Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problems 1.11-12.
This exercise in continuous probability distributions actually precedes the problem on Buffon’s needle, so it uses the same logic.
Suppose we have a needle mounted on a pivot so that the needle is free to swing anywhere in the top semicircle, so that when it comes to rest, its angular coordinate is equally likely to be any value between 0 and . In that case, the probability density is a constant in this range, and zero outside it. That is
From normalization, we must have
The statistics of the distribution are
We now want the probability that the projection of the needle onto the axis lies between and . If the needle is at angle , then its coordinate is (where is the length of the needle). If the angle changes by , its coordinate changes by so for the probability density, we take absolute values and get
As a check:
Since , we can get and from 8 and 9, but we can also calculate it the hard way, using :
Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.10.
Here are a few statistical properties of the first 25 digits of (if you want more digits, here’s a link to the first million digits):
The frequency of each digit and the probability of getting each one are:
The most probable digit is 3, the median is 4 (there are 10 digits and 12 digits so that’s as close as we can get to dividing the distribution equally) and the average is 4.72.
We can get the variance by calculating , so we get ; . The standard deviation is
We’d need to use quite a few more digits to get a properly random collection of numbers.
Required math: algebra, calculus (partial derivatives and integration by parts), complex numbers
Required physics: Schrödinger equation, probability density
Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.9.
Suppose a particle is in the quantum state
where is the normalization constant and is a constant with dimensions of 1/time. We can find from normalization:
The spatial component of the wave function is
and it must satisfy the time-independent Schrödinger equation in one dimension
The energy can be found from the time equation:
From 7 we have
This is the harmonic oscillator potential, and the wave function is actually the ground state of that potential.
We can work out a few average values:
since is even.
The standard deviations are
and the uncertainty principle is
so in this case, the uncertainty is the minimum possible.