Heisenberg's Uncertainty Principle — Question 1

Question 1

A real Gaussian saturates the position-momentum uncertainty bound

At one instant, $\psi(x)=A\exp[-x^2/(4a^2)]$ on the full real line, where $a>0$ and A is positive. Use $\int e^{-x^2/(2a^2)}dx=\sqrt{2\pi}a$ and $\int x^2e^{-x^2/(2a^2)}dx=\sqrt{2\pi}a^3$, with both integrals over all x. For this state $\langle p^2\rangle=\hbar^2\int|\psi\prime|^2dx$. All uncertainties here are standard deviations, with $\sigma_x\sigma_p\ge\hbar/2$.

1. Normalize the state and find $\sigma_x$.

2. Find $\sigma_p$ and the uncertainty product.

Question 1 - Solution

A real Gaussian saturates the position-momentum uncertainty bound

(a) Method and calculation

$$1=A^2\sqrt{2\pi}a\quad\Rightarrow\quad\boxed{A=(2\pi a^2)^{-1/4}}.$$$$\langle x\rangle=0,\quad\langle x^2\rangle=a^2\quad\Rightarrow\quad\boxed{\sigma_x=a}.$$

(b) Interpretation and further calculation

$$\psi\prime=-\frac{x}{2a^2}\psi,\quad\langle p\rangle=0,\quad\langle p^2\rangle=\frac{\hbar^2}{4a^4}\langle x^2\rangle=\frac{\hbar^2}{4a^2}.$$$$\boxed{\sigma_p=\frac{\hbar}{2a}},\qquad\boxed{\sigma_x\sigma_p=\frac\hbar2}.$$

Physical check

The parameter a is the standard deviation of the probability density, not of the amplitude profile. Confusing those profiles produces a factor-of-square-root-of-two error.