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Consider the following hamiltonian

$$H=-h\sum_{i=1}^N\sigma_i$$

where $\sigma_i=\pm1$ and $h$ is the magnetization.

Let us assume that the system is equilibrated with a bath at temperature $T$ with $h=0$. Then the field is ramped up to $h=h_0$, and the system equilibrate again with a bath at temperature $T$.

what is the work performed in this process?

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closed as off-topic by By Symmetry, Jon Custer, Kyle Kanos, AccidentalFourierTransform, John Rennie Aug 4 '18 at 5:34

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  • $\begingroup$ Your use of the word “equilibrate” is very unclear to me. Do you perhaps mean: “put into thermal equilibrium with...” and “the system in equilibrium” resp. instead? $\endgroup$ – Antaios Jul 29 '18 at 7:50
  • $\begingroup$ @Antaios. the system equilibrate again with a bath at temperture $T$. corrected in the post. $\endgroup$ – jarhead Jul 29 '18 at 7:53
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By work, I presume you mean macroscopic work. This is the work done by an external agent in sustaining a magnetic field $h_0$ until a Magnetisation $M$ is induced.

Macroscopic work done in changing the magnetisation from $0$ to $M$ is given by a by the expression(See Heat and Thermodynamics, Zemansky and Dittman, Seventh Edition, Eq 3.11): $$W=\int_0^M hdM$$ Assuming the field is flipped to $h_0$ instantaneously $$W=h_0\int_0^M dM=h_0 M$$ This is numerically equal to the average energy with a negative sign. This is a standard exercise in statistical mechanics that requires the definition of a partition function: $$Z=2^Ncosh^N \beta h_0$$ Where $\beta=\frac{1}{KT}$.

Average Energy is given by: $$U=-\frac{\partial \log Z}{\partial \beta}$$

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