Answered: Calculate the uncertainties dr = V(x2)…

Calculate the uncertainties dr = V(x2) and op = V(p²) for %3D a particle confined in the region -a < a < a. The wave function of the particle is: 1 (x) = COS -a a, r<-a. %3D

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Question

Calculate the uncertainties dr = V(x2) and op = V(p²) for
%3D
a particle confined in the region -a < a < a. The wave function of the
particle is:
1
(x) =
COS
-a <r < a,
Va
2a
(x) = 0,
a > a,
r<-a.
%3D

Expert Solution

Step 1

Given:

The wavefunction of the particle is:

Introduction:

The uncertainty principle is a mathematical inequality that asserts a fundamental limit to the precision with which the values for certain pairs of physical quantities of a particle. Such pairs of variables are known as canonically conjugate pairs.

Step 2

Calculation:

Calculate the value of $<x^{2}>$ .

$<x^{2}> = \int_{- a}^{a} x^{2} \frac{1}{a} \cos^{2} (\frac{πx}{2 a}) d x <x^{2}> = \frac{1}{2 a} \int_{- a}^{a} x^{2} (1 + \cos (\frac{πx}{a})) d x <x^{2}> = \frac{1}{2 a} \int_{- a}^{a} x^{2} (1 + \cos (\frac{πx}{a})) d x <x^{2}> = \frac{1}{2 a} {[\frac{x^{3}}{3}]}_{- a}^{a} + \frac{1}{2 a} \int_{- a}^{a} x^{2} (\cos (\frac{πx}{a})) d x <x^{2}> = \frac{1}{2 a} [\frac{2 a^{3}}{3}] + \frac{1}{2 a} x^{2} \int_{- a}^{a} (\cos (\frac{πx}{a})) d x - \frac{2 a}{2 a π} \int_{- a}^{a} x (\sin (\frac{πx}{a})) d x <x^{2}> = \frac{1}{2 a} [\frac{2 a^{3}}{3}] + \frac{1}{2 a} x^{2} {(\sin (\frac{πx}{a}))}_{- a}^{a} + \frac{1}{π} x {(\cos (\frac{πx}{a}))}_{- a}^{a} <x^{2}> = \frac{1}{2 a} [\frac{2 a^{3}}{3}] + \frac{1}{π} x {(\cos (\frac{πx}{a}))}_{- a}^{a} <x^{2}> = [\frac{a^{2}}{3}]$

Calculate the value of $\sqrt{<x^{2}>}$ .

$\sqrt{<x^{2}>} = \sqrt{\frac{a^{2}}{3}}$

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