Spontaneous symmetry breaking. Ferromagnet.


by LagrangeEuler
Tags: ferromagnet, spontaneous, symmetry
LagrangeEuler
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#1
Feb16-13, 10:31 AM
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I'm not sure what people meant about this. Heisenberg hamiltonian is ##O(3)## invariant.
[tex]H=-J\sum_{\langle i,j \rangle} \vec{S}_i \cdot \vec{S}_j[/tex]
##\langle \rangle## denotes nearest neighbors.
It has ##O(3)## symmetry. If I understand well ground state is infinitely degenerate. But system must choose some ground state. Now where is symmetry breaking here?
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Vanadium 50
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#2
Feb16-13, 10:42 AM
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The solutions to the ground state do not have O(3) symmetry, do they? (The sum of solutions does, but an individual solution does not)
LagrangeEuler
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#3
Feb16-13, 10:51 AM
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I don't understand what do you say. Could you be more explicit? What is spontaneous symmetry breaking?

atyy
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#4
Feb16-13, 11:48 AM
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Spontaneous symmetry breaking. Ferromagnet.


Roughly, it means that a ground state does not have the rotational symmetry. Rotating brings you from one ground state to another. But how about the superposition of ground states which is also a ground state? I found Eq 9 and the introduction of http://arxiv.org/abs/physics/0609177v1 helpful.

"The general idea behind spontaneous symmetry breaking is easily formulated: as a collection of quantum mechanical particles grows larger, the object as a whole becomes ever more unstable against small perturbations. In the end even an infinitesimal perturbation is enough to cause the collective system to break the underlying symmetry of the Hamiltonian. The fact that the symmetry breaking can happen spontaneously is then signaled by a set of non-commuting limits: In the complete absence of perturbations even a macroscopic system should conform to the symmetry of the Hamiltonian. If on the other hand the system is allowed to grow to macroscopic size in the presence of even just an infinitesimal perturbation, then it will be able to break the symmetry and end up in a classical state."
Timo
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#5
Feb16-13, 11:53 AM
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Quote Quote by Vanadium 50 View Post
The solutions to the ground state do not have O(3) symmetry, do they? (The sum of solutions does, but an individual solution does not)
The macroscopic ground state is when all spins point in the same direction S. This is realized by an infinite amount of microstates, one of each possible S. These microstates can be mapped onto each other via a rotation or reflection. Similarly, applying a rotation to the set of all these microstates creates a new set that is identical to the original one. In that sense, the macroscopic ground state has an O(3) symmetry - assuming I remember the posh-naming convention of symmetries correctly.

To answer the original question: The breaking of this "macroscopic" symmetry resulting in a stable permanent magnetization occurs in practice, not in the abstract math model. Real materials have impurities and anisotropies.
LagrangeEuler
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#6
Feb16-13, 12:20 PM
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I'm interesting in symmetry breaking in classical physics. Why I can't say that with rotation of ground state I get same ground state. Also is there symmetry breaking for finite lattice?
atyy
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#7
Feb16-13, 12:25 PM
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Quote Quote by LagrangeEuler View Post
I'm interesting in symmetry breaking in classical physics. Why I can't say that with rotation of ground state I get same ground state.
In classical physics, the idea is that you set up an "inertial frame" using "fixed stars" in the universe or lab external to the system which don't have this symmetry. When you rotate the ground state, the rest of the universe is fixed, so rotating the ground state brings you to a different position relative to the "fixed stars" in the universe.

Quote Quote by LagrangeEuler View Post
Also is there symmetry breaking for finite lattice?
Rod Munoz's comment linked to an interesting article about this at http://physics.stackexchange.com/que...uantum-systems.
LagrangeEuler
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#8
Feb16-13, 01:59 PM
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Ok. But what is in that case SPONTANEOUSLY SYMMETRY BREAKING? States are not equivalent. Ok. Let in T=0 all spins are in direction of z-axis. And what now?
Timo
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#9
Feb16-13, 02:34 PM
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You have to admit that it is you who claimed that there was spontaneous symmetry breaking, so it's somewhat strange to ask us where it is. Example origins of symmetry breaking would be an external magnetic field in the z-direction or impurities with net magnetic moment in the z-direction.
DrDu
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#10
Feb17-13, 03:17 AM
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In the limit of an infinite crystal, e.g. the state where all spins are pointing in z-direction is orthogonal to the state were all spins point in x direction. Furthermore it requires changing the direction of an infinity of spins to move one state into the other. This is not possible in one Hilbert space. Hence the operation which rotates one ground state in the other cannot be implemented in one Hilbert space. So the O(3) symmetry has really ceased to exist on a quantum level.
LagrangeEuler
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#11
Feb17-13, 03:52 AM
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Thanks you all for discussion. If I understand it you need infinite energy to flip infinite many spins. What if lattice is finite? For example we have 24x24x24 lattice sites which all pointing up in the z- direction. I don't need infinite energy to flip them all. Right?

I have a problem to understand this problem in classical level. What will you say that the spontaneous symmetry breaking is?
DrDu
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#12
Feb17-13, 04:05 AM
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Quote Quote by LagrangeEuler View Post
What if lattice is finite?
Symmetry breaking strictly speaking only occurs in infinite systems, like other phase transitions.
So it's an idealisation.
Maybe more physical is the characterisation in terms of long range order, as this is directly observable.
I regard spontaneous symmetry breaking as mainly a quantum mechanical phenomenon.
However on scholarpedia, there are two articles, one for symmetry breaking in classical and one for breaking in quantum systems:
http://www.scholarpedia.org/article/...ssical_systems
http://www.scholarpedia.org/article/...uantum_systems
DrDu
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#13
Feb17-13, 04:07 AM
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Quote Quote by LagrangeEuler View Post
If I understand it you need infinite energy to flip infinite many spins.
No, you don't need infinite energy. You can rotate a magnet using arbitrary little energy.
The point is that you cannot represent this rotation as a quantum mechanical operator.
LagrangeEuler
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#14
Feb17-13, 04:15 AM
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Sorry but what quantum mechanics operator if I talk about classical spins systems?
DrDu
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#15
Feb17-13, 05:10 AM
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I didn't understand from your OP that you are interested specifically in classical systems.
The point is that, if you were living in an infinite ferromagnet, you would perceive the magnetization axis as singled out and you would not have any means to rotate it. So although your underlying hamiltonian is invariant under rotations, you do not have the means to implement these rotations.
That's what is meant with spontaneous symmetry breaking.


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