Suppose we have a gas of $N$ diatomic molecules (ex $O_2$) with one-molecule hamiltonian being: $$\mathcal{H} = \frac{\vec{p}_1^2}{2m}+\frac{\vec{p}_2^2}{2m} + V(r_{rel}) $$ Where $r_{rel}$ is the relative distance between the two atoms. Say the potential is something like $V= \frac12 m ω^2 r_{rel}^2 = \frac12 k r_{rel}^2$ Thus the one body partition function is going to be:  $$ Z_1 = \frac1{h^6} \int d^3r_1 \int d^3r_2 \int d^3p_1 \int d^3p_2 e^{-β\mathcal{H}}$$   or more conveniently-using center of mass coordinates  : $$ Z_1 = \frac1{h^6} \int d^3r_{cm} \int d^3r_{rel} \int d^3p_1 \int d^3p_2 e^{-β\mathcal{H}}$$    (something tells me I got something wrong here?)  It's easy to see the first integral gives us the volume $V$, while the last two are the same so we'll get a term $ (\frac{2mπ}{β})^{3}$.   Now the point were I'm getting confused is the integral of the relative distance: I'm reading in notes/books that the relative distance is one extra quadratic term in the hamiltonian and will thus give us an extra term  $\sim (\frac{1}{β})^{1/2}$. However I'm thinking of it as three extra terms, since $\vec{r}_{rel}= \vec{r}_1 - \vec{r}_2$ is a vector, same as $\vec{p}_{1,2}$, right? Indeed doing the integral I get : $$ \int d^3r_{rel} exp(-\frac{βk}{2} r_{rel}^2) =4π \int dr r^2 exp(-\frac{βk}{2} r^2) \sim (\frac{1}{β})^{3/2}$$ So I'd get a heat capacity $9/2 k N$ instead of $7/2 k N$.   What am I getting wrong? Why are the momenta  considered as 3 quadratic terms each, while the relative distance in the potential as 1 and how does this become apparent when integrating with respect to center of mass coordinates?
Question  I calculated the classical heat capacity of a diatomic gas as $C_V = (9/2)Nk_B$, however the accepted value is $C_V = (7/2)Nk_B$.  I assumed the classical Hamiltonian of two identical atoms bound together as $$ H = \dfrac{1}{2m}( |\bar{p}_2|^2 + |\bar{p}_2|^2)+ \dfrac{\alpha}{2} |\bar{q}_1-\bar{q}_2|^2. $$ I calculated the partition function of $N$ particles as $$ Z = \left( \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} e^{-\beta H} ~d^3q_1~d^3p_1~d^3q_2~d^3p_2 \right)^N \propto V^N T^{(9/2)N}. $$ I calcuated the heat capacity as $$ C_V = \dfrac{\partial }{\partial T} \left( k_B T^2 \dfrac{\partial \ln(Z)}{\partial T} \right) = \dfrac{9}{2}k_BN. $$  Why does the classical argument fail?  Classical Derivation  The partition function is \begin{align} Z &amp;=&amp; \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta H(\bar{q}_1,\bar{q}_2,\bar{p}_1,\bar{p}_2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1   ~d^{3}p_2 \right)^N \\&amp;=&amp; \left( \frac{1}{h^6} \int \mathrm{e}^{- \beta ((|\bar{p}_1|^2+|\bar{p}_2|^2)/(2m)+\alpha |\bar{q}_1-\bar{q}_2|^2/2)} ~d^{3}q_1 ~d^{3}q_2 ~d^{3}p_1   ~d^{3}p_2 \right)^N \end{align} A useful gaussian integral \begin{align} \int_{-\infty}^{\infty} e^{-\gamma (x-x_0)^2}dx = \sqrt{\dfrac{\pi}{\gamma}} \end{align} The partition function can be evaluated using separated integrals \begin{align} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_1|^2} ~d^{3}p_1 = \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta |\bar{p}_2|^2} ~d^{3}p_2 = \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \end{align} and \begin{align} \iiint_{-\infty}^{\infty} \iiint_{-\infty}^{\infty} \mathrm{e}^{- \beta \alpha |\bar{q}_1-\bar{q}_2|^2/2 } ~d^{3}q_1 ~d^{3}q_2 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 \iiint_{-\infty}^{\infty}  ~d^{3}q_1 = \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \end{align} The last set of integrals are improper integrals. One has to take the limit as the space approaches infinite containment. In that limit, integrating one set of variables $d^3q_2$ approaches the limit of a finite Gaussian term, while the other $d^3q_1$ approaches the diverging value of the total volume of the gas.  The partition function is \begin{align} Z &amp;=&amp; \left( h^{-6} \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left(\sqrt{\dfrac{\pi}{\beta}}\right)^3 \left( \sqrt{\dfrac{\pi}{\beta \alpha/2}} \right)^3 V \right)^N \\&amp;=&amp; \left( h^{-6} \left(k_B T \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} V \right)^N \\&amp;=&amp; \left( h^{-6} \left(k_B \pi\right)^{9/2} \left( \dfrac{2}{\alpha} \right)^{3/2} \right)^N V^N T^{9N/2} \end{align}
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