2
$\begingroup$

Is the following intuition correct?

Consider a 1D Ising model of length $N$ in a heat bath with inverse temperature $\beta$ (i.e. a canonical ensemble). The Boltzmann factor is given by $$ p(E_i) = \frac{g_i e^{-\beta E_i}}{\sum_i g_i e^{-\beta E_i}}, $$

where the factor of $g_i$ accounts for the multiplicity of macrostate with energy $E_i$ (i.e. its degeneracy).

To further study $p(E_i)$, one can diagonalise the Hamiltonian with an eigenbasis given by the tensor product of the eigenvectors of the $\sigma_z$ operator $\{|\uparrow \rangle, |\downarrow \rangle \}$, i.e. with $$ \{\underbrace{|\uparrow \rangle\otimes|\uparrow \rangle \otimes\cdots \otimes|\uparrow \rangle}_\text{$N$ times}, \ |\downarrow \rangle\otimes|\uparrow \rangle \otimes\cdots \otimes|\uparrow \rangle,\ \cdots, \ |\downarrow \rangle\otimes|\downarrow \rangle \otimes\cdots \otimes|\downarrow \rangle \}. $$ It is easy to see, that for a given total magnetisation $M_T = \langle S^z_T \rangle \equiv \langle \sum_i S^z_i \rangle$, when $M_T=0$ the multiplicity of this macrostate is the largest, given by $g={N \choose N/2}$ microstates.

Then, $e^{-\beta E_i}$ in the limit $\lim_{β\rightarrow 0}e^{-\beta E_i}=1$ and the Boltzmann factor is given by $$ p(E_i, \beta =0)= \frac{g_i}{\sum_i g_i}. $$ Thus in the thermodynamic limit, the state with the largest multiplicity $g_i$ dominates i.e. we have $$ p(E_i,\beta=0) \sim \left\{\begin{aligned} 1, \quad M_T=0\\ 0, \quad M_T\neq0 \end{aligned}\right. $$ In this particular example, the statistics are dominated by the eigenstates that lie in the middle of the spectrum but whereas this is true for this example, in general, it might not be true in other cases. As @Jahan Claes pointed out, it depends on the shape of the density of states as shown above.

$\endgroup$
1
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Dec 8 '20 at 18:50
1
$\begingroup$

Your intuition is sound. Since $e^{-\beta E_i}\rightarrow 1$ in the limit of high temperature, the most probable macrostate is the one with the highest multiplicity, so the equilibrium configuration of the system will be the one which maximizes the density of states.

For a system with bounded energy like the Ising model (whose density of states is peaked at $m=0\iff E = \frac{E_{max}+E_{min}}{2}$), this is physically meaningful and corresponds to a configuration with zero net magnetization. For a system with unbounded energy like the ideal gas, the density of states will generally not have any maxima so the $T\rightarrow \infty$ limit is not really well-defined.

$\endgroup$
1
  • $\begingroup$ You are right in your first coment, I got the terms mixed up! I have edited it. $\endgroup$ – FriendlyLagrangian Dec 6 '20 at 17:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.