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Chapter 15: Q2P (page 765)

Do Problem 6.4 with f(x) in (8.1).

Short Answer

Expert verified

The mean is equal to 0.

The variance is equal to $蟽^{2}$

Step by step solution

01

Given information

A function that is proportional to $e^{- a x^{2}}$ .

02

Definition of variance

A statistical measurement of the dispersion between values in a data collection is known as variance.

03

The density of harmonic oscillator

The probability density functions of a harmonic oscillator $f (x) = c (a) . e^{- a x^{2}}$ .

The density function is $g (x) = \frac{1}{\sqrt{2 蟺蟽^{2}}} e^{\frac{- x^{2}}{(2 蟽^{2})}}$

It is the only symmetric function proportional to $e^{- a x^{2}}$ that integrates to 1.

$\begin{array}{rcl} a & = & \frac{1}{2 蟽^{2}} c (a) \\ = & \frac{1}{\sqrt{2 蟺蟽^{2}}} \end{array}$

Hence the mean is equal to 0 and the variance is equal to $\begin{array}{rcl} 蟽^{2} \end{array}$

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Most popular questions from this chapter

By expanding $w (x, y, z)$ in a three-variable power series similarto , $(10.10)$ show that

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