I understand that the number of microstates in a two-state system is $W = 2^N$. However, if the energy of the system is fixed at zero (for example, exactly half of a paramagnet's N dipoles point up and the other point down), are all $2^N$ microstates accessible to the system given an infinite length of time? Will it be able to change microstates at all?
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$\begingroup$ I assume that you're in a system where the two energies are $\pm\epsilon$? In that case, $E=0$ means that there are an equal number of ups as downs. This is not the full $2^N$-sized state-space. Rather, there are $N$-choose-$N/2$ such microstates. This does turn out to be the macro-state of highest multiplicity, but of course fixing the energy to be a particular value necessarily means that you are only sampling a subset of the full state-space. $\endgroup$– marchCommented Sep 26 at 18:03
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$\begingroup$ @march This answers my question perfectly. Thank you! $\endgroup$– umntcCommented Sep 26 at 20:38
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I assume that the OP is talking about a sysetm where the two energies are $\pm \epsilon$. In that case, $E = 0$ means that there are an equal number of ups as downs. Such states do not fill out the entire $2^N$-sized state-space. Rather, there are $N$-choose-$N/2$ such microstates. This does turn out to be the macro-state of highest multiplicity, but of course fixing the energy to be a particular value necessarily means that you are only sampling a subset of the full state-space.
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If the energy is fixed as $E=0$ then only the microstates of zero energy are available (which, in a two state system, would correspond to the largest set of microstates). In reality, this never happens because the energy is never truly fixed. Even if a system is isolated, the state of each individual component of the system is not classical, it's a quantum state that may spontaniously acquire a higher energy.
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$\begingroup$ What do you mean by the largest set of microstates? Do you mean the modal macrostate? $\endgroup$– umntcCommented Sep 26 at 17:50
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$\begingroup$ If the two states correspond to energies $+1$ and $-1$, then the zero energy state has the largest set of microstates and is therefore the most probable state. This is equivalent to throwing millions of coins, labeling heads as $+1$ and tails as $-1$, and then expecting that the net result will be zero. $\endgroup$– agaminonCommented Sep 26 at 18:00