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A simple example - the quantum harmonic oscillator

  • Page ID
    5259
  • As a simple example of the trace procedure, let us consider the quantum harmonic oscillator. The Hamiltonian is given by

    \[ H = {P^2 \over 2m} + {1 \over 2}m\omega^2 X^2\]

    and the eigenvalues of \(H\) are

    \[ E_n = \left(n + {1 \over 2}\right)\hbar\omega,\;\;\;\;\;\;\;\;\;\;n=0,1,2,...\]

    Thus, the canonical partition function is

    \[Q(\beta) = \sum_{n=0}^{\infty} e^{-\beta (n+1/2)\hbar\omega} = e^{-\beta \hbar \omega/2}\sum_{n=0}^{\infty}\left(e^{-\beta\hbar\omega}\right)^n\]

    This is a geometric series, which can be summed analytically, giving

    \[ Q(\beta) = {e^{-\beta \hbar\omega/2} \over 1-e^{-\beta \hbar \omega}} = {1 \over e^{\beta \hbar \omega / 2} - e^{-\beta \hbar\omega/2}}= {1 \over 2}csch(\beta\hbar\omega/2)\]

    The thermodynamics derived from it as as follows:

    1.
    Free energy:

    The free energy is

    \[A = -{1 \over \beta}\ln Q(\beta) = {\hbar\omega \over 2} +{1 \over \beta}\ln \left(1-e^{-\beta \hbar \omega}\right)\]

    2.
    Average energy:

    The average energy \(E = \langle H \rangle \) is

    \[ E = -{\partial \over \partial \beta}\ln Q(\beta)= {\hbar\omega \over 2} + {\hbar \omega e^{-\beta \hbar \omega}}= \left({1 \over 2} + \langle n \rangle \right)\hbar\omega\]

    3.
    Entropy

    The entropy is given by

    \[ S = k\ln Q(\beta) + {E \over T} =-k\ln \left(1-e^{-\beta \hbar \omega} \right ) + {\hbar \omega \over T}{e^{-\beta \hbar \omega}\over 1-e^{-\beta \hbar \omega}}\]

    Now consider the classical expressions. Recall that the partition function is given by

    \[ Q(\beta) = {1 \over h} \int dp dx e^{-\beta \left({p^2 \over 2m} + {1 \over 2} m\omega ^2 x^2 \right )} = {1 \over h} \left ( {2 \pi m \over \beta} \right )^{1/2} = {2\pi \over \beta \omega h} ={1 \over \beta \hbar \omega}\]

    Thus, the classical free energy is

    \[A_{\rm cl} = {1 \over \beta}\ln(\beta \hbar \omega)\]

    In the classical limit, we may take \(\hbar \) to be small. Thus, the quantum expression for \(A\) becomes, approximately, in this limit:

    \[A_{\rm Q} \longrightarrow {\hbar \omega \over 2} + {1 \over \beta}\ln (\beta \hbar \omega)\]

    and we see that

    \[A_{\rm Q} - A_{\rm cl} \longrightarrow {\hbar \omega \over 2}\]

    The residual \({\hbar \omega \over 2} \) (which truly vanishes when \(\hbar \rightarrow 0 \)) is known as the quantum zero point energy. It is a pure quantum effect and is present because the lowest energy quantum mechanically is not \(E = 0 \) but the ground state energy \(E={\hbar\omega \over 2} \).