Question 23.3: (a) Calculate the vibrational partition function of a diatom...

In this problem, the main goal is to compare the classical and quantum mechanical representations of q and <\varepsilon>. At some temperature T, the classical and the quantum mechanical values approach each other. Although we are asked to calculate the vibrational contribution, similar calculations could be carried out for the translational or the rotational contributions. In Part III, we discussed the vibrational partition function in detail and provided an introductory exposition to the classical partition function.

Part(a): Calculate the Classical Vibrational Partition Function

We have already calculated the quantum mechanical vibrational contribution to the partition function for a diatomic molecule. Let us recall the results:

(i) The energy levels of the harmonic oscillator are quantized according to:

\varepsilon_{n}=\left(n+\frac{1}{2}\right) h \nu, n=0,1,2 \ldots                           (36)

(ii) Results presented in Part III indicate that:

q_{v i b}^{Q M}=\frac{\exp \left(-\theta_{v i b} / 2 T\right)}{1-\exp \left(-\theta_{v i b} / T\right)}, \theta_{v i b}=h \nu / k                           (37)

<\varepsilon_{v i b}>{ }^{Q M}=k T\left\{\frac{\theta_{v i b}}{2 T}+\frac{\frac{\theta_{v i b}}{T} \exp \left(\frac{-\theta_{v i b}}{T}\right)}{1-\exp \left(\frac{-\theta_{v ib} }{T}\right)}\right\}                                (38)

where QM stands for quantum mechanics.
In this problem, we are asked to calculate q _{v i b} \text { and }<\varepsilon_{v i b}> treating the vibrational modes classically. We therefore consider the energy of each possible vibrational state as a continuous variable. We consider the classical Hamiltonian description of the energy of a diatomic molecule in terms of its position, \vec{q}, and momentum, \vec{p} : Here, we envision a diatomic molecule as made up of two atoms connected by a Hookean spring. In this case, we have two degrees of freedom: the displacement, x, and the momentum associated with that displacement, p. We can write the classical Hamiltonian describing the motion of the two atoms and their interactions (through the stretch of the spring) as follows (see Part III):

H=\frac{p^{2}}{2 m}+\frac{1}{2} k_{s} x^{2}

where m is the reduced mass of the diatomic molecule and k _{s} is the spring constant.
As shown in Part III, the spring constant is given by

k_{s}=4 \pi^{2} \nu^{2} m                           (39)

where \nu is the characteristic vibrational frequency of the system.
We can also use a result derived in Part III to calculate q_{v i b}^{C}, where C stands for classical. Specifically:

q_{v i b}^{c}=\frac{1}{h} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\beta H(p, x)} d p d x                            (40)

Using the expression for H above in Eq. (40) and separating the integrals over p and x, we obtain:

q_{v i b}^{c}=\frac{1}{h}\left(\int_{-\infty}^{\infty} e^{-\frac{\beta p^{2}}{2 m}} d p\right)\left(\int_{-\infty}^{\infty} e^{-\frac{\beta}{2} k_{s} x^{2}} d x\right)                          (41)

where the two integrals in Eq. (41) are Gaussian integrals. In Part III, we showed that:

\int_{-\infty}^{\infty} e^{-a^{2} y^{2}} d y=\frac{\sqrt{\pi}}{a}                           (42)

Using Eq. (42) in Eq. (41), we obtain:

q_{v i b}^{c}=\frac{1}{h}\left(\frac{2 \pi m}{\beta}\right)^{1 / 2}\left(\frac{2 \pi}{\beta k_{s}}\right)^{1 / 2}=\frac{2 \pi m^{1 / 2}}{h \beta k_{s}^{1 / 2}}                             (43)

or

q_{v i b}^{c}=\left(\frac{2 \pi m^{1 / 2}}{h k_{s}^{1 / 2}}\right) k_{B} T                         (44)

Part (b): Calculate the Average Vibrational Energy

To calculate the average vibrational energy, we use the following equation derived in Part III:

<\varepsilon_{v i b}>^{c}=k_{B} T^{2}\left(\frac{\partial \ln q_{v i b}^{c}}{\partial T}\right)_{\underline{ V }}                             (45)

Using Eq. (44) in Eq. (45) yields:

<\varepsilon_{v i b}>^{c}=k_{B} T^{2}\left(\frac{\partial}{\partial T}(\ln T+\text { terms that do not contain } T)\right)_{\underline{ V }}                          (46)

<\varepsilon_{v i b}>^{c}=k_{B} T                           (47)

Part (c): Compare the Classical and the Quantum Mechanical Limits

An examination of Eq. (37) shows that if T \gg \theta_{v i b}, we obtain the following result (see Part III):

\lim T \gg \theta_{v i b}, q_{v i b}^{Q M}=\frac{1-\theta_{v i b} / 2 T}{1-\left(1+\left(\theta_{v i b} / T\right)\right)}=\frac{T}{\theta_{v i b}}=\frac{T}{\frac{h \nu}{k_{B}}}=\frac{k_{B} T}{h \nu}                                (48)

According to Eq. (44):

q_{v i b}^{c}=\left(\frac{2 \pi m^{1 / 2}}{h k_{s}^{1 / 2}}\right) k_{B} T

Using the expression for k _{s}, as well as Eq. (56), in Eq. (61) yields:

q_{v i b}^{c}=\frac{k_{B} T}{h \nu}                           (49)

A comparison of Eq. (66) and Eq. (65) shows that when T \gg \theta_{v i b}, q_{v i b}^{c}=q_{v i b}^{Q M}.
Next, let us consider the average energy in the quantum mechanical representation. Recall that for the QM case:

<\varepsilon_{v i b}>^{Q M}=k T\left\{\frac{\theta_{v i b}}{2 T}+\frac{\frac{\theta_{v i b}}{T} \exp \left(\frac{-\theta_{v i b}}{T}\right)}{1-\exp \left(\frac{-\theta_{v i b}}{T}\right)}\right\}                           (50)

In the limit T \gg \theta_{v i b}, Eq. (67) reduces to:

<\varepsilon_{v i b}>^{Q M}=k T\left\{\frac{\theta_{v i b}}{2 T}+\frac{\frac{\theta_{v i b}}{T}\left(1-\frac{\theta_{v i b}}{T}\right)}{1-1+\frac{\theta_{v i b}}{T}}\right\}=k_{B} T                           (51)

A comparison of Eq. (51) and Eq. (47) shows that in the limit T \gg \theta_{v i b}, <\varepsilon_{v i b}>^{C}=<\varepsilon_{v i b}>^{Q M}.