When computing the mean magnetic moment of a system with Hamiltonian $$ \mathcal{H} = -\mu \vec{H} \sum_{i=1}^{N} \cos{\theta_i} $$ and external field $\vec{H}=H\hat{e}_z$ we first evaluate the canonical partition function of the system which equals to $$ Z_N = Z_1^N = \left( \frac{4\pi R^3}{3h^3 \beta \mu H} \right)^N. $$ But why, when computing the ensemble average $$ \langle\mu_z\rangle = \frac{\displaystyle \int \prod_i \mathrm{d}^3q_i \exp{\left(-\beta \mu H \cos\theta_i \right) } \cos\theta_i |\vec{\mu}|}{\displaystyle \int \exp\left(-\beta \mu H \sum_j \cos \theta_j\right)} $$ do we get $$ \langle \mu_z \rangle = \left[ \mu \left( \coth(\mu\beta H) - \frac{1}{\mu\beta H} \right) \right] ^N $$ where this is just different from the one-particle average (without the sum) due to the $(...)^N$? Shouldn't this be the same?

Or is my conception about ensemble averages completely flawed?

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    $\begingroup$ I'm pretty sure you've calculated the ensemble average wrong. I haven't followed the details but your final expression is not dimensionally consistent (taking $\mu$ to be dimensionful). I suspect there's meant to be an $N$ in front and no power of $N$. $\endgroup$ – jacob1729 Jun 16 at 12:46
  • $\begingroup$ How would you go about calculating the ensemble average? $\endgroup$ – samox Jun 16 at 14:31
  • $\begingroup$ I haven't checked how to do that integral specifically. I would use some relation like $\langle \mu_z \rangle = \frac{1}{H} \partial_\beta ln(Z^N)$. I guess you get this by noticing the numerator of your integral looks like a partial derivative of exp(...). $\endgroup$ – jacob1729 Jun 16 at 14:51

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