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Statistical Mechanics - Physics at Oregon State University

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A.1. SOLUTIONS FOR CHAPTER 1. 235<br />

The energy eigenst<strong>at</strong>es of a harmonic oscill<strong>at</strong>or are ɛn = ω(n + 1<br />

2 ) for n =<br />

0, 1, 2, · · · Consider a system of N such oscill<strong>at</strong>ors. The total energy of this<br />

system in the st<strong>at</strong>e {n1, n2, · · · , nN} is<br />

where we have defined<br />

U =<br />

N<br />

i=1<br />

ɛni<br />

M =<br />

= (M + 1<br />

2 N)ω<br />

Calcul<strong>at</strong>e the multiplicity function g(M, N). Hint: rel<strong>at</strong>e g(M, N) to g(M, N +<br />

1) and use the identity<br />

N<br />

i=1<br />

ni<br />

m<br />

<br />

n + k n + 1 + m<br />

=<br />

n n + 1<br />

k=0<br />

Show th<strong>at</strong> for large integers g(M, N) is a narrow Gaussian distribution in x =<br />

M<br />

N .<br />

It is easy to show th<strong>at</strong><br />

g(M, N) =<br />

M<br />

g(m, n)g(M − m, N − n)<br />

m=0<br />

because the st<strong>at</strong>es of parts of the system are independent. We also know th<strong>at</strong><br />

g(M, 1) = 1<br />

because for one particle there is only one way of obtaining M units of energy.<br />

Therefore<br />

g(M, N + 1) =<br />

M<br />

g(m, N)<br />

m=0<br />

Comparing with the hint we see th<strong>at</strong><br />

<br />

n + M<br />

g(M, N) =<br />

n<br />

for some value of n linearly rel<strong>at</strong>ed to N, or n = N + n0. Because g(M, 1) = 1<br />

and 0+M<br />

0 = 1 we find th<strong>at</strong> n = N − 1 and hence<br />

<br />

N − 1 + M<br />

g(M, N) =<br />

N − 1<br />

Next we take the logarithms:

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