Solving the 1D Ising Model with Transfer Matrices

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Hello! Welcome back to our exploration of the Ising model.

In the last lesson, we formulated the Ising Hamiltonian, which gives us the energy for any microscopic arrangement of spins. Our ultimate goal is to connect this microscopic description to macroscopic thermodynamic properties like magnetization and specific heat. The bridge between these two worlds is the partition function, $Z$ .

Today's learning outcome is to solve the one-dimensional Ising model exactly using the transfer matrix method. This is a classic and beautiful result in statistical mechanics. It not only provides a complete solution for the 1D system but also introduces a powerful technique that is central to many areas of theoretical physics.

From Hamiltonian to Partition Function

Let's briefly recall the setup from last lesson. For a one-dimensional chain of $N$ spins, the Hamiltonian is:
$H = - J \sum_{i = 1}^{N} σ_{i} σ_{i + 1} - B \sum_{i = 1}^{N} σ_{i}$
To eliminate boundary effects, we'll consider the spins arranged on a ring, which imposes periodic boundary conditions: $σ_{N + 1} = σ_{1}$ .

The partition function is the sum over all $2^{N}$ possible spin configurations:
$Z = \sum_{{σ}} e^{- β H} = \sum_{σ_{1} = \pm 1} \dots \sum_{σ_{N} = \pm 1} exp (β J \sum_{i = 1}^{N} σ_{i} σ_{i + 1} + βB \sum_{i = 1}^{N} σ_{i})$
Directly evaluating this sum is impossible for large $N$ . The challenge lies in the coupling term $σ_{i} σ_{i + 1}$ , which prevents the sum from being separated into a product of independent sums for each spin. The transfer matrix method is a clever way to handle exactly this kind of nearest-neighbor coupling.

The Transfer Matrix Method

The core idea is to rewrite the exponential of the sum as a product of exponentials. This allows us to view the partition function calculation as a sequence of matrix multiplications.

To make this work, it's convenient to split the external field term symmetrically between adjacent sites. The total energy remains the same under the sum, but the form becomes more useful:
$H = - \sum_{i = 1}^{N} (J σ_{i} σ_{i + 1} + \frac{B}{2} (σ_{i} + σ_{i + 1}))$
Now, the partition function can be written as:
$Z = \sum_{σ_{1}, \dots, σ_{N}} \prod_{i = 1}^{N} exp (β J σ_{i} σ_{i + 1} + \frac{βB}{2} (σ_{i} + σ_{i + 1}))$

Let's define a quantity that depends only on two adjacent spins, $σ_{i}$ and $σ_{i + 1}$ . We can think of this as the "link" between site $i$ and site $i + 1$ . Let's organize these quantities into a 2x2 matrix, which we'll call the transfer matrix, $T$ . The elements of this matrix are indexed by the spin values $σ = \pm 1$ and $σ^{'} = \pm 1$ :
$T_{σ, σ^{'}} = exp (β J σ σ^{'} + \frac{βB}{2} (σ + σ^{'}))$

Let's write this matrix out explicitly:

$T_{+ 1, + 1} = exp (β J + βB)$
$T_{- 1, - 1} = exp (β J - βB)$
$T_{+ 1, - 1} = exp (- β J)$
$T_{- 1, + 1} = exp (- β J)$

So, the transfer matrix is:
$T = (e^{β (J + B)} e^{- β J} e^{- β J} e^{β (J - B)})$

This matrix $T$ contains all the information about the interaction between one spin and its neighbor.

Solving the 1D Ising Model

The following notes from a course by Prof. Steven Kivelson at Stanford provide an exceptionally clear walkthrough of the transfer matrix method. We will follow its logic closely.

Please read the sections 'Writing the Hamiltonian as a sum over bonds' and 'Defining the transfer matrix'. The text clearly shows how to rewrite the Hamiltonian and define the matrix elements of T. Note the use of bra-ket notation, which might feel familiar from your quantum mechanics courses and can be a helpful way to think about the spin states.

The Partition Function as a Trace

Now, let's see how this matrix simplifies the partition function. We can rewrite the sum as:
$Z = \sum_{σ_{1} = \pm 1} \dots \sum_{σ_{N} = \pm 1} T_{σ_{1}, σ_{2}} T_{σ_{2}, σ_{3}} \dots T_{σ_{N - 1}, σ_{N}} T_{σ_{N}, σ_{1}}$
Let's look at the sum over $σ_{2}$ : $\sum_{σ_{2}} T_{σ_{1}, σ_{2}} T_{σ_{2}, σ_{3}}$ . This is precisely the definition of the matrix product $(T^{2})_{σ_{1}, σ_{3}}$ . If we perform the sum over $σ_{3}$ , we get $(T^{3})_{σ_{1}, σ_{4}}$ , and so on.

After summing over all intermediate spins $σ_{2}, \dots, σ_{N}$ , we are left with:
$Z = \sum_{σ_{1} = \pm 1} (T^{N})_{σ_{1}, σ_{1}}$
This is the sum of the diagonal elements of the matrix $T^{N}$ , which is, by definition, the trace of $T^{N}$ .
$Z = Tr (T^{N})$
This is a remarkable result. The complicated sum over $2^{N}$ configurations has been reduced to computing the $N$ -th power of a 2x2 matrix.

Solving the 1D Ising Model

Let's continue with the same notes to see how the partition function is formally written as a trace.

Read the section 'Rewriting the partition function as a trace'. It explains the 'key trick' of how summing over a shared spin index is equivalent to matrix multiplication. The analogy to the path integral propagator at the end is a nice connection you might appreciate.

Calculating the Trace via Eigenvalues

The trace of a matrix is independent of the basis in which it's calculated. The most convenient basis is the eigenbasis of $T$ . Let the eigenvalues of $T$ be $λ_{1}$ and $λ_{2}$ . In its eigenbasis, $T$ is diagonal:
$T_{diag} = (λ_{1} 0 0 λ_{2})$
Then $T^{N}$ is simply:
$(T^{N})_{diag} = (λ_{1}^{N} 0 0 λ_{2}^{N})$
The trace is the sum of these diagonal elements:
$Z = Tr (T^{N}) = λ_{1}^{N} + λ_{2}^{N}$

This gives us the exact partition function for any finite $N$ .

The Thermodynamic Limit

We are typically interested in macroscopic systems, which corresponds to the thermodynamic limit $N \to \infty$ . Let's order the eigenvalues such that $λ_{1} > λ_{2}$ . Then we can write:
$Z = λ_{1}^{N} (1 + (\frac{λ _{2}}{λ _{1}})^{N})$
Since $∣ λ_{2} / λ_{1} ∣ < 1$ , the second term vanishes as $N \to \infty$ . Therefore, in the thermodynamic limit, the partition function is dominated entirely by the largest eigenvalue:
$Z \approx λ_{1}^{N} (N \to \infty)$
The Helmholtz free energy per spin, $f = F / N$ , is then:
$f = - \frac{1}{N} k_{B} T ln Z \approx - \frac{1}{N} k_{B} T ln (λ_{1}^{N}) = - k_{B} T ln λ_{1}$
This is the central result of the transfer matrix method: all the thermodynamic properties of the infinite 1D Ising chain are determined by the largest eigenvalue of a simple 2x2 matrix!

Finding the Eigenvalues

The final step is to find the eigenvalues of $T$ by solving the characteristic equation $det (T - λ I) = 0$ :
$det (e^{β (J + B)} - λ e^{- β J} e^{- β J} e^{β (J - B)} - λ) = 0$
$(e^{β (J + B)} - λ) (e^{β (J - B)} - λ) - e^{- 2 β J} = 0$
This is a quadratic equation for $λ$ . After some algebraic manipulation (involving identities for hyperbolic functions), the two solutions are found to be:
$λ_{1, 2} = e^{β J} cosh (βB) \pm e^{2 β J} sinh^{2} (βB) + e^{- 2 β J}$
The larger eigenvalue, $λ_{1}$ , is the one with the plus sign.

The Ising model

The following video provides a walkthrough of the derivation, including the explicit calculation of the eigenvalues. It offers a good dynamic complement to the textual resources.

Watch from 8:51 to 12:21 to see how the partition function is expressed as the trace of the transfer matrix power. Then, watch from 16:29 to 20:19, which details the process of finding the eigenvalues by solving the characteristic equation. The video focuses on the B=0 case for simplicity at the end, which we will discuss next.

Special Case: Zero External Field ( $B = 0$ )

The physics is often clearest in the absence of an external field. Setting $B = 0$ , we have $cosh (0) = 1$ and $sinh (0) = 0$ . The eigenvalues simplify significantly:
$λ_{1, 2} = e^{β J} \pm e^{- 2 β J} = e^{β J} \pm e^{- β J}$
This gives us:

$λ_{1} = 2 cosh (β J)$
$λ_{2} = 2 sinh (β J)$

So, for $B = 0$ , the exact partition function for a 1D Ising ring is $Z = (2 cosh (β J))^{N} + (2 sinh (β J))^{N}$ , and the free energy per spin in the thermodynamic limit is $f = - k_{B} T ln (2 cosh (β J))$ .

Test your understanding!

Using the zero-field result for the free energy per spin, $f = - k_{B} T ln (2 cosh (β J))$ , calculate the internal energy per spin, $u = U / N$ .

Recall that $U = (\frac{\partial ( βF )}{\partial β})_{V}$ , which means $u = \frac{\partial ( β f )}{\partial β}$ . What do you get for the energy in the high temperature ( $T \to \infty$ or $β \to 0$ ) and low temperature ( $T \to 0$ or $β \to \infty$ ) limits? Do these limits make physical sense?

Show answer

First, let's calculate the internal energy per spin, $u$ .
$β f = - ln (2 cosh (β J))$
$u = \frac{\partial ( β f )}{\partial β} = - \frac{\partial}{\partial β} ln (2 cosh (β J)) = - \frac{1}{2 c o s h ( β J )} \cdot 2 sinh (β J) \cdot J = - J tanh (β J)$
This is the exact internal energy per spin for the 1D Ising model at zero field.

Now let's check the limits:

High Temperature ( $T \to \infty ⟹ β \to 0$ ):
As $β \to 0$ , $tanh (β J) \approx β J$ . So, $u \approx - J (β J) = - J^{2} / k_{B} T$ . As $T \to \infty$ , $u \to 0$ . This makes sense: at infinite temperature, the spins are completely random, so the average value of $σ_{i} σ_{i + 1}$ is zero, and the interaction energy vanishes.
Low Temperature ( $T \to 0 ⟹ β \to \infty$ ):
As $β \to \infty$ , $tanh (β J) \to 1$ (for ferromagnetic $J > 0$ ). So, $u \to - J$ . This also makes sense: at zero temperature, the system is in its ground state. For a ferromagnet, all spins are aligned, so every $σ_{i} σ_{i + 1} = 1$ . The energy per bond is $- J$ , so the energy per spin is also $- J$ (since each spin is part of two bonds on a ring, but each bond is shared by two spins).

An Alternative View: Bond Variables

Interestingly, there is another elegant way to solve the 1D Ising model, developed by Hendrik Kramers and Gregory Wannier (and showcased by Leonard Susskind in the video below). Instead of focusing on the spins on the sites, you can redefine your variables to describe the bonds between the sites. This transforms the problem of interacting spins into a problem of non-interacting bonds, which is trivial to solve.

Statistical Mechanics Lecture 9

For a different and highly intuitive perspective on the same problem, watch this excerpt from a Stanford lecture by Leonard Susskind. He demonstrates a clever 'duality' trick to solve the 1D model without matrices.

Watch the segment from 25:43 to 35:40. Pay close attention to how he defines the 'bond' variables \mu_i = \sigma_i \sigma_{i+1} and how this simplifies the energy expression, effectively decoupling the problem.

While the transfer matrix method is more general and can be extended to 2D (though not solved as simply), this alternative approach is a beautiful example of physical reasoning and highlights the concept of duality, which is a deep and recurring theme in theoretical physics.

Conclusion

Today we have accomplished a major goal: the exact solution of the 1D Ising model. We have seen how a seemingly intractable sum over an exponential number of states can be tamed using the transfer matrix method.

Key Takeaways:

The partition function $Z$ of the 1D Ising model on a ring can be expressed as the trace of the $N$ -th power of a 2x2 transfer matrix $T$ : $Z = Tr (T^{N})$ .
In the thermodynamic limit ( $N \to \infty$ ), the free energy per spin depends only on the largest eigenvalue $λ_{1}$ of the transfer matrix: $f = - k_{B} T ln λ_{1}$ .
By finding the eigenvalues of $T$ , we obtained a closed-form expression for the free energy, which contains all thermodynamic information about the system. For $B = 0$ , $f = - k_{B} T ln (2 cosh (β J))$ .

Preview of the Next Lesson:
We now have an exact expression for the free energy. What does this tell us about phase transitions? A phase transition is marked by a non-analyticity in the free energy or its derivatives. In the next lesson, we will analyze our solution for $f$ and use it to demonstrate the absence of a finite-temperature phase transition in the one-dimensional Ising model, a foundational result with profound implications.