Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In the sigma-model of spontaneous symmetry breaking, we have degenerate vacuum states. But if we don't pick up a particular value of VEV, we won't have any symmetry breaking. As I read from a book, in field theory at finite volume, there would be no problem; but at infinite volume, we are forced to choose a value of the $\pi$ field. But why?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

This is the superselection phenomenon. If you have a particle in a potential well which has a symmetric minimum, like a particle in 2d with the potential

$$ V(x,y) = a(x^2+y^2)^2 - b(x^2+y^2) $$

With $a>0$ and $b>0$, the ground state is symmetric under rotations--- the ground state wavefunction will be rotationally invariant. This can be proved rigorously, but it is also clear from the relation of the path integral to a biased random walk--- the random walk will wander around the sphere.

When you have a field theory in finite volume (on a lattice), the field value center of mass will also random walk around the circle. If you have two scalar fields, X and Y, the values are X(p) Y(p), where p ranges over the points of the lattice. There are a large but finite number of points p, the field operators are like the X,Y position of the point p, and the particles can all move thier center of mass, which will fluctuate until it is all around the circle.

But if you take the infinite lattice limit, the center of mass degree of freedom is infinitely massive, and has no fluctuation. This means that no local operator will have matrix elements between states with different values for the center of mass position--- the local operators just shift the value at a few nearby lattice sites, and the mean value is defined by infinitely many lattice sites far away.

The result is that every different value of the center of mass position is a different vacuum, completely unreachable from any other. Under this condition, there is no sense is saying that the vacuum is superposed over all orientations, because any local observer will never see more than one orientation--- the vacuum is frozen at some center of mass value which can be determined by local measurements, and the value never quantum fluctuates over all space.

Such a situation was called a "superselection sector" by Wigner. The terminology is unfortunate, because it is not much like a selection rule, and this was the analogy. A better name might be a "macroscopic variable".

The same thing happens in statistical mechanics. If you have a solid crystal on a table, the canonical ensemble sums over all possible positions of the solid on the table--- but once you see where the solid is, it is never going to move, because a coordinated center of mass motion that moves all the points of a solid in the same direction is infinitely improbable in the large system limit.

There are two limits involved in a field theory--- the infrared infinite lattice limit, and the ultraviolet fine-lattice limit. In the ultraviolet limit, the m term goes to zero by scaling, so if you make the lattice too fine, you will find that the field fluctuations do not pick out a direction at small distances, the field values are random from point to point. But over a coarse lattice, the same theory on this coarse lattice has a negative mass. This means that the fields X,Y will tend to point in some direction, and nearby lattice points will tend to point in the same direction.

The vanishing of the m term at short distances means that any finite volume field theory is effectively like a finite number of particles--- no superselection rule. In the infinite volume limit, the superselection rule emerges, and the field vacuum is one or another direction, not all of them in coherent superposition. The reason is that the local operators cannot knock you from one vacuum to another.

share|improve this answer
    
Is this related to the "Wagner-Mermin" theorem in some way? –  QuantumDot Sep 12 '12 at 20:25

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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