Heisenberg's Uncertainty Principle — Question 5

Question 5

An exponential position state obeys the bound without saturating it

A normalized state is $\psi(x)=\sqrt\kappa e^{-\kappa|x|}$, where $\kappa>0$. Use $\int_0^\infty x^2e^{-bx}dx=2/b^3$ and the momentum second-moment form $\langle p^2\rangle=\hbar^2\int|\psi\prime|^2dx$. Both means vanish by symmetry. All uncertainties here are standard deviations, with $\sigma_x\sigma_p\ge\hbar/2$.

1. Find the position variance.

2. Find the momentum variance and compare the uncertainty product with the Gaussian limit.

Question 5 - Solution

An exponential position state obeys the bound without saturating it

(a) Method and calculation

$$\sigma_x^2=2\kappa\int_0^\infty x^2e^{-2\kappa x}dx=\frac1{2\kappa^2}.$$$$\boxed{\sigma_x=1/(\sqrt2\kappa)}.$$

(b) Interpretation and further calculation

$$|\psi\prime|^2=\kappa^2|\psi|^2\quad\text{away from }0\quad\Rightarrow\quad\boxed{\sigma_p=\hbar\kappa}.$$$$\boxed{\sigma_x\sigma_p=\hbar/\sqrt2>\hbar/2}.$$

Physical check

The cusp has a square-integrable first derivative, so the momentum second moment is finite. It must not be confused with a discontinuous top-hat wave function.