The Mermin-Wagner Theorem

Recommended resources

The freely available resources this lesson will direct you to.

Hello! Welcome to the first lesson in our module on Advanced Topics in Critical Phenomena.

In our previous module, we saw how the renormalization group and the epsilon expansion provide a comprehensive framework for understanding continuous phase transitions. We explored the O(n) model and found that its critical behavior depends on both dimension $d$ and the symmetry of the order parameter, specified by $n$ . This naturally leads to a crucial question: What happens in low dimensions? For instance, the 2D Ising model ( $n = 1$ , discrete symmetry) famously exhibits a phase transition, yet we've seen hints (e.g., from the spherical model, $n \to \infty$ ) that models with continuous symmetry ( $n \geq 2$ ) do not have a finite-temperature transition in $d = 2$ .

Today, we will address this question head-on. Our learning goal is to state the Mermin-Wagner theorem and explain why continuous symmetries cannot be spontaneously broken in d≤2. This is a cornerstone result in statistical physics that places a fundamental constraint on the existence of long-range order. We will explore this theorem from three angles: a physical, intuitive argument; a concrete calculational example; and a sketch of a rigorous proof.

1. Spontaneous Symmetry Breaking and Goldstone Modes

The key distinction between models like the Ising model and those like the XY or Heisenberg models is the nature of their symmetry.

Ising Model: Has a discrete $Z_{2}$ symmetry ( $σ_{i} \to - σ_{i}$ ). To flip a domain of ordered spins, you must overcome a finite energy barrier, the energy of the domain wall.
XY/Heisenberg Models: Have a continuous rotational symmetry (U(1) or SO(3)). In the ordered state, all spins align in a specific direction, spontaneously breaking this symmetry. However, one can create a very-long-wavelength twist in the spin direction. As the wavelength of this twist goes to infinity, the energy cost to create it goes to zero.

These zero-energy (or massless) excitations associated with the breaking of a continuous symmetry are known as Goldstone modes. In condensed matter systems, they manifest as familiar excitations:

Crystals: Breaking continuous translational symmetry gives rise to massless phonons.
Ferromagnets: Breaking continuous rotational (spin) symmetry gives rise to massless magnons (spin waves).

This concept is central to understanding the Mermin-Wagner theorem. The following video provides an excellent physical explanation of Goldstone modes and their direct connection to the absence of magnetic order in low dimensions.

Symmetry Breaking and Magnetism - Prof Stephen Blundell - OUPS Lecture

In this segment, Professor Stephen Blundell explains the concept of Goldstone's theorem and then uses it to provide a clear, physical argument for the Mermin-Wagner theorem. Focus on the idea of massless excitations and how their proliferation destroys long-range order.

Please watch the segment from 40:20 to 48:27. Pay close attention to the link between breaking a continuous symmetry, the appearance of Goldstone modes (like magnons), and the argument for why these modes prevent ordering in 1D and 2D.

The physical argument, as you saw in the video, can be summarized as follows:

Spontaneously breaking a continuous symmetry requires the existence of massless Goldstone modes.
At any finite temperature $T > 0$ , thermal energy will excite these modes.
The crucial question is: How many of these modes are excited? To find out, we must integrate the number of modes at a given energy (given by the Bose-Einstein distribution for bosonic excitations like magnons) over all possible wavevectors $k$ .
For a $d$ -dimensional system, the volume element in k-space is $d^{d} k \sim k^{d - 1} d k$ . For low-energy magnons, the dispersion relation is typically $ω (k) \propto k^{2}$ .
The total number of thermally excited magnons is proportional to the integral $\int \frac{k ^{d - 1} d k}{e ^{β ω (k)} - 1}$ . At low $k$ , this behaves like $\int \frac{k ^{d - 1} d k}{k ^{2}} = \int k^{d - 3} d k$ .
This integral diverges at the lower limit ( $k \to 0$ ) for $d \leq 2$ . This divergence means that at any non-zero temperature, an infinite number of spin-flipping magnons are created, which completely scramble any preferred orientation and destroy long-range magnetic order.

2. A Concrete Calculation: The Classical XY Model

The physical argument is powerful, but we can make it more concrete by calculating the correlation function for a specific model. The classical XY model is a perfect test case, as it has a continuous U(1) symmetry. The Hamiltonian is:

H = - J ⟨ i, j ⟩ \sum s_{i} \cdot s_{j} = - J ⟨ i, j ⟩ \sum cos (θ_{i} - θ_{j})

where $s_{i}$ are 2D unit vectors represented by angles $θ_{i}$ .

If long-range order existed, all spins would, on average, point in the same direction, and the correlation function $C (r) = ⟨ s (r) \cdot s (0)⟩$ would approach a non-zero constant as $r \to \infty$ . Let's see if this holds up to scrutiny.

The following tutorial provides a clear, step-by-step calculation. We will assume order exists (meaning fluctuations $θ_{i} - θ_{j}$ are small), approximate the Hamiltonian, and calculate the correlation function to check for self-consistency.

[PDF] Tutorial III The Mermin-Wagner theorem

This tutorial from the Weizmann Institute, titled 'The Mermin-Wagner theorem', walks through the classic calculation for the XY model. It demonstrates mathematically how thermal fluctuations destroy long-range order in dimensions one and two.

Please read sections 2 and 3 of the document (pages 1-4). Section 2 introduces the XY model and the spin-wave approximation. The key part is Section 3, which calculates the correlation function. Focus on how the mean-square fluctuation (\langle[ heta(r)- heta(0)]^2 angle) is calculated and why its dependence on distance r is different for d=1, 2, and 3. The final result in equation (16) is the punchline.

As the tutorial shows, the long-range order is destroyed by the divergence of fluctuations. The correlation function is $C (r) = ⟨ cos (θ (r) - θ (0))⟩ \approx e^{- \frac{1}{2} ⟨(θ (r) - θ (0))^{2} ⟩}$ for small fluctuations. The calculation reveals:

In d=3, $⟨(θ (r) - θ (0))^{2} ⟩$ approaches a constant as $r \to \infty$ , so $C (r)$ asymptotes to a non-zero value. Long-range order is possible.
In d=2, $⟨(θ (r) - θ (0))^{2} ⟩ \propto lo g (r)$ , which diverges as $r \to \infty$ . This causes $C (r) \propto r^{- η (T)}$ to decay to zero as a power law. No true long-range order, but a "quasi-long-range" order.
In d=1, $⟨(θ (r) - θ (0))^{2} ⟩ \propto r$ , which diverges even more strongly. This causes $C (r) \propto e^{- r / ξ}$ to decay exponentially to zero. Only short-range order exists.

This calculation confirms our physical intuition: the long-wavelength fluctuations are too strong in low dimensions and wash out any potential long-range order.

3. The Formal Theorem and a Rigorous Proof

We can now state the theorem formally.

The Mermin-Wagner-Hohenberg Theorem:

In systems with short-range interactions, spontaneous breaking of a continuous symmetry is forbidden at any non-zero temperature for spatial dimensions $d \leq 2$ . Consequently, the expectation value of any order parameter that is not invariant under the continuous symmetry group must be zero.

This theorem is very general and applies to quantum systems as well. The original proof by Mermin and Wagner for the Heisenberg model uses a powerful mathematical tool called the Bogoliubov inequality. Given your background, it's worth seeing the structure of this rigorous proof.

The Bogoliubov inequality states:

\frac{1}{2} ⟨{A, A^{†}}⟩ \geq k_{B} T \frac{∣ ⟨[ C , A ]⟩ ∣ ^{2}}{⟨[ C , [ C ^{†} , H ]]⟩}

where $A$ and $C$ are any operators.

The strategy of the proof is to choose $A$ and $C$ cleverly. For the Heisenberg model, one chooses them to be Fourier components of the spin raising/lowering operators. The proof then proceeds as follows:

The term $⟨[C, A]⟩$ on the right-hand side becomes proportional to the magnetization $M$ .
The double commutator in the denominator, after some algebra and taking a $k \to 0$ limit, becomes proportional to $k^{2}$ .
This results in an inequality of the form $(const.) \geq M^{2} \int \frac{d ^{d} k}{a k ^{2} + b M}$ , where $a$ and $b$ are constants.
For $d \leq 2$ , the integral on the right-hand side diverges as $M \to 0$ . For the inequality to hold, the magnetization $M$ must be strictly zero for any $T > 0$ .

The following resource provides the details of this argument.

[PDF] The Mermin-Wagner Theorem

These presentation slides by Andreas Werner, titled 'The Mermin-Wagner Theorem', outline the rigorous proof using the Bogoliubov inequality for the isotropic Heisenberg model.

Please review the slides covering the 'Proof of the Mermin-Wagner Theorem' (starting around slide 9). You don't need to re-derive every commutator, but focus on the logical flow: Note the specific choice of operators A and C on slide 14. See how the terms of the Bogoliubov inequality are evaluated, leading to the final inequality on slide 22. Understand why the integral on slide 22 diverges for d=1 and d=2, forcing the spontaneous magnetization to be zero (slides 23-25).

This rigorous proof solidifies our understanding, showing that the absence of long-range order for continuous symmetries in low dimensions is not an artifact of a specific model or approximation, but a fundamental consequence of the principles of statistical mechanics.

Test your understanding!

The Mermin-Wagner theorem relies on the symmetry being continuous. Consider the 2D ANNNI (Anisotropic Next-Nearest-Neighbor Ising) model. It's an Ising model, so spins are $\sigma_i = \pm 1$. However, at certain parameters, it can form phases where the spins order in striped patterns (e.g., `...++--++--...`). This striped phase breaks translational symmetry. Is the Mermin-Wagner theorem applicable to this transition? Why or why not?

Show answer

The Mermin-Wagner theorem is not directly applicable, but its principle is. The order in the striped phase breaks the *discrete* translational symmetry of the underlying lattice (e.g., a stripe of width 2 breaks symmetry under one-lattice-site translations). However, the *phase* of the stripes (their overall position) can be shifted continuously. This emergent U(1) "phason" mode is a Goldstone mode associated with breaking translational symmetry. Therefore, long-wavelength fluctuations of the stripe positions will destroy true long-range positional order of the stripes, in accordance with the Mermin-Wagner principle. The system will have algebraic (power-law) correlations of the stripe positions, similar to the 2D XY model.

Conclusion

In this lesson, we have established a crucial result limiting the possibility of long-range order in the physical world.

Key Takeaways:

Mermin-Wagner Theorem: Prohibits the spontaneous breaking of any continuous symmetry for systems in dimensions $d \leq 2$ at any finite temperature ( $T > 0$ ), provided the interactions are sufficiently short-ranged.
Physical Mechanism: The breaking of a continuous symmetry would imply the existence of massless Goldstone modes. In low dimensions, thermal fluctuations populate these long-wavelength modes so abundantly that they destroy any potential long-range order.
Manifestation: This destruction of order is seen in the decay of correlation functions. For $d = 2$ , the decay is power-law ("quasi-long-range order"), and for $d = 1$ , it is exponential (short-range order).
Rigorous Foundation: The theorem can be proven formally for quantum systems using the Bogoliubov inequality.

Preview of the Next Lesson:

The Mermin-Wagner theorem seems to imply that 2D systems with continuous symmetries are "boring" at finite temperatures. This is not true! While they can't have true long-range order, the 2D XY model exhibits a very peculiar phase transition without an order parameter, known as the Kosterlitz-Thouless (KT) transition. In our next lesson, we will explore this fascinating phenomenon, which is driven by the binding and unbinding of topological defects called vortices.