= Solution
The density $|\Psi_0|^2=\sqrt{\alpha/\pi}\,e^{-\alpha x^2}$ is even. Hence $\langle x\rangle=0$; also $\Psi_0'=-\alpha x\Psi_0$ gives $\langle p_x\rangle=i\hbar\alpha\langle x\rangle=0$.
To evaluate the second position moment, differentiate the <Gaussian integral> with respect to its positive parameter:
$$
\begin{aligned}
G(\alpha)&=\int_{-\infty}^{\infty}e^{-\alpha x^2}\,\mathrm dx=\sqrt\pi\,\alpha^{-1/2},\\
\int_{-\infty}^{\infty}x^2e^{-\alpha x^2}\,\mathrm dx
&=-G'(\alpha)=\frac{\sqrt\pi}{2\alpha^{3/2}},\\
\langle x^2\rangle&=\sqrt{\frac\alpha\pi}\frac{\sqrt\pi}{2\alpha^{3/2}}
=\frac1{2\alpha}=\frac{\hbar}{2m\omega}.
\end{aligned}
$$
For the momentum second moment, apply the <momentum operator> twice:
$$
\langle p_x^2\rangle=-\hbar^2\int\Psi_0^*\Psi_0''\,\mathrm dx
=\hbar^2\left(\alpha-\alpha^2\langle x^2\rangle\right)
=\frac{\hbar^2\alpha}{2}=\frac{m\hbar\omega}{2}.
$$
Both means vanish, so
$$
\boxed{\Delta x=\sqrt{\frac{\hbar}{2m\omega}},\qquad
\Delta p_x=\sqrt{\frac{m\hbar\omega}{2}},\qquad
\Delta x\,\Delta p_x=\frac\hbar2.}
$$
The <ground state> attains the bound proved above and is therefore a <minimum uncertainty state>. A wider Gaussian reduces momentum spread while increasing position spread: the product remains at this lower limit. Other states, including excited oscillator eigenstates, generally give a larger product.
Back to article page
