1D Ising Model: No Finite-Temperature Phase Transition

Hello! In our last lesson, we achieved a significant milestone by solving the one-dimensional Ising model exactly using the transfer matrix method. We found that in the thermodynamic limit, the Helmholtz free energy per spin, $f$, is determined entirely by the largest eigenvalue, ${\lambda }_1$, of the transfer matrix:
$f=-k_{B}T\ln {\lambda }_1$
where
${\lambda }_1=e^{\beta J}\cosh (\beta B)+\sqrt{e^{2\beta J}{\sinh }^2(\beta B)+e^{-2\beta J}}$
Today, we will use this exact result to demonstrate a fundamental feature of one-dimensional systems: the absence of a finite-temperature phase transition in the 1D Ising model.

A phase transition is mathematically signaled by a non-analyticity in the free energy or one of its derivatives as a function of temperature or an external field. The most prominent example is the emergence of a spontaneous magnetization below a critical temperature, $T_c$. We will investigate our solution from three complementary perspectives to show why this never happens for $T>0$ in one dimension:

1. A direct calculation showing the absence of spontaneous magnetization.
2. A physical argument involving the energy and entropy of domain walls.
3. A rigorous mathematical argument based on the properties of the transfer matrix.

Argument 1: The Absence of Spontaneous Magnetization

The defining characteristic of a ferromagnet is its ability to maintain a net magnetization even after the external magnetic field is removed. This spontaneous magnetization, $m_0$, is the order parameter for the ferromagnetic transition. We can find it by calculating the magnetization $m(T,B)=-\partial f/\partial B$ and then taking the limit as the field approaches zero:
$m_0(T)={\lim }_{B\to 0^{+}}m(T,B)$
If $m_0(T)$ is non-zero for some range of temperatures, a phase transition exists. Let's see what our exact solution tells us.

The video below directly calculates the magnetization per spin from the free energy and examines its behavior as the external field $H$ (or $B$ in our notation) is turned off.

As the video shows, the exact magnetization per spin is given by:
$m(T,B)=\frac{\sinh (\beta B)}{\sqrt{{\sinh }^2(\beta B)+e^{-4\beta J}}}$
When we examine the limit as $B\to 0$ for any finite temperature ($T>0⟹\beta <\infty$), we find:

• The numerator, $\sinh (\beta B)$, goes to 0.
• The denominator goes to $\sqrt{0+e^{-4\beta J}}=e^{-2\beta J}$, which is a non-zero constant.

Therefore,
$m_0(T)={\lim }_{B\to 0}m(T,B)=\frac{0}{e^{-2\beta J}}=0\text{for all }T>0.$

The spontaneous magnetization is zero at all finite temperatures. The system is only magnetized when an external field is present. Once the field is removed, thermal fluctuations are always strong enough to destroy the long-range order, and the net magnetization vanishes. This is the mathematical proof that the 1D Ising model does not have a finite-temperature ferromagnetic phase transition.

Argument 2: The Physical Picture of Domain Walls

Why is it that thermal fluctuations are always dominant in 1D? A powerful physical argument, first formulated by Rudolf Peierls, provides the intuition. It involves analyzing the thermodynamic stability of the perfectly ordered state against defects.

Imagine the system at $T=0$. For a ferromagnet ($J>0$), the ground state consists of all spins perfectly aligned, for instance, all up:
↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑
The energy of each of the $N$ bonds is $-J$, so the total energy is $-NJ$. This is a state of perfect order, so its entropy is zero.

Now, let's consider the system at a very low but non-zero temperature, $T>0$. Let's introduce a single defect—a domain wall—by flipping all spins to the right of some point:
↑ ↑ ↑ ↑ | ↓ ↓ ↓ ↓

Let's analyze the change in the free energy, $\Delta F=\Delta E-T\Delta S$, from creating this single wall.

• Energy Cost ($\Delta E$): The single bond ↑|↓ now has energy $+J$ instead of $-J$. All other bonds are unaffected. The energy cost to create this wall is therefore $\Delta E=J-(-J)=2J$. This is a finite, constant energy cost.

• Entropy Gain ($\Delta S$): The domain wall can be placed between any of the $N$ pairs of spins in the chain. Thus, creating one wall introduces $N$ possible microstates for the system. Using Boltzmann's entropy formula, the entropy gain is $\Delta S=k_B\ln N$.

The change in free energy is:
$\Delta F=2J-k_{B}T\ln N$
In the thermodynamic limit ($N\to \infty$), the entropic term $-k_{B}T\ln N$ becomes infinitely negative for any $T>0$, no matter how small. This term will always overwhelm the constant energy cost $2J$.

This means it is always thermodynamically favorable to introduce domain walls. The creation of one wall lowers the free energy, which encourages the creation of more, until the system is filled with a random arrangement of up and down domains. This completely destroys the long-range order of the ground state.

This elegant argument highlights the special nature of one dimension: a single defect is enough to cut the entire system in two, and the entropy associated with placing this defect anywhere along the infinite chain is overpowering.

Argument 3: The Rigorous Mathematical Foundation

The fact that we could write down a simple, well-behaved (analytic) expression for the magnetization stems from a deeper mathematical property of the transfer matrix solution. Phase transitions are points of non-analyticity. Their absence in the 1D model is directly related to the analyticity of the free energy, $f=-k_{B}T\ln {\lambda }_1$.

The function $\ln (x)$ is analytic as long as its argument $x$ is positive and finite. The largest eigenvalue, ${\lambda }_1$, is a sum and square root of exponential and hyperbolic functions. These are all analytic. The only place a non-analyticity could arise is if the term inside the square root becomes zero or negative, leading to a branch point singularity.
$\text{Argument of square root}=e^{2\beta J}{\sinh }^2(\beta B)+e^{-2\beta J}$
Since ${\sinh }^2(\beta B)\ge 0$ and $e^{-2\beta J}>0$ for any finite $\beta$, the entire expression is strictly positive. Therefore, the square root is always real and well-defined, and ${\lambda }_1$ is an analytic function of both $T$ and $B$ for all $T>0$. Consequently, the free energy $f$ is also analytic. No non-analyticity means no phase transition.

This property is quite general for 1D systems with short-range interactions and is formalized by the Perron-Frobenius theorem.

In essence, the Perron-Frobenius theorem provides the rigorous justification for our observation: the largest eigenvalue ${\lambda }_1$ is always strictly greater than ${\lambda }_2$ for $T>0$, meaning they never cross. This absence of eigenvalue degeneracy is what guarantees the analyticity of the free energy.

Conclusion

We have established a cornerstone result of statistical mechanics: the one-dimensional Ising model with short-range interactions does not exhibit a phase transition at any finite temperature.

Key Takeaways:

• Direct Calculation: The exact expression for magnetization shows that spontaneous magnetization is zero for all $T>0$.
• Physical Argument: The entropy gained by creating a domain wall anywhere along the infinite 1D chain always overcomes the finite energy cost, destroying long-range order for any $T>0$.
• Mathematical Rigor: The Perron-Frobenius theorem ensures that the largest eigenvalue of the transfer matrix is unique and analytic, leading to an analytic free energy, which precludes the possibility of a phase transition.

This result is not just a mathematical curiosity; it highlights the crucial role of dimensionality in collective phenomena. As our mini-exercise hinted, the situation changes dramatically when we move to two dimensions.

Preview of the Next Lesson:
We will move into the fascinating world of the two-dimensional Ising model. While its solution is far more complex, we will state the celebrated exact result for the free energy found by Lars Onsager. We will see that it does indeed contain a non-analyticity, signaling a genuine phase transition, and we will begin to explore the nature of this critical point.