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Consider $N$ particles in a volume $V$ at temperature $T$. Given the Hamiltonian $$\sum_{i=1}^\infty \frac{(p_{x,i}+\frac{q}{2}y_i B)^2+(p_{y,i}-\frac{q}{2}x_i B)^2+p_{z,i}^2}{2m}$$ find the partition function $Z$.

My problem is that I’m facing an integral in $dxdydzdp_x dp_y dp_z$ of the function $e^{\frac{-\beta}{2m}(p_{x,i}+\frac{q}{2}y_i B)^2}$. And I don't know what to do, if I integrate with respect $p_x$ in $\mathbb{R}$ I get something independent of $y$ and $B$ which seems odd.

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closed as off-topic by sammy gerbil, Kyle Kanos, ZeroTheHero, glS, Sebastian Riese May 10 '18 at 12:10

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  • $\begingroup$ You are being sloppy in the integration. For a clearer picture, try using the coordinate transformation : $p'_x = p_x +q/2yB$, $p'_y = p_y - q/2xB$, and then integrate. The partition function won't be independent of $B$. $\endgroup$ – Bruce Lee Apr 30 '18 at 11:43
  • $\begingroup$ I still get the integral of $e^{\frac{-\beta}{2m}(p'_x)^2 dp'_x}$ from $-\infty$ to $\infty$. And that's just $(\frac{2\pi m}{\beta})^{1/2}$ $\endgroup$ – Rafa Apr 30 '18 at 12:50
  • $\begingroup$ This seems to be a primitive version of the van Leeuwen theorem. $\endgroup$ – AccidentalFourierTransform Apr 30 '18 at 14:26