Symmetry breaking

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How does symmetry breaking occur?
 

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  • #2
vanesch
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In general, when you have a problem that is symmetrical under a symmetry group G, it means that your solutions to the problem are representations of that group G. It DOESN'T mean that each individual solution is mapped onto itself (is symmetrical) under G, but only that when you have a solution S, that when you take a symmetry operation g in G, then g S = T and T is an "equivalent" solution.

Let us take an example: consider the problem in euclidean plane geometry: point a and b are separated by a positive integer number of centimeters. This problem is translation invariant, and rotation invariant, so the problem is invariant under the euclidean group in 2 dimensions, call it E(2). Each pair of points {u,v} that satisfies the problem definition is a solution, but of course each pair of points by itself is NOT rotation invariant or translation invariant !

You see that the total solution set is in fact the union of different sets: S1 is the set of all pairs of points that are 1 cm separated from eachother, S2 is the set of all pairs of points that are 2 cm separated from eachother etc...
We will consider that S0, the set of couples {a,a}, is also part of the solution.

Now, it can be that for some reason, we have to choose ONE SINGLE solution. Imagine it is a solution from S0. In that case, we have to choose a specific point O, and the solution we take is {O,O}. We notice that this specific solution is still invariant under rotations about O, but not under translations. Everything we will do with this particular solution HAS BROKEN THE TRANSLATION SYMMETRY of the problem.

But there is more. Imagine, now, that we have to choose a solution from S1. We take two points, A and B, 1 cm from eachother {A,B}. This solution has not only no translation symmetry, it also lost its rotational symmetry. So the symmetry of the original problem is completely gone when we use this solution. We can say that we have full symmetry breaking.

The problem in physics usually comes down to the following situation: usually we have a problem with a certain symmetry (G), and we have to choose a solution (ground state, say) to work with (to have, say, perturbation expansions or so). In many cases, the ground state makes up a solution space of its own ; in that case, the ground state has the full symmetry G. We are so used to this particular case that we often forget that it is not necessary for this to happen. When the ground state solution space contains many elements, however, only the full SET is symmetrical under G and not each element individually. If we are forced to choose one, we BREAK the original symmetry of the problem, and our perturbation series around that ground state will not be invariant under group G.

cheers,
Patrick.
 
  • #3
Haelfix
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There are two ways to break symmetry that we know off.

One is by explicitly writing down symmetry breaking terms in the lagrangian. This is not very interesting, but does the trick just fine.

The other is what Patrick wrote above, and is called spontaneous symmetry breaking..
 
  • #4
turin
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vanesch said:
... when you take a symmetry operation g in G, then g S = T and T is an "equivalent" solution.
What does it mean to be an "equivalent" solution. Does it just mean that both S and T are members of some solution set with no preference of selecting one over the other?




vanesch said:
Now, it can be that for some reason, we have to choose ONE SINGLE solution. Imagine it is a solution from S0. In that case, we have to choose a specific point O, and the solution we take is {O,O}.
I'm a little confused by the term "solution." What are we solving?




vanesch said:
Everything we will do with this particular solution HAS BROKEN THE TRANSLATION SYMMETRY of the problem.
How is that? If we translate the origin, the distance between the points is still zero, isn't it?




Can you give a more physical example? I am lost in the abstraction.
 
  • #5
Haelfix
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Think of the Mexican hat potential. If you pick a particular ground state, and perturb around it, you have locally broken the theories symmetry (say the SO(2) rotational symmetry)

This problem is not manifest in a nonperturbative theory of course, but thats life in QFT.
 
  • #6
vanesch
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turin said:
1) I'm a little confused by the term "solution." What are we solving?

2) How is that? If we translate the origin, the distance between the points is still zero, isn't it?

3) Can you give a more physical example? I am lost in the abstraction.
Allright, I wanted to take a toy problem and apparently things got more confused for you now. Sorry.

I'll first treat 3). Let's take the following surface:

z(x,y) = b (x^2+y^2)^2 - x^2 - y^2

clearly, z is only a function of x^2 + y^2, so is invariant under a rotation around the origin: z(r) = b r^4 - r^2

Now let us consider that we put a marble on that surface, and we would like to calculate its natural oscillation frequency for small movements.
The problem is that if we put the marble on the surface, it will roll down towards a "ground state" with lowest z, and that we will have to do a series expansion around that position (that's what I called choosing our solution).

Calculating the equilibrium points:
z'(r) = 4 b r^3 - 2 r = 0 gives r=0 and r = \sqrt{1/(2 b)}
The first solution is unique (r=0) and hence is also rotationally symmetrical, but unfortunately, it is an unstable equilibrium.

The second solution, r = 1/sqrt{2b} is in fact a whole set of solutions, namely a circle with radius 1/sqrt{2b}. But the marble will pick out ONE of these points on the circle ; we take it to be the point P = {x = 1/sqrt{2b}, 0}.
Clearly, although the CIRCLE (set of solutions) is rotationally invariant, the point we took by itself isn't of course.
We will recenter ourselves on P, with x = u+1/sqrt[2b].
If we do now a Taylor expansion of z(u+1/\sqrt{2b},y) around P, we find:

z[x,y] ~ 2 u^2-1/(4b) + higher orders

So we see that in this expansion, y doesn't appear until higher orders, and we have a parabola in u. No rotational symmetry anymore between u and y.
Nevertheless, the above is the correct potential for small movements of our marble: along the u direction (= shifted x) there is a local parabolic minimum, while along the y direction, the movement is indifferent (indeed, in first order it is moving along the circle of minima).
The original symmetry has been broken because our marble HAD to choose one of the different equivalent ground states.

The problem with this example is that it is SO TERRIBLY trivial. It is only after having seen this exact application to QFT that one realizes what it implies.

Coming back to 1) and 2), I tried to present a toy problem. Imagine that we have these magetic sticks, of 1 cm long, which can be concatenated into sticks of n cm when you put n of these elementary sticks together. The 'solution' to the problem of all possible stick positions in the plane was discussed, where we consider a "stick position" as uniquely described by the two end points of the stick.
Clearly, the problem of "find a stick position" was described by my set of couples of points {a,b}, n cm apart, and has euclidean symmetry (rotations and translations).
This was the "problem" to be solved.

2) YES, you stated the symmetry of the original problem. However, the set with one point, {O}, is of course NOT translationally invariant because O will be mapped onto another point which is not in the set. However, {O} IS rotationally invariant wrt a rotation around O: indeed, O is mapped onto O by such a rotation.
This is not true anymore for the set { (a,b)}.

cheers,
patrick.
 
  • #7
turin
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For the particle in the cylindrically symmetric potential, what is your justification for the particle restricting itself to radial displacements? It seems like any initial azimuthal velocity would result in the paticle periodically orbiting the origin in 2-D (assuming small perturbations about the stable equilibrium, of course). In effect, doesn't this problem just become a 1-D problem with an effective potential that is essentially original radial potential plus the centrifugal potential? In that perspective, I can see that the 1-D effective potential is not symmetric about the stable equilibrium point. So is there always this effective reduction in (non-trivial) degrees of freedom in the problem when symmetry breaking is induced?
 
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  • #8
vanesch
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turin said:
So is there always this effective reduction in (non-trivial) degrees of freedom in the problem when symmetry breaking is induced?
Yes ! And this is a very important observation. With each continuous broken symmetry corresponds a degree of freedom that you can "neglect" in a certain way ; it is in fact the remnant of the symmetry which is hidden now.
This has VERY important consequences in the standard model. In fact, the "broken" degrees of freedom correspond to fields called Goldstone bosons.

cheers,
patrick.
 
  • #9
arivero
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Time ago, a prototypical example was Curie's temperature.
 
  • #10
turin
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Thanks, vanesch. I have been elevated to a new level of understanding.
 
  • #11
Haelfix
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In fact, this loss of degrees of freedom is tantamount to the description of the standard model. Its sort of the central feature, that gives particles their mass via the Higgs mechanism.

In that case the spurious degree of freedom (usually associated with radial fluctuations) called the Goldstone Boson, eats a free, and ackward Gauge term. And presto, a new mass term comes into being.

In fact, thats pretty much the only mechanism we know off that gives particles a mass
 
  • #12
turin
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Does the formalism explain how this mass is related to the energy of the particle (exitation of the field)?
 

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