91Ó°ÊÓ

In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H₂ molecule, for example, are more accurately described by the approximate formula

$E_n≈\mathrm{Ï\mu }\left(1.03n-0.03n^2\right),n=0,1,2,…$

where $\mathrm{Ï\mu }$ is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for slightly higher n it would say that E_n decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n$≈$15.)

In actual life, most oscillators are not perfect harmonic oscillators; the energy of real oscillators is provided by:

$E_n=\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)$

Where

$\mathrm{Ï\mu }$is the energy difference between the first two levels.

The partition function is:

$Z=\underset{n}{∑}{e}^{-\mathrm{β}{E}_n}\left(1\right)$

Substitute with energy

localid="1647451366905">Z=∑ne-βϵ1.03n-0.03n2(2)

Using python the summation is found with the code:

The average energy is given as:

$\stackrel{¯}{E}=-\frac{1}{Z}\frac{∂Z}{∂\mathrm{β}}$

Substitute from (1),

role="math" localid="1647451769495" $$\begin{gathered} \stackrel{¯}{E}=-\left(1-e^{-\mathrm{β}{E}_n}\right)\frac{∂}{∂\mathrm{β}}\left(\frac{1}{1-e^{-\mathrm{β}{E}_n}}\right) \\ \stackrel{¯}{E}=\left(1-e^{-\mathrm{β}{E}_n}\right)\frac{E_n{e}^{-\mathrm{β}{E}_n}}{{\left(1-e^{-\mathrm{β}{E}_n}\right)}^2} \\ \stackrel{¯}{E}=\frac{E_n{e}^{-\mathrm{β}{E}_n}}{1-e^{-\mathrm{β}{E}_n}} \\ \stackrel{¯}{E}=\frac{E_n}{e^{\mathrm{β}{E}_n}-1} \end{gathered}$$

Substituting the energy,

$\stackrel{¯}{E}=\underset{n}{∑}\frac{\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)}{e^{\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)}-1}$

The partial derivative of total energy with respect to temperature equals the heat capacity, which is:

$$\begin{gathered} C=\frac{∂U}{∂T}=\frac{∂\mathrm{β}}{∂T}\frac{∂U}{∂\mathrm{β}} \\ C=\frac{∂\mathrm{β}}{∂T}\frac{∂}{∂\mathrm{β}}\left(\frac{N{E}_n}{e^{\mathrm{β}{E}_n}-1}\right) \\ C={\left(\frac{∂T}{∂\mathrm{β}}\right)}^{-1}\left(-\frac{N{\left(E_n\right)}^2{e}^{\mathrm{β}{E}_n}}{{\left(e^{\mathrm{β}{E}_n}-1\right)}^2}\right) \end{gathered}$$

But $\mathrm{β}=1/kT→T=1/k\mathrm{β}$

${\left(\frac{∂T}{∂\mathrm{β}}\right)}^{-1}=-k{\mathrm{β}}^2$

Therefore,

$C=\frac{Nk{\left(\mathrm{β}{E}_n\right)}^2{e}^{\mathrm{β}{E}_n}}{{\left(e^{\mathrm{β}{E}_n}-1\right)}^2}$

Substitute the energy:

role="math" localid="1647452008304" $$\begin{gathered} C=\underset{n}{∑}\frac{Nk{\left(\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)\right)}^2{e}^{\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)}}{{\left(e^{\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)}-1\right)}^2} \\ \frac{C}{Nk}=\underset{n}{∑}\frac{{\left(\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)\right)}^2{e}^{\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)}}{{\left(e^{\mathrm{β}\mathrm{Ï\mu }\left(1.03n-0.03n^2\right)}-1\right)}^2} \end{gathered}$$

If $x=1/\mathrm{β}\mathrm{Ï\mu }$, then

$\frac{C}{Nk}=\underset{n}{∑}\frac{{\left(\left(1.03n-0.03n^2\right)/x\right)}^2{e}^{\left(1.03n-0.03n^2\right)/x}}{{\left(e^{\left(1.03n-0.03n^2\right)/x}-1\right)}^2}$

Using python to plot the function for different values of $n_{\max }$. The code is given below:

In the graph below, the red curve is for harmonic oscillator: