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As it is explained in this answer (and nicely so!), the second derivative of a thermodynamic potential to an intensive quantity, for example pressure or magnetic field strength (or temperature) will give the variance of the according extensive quantity in the given ensemble.

This had me thinking wether one could extend this interpretation to the thermodynamic limit. However, in the thermodynamic limit, all extensive quantities are sharply peaked - wouldn't that mean that the variance, and hence the susceptibilities, are all zero?

Or am I wrong and there is still some uncertainty left in the thermodynamic limit?

Quantumwhisp
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In the thermodynamic limit both the extensive quantity (e.g., the total magnetization) and its variance (i.e., the magnetic susceptibility) scale proportionally to the size of the system: $$ \langle M\rangle \sim N,\\\sigma_M^2=\langle M^2\rangle - \langle M\rangle^2\sim N $$ This means that the relative fluctuations of the quantity in question are decreasing as $1/\sqrt{N}$: $$ \frac{\sigma_M}{\langle M\rangle} \sim \frac{1}{\sqrt{N}}, $$ in other words the standard deviation grows as $\sim \sqrt{N}$, i.e., much slower the quantity itself. However, the variance, which is the square of the standard deviation, is still growing proportionally to $N$, and therefore has finite value when taken per number of particles/moles/mass/volume/etc. See Volume susceptibility.

Roger V.
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