The question is I was given the Hamiltonian of two classical particles with masses $m$ with which I must calculate Partition function $$H =\frac{P_1^2}{2m} +\frac{P_2^2}{2m} + q_1^2 + q_2^2 +\frac{q_1q_2}{4}$$

To calculate Partition function $$Z = \int^∞_{-∞} \exp\left[ -\beta\left(\frac{P_1^2}{2m} +\frac{P_2^2}{2m} + q_1^2 + q_2^2 +\frac{q_1q_2}{4}\right) \right] dp_1 dp_2 dq_1 dq_2$$ I cn see that the first two terms of the Hamiltonian in the integral can be calculated by the basic integrals formula involving gamma functions but the last three terms of the Hamiltonian challenges me. help me solve this!


1 Answer 1


The two integrals over the momentum are Gaussian $$\int_{-\infty}^{+\infty} e^{-\beta{p^2\over 2m}}dp =\big(2\pi mk_BT\big)^{1/2}$$ The two integrals over $q_1$ and $q_2$ can be computed separately. Let us try to introduce the change of variable $$q_3=q_2+\alpha q_1$$ Since $q_3^2=q_2^2+\alpha_2^2q_1^2+2\alpha q_1q_2$, we see that the integrals are decoupled when choosing $\alpha=1/8$. The determinant of the Jacobian is $1$ so the integral becomes $$\eqalign{ \int e^{-\beta(q_1^2+q_2^2+{1\over 4}q_1q_2)}dq_1dq_2 &=\int e^{-\beta(q_1^2+q_3^2-\alpha^2q_1^2)}dq_1dq_3 \cr &=\int e^{-\beta(1-\alpha^2)q_1^2}dq_1\int e^{-\beta q_3^2}dq_3\cr &=\sqrt{\pi k_BT\over 1-\alpha^2}.\sqrt{\pi k_BT} \cr &={k_BT\over\sqrt{1-\alpha^2}} }$$

  • $\begingroup$ thank you @Christophe. Why have you mentioned determinant of jacobi is 1? what does it mean? $\endgroup$
    – kayalvizhi
    Commented Oct 15, 2022 at 10:19
  • 1
    $\begingroup$ Since Christophe changed the variable of integration, there can be change of measure. One uses Jacobian to determine how the measure should change. $\endgroup$
    – Andy Chen
    Commented Oct 15, 2022 at 11:56

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