Download Statistical Physics Problems: Gaussian Distribution and Harmonic Oscillator and more Assignments Physics in PDF only on Docsity!
These problems will serve the purpose of letting you carry out some statistical
physics calculations for yourself. Hopefully you will realize how powerful
these methods can be once you get over your natural fear of all those
integrals!
Problem 1.
One very commonly encountered probability density is the Gaussian or
normal distribution given by
w ( x ) = C e
!
x
2
2 "
2
.
Suppose we are measuring the x-position of a particle and the results scatter
around according to a Gaussian distribution, with any value of x between
negative infinity and positive infinity being possible, but those near where the
distribution is maximum (i.e near zero) being the most likely. Remember
what w ( x ) actually means. The probability of finding the particle to lie
between position x and position x + dx is just given by P ( x , x + dx ) = w ( x ) dx.
(a.) To become familiar with this function, calculate the value of the
normalization constant C^. If you look up the integral you will find that
e
! a
2 x
2
dx =
"
a
!#
(b.) Show that the average value of x is 0, i.e x = 0 , as you would expect.
(c.) Show that x
2 =!
2
. The fact that x
2
!"
"
e
!
x
2
2 $
2
dx =
2 a
3
should be
useful. You can get this from the first integral by taking the derivative of
both sides with respect to a , which brings down a factor of x
2 , inside the
integral, which is a nifty trick!
(d.) Show that the root-mean-square spread in the measured values of x will
just equal!. Root-mean-square spread means the square root of the average
of the quantity ( x! x )
2 , so in this case you already know the answer!
Problem 2.
Probably the most fundamental and useful result from all of statistical physics
is that the probability of finding a system in a particular state (called it state #
i ) is proportional to e
!
Ei
kBT , where E i
is the energy of the system when it is in
state i. This theorem applies to any system that is in thermal equilibrium at
temperature T.
Of course, to find the actual probability we must impose the requirement that
the sum of the probabilities for all possible states must be exactly 1.000,
because the system must be in one of its states.
For quantum systems the idea of a state is naturally well-defined. For
example, a one-dimensional harmonic oscillator has states that are defined by
a single integer, n. When the oscillator is in state # n , it has energy
E
n
= ( n + 1 / 2 )!! , where! is the natural angular frequency of the oscillator,
and! is Planck’s constant divided by 2! , which is an exceedingly small
number (1.05! 10
" 34 Joule seconds). The fact that! is so small explains
why we never notice that we can’t give a harmonic oscillator whatever energy
we like, but that actually we can only change its energy by one unit at a time.
The lowest possible energy an oscillator can have is not zero as you might
have expected but is instead equal to half the energy difference between
levels.
(a.) Using the fact that the oscillator must be in one of its states, i.e. the sum
of all the probabilities must be exactly 1.000, show that the probability of
finding an oscillator (that is in thermal equilibrium at temperature T ) to be in
state # n is given by P ( n ) = 1! e
!
! w
kBT
e
!
n! (
kBT
. (Hint: You are trying to sum a
geometric series of the form 1 + x + x
2
(b.) Using the result of part (a.) find the average energy the oscillator has at
temperature T. (You should get
!! e
"
!!
kBT
1 " e
"
!!
kBT
, if you are sufficiently
careful!) You will find it useful to know that n e
! nx
n = 0
n ="
e
! x
1! e
! x $ %
2
. If you
are mathematically inclined, you can prove this by taking the derivative of
