The origin of my doubt lays on the Peierls argument to show that there is no phase transition in the 1D Ising Model.

I understand that the Helmholtz Potential is the partial Lengendre transform of $U$, and therefore can be written as \begin{equation} F = U - TS \end{equation}

Now, making use of the previous expression we can easily compute $F$ for an unidimensional Ising system with its all $N$ spins pointing in the same direction: \begin{equation} F_{1} = -NJ - k_{B}T\ln 2 \end{equation}

where $J$ is the interaction factor and I calculated the entropy as $k_{B}\ln\Omega$ being $\Omega$ the number of possible microstates with $U=-NJ$.

If we invert a strip of spins of undefined lenght the new Helmholtz Free Energy is the following: \begin{equation} F_{2} = -NJ + 4J -k_{B}T\ln [N(N-1)] \end{equation} where $4J$ is the contribution due to the walls and again I calculated the entropy counting the number of possible states.

It is quite clear that the Helmholtz Potential varies, and I don't find it unphysical at all: the energy changes and so does the configurational entropy.

However, as far as I know, the Helmholtz Potencial is a function of the temperature, volume and number of elements, $F = F(T,V,N)$ in its standard form, and those do not depend on the spins configuration. Following this reasoning it appears that $F$ should be always the same, no matter the pointing direction of the spins. Also, when we define $F = -k_{B}T\ln Z$ in the canonical ensemble it doesn't look like it will vary with the possible configurations.

So, where is the mistake in my last reasoning?

  • 1
    $\begingroup$ Just a remark: this is not what is known as Peierls' argument. The latter is about the existence of phase transition in the 2d Ising model (as developed in his famous 1936 paper). The argument you sketch is usually attributed to Landau and appears in his statistical physics book with Lifshitz. $\endgroup$ Commented May 27, 2018 at 7:15
  • $\begingroup$ Isn't the answer to why the free energy ordinarily does not depend on configuration just the fact that what we are usually considering is the free energy of the minimum configuration in which case F=-TlnZ? In this case we are comparing F of different configurations to find out which one is smaller. $\endgroup$
    – Hmmmm
    Commented Oct 23, 2020 at 1:40

1 Answer 1


As defined in standard thermodynamics, the Helmholtz free energy $F$ is a function of the macroscopic variables $T$, $V$, and $N$. It does not depend on the configuration.

However, there are related "free energies" that do depend on some features of the configuration. Note that $F$ is defined in statistical mechanics as $$e^{-\beta F} = \sum_{\{s_i\}} e^{-\beta H[s_i]}$$ where we sum over all configurations. But then too little information remains in $F$ to do the Peierls argument, which involves counting domain walls in particular configurations.

Instead, we can perform the sum "halfway". We let $$e^{-\beta F} = \sum_n e^{-\beta F'(n)}, \quad e^{-\beta F'(n)} = \sum_{n \, \text{domain walls}} e^{-\beta H[s_i]}.$$ The notation is a bit clunky, but the point is that while the thermodynamic free energy $F$ sums over all configurations, the quantity $F'(n)$ sums over only configurations with $n$ domain walls. Your quantities $F_1$ and $F_2$ are $F'(0)$ and $F'(1)$.

The probability of having $n$ domain walls is proportional to $e^{-\beta F'(n)}$, so your calculation shows that having one domain wall is much more likely than having none. That's the Peierls argument.

From a deeper perspective, $F'(n)$ is a Wilsonian effective Hamiltonian obtained by integrating out all degrees of freedom except for the number of domain walls. In general, such an object is called the Landau free energy or the Landau-Ginzburg Hamiltonian, as I explain here. The principle is that we should sum over the degrees of freedom we don't care about to simplify the problem. The Peierls argument is an extreme example, summing over almost absolutely everything.

  • $\begingroup$ Thank you for the detailed explanation and the further reading link, I appreciate. $\endgroup$
    – Mat
    Commented May 26, 2018 at 17:51

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