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GR0877 Problem 100

PROBLEM STATEMENT: The partition function Z in statistical mechanics can be written as

\displaystyle Z = \sum_{r} e^{- E_r /kT},

where the index r ranges over all possible microstates of a system and E_r is the energy of microstate r. For a single quantum mechanical harmonic oscillator with energies

E_{n} = \left( n + \frac{1}{2} \right)\hbar \omega, where n = 0, 1, 2,\dots,

the partition function Z is given by which of the following?
(A) \displaystyle Z = e^{- \frac{1}{2}\hbar \omega /kT}

(B) \displaystyle Z = e^{\frac{1}{2}\hbar \omega /kT}

(C) \displaystyle Z = e^{\frac{1}{2}\hbar \omega /kT} - 1

(D) \displaystyle Z = e^{\frac{1}{2}\hbar \omega /kT} + 1

(E) \displaystyle Z = \frac{e^{\frac{1}{2}\hbar \omega /kT}}{e^{\hbar \omega /kT}-1}

SOLUTION: (E) The partition function for a single QM harmonic oscillator is:
\displaystyle Z = \sum_{n=0}^{\infty} \exp\left( -\frac{E_n}{kT}\right) = \exp\left( -\frac{\hbar \omega}{2kT}\right) \sum_{n=0}^{\infty} \exp\left( -\frac{n \hbar \omega}{kT}\right).
\displaystyle \sum_{n=0}^{\infty} \exp\left( -\frac{n \hbar \omega}{kT}\right) = 1 + \exp\left( -\frac{\hbar \omega}{kT}\right) + \exp\left( -\frac{2 \hbar \omega}{kT}\right) + \dots which is an infinite geometric series:
\displaystyle \sum_{n=0}^{\infty} \exp\left( -\frac{n \hbar \omega}{kT}\right) = \frac{1}{1 - \exp\left( -\frac{\hbar \omega}{kT}\right)}. Thus, \displaystyle Z = \frac{\exp\left( -\frac{\hbar \omega}{2kT}\right)}{1 - \exp\left( -\frac{\hbar \omega}{kT}\right)} = \frac{\exp\left( \frac{\hbar \omega}{2kT}\right)}{\exp\left( \frac{\hbar \omega}{kT}\right) -1}.

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  1. Eric
    09.10.2012 at 02:18

    I just wanted to say thank you for your solutions. They have helped so much! I take the PGRE in 5 days, and your solutions to this exam were so, so helpful!!! I wish you all the best.

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