# Heisenberg ferromagnet -- Spontaneous symmetry breaking

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• LagrangeEuler
In the thermodynamic limit, the ground state will spontaneously break symmetry and the Cooper pair will be forced into a local pairing.In summary, in type I superconductors, spontaneous symmetry breaking means that the ground state breaks symmetry even though the Hamiltonian does not.

#### LagrangeEuler

Heisenberg model of ferromagnet is defined by

$$\hat{H}=-J \sum_{\langle i,j \rangle} \vec{S}_i \cdot \vec{S}_j$$
where ##J>0## and summation is between nearest neighbours. Hamiltonian is perfectly rotational symmetric. However, the ground state “spontaneously” chooses a particular orientation ##|\psi\rangle=|\uparrow...\uparrow...\uparrow \rangle## and hence is not invariant under the symmetry (rotation).

I do not understand part the ground state “spontaneously” chooses a particular orientation. As far as I understand ground state is the state with minimal energy. So state ##|\psi_2\rangle=|\downarrow...\downarrow...\downarrow \rangle## also corresponds to same energy. And state ##|\psi_3\rangle=|←...←...← \rangle## coresponds to same energy. Why then is the part the ground state “spontaneously” chooses a particular orientation so important?

If you have spontaneous symmetry breaking (for a global symmetry), the ground state is necessarily degenerate. Any state, with all spins pointing in one direction has the same minimal energy. Which ground state the system takes is indetermined. You either need some breaking of the symmetry, e.g., by introducing an external magnetic field with ##\hat{H}' \propto -\sum_i \vec{S}_i \cdot \vec{B}##, then your magnetization in the ground state will be in direction of the magnetic field, or by some initial fluctuation the system "spontaneously" chooses some direction along which the spins orient themselves in the ground state.

Thanks a lot for the answer. So I could say that in this case symmetry of the Hamiltonian larger then the symmetry of the ground state. Does ground state have symmetry in that case?

A symmetry can only be spontaneously broken if the system is infinite. In this case, the operation which changes one ground state into another one is no longer a unitary transformation and the different ground states span different Hilbert spaces. This is the mathematical definition of a broken symmetry. Actually, it is irrelevant whether a special ground state really has the symmetry of the Hamiltonian or not. E.g. in case of a ferromagnet, there is a ground state which is a superposition of all possible orientations with equal sign which is symmetric. The real physical feature which defines a spontaneously broken symmetry is the appearance of long range correlations. E.g. while the expectation value ##<\vec{S}>## may or may not be zero for a given ground state, ##<\vec{S}(r) \cdot \vec{S}(r')>## will never vanish.

• vanhees71
There is no need for a symmetry breaking field. In the case of e.g. superconductivity, it does not even exist.

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What I meant is that taking a single piece of metal and cooling it, it will become superconductive although electron number is evidently fixed and won't fluctuate. I would have to re-read the articles you cited. I didn't consider them very useful when I tried to understand superconductivity, but the "thin spectrum" they discuss maybe potentially helpful in some specialized situations.

DrDu said:
What I meant is that taking a single piece of metal and cooling it, it will become superconductive although electron number is evidently fixed and won't fluctuate. I would have to re-read the articles you cited. I didn't consider them very useful when I tried to understand superconductivity, but the "thin spectrum" they discuss maybe potentially helpful in some specialized situations.

Yes, but by that argument even an infinite size system is not needed since real superconductors are not infinite in size.

Spontaneous symmetry breaking implies that you have a nonzero vacuum expectation value of the order parameter without a source. This means that the ground state breaks symmetry even though your Hamiltonian does not. In the case of a ferromagnet, if you start in the spin up ordered phase, you will never be able to get to the spin down ordered phase in an infinite system through local fluctuations even though technically spin up and spin down have the same energy in zero field. So once you choose a ground state you will remain in that ground state super selection sector. You can only go between these super selection sectors through global operations on the system so they only make sense in the thermodynamic limit. If you had a finite system you could technically get between ground states in a finite time flipping spins. However this time grows exponentially with the system size where you measure lengths in terms of L/ξ. Superselection sectors come up often when discussing topologically ordered states.

For type I superconductors the order parameter would be the expectation value of the complex pairing field and you can find the critical temperature and critical B field in Ginzburg-Landau theory by creating an effective action for the cooper pair field which can be allowed to vary spatially.

• vanhees71
However this time grows exponentially with the system size where you measure lengths in terms of L/ξ. Superselection sectors come up often when discussing topologically ordered states.

But a difference is that it seems that topological order doesn't require infinite size (kitaev?), whereas the known examples of symmetry breaking do?

Whenever you talk about ground state degeneracies in topologically ordered systems (for this purpose I will not mention SPTs), you are talking about the thermodynamic limit and usually consider periodic boundary or anti-periodic boundary conditions. The definition of topological order is that it is robust against all local perturbations. This can be seen by observing that corrections to the energy fall off exponentially with the size of the system. You can see this in the toric code. So the ground state degeneracy really is only robust in the thermodynamic limit.

Even if you don't have an infinite system, you can still study phase transitions by looking at finite size effects. For example, the critical point of the Ising model is described by the simplest CFT of massless free fermions. However, in a finite size system, conformal invariance has been broken by introducing a length scale. This will show up in 1/L corrections to the free energy which are proportional to the central charge.

Another thing to consider is that these systems are sensitive to boundary conditions. For example, the finite size corrections will be different for PBCs and APBCs conditions. If you look at the quantum Ising model/Kitaev chain you will see the choice of BCs reflected in the Kramer's degeneracy (whether or not you have a zero state).

A last comment is that every finite size system is gapped. Again, you can tell whether the system is gapped or gapless in the thermodynamic limit by finite size scaling.