# A Heisenberg ferromagnet -- Spontaneous symmetry breaking

1. May 19, 2016

### 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?

2. May 19, 2016

### vanhees71

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.

3. May 19, 2016

### LagrangeEuler

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?

4. May 19, 2016

### DrDu

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.

5. May 19, 2016

### atyy

http://arxiv.org/abs/physics/0609177
Spontaneous Symmetry Breaking in Quantum Mechanics
Jasper van Wezel, Jeroen van den Brink

The article explains two things that have been mentioned by vanhees71 and Dr Du:
1) the need for a symmetry breaking external field
2) the need for the limit of an infinitely large system

6. May 19, 2016

### DrDu

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

7. May 19, 2016

### atyy

Last edited: May 19, 2016
8. May 19, 2016

### DrDu

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.

9. May 19, 2016

### atyy

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

10. May 19, 2016

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.

11. May 20, 2016

### atyy

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

12. May 20, 2016