$$pa/(2\hbar)=\pi\quad\Rightarrow\quad\boxed{p_{\mathrm{zero}}=2\pi\hbar/a=h/a}.$$

(b) Interpretation and further calculation

$$p^2|\widetilde\psi(p)|^2=\frac{2\hbar}{\pi a}\sin^2\!\left(\frac{pa}{2\hbar}\right).$$$$\int_{-\infty}^{\infty}p^2|\widetilde\psi(p)|^2dp=\infty\quad\Rightarrow\quad\boxed{\text{no finite momentum variance}}.$$

Physical check

The first diffraction zero is a useful width scale but is not a standard deviation. An ideal discontinuous aperture produces long momentum tails; smoothing its edges changes those tails.