A symmetry is spontaneously broken in a system with infinite number of degrees of freedom (DOF), when the system finds itself in the ground state that breaks the symmetry of the Hamiltonian. For example, the $SO(3)$ symmetry of the Hamiltonian in Heisenberg model, is spontaneously broken in paramagnetic to ferromagnetic transition.

For systems with infinite number of degrees of freedom (DOF), thermal fluctuations cannot restore the symmetry of the ground state i.e., the ground state does not share the symmetry of the Hamiltonian. However, with finite DOF, symmetry can be restored via thermal fluctuations.

How does this happen? In other words, how and why is it that thermal fluctuations can restore the invariance of the ground state under the symmetry (of the Hamiltonian) when the system has finite DOF but fails to do so when the system has infinite DOF?


There seems to be some confusion in your question between thermal fluctuations and quantum fluctuations, so I will try to address both of them in my answer.

Spontaneous symmetry breaking occurs when the ground state of the Hamiltonian does not exhibit the full symmetries of the Hamiltonian. In other words, there are multiple degenerate configurations with the lowest energy. Whether the symmetry can be restored by fluctuations depends on the "size" of the manifold of minimum energy states.

The way the restoration occurs is that fluctuations can carry the system from one minimum-energy configuration to another. If the fluctuations are strong enough, and the "number" of minimum-energy states is small enough, the full symmetry of the system is restored. For example, in the classical system of one a particle in a potential with two equally deep wells, if the temperature is high enough (bigger than the barrier between the wells, roughly), the symmetry is restored, because the particle can move freely back and forth between the wells, thanks to the thermal energy it possesses.

Quantum fluctuations can work the same way. For the same particle in a double well potential, the quantum ground state is actually equally distributed between the two wells. The quantum fluctuations, which lead to tunneling through the barrier, restore the symmetry. As the "size" of the manifold of ground states increases, it becomes harder to restore the symmetry. In a field theory system with one spatial dimension, the quantum fluctuations always restore the symmetry; the thermal fluctuations do so as well, provided $T>0$.

As the dimension of the field theory increases, it becomes harder to restore the symmetry. The restorration also depends on the nature of the states. The two-dimensional Ising model, with a discrete state space, has broken symmetry below a threshold temperature. However, a system with a continuous state space, such as a Heisenberg model, does not exhibit symmetry breaking in two dimensions.

The key difference between quantum and thermal fluctuations is that at high enough temperature, thermal fluctuations can always restore a symmetry. If the thermal energy is vast compared with the energy barriers, the system is free to move between the ground states, and the dynamics are symmetric. Contrary to what you suggest in the question, the rotation symmetry of a ferromagnet will be restored at high temperature.

  • $\begingroup$ @ Buzz-But isn't it that without infinite DOF there can be no spontaneous symmetry breaking (SSB)? If I understood it correct, your answer pretends that whether SSB can happen depends only upon the dimensionality of the system and manifold of degenerate vacua. What about Heisenberg model in 3D for a finite system and an infinite system? Can SSB occur in the former case? $\endgroup$
    – SRS
    Dec 22 '16 at 16:45
  • 2
    $\begingroup$ The fact that quantum corrections restore the symmetry for a fixed number of particle in a potential (a system with finite degrees of freedom) is equivalent to the statement that there can only be SSB in a system with an infinite number of degrees of freedom. That means that, technically, there is no SSB for any finite system. However, in practice, the time to tunnel between the degenerate ground states becomes exponentially long in the number of degrees of freedom, so there effectively is SSB. This is just like the thermodynamic limit for large but not truly infinite systems. $\endgroup$
    – Buzz
    Dec 22 '16 at 16:49
  • $\begingroup$ @Buzz- Does the restoration of symmetry imply that the actual ground state of the system becomes a quantum superposition of all degenerate ground states? $\endgroup$
    – SRS
    Dec 22 '16 at 17:07
  • 1
    $\begingroup$ @SRS For the quantum restoration, yes. The ground state is a superposition of states centered around the separate classical ground states. For the thermal restoration, the ground state doesn't change, but the ground state is no longer the resting/equilibrium configuration of the system. The system is in a mixed state that includes equal admixtures of all the classical ground states, as well as an amount of higher-energy states that is determined by the available thermal energy. $\endgroup$
    – Buzz
    Dec 22 '16 at 17:10
  • 1
    $\begingroup$ @SRS I don't know of one off the top of my head, but I will see what I can find. It may be a some time though, since I am out of my office for the holiday break. $\endgroup$
    – Buzz
    Dec 22 '16 at 17:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.