The GATE Grind

GATE 2021 PH – Question 46

Thermodynamics and Statistical Mechanics · Partition function, free energy, calculation of thermodynamic quantities · 2 marks · Multiple choice

A one-dimensional harmonic oscillator of angular frequency $\omega$ is in equilibrium at $T=(k_B\beta)^{-1}$. Its non-degenerate levels have energies $E_n=(n+1/2)\hbar\omega$. The source describes the states as having “equal probability”. Choose its Helmholtz free energy.

  1. $\frac{\hbar\omega}{2}+\beta^{-1}\ln[1-\exp(\beta\hbar\omega)]$
  2. $\frac{\hbar\omega}{2}+\beta^{-1}\ln[1-\exp(-\beta\hbar\omega)]$
  3. $\frac{\hbar\omega}{2}+\beta^{-1}\ln[1+\exp(-\beta\hbar\omega)]$
  4. $\beta^{-1}\ln[1-\exp(-\beta\hbar\omega)]$

Practise this question in The GATE Grind →

Show answer and explanation

Correct answer: (B) $\frac{\hbar\omega}{2}+\beta^{-1}\ln[1-\exp(-\beta\hbar\omega)]$

Explanation

Thermal equilibrium at fixed temperature uses Boltzmann probabilities, not equal probabilities over infinitely many levels. Under that intended canonical interpretation, $Z=\sum_{n=0}^\infty e^{-\beta\hbar\omega(n+1/2)}=e^{-\beta\hbar\omega/2}/(1-e^{-\beta\hbar\omega})$. Thus $F=-\beta^{-1}\ln Z=\hbar\omega/2+\beta^{-1}\ln(1-e^{-\beta\hbar\omega})$, which is B. Literal equal probabilities would not define a normalized equilibrium distribution for this infinite spectrum.

More Partition function, free energy, calculation of thermodynamic quantities questions

All Partition function, free energy, calculation of thermodynamic quantities questions →