# Chemistry - What is the classical partition function for a system of anharmonic oscillators?

## Solution 1:

Here follows a complete mathematical derivation of the expressions for the internal energy and isochoric heat capacity. I am not sure why taking the high-temperature limit is unphysical, but maybe someone else knows.

The potential is

$$ V(x) = \frac{k_0x^2}{2} + \alpha x^4 $$

for one dimensional, *localized* oscillators. The Hamiltonian then becomes

$$ H(p,x) = \frac{k_0x^2}{2} + \alpha x^4 + \frac{p^2}{2m} $$

for each particle. Since we have a system of $N$ particles, we sum over all particles to get the full Hamiltonian for the entire system (assuming all oscillators have the same mass)

$$ H(p,x) = \sum\limits_{j=1}^{N} \frac{k_0x_j^2}{2} + \alpha x_j^4 + \frac{p_j^2}{2m} $$

The general expression for the classical canonical partition function is

\begin{align} Q_{\text{N,V,T}} &= \frac{1}{N! h^{3N}} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}e^{-H(x,p)/kT} \text{d}x\text{d}p \\ \end{align}

The text says that the oscillators are localized, so we should take away the N! in the expression for Q, since we are dealing with distinguishable particles. Plugging in the Hamiltonian, we then get for the partition function {dropping the subscript from now on}

\begin{align} Q = \frac{1}{h^{3N}} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}e^{-\left( \sum\limits_{j=1}^{N} \frac{k_0 x_j^2}{2} + \alpha x_j^4 + \frac{p_j^2}{2m} \right)\frac{1}{kT}} \text{d}x\text{d}p \end{align}

We know that an exponential where the exponent is a sum we can rewrite into a product of exponentials $e^{a+b+c} = e^ae^be^c$

\begin{align} Q &= \frac{1}{h^{3N}} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \prod\limits_{j=1}^{N} e^{-\frac{k_0x_j^2}{2kT}} e^{-\frac{\alpha x_j^4}{kT}} e^{-\frac{p_j^2}{2mkT}} \text{d}x\text{d}p \\ &= \frac{1}{h^{3N}} \prod\limits_{j=1}^{N} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x_j^2}{2kT}} e^{-\frac{\alpha x_j^4}{kT}} e^{-\frac{p_j^2}{2mkT}} \text{d}x\text{d}p \\ &= \frac{1}{h^{3N}} \prod\limits_{j=1}^{N} \int\limits_{-\infty}^{\infty} e^{-\frac{p_j^2}{2mkT}} \text{d}p \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x_j^2}{2kT}} e^{-\frac{\alpha x_j^4}{kT}} \text{d}x \\ &= \frac{1}{h^{3N}} \left( \int\limits_{-\infty}^{\infty}e^{-\frac{p^2}{2mkT}} \text{d}p \right)^N \left( \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \right)^N \end{align}

where in the last step we saw that the integrand will be the same for all values of $j$. The integral over the momenta are easily evaluated; just look up any table of definite exponential integrals to find the expression. We then get

$$ Q = \left( \frac{2mkT\pi}{h^2}\right)^{3N/2} \left( \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \right)^N $$

The remaining integral is too complicated to evaluate analytically, so we perform a Taylor series to first order around $\alpha = 0$ (first order Maclaurin series in $\alpha$) for the anharmonic exponential. This is justified since, as the text explains, $a > x^4 >> kT$. Doing the expansion:

\begin{align} e^{-\frac{\alpha x^4}{kT}} &\approx e^0 + a \left( \frac{\text{d}}{\text{d}a} e^{-\frac{\alpha x^4}{kT}} \right)_{a=0} \\ &\approx 1 + a\left( e^{-\frac{\alpha x^4}{kT}} \left( -\frac{x^4}{kT} \right) \right)_{a=0} \\ &\approx 1 - \frac{ax^4}{kT} \end{align}

We now use this expression instead of the exponential in the integral. Lets just evaluate the integral for now, and then plug in for $Q$ once we have an expression for the integral. Calling the integral $I$, we get

\begin{align} I &= \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \\ &= \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} \left( 1 - \frac{ax^4}{kT} \right) \text{d}x \\ &= \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} \text{d}x - \frac{a}{kT} \int\limits_{-\infty}^{\infty} x^4 e^{-\frac{k_0x^2}{2kT}} \text{d}x \\ &= \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} - \frac{a}{kT} \int\limits_{-\infty}^{\infty} x^4 e^{-\frac{k_0x^2}{2kT}} \text{d}x \end{align}

While this integral looks intimidating, we can look it up in standard tables. It has the following solution

$$ \int\limits_{0}^{\infty} x^{2n} e^{-cx^2} \text{d}x = \left( \frac{\pi}{c} \right)^{1/2} \frac{1\cdot 3 \cdots (2n-1)}{2^{n+1}c^n} $$

Since our limits go from $-\infty$ instead of $0$, we multiply with a factor of 2 to get the correct results for the integral. Using this, we obtain

\begin{align} I = \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} - \frac{2a}{kT} \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} \frac{3}{8} \left( \frac{2kT}{k_0^2} \right)^2 \end{align}

We can now cancel the 8 in the denominator with the $2^2$ and 2 in front of $a$, cancel a factor of $kT$ and then factorize by moving the square root out

\begin{align} I &= \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} \left(1 - \frac{3akT}{k_0^4} \right) \end{align}

We can now insert this into our expression for the partition function:

$$ Q = \left( \frac{2mkT\pi}{h^2}\right)^{3N/2} \left( \frac{2kT\pi}{k_0^2} \right)^{N/2} \left(1 - \frac{3akT}{k_0^4} \right)^N $$

In statistical thermodynamics, all thermodynamic properties are related to the natural logarithm of the partition function, so we now take the logarithm. This is the reason for factorizing the expressions; it just makes taking the logarithm much easier.

$$ \ln Q = \frac{3N}{2} \ln \left( \frac{2mkT\pi}{h^2} \right) + \frac{N}{2}\ln\left( \frac{2kT\pi}{k_0^2} \right) + N\ln \left( 1 - \frac{3akT}{k_0^4} \right) $$

From here we can start to get expressions for the internal energy and isochoric heat capacity. First, internal energy

\begin{align} U &= kT^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{N,V} \\ &= kT^2 \left( \frac{3N}{2} \frac{1}{T} + \frac{N}{2} \frac{1}{T} - N \frac{3ak}{k_0^4 - 3akT} \right) \\ &= \frac{3NkT}{2} + \frac{NkT}{2} - \frac{3Nak}{k_0^4 - 3akT} \end{align}

The isochoric heat capacity is

\begin{align} C_V &= \left( \frac{\partial U}{\partial T} \right)_{N,V} \\ &= \frac{3Nk}{2} + \frac{Nk}{2} - N\left( \frac{3ak}{k_0^4 - 3akT} \right)^2 \end{align}

The first term in $C_V$ comes from the kinetic energy, and is exactly what the equipartition theorem predicts. The second term is related to the potential energy, while the last term clearly is related to the anharmonic pertubation.

The exam question now asks why taking the high-temperature limit for $C_V$ is unphysical. This I am unable to answer. Mathematically, the limit seems to diverge, but I'm not sure what kind of answer is expected here. But at least I think the mathematical derivation is correct.

## Solution 2:

The first answer has one problem, you have N oscilators not N particles. So we have $$(\frac{2mkTπ}{h^{2}})^{N}$$ instead of $$(\frac{2mkTπ}{h^{2}})^{\frac{3N}{2}}$$. With this correction your internal energy becomes: $$U=NK_{b}T+\frac{3\alpha NK_{b}T}{3\alpha-\frac{k^{2}}{K_{b}T}} $$. Note that if $\alpha=0$ the energy becomes: $$U=NK_{b}T$$ which is the internal energy of N harmonic oscillators as expected.