Symmetry breaking Suppose the system had SU(3) symmetry initially. - - PowerPoint PPT Presentation

Symmetry breaking

• Suppose the system had SU(3) symmetry
• initially. Let perturbation break the symmetry

to SU(2).

• In this case (u,d,s) which describe 3-

dimensional irrep of SU(3) will break into

• Two irreps of SU(2) (2 1)

Dynamical Symmetry

• Suppose symmetry of the system has no geometrical

interpretation, we say that the symmetry is dynamical symmetry

• Recall Kepler laws of planetary

motion/charged particle in a Coulombic potential

• Hamiltonian

commutes with angular momentum. This system has rotational and time translation symmetry- L and E are conserved

• Runge-Lenz vector M is another conserved quantity

Dynamical Symmetry (contd)

• Classically M satisfies the following constraints:
• Quantum mechanical operator M is
• Classical relations modifies to

Hydrogen atom- Lie algebra

• The conserved quantities which commutes

with Hamiltonian are L and M. Lie algebra involving them:

• Rescale so that we obtain so(4)

algebra- determine `a’

Hydrogen atom- Lie algebra

• The rescale factor `a’ in will be
• The last equation acting on a state with

energy E will resemble so(4) if

Hydrogen atom- so(4) algebra

• The generators are
• Hence we add a fourth fictious coordinate ω

so that the rotation in 4-dimensional space is given by

• Recall the energy levels are

En = -13.6/n2 eV.

• Can we exploit SO(4) symmetry to obtain this

energy?

SO(4)= SU(2) xSU(2)

• We can rewrite so(4) generators such that the

algebra resembles product of two su(2) algebras

• That is.,
• Clearly,
• Hence we can write states which are

simultaneous eigenstates of H, Iz , Kz

SO(4)= SU(2) xSU(2)

• The simultaneous eigenstate is denoted as
• We can construct two Casimir operators which

commutes with all so(4) generatos and H:

• Using the form of M and L, you can check that C2 is a

null operator

• This implies that i and k are constrained to be same
• Hence

where

SO(4)= SU(2) xSU(2)

• From the two equations:
• E is determined by equating
• The exact expression is

SO(4)= SU(2) xSU(2)

• Replace (2i+1) = n in the energy eigenvalues
• The eigenvalues are the familiar hydrogen atom

energy levels:

• We know that the energy levels are n2 fold

degenerate ignoring spin quantum number

• We can see this from the su(2) X su(2) algebra

Physical orbital angular momentum L

• Note that the orbital angular momentum L is

given by addition of the two su(2) generators :

• The eigenstates of Lz
• Where the range of will be from
• With the condition i=k ,
• in terms of n=2i+1, we get = 0,1,2,…n-1

giving degeneracy