# Uncertainty in Operators | Quantum Mechanics

Quantum Mechanics
Basic Quantum Mechanics

Uncertainty in Operators | Quantum Mechanics

$\triangle$ A = uncertainty in operator A, and it is calculated as followes

$\triangle A = \sqrt{<A^{2}>-<A>^{2}}$

Problem:

Calculate the expectation value of position and momentum of a particle represented by the wavefunction:

$\psi(x) = \left\{ \begin{array}{c} \frac{1}{\sqrt{L}}sin(\frac{\pi x}{L}),\,0<x<L\\ 0,\,otherwise \end{array}\right\}$

Solution:

$<\hat{x}>$ = $\frac{\intop_{-\infty}^{\infty}\psi^{*}(x\psi)dx}{\intop_{-\infty}^{\infty}\psi^{*}\psi dx}$

= $\frac{\intop_{-\infty}^{0}\psi^{*}(x\psi)dx\,+\,\intop_{0}^{L}\psi^{*}(x\psi)dx\,+\intop_{L}^{\infty}\psi^{*}(x\psi)dx\,}{\intop_{-\infty}^{\infty}\psi^{*}\psi dx}$

= $\frac{\frac{1}{L}\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})xdx}{\frac{1}{L}\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$

= $\frac{\frac{1}{2}\intop_{0}^{L}(1-cos\frac{2\pi x}{L})xdx}{\frac{1}{2}\intop_{0}^{L}(1-cos\frac{2\pi x}{L})dx}$

= $\frac{[\frac{x^{2}}{2}]{}_{0}^{L}-\intop_{0}^{L}(xcos\frac{2\pi x}{L})dx}{[x-\frac{L}{2\pi}sin\frac{2\pi x}{L}]_{0}^{L}}$

= $\frac{\frac{L^{2}}{2}-\{[x\frac{L}{2\pi}sin\frac{2\pi x}{L}]_{0}^{L}\,-\,\intop_{0}^{L}[\frac{L}{2\pi}sin\frac{2\pi x}{L}dx]\}}{L}$

= $\frac{\frac{L^{2}}{2}-\{0-\frac{L}{2\pi}\frac{L}{2\pi}[cos\frac{2\pi x}{L}]_{0}^{L}\,\}}{L}$

= $\frac{\frac{L^{2}}{2}-\frac{L^{2}}{4\pi}(1-1)}{L}$

=  $\frac{L}{2}$

$<\hat{x}>$ = $\frac{L}{2}$

Now, Expectation Value of Momentum Operator,

$<\hat{p}>$ = $\frac{\intop_{-\infty}^{\infty}\psi^{*}(p\psi)dx}{\intop_{-\infty}^{\infty}|\psi|^{2}dx}$

= $\frac{\frac{1}{L}\intop_{0}^{L}sin(\frac{\pi x}{L})\{-i\hbar\frac{d}{dx}sin(\frac{\pi x}{L})\}dx}{\frac{1}{L}\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$

= $\frac{\frac{1}{L}\intop_{0}^{L}sin(\frac{\pi x}{L})\{-i\hbar\frac{d}{dx}sin(\frac{\pi x}{L})\}dx}{\frac{1}{L}\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$

= $\frac{\frac{-i\hbar\pi}{L}\intop_{0}^{L}sin(\frac{\pi x}{L})cos(\frac{\pi x}{L})dx}{\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$

= $\frac{\frac{-i\hbar\pi}{2L}\intop_{0}^{L}sin^{2}(\frac{2\pi x}{L})dx}{\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$

= $\frac{\frac{-i\hbar\pi}{2L}\left[\frac{-cos(\frac{2\pi x}{L})}{\frac{2\pi}{L}}\right]_{0}^{L}}{\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$

= $\frac{\frac{i\hbar\pi}{2L}(cos2\pi-cos0)}{\intop_{0}^{L}sin^{2}(\frac{\pi x}{L})dx}$ = 0