Question

Consider a single one-dimensional harmonic oscillator of angular frequency \( \omega \), in equilibrium at temperature \( T = \left( k_B \beta \right)^{-1 }\). The states of the harmonic oscillator are all non-degenerate having energy \( E_n = \left( n + \frac{1}{2} \right) \hbar \omega \) with equal probability, where \( n \) is the quantum number. The Helmholtz free energy of the oscillator is
 

• A. \( \frac{\hbar \omega}{2} + \beta^{-1} \ln \left[ 1 - \exp(\beta \hbar \omega) \right] \)
• B. \( \frac{\hbar \omega}{2} + \beta^{-1} \ln \left[ 1 - \exp(-\beta \hbar \omega) \right] \)
• C. \( \frac{\hbar \omega}{2} + \beta^{-1} \ln \left[ 1 + \exp(-\beta \hbar \omega) \right] \)
• D. \( \beta^{-1} \ln \left[ 1 - \exp(-\beta \hbar \omega) \right] \)

Hint:

For a quantum harmonic oscillator, the Helmholtz free energy can be derived using the partition function and involves a geometric series sum.

Answer 1

The Correct Option is B

The Helmholtz free energy is given by the formula: \[ F = -k_B T \ln Z, \] where \( Z \) is the partition function, \( T \) is the temperature, and \( k_B \) is the Boltzmann constant. For a quantum harmonic oscillator, the partition function is given by: \[ Z = \sum_{n=0}^{\infty} \exp \left( - \beta E_n \right), \] where \( E_n = \left( n + \frac{1}{2} \right) \hbar \omega \) is the energy of the nth state, and \( \beta = \frac{1}{k_B T} \). Substituting the energy expression into the partition function: \[ Z = \sum_{n=0}^{\infty} \exp \left( - \beta \left( n + \frac{1}{2} \right) \hbar \omega \right) = \exp \left( - \frac{\beta \hbar \omega}{2} \right) \sum_{n=0}^{\infty} \exp \left( - \beta n \hbar \omega \right). \] This sum is a geometric series, and can be summed as: \[ Z = \frac{1}{1 - \exp(-\beta \hbar \omega)}. \] Now, the Helmholtz free energy is: \[ F = -k_B T \ln Z = -k_B T \ln \left( \frac{1}{1 - \exp(-\beta \hbar \omega)} \right). \] This simplifies to: \[ F = \frac{\hbar \omega}{2} + \beta^{-1} \ln \left[ 1 - \exp(-\beta \hbar \omega) \right]. \] Thus, the correct expression for the Helmholtz free energy is given by option (B).