# Standard deviation problem

## Homework Statement

Problem 1.2 from the book Introduction to Quantum Mechanics (2e) by Griffiths:

Suppose a rock is dropped off a cliff of height h. As it

falls, a million photos are snapped, at random intervals. On each picture the distance the rock has fallen is measured.

a) Find the standard deviation of the distribution.

## The Attempt at a Solution

Starting with the equation of motion (assuming it falls in the positive x direction):

$$x(t)=x_0+v_0(t)+\frac{1}{2}at^2=\frac{1}{2}gt^2$$

The time T it takes to fall:

$$h=\frac{1}{2}gt^2$$
$$T=\sqrt{\frac{2h}{g}}$$

Probability a picture is taken in interval dt:

$$\frac{dt}{T}=dt \sqrt{\frac{g}{2h}}=\frac{dx}{gt}\sqrt{\frac{g}{2h}}=\frac{1}{2h}\frac{1}{\sqrt{gt}}dx=\frac{1}{2\sqrt{hx}}dx$$

So the probability density is:

$$\rho(x)=\frac{1}{2\sqrt{hx}}$$

Expectation Value:

$$\mu=\int_0^h x \rho(x)dx=\int_0^h x \frac{1}{2\sqrt{hx}}dx=\frac{1}{2\sqrt{h}}\frac{2}{3}x^\frac{3}{2}=\frac{h}{3}$$

Standard deviation:

$$\sigma^2=\int_0^h (x-\mu)^2 \rho(x)dx$$
$$\sigma^2=\int_0^h (x-\frac{h}{3})^2 \frac{1}{2\sqrt{hx}}dx=\frac{4}{45}h^2$$
$$\sigma=\frac{2h}{\sqrt{45}}$$

Look right?

Consider the Gaussian distribution $$\rho(x)=Ae^{-\lamba(x-a)^2}$$ where A, α, and λ are positive real constants.

Find <x>, <x²> and std dev.

$$<x>=\int_{-\infty}^{\infty} xAe^{-\lamba(x-a)^2}dx$$

$$<x^2>=\int_{-\infty}^{\infty} x^2Ae^{-\lamba(x-a)^2}dx$$

$$\sigma^2=<x^2>-<x>^2$$

I tried using integration by parts to evaluate the integrals, but I didn't get anywhere. I also tried looking them up on a table of integrals - also no luck.

I'm guessing <x>=a because a is the peak in a Gaussian distribution. But how do I show it?

Dick