In explicit symmetry breaking, the equations of motion of a physical system are variant under the broken symmetry; by contrast, for spontaneous symmetry breaking (SSB), these equations are invariant, but the entire system is not because its vacuum (background) is non-invariant. Further use for the SSB characteristic nonlinear realizations (Goldstone mode), and the group theoretical patterns involved.
Questions tagged [symmetry-breaking]
1008 questions
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Is this an example of spontaneous symmetry breaking?
Consider a pencil standing in a (ideally) perfectly vertical position.
The gravitational field will the same no matter the (angular) direction it will fall in.
But it will end up falling in a particular direction.
Is this an example of spontaneous…
SuperCiocia
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Spontaneous symmetry breaking of SHO
Spontaneous symmetry breaking refers to the solution of a system loses some symmetry in its Lagrangian. Consider a Simple Harmonic Oscillator, its lagrangian is time translationally invariant but its solution is periodic in time, thus not…
Shadumu
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Why is $\rm SU(2)\times SU(2)$ spontaneously broken?
Following the approach of Weinberg's book to discuss the chiral symmetry, at a certain point he says
If the $\rm SU(2)\times SU(2)$ symmetry is exact and unbroken, then this
would require any one-hadron state $|h\rangle$ to be degenerate with…
apt45
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Spatial symmetry breaking and locality
I consider a system described by a state $\Psi(\mathbf{r})$, where $\mathbf{r}$ are the spatial coordinates. The energy of the system is a functional $E[\Psi]$.
An usual analysis of a phase transition, consists first in finding the group $G$ of the…
user3719401
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Why does the symmetry breaking parameter `$v$' have dimensions of mass
A symmetry is spontaneously broken by an operator $\hat{O}$ which acquires a non-zero vacuum expectation value. This is expressed as
$$ \langle 0 | \hat{O} | 0 \rangle = v$$
The parameter $v$ is said to have dimensions of mass. Can someone…
user46837
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Non-abelian Goldstone bosons
Suppose we have a theory with a non-abelian symmetry group $G$ that is spontaneously broken to the subgroup $H\subset G$--this is a global symmetry, $not$ a gauge redundancy. Let $X^a$ be the generators of the unbroken subgroup $H$ and $Y^m$ be the…
user105620
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