# The Wave Function Exercises 01

Exercise 1

• ${^2=21^2=441}$

${=1/N\sum j^2N(J)=\dfrac{6434}{14}=459.6}$

• Calculating for each ${\Delta j}$
 ${j}$ ${\Delta j=j-}$ 14 ${14-21=-7}$ 15 ${15-21=-6}$ 16 ${16-21=-5}$ 22 ${22-21=1}$ 24 ${24-21=3}$ 25 ${25-21=4}$

Hence for the variance it follows

${\sigma ^2=1/N\sum (\Delta j)^2N(j)=\dfrac{260}{14}=18.6}$

Hence the standard deviation is

$\displaystyle \sigma =\sqrt{18.6}=4.3$

• ${\sigma^2=-^2=459.6-441=18.6}$

And for the standard deviation it is

$\displaystyle \sigma =\sqrt{18.6}=4.3$

Which confirms the second equation for the standard deviation.

Exercise 2 Consider the first ${25}$ digits in the decimal expansion of ${\pi}$.

• What is the probability of getting each of the 10 digits assuming that one selects a digit at random.

The first 25 digits of the decimal expansion of ${\pi}$ are

$\displaystyle \{3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3\}$

Hence for the digits it is

 ${N(0)=0}$ ${P(0)=0}$ ${N(1)=2}$ ${P(1)=2/25}$ ${N(2)=3}$ ${P(2)=3/25}$ ${N(3)=5}$ ${P(3)=1/5}$ ${N(4)=3}$ ${P(4)=3/25}$ ${N(5)=3}$ ${P(5)=3/25}$ ${N(6)=3}$ ${P(6)=3/25}$ ${N(7)=1}$ ${P(7)=1/25}$ ${N(8)=2}$ ${P(8)=2/25}$ ${N(9)=3}$ ${P(9)=3/25}$
• The most probable digit is ${5}$. The median digit is ${4}$. The average is ${\sum P(i)N(i)=4.72}$.

• ${\sigma=2.47}$

 Exercise 3 The needle on a broken car is free to swing, and bounces perfectly off the pins on either end, so that if you give it a flick it is equally likely to come to rest at any angle between ${0}$ and ${\pi}$. Along the ${\left[0,\pi\right]}$ interval the probability of the needle flicking an angle ${d\theta}$ is ${d\theta/\pi}$. Given the definition of probability density it is ${\rho(\theta)=1/\pi}$. Additionally the probability density also needs to be normalized. $\displaystyle \int_0^\pi \rho(\theta)d\theta=1\Leftrightarrow\int_0^\pi 1/\pi d\theta=1$ which is trivially true. The plot for the probability density is Compute ${\left\langle\theta \right\rangle}$, ${\left\langle\theta^2 \right\rangle}$ and ${\sigma}$. {\begin{aligned} \left\langle\theta \right\rangle &= \int_0^\pi\frac{\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\theta d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^2}{2} \right]_0^\pi\\ &= \frac{\pi}{2} \end{aligned}} For ${\left\langle\theta^2 \right\rangle}$ it is {\begin{aligned} \left\langle\theta^2 \right\rangle &= \int_0^\pi\frac{\theta^2}{\pi}d\theta\\ &= \frac{1}{\pi} \left[ \frac{\theta^3}{3} \right]_0^\pi\\ &= \frac{\pi^2}{3} \end{aligned}} The variance is ${\sigma^2=\left\langle\theta^2 \right\rangle-\left\langle\theta\right\rangle^2 =\dfrac{\pi^2}{3}-\dfrac{\pi^2}{4}=\dfrac{\pi^22}{12}}$. And the standard deviation is ${\sigma=\dfrac{\pi}{2\sqrt{3}}}$. Compute ${\left\langle\sin\theta\right\rangle}$, ${\left\langle\cos\theta\right\rangle}$ and ${\left\langle\cos^2\theta\right\rangle}$. {\begin{aligned} \left\langle\sin\theta \right\rangle &= \int_0^\pi\frac{\sin\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\sin\theta d\theta\\ &= \frac{1}{\pi} \left[ -\cos\theta \right]_0^\pi\\ &= \frac{2}{\pi} \end{aligned}} and {\begin{aligned} \left\langle\cos\theta \right\rangle &= \int_0^\pi\frac{\cos\theta}{\pi}d\theta\\ &= \frac{1}{\pi}\int_0^\pi\cos\theta d\theta\\ &= \frac{1}{\pi} \left[ \sin\theta \right]_0^\pi\\ &= 0 \end{aligned}} We’ll leave ${\left\langle\cos\theta^2 \right\rangle}$ as an exercise for the reader. As a hint remember that ${\cos^2\theta=\dfrac{1+\cos(2\theta)}{2}}$.

 Exercise 4 In exercise ${1.1}$ it was shown that the the probability density is $\displaystyle \rho(x)=\frac{1}{2\sqrt{hx}}$ Hence the mean value of ${x}$ is {\begin{aligned} \left\langle x \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{h}{3} \end{aligned}} For ${\left\langle x^2 \right\rangle}$ it is {\begin{aligned} \left\langle x^2 \right\rangle &= \int_0^h\frac{x}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\int_0^h x^{3/2}dx\\ &= \frac{1}{2\sqrt{h}}\left[\frac{2}{5}x^{5/2} \right]_0^h\\ &= \frac{h^2}{5} \end{aligned}} Hence the variance is $\displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{h^2}{5}-\frac{h^2}{9}=\frac{4}{45}h^2$ and the standard deviation is $\displaystyle \sigma=\frac{2h}{3\sqrt{5}}$ For the distance to the mean to be more than one standard deviation away from the average we have two alternatives. The first is the interval ${\left[0,\left\langle x \right\rangle+\sigma\right]}$ and the second is ${\left[\left\langle x \right\rangle+\sigma,h\right]}$. Hence the total probability is the sum of these two probabilities. Let ${P_1}$ denote the probability of the first interval and ${P_2}$ denote the probability of the second interval. {\begin{aligned} P_1 &= \int_0^{\left\langle x \right\rangle-\sigma}\frac{1}{2\sqrt{hx}}dx\\ &= \frac{1}{2\sqrt{h}}\left[2x^{1/2} \right]_0^{\left\langle x \right\rangle-\sigma}\\ &= \frac{1}{\sqrt{h}}\sqrt{\frac{h}{3}-\frac{2h}{3\sqrt{5}}}\\ &=\sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}} \end{aligned}} Now for the second interval it is {\begin{aligned} P_2 &= \int_{\left\langle x \right\rangle+\sigma}^h\frac{1}{2\sqrt{hx}}dx\\ &= \ldots\\ &=1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}} \end{aligned}} Hence the total probability ${P}$ is ${P=P_1+P_2}$ {\begin{aligned} P&=P_1+P_2\\ &= \sqrt{\frac{1}{3}-\frac{2}{3\sqrt{5}}}+1-\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}}\\ &\approx 0.3929 \end{aligned}}

 Exercise 5 The probability density is ${\rho(x)=Ae^{-\lambda(x-a)^2}}$ Determine ${A}$. Making the change of variable ${u=x-a}$ (${dx=du}$) the normalization condition is {\begin{aligned} 1 &= A\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &= A\sqrt{\frac{\pi}{\lambda}} \end{aligned}} Hence for ${A}$ it is $\displaystyle A=\sqrt{\frac{\lambda}{\pi}}$ Find ${\left\langle x \right\rangle}$, ${\left\langle x^2 \right\rangle}$ and ${\sigma}$. {\begin{aligned} \left\langle x \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty ue^{-\lambda u^2}du+a\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &=\sqrt{\frac{\lambda}{\pi}}\left( 0+a\sqrt{\frac{\pi}{\lambda}} \right)\\ &= a \end{aligned}} If you don’t see why ${\displaystyle\int_{-\infty}^\infty ue^{-\lambda u^2}du=0}$ check this post on my other blog. For ${\left\langle x^2 \right\rangle}$ it is {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right) \end{aligned}} Now ${\displaystyle 2a\int_{-\infty}^\infty u e^{-\lambda u^2}du=0}$ as in the previous calculation. For the third term it is ${\displaystyle a^2\int_{-\infty}^\infty e^{-\lambda u^2}du=a^2\sqrt{\frac{\pi}{\lambda}}}$. The first integral is the hard one and a special technique can be employed to evaluate it. {\begin{aligned} \int_{-\infty}^\infty u^2e^{-\lambda u^2}du &= \int_{-\infty}^\infty-\frac{d}{d\lambda}\left( e^{-\lambda u^2} \right)du\\ &= -\frac{d}{d\lambda}\int_{-\infty}^\infty e^{-\lambda u^2}du\\ &=-\frac{d}{d\lambda}\sqrt{\frac{\pi}{\lambda}}\\ &=\frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}} \end{aligned}} Hence it is {\begin{aligned} \left\langle x^2 \right\rangle &= \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^\infty (u+a)^2e^{-\lambda u^2}du\\ &= \sqrt{\frac{\lambda}{\pi}}\left(\int_{-\infty}^\infty u^2e^{-\lambda u^2}du+2a\int_{-\infty}^\infty u e^{-\lambda u^2}du+a^2\int_{-\infty}^\infty e^{-\lambda u^2}du \right)\\ &= \sqrt{\frac{\lambda}{\pi}}\left( \frac{1}{2}\sqrt{\frac{\pi}{\lambda^3}}+0+a^2\sqrt{\frac{\pi}{\lambda}} \right)\\ &=a^2+\frac{1}{2\lambda} \end{aligned}} The variance is $\displaystyle \sigma^2=\left\langle x^2 \right\rangle-\left\langle x \right\rangle^2=\frac{1}{2\lambda}$ Hence the standard deviation is $\displaystyle \sigma=\frac{1}{\sqrt{2\lambda}}$

— Mathematica file —

The resolution of exercise 2 was done using some basic Mathematica code which I’ll post here hoping that it can be helpful to the readers of this blog.

// N[Pi, 25]

piexpansion = IntegerDigits[3141592653589793238462643]

digitcount = {}

For[i = 0, i <= 9, i++, AppendTo[digitcount, Count[A, i]]]

digitcount

digitprobability = {}

For[i = 0, i <= 9, i++, AppendTo[digitprobability, Count[A, i]/25]]

digitprobability

digits = {}

For[i = 0, i <= 9, i++, AppendTo[digits, i]]

digits

j = N[digits.digitprobability]

digitssquared = {}

For[i = 0, i <= 9, i++, AppendTo[digitssquared, i^2]]

digitssquared

jsquared = N[digitssquared.digitprobability]

sigmasquared = jsquared - j^2

std = Sqrt[sigmasquared]

deviations = {}

deviations = piexpansion - j

deviationssquared = (piexpansion - j)^2

variance = Mean[deviationssquared]

standarddeviation = Sqrt[variance]