Spontaneous Symmetry Breaking in Ferromagnets

The Ferromagnetic Phase Transition: A Tale of Two Symmetries

A ferromagnet is a material that can exhibit a net magnetic moment even in the absence of an external magnetic field. However, this property is temperature-dependent.

• Above the Curie Temperature ($T>T_c$): The material is in a paramagnetic phase. Thermal energy is high enough that the microscopic magnetic moments (spins) of the atoms are randomly oriented. There is no net magnetization, $\vec{M}=0$, and the system as a whole is isotropic—it looks the same in all directions.
• Below the Curie Temperature ($T<T_c$): The material enters the ferromagnetic phase. It becomes energetically favorable for the spins to align, creating a net spontaneous magnetization, $\vec{M}\ne 0$. Although the underlying physical laws governing the spin interactions are still isotropic, the system's ground state is not. It has "chosen" a specific direction in space for its magnetization.

A crucial, and perhaps counter-intuitive, point is identifying which phase is more symmetric. The following video by Prof. Stephen Blundell addresses this question directly.

As Prof. Blundell explains, the paramagnetic state possesses the full rotational symmetry of the underlying laws because, on average, no direction is preferred. The ferromagnetic state, by selecting a specific axis of magnetization, has spontaneously broken this rotational symmetry.

This phenomenon is captured by an order parameter, a macroscopic quantity that is zero in the symmetric phase and non-zero in the broken-symmetry phase. For a ferromagnet, the order parameter is the net magnetization vector, $\vec{M}$. The transition from $\vec{M}=0$ to $\vec{M}\ne 0$ as the temperature drops below $T_c$ signals the occurrence of SSB.

The Microscopic Picture: The Heisenberg Model

To analyze this more rigorously, we can use a microscopic model. The Heisenberg model provides an excellent description for many magnetic insulators. The Hamiltonian is given by:

where ${\vec{S}}_i$ is the spin operator at lattice site $i$, and $J_{ij}$ is the exchange coupling between sites $i$ and $j$. We will consider the ferromagnetic case, where $J_{ij}>0$.

The crucial feature of this Hamiltonian is its symmetry. The scalar product ${\vec{S}}_i\cdot {\vec{S}}_j$ is invariant under a simultaneous rotation of all spins. This means the Hamiltonian has a global SU(2) symmetry (or SO(3) if we treat the spins as classical vectors).

However, the ground state of the system does not share this symmetry. To minimize the energy, all spins must align parallel to each other. But in which direction should they align? Any direction in space is equally good. This means there is a continuous manifold of degenerate ground states, each corresponding to a different direction of total magnetization. This manifold is the two-sphere, $S^2$, since each direction can be specified by a unit vector.

When the system cools below $T_c$, it settles into one of these ground states, for instance, with all spins pointing along the $z$-axis. This choice spontaneously breaks the SU(2) symmetry. The state is now only symmetric under rotations about the chosen $z$-axis. This residual symmetry is a U(1) subgroup of SU(2).

So, the symmetry breaking pattern is:

For a more detailed look at the structure of the Heisenberg model and its ground state, the following text is highly recommended.

Goldstone's Theorem and Magnons

Now we can apply Goldstone's theorem from our previous lesson. The number of broken generators is:

Therefore, we expect two massless Goldstone bosons. What are these excitations? They are long-wavelength, low-energy fluctuations of the order parameter—in this case, slow variations in the direction of the magnetization vector. These collective excitations are known as spin waves, and their quanta are called magnons.

The paper you just read continues to derive the dispersion relation for these magnons. It finds that the energy $E_k$ of a magnon with momentum $k$ behaves as:

This presents a puzzle. We have a quadratic dispersion relation, not the linear $E_k\propto k$ typical of a massless relativistic particle. Furthermore, we seem to have found only one mode, but Goldstone's theorem predicted two.

This is not a contradiction but a vital subtlety of Goldstone's theorem in non-relativistic systems. The theorem, in its most general form, guarantees the existence of gapless modes but does not mandate a linear dispersion. The counting rule is also more nuanced.

• Type-I Goldstone Bosons: These are the "standard" bosons with a linear dispersion, $E_k\propto k$. They arise when the charge density of the broken generator is zero in the ground state.
• Type-II Goldstone Bosons: These have a quadratic dispersion, $E_k\propto k^2$. They can arise when the ground state has a non-zero density of a conserved charge corresponding to a broken generator.

In the ferromagnet, the broken generators are related to spin rotations away from the magnetization axis (e.g., $S_x$ and $S_y$ if $\vec{M}$ is along $z$). The ground state has a non-zero density of $S_z$. This coupling between the broken generators and a non-zero charge density leads to the appearance of a single Type-II Goldstone boson (the magnon) instead of two Type-I bosons.

This distinction is made clearly in the following resource.

The Phenomenological View: Landau Theory

Finally, let's connect this microscopic picture to the "Mexican hat" potential we've used as our canonical model for SSB. This is done through Landau's theory of phase transitions. We write down the Helmholtz free energy $F$ as a power series in the order parameter $M$. Since the energy must be invariant if we flip the direction of magnetization ($\vec{M}\to -\vec{M}$), the expansion can only contain even powers of $M$:

The key insight of Landau theory is that near the critical temperature $T_c$, the coefficient $\alpha (T)$ changes sign. A simple model is $\alpha (T)=a(T-T_c)$ with $a>0$, while $\beta (T)\approx b>0$.

• For $T>T_c$, $\alpha >0$. The minimum of $F$ is at $M=0$.
• For $T<T_c$, $\alpha <0$. The $M^2$ term is now negative, creating an unstable maximum at $M=0$. The minimum of $F$ moves to a non-zero value $M_0=\sqrt{-\alpha /(2\beta )}$.

This is precisely the behavior of the Mexican hat potential. The video by Prof. Blundell provides a clear animation of this process.

Conclusion

Today, we have thoroughly examined the ferromagnet as a physical realization of spontaneous symmetry breaking. By connecting the microscopic Heisenberg model to the phenomenological Landau theory and applying the powerful framework of Goldstone's theorem, we have seen how a phase transition can be understood as a change in the symmetry of the system's ground state.

Key Takeaways:

• A ferromagnet spontaneously breaks continuous SU(2) rotational symmetry down to a U(1) symmetry below the Curie temperature $T_c$.
• The order parameter for this transition is the net magnetization $\vec{M}$.
• The broken symmetry gives rise to gapless Goldstone bosons, which in this system are magnons (spin waves).
• Because the ferromagnetic ground state has a non-zero "charge density" (the magnetization itself), the system exhibits a Type-II Goldstone boson with a quadratic dispersion relation ($E_k\propto k^2$).
• The Landau theory of phase transitions provides a phenomenological description that maps directly onto the "Mexican hat" potential model of SSB.

Preview of the Next Lesson:

Having explored SSB in a magnetic system with broken rotational symmetry, we will turn our attention to another classic condensed matter system: a superfluid. We will describe the spontaneous breaking of a global U(1) phase symmetry, which gives rise to a Type-I Goldstone boson—the phonon. This will provide another concrete application of the theorem and serve as a crucial stepping stone towards understanding the Higgs mechanism, where the symmetry is local rather than global.