Natural Science Forum / Physics / Research / January 2005

Thanks. Two references are

E.P. Wigner and E. Inönü,
Proc. Natl. Acad . Sci 39, 510 (1953).

E. Inönü and E. Wigner,
Nuovo cimento 9, 705 (1952).

Arnold Neumaier

I'll resolve the matter here, following up on both your comments and
my other articles in this thread.

A projective representation of a Lie group is equivalent to an
ordinary representation of a central extension of the group.  A
central extension is a U(1) bundle over the group, and its Lie
algebra is obtained by adding an abelian unit 1 to the algebra.

So the central extension of the Lie algebra I listed before would
look like:

[J_a, J_b] = e^c_{ab} J_c + A_{ab} 1
[J_a, K_b] = e^c_{ab} K_c + B_{ab} 1
[J_a, P_b] = e^c_{ab} P_c + C_{ab} 1
[J_a, E] = D_a 1
[K_a, K_b] = E_{ab} 1
[K_a, P_b] = F_{ab} 1
[K_a, E] = P_a + G_a 1
[P_a, P_b] = H_{ab} 1
[P_a, E] = I_a 1,
and of course:
[X,1] = 0 for all X.
where
      e^1_{23} = 1 = -e^1_{32}, with cyclic permutations
      e^a_{bc} = 0 if a=b,b=c or c=a.

The A's, E's and H's are all anti-symmetric.  In particular,
the A's can be written as A_{ab} = e^c_{ab} A_c, and can be
absorbed into a redefinition of J, without changing the other
commutators.

The Jacobi identity further limits the range of possibilities.
The identities on the J-J-K and J-J-P combinations show that the
B's and C's are anti-symmetric.  So, these likewise can be
reabsorbed into redefinitions of K, P (and G).  The Jacobi
identities on the J-J-E combinations show that the D's are 0.
Applying it to the J-K-K and J-P-P combinations shows that
the E's and H's are all 0, too.  Applying it to the J-K-E
combinations show that the G's are all 0; on the J-P-E
combinations, you find the I's are all 0; and finally on the
J-K-P's you find that the F's are diagonal: F_{ab} = m delta_{ab}.

So, the most general non-trivial central extension of the
Galilei group has the Lie algebra given by:

[J_a, J_b] = e^c_{ab} J_c
[J_a, K_b] = e^c_{ab} K_c
[J_a, P_b] = e^c_{ab} P_c
[K_a, P_b] = m delta_{ab}
[K_a, E] = P_a
all other commutators 0.

For non-zero m, neither P^2 nor |KxP|^2 are invariant any longer.
In particular, you find that:
          [K_a, P^2] = 2 m P_a,
so that instead of P^2, you have E0 = E - P^2/(2m) invariant.

Defining X = K/m, and E0 = E - P^2/(2m), you find that:

[J_a, J_b] = e^c_{ab} J_c
[J_a, X_b] = e^c_{ab} X_c
[J_a, P_b] = e^c_{ab} P_c
[X_a, P_b] = delta_{ab}
[E0,x] = 0 for all x.

So the algebra splits into the parts <E0> x <J,X,P>.

One can define L = X x P, and finds that
[X_a,L_b] = e^c_{ab} X_c
[P_a,L_b] = e^c_{ab} P_c
[L_a,L_b] = e^c_{ab} L_c
[J_a,L_b] = e^c_{ab} L_c.
So, defining S = J - L, one gets:
[X_a,S_b] = [P_a,S_b] = [J_a,S_b] = 0
[S_a,S_b] = e^c_{ab} S_c.

Therefore, the algebra splits into <S> x <X,P>.  The subalgebra
corresponding to <X,P> is just the Heisenberg algebra; that
corresponding to <S> is just SU(2).

Finally, in place of |KxP|^2 is the invariant
                        |S|^2 = |J-(KxP)/m|^2.
So, the eigenstates of an irreducible representation have the
form
         |f,l,m,e> = f_{e m}(x) |l,n>
where
    E |f,l,n,m,e> = e |f,l,n,m,e>
    P |f,l,n,m,e> = d/dx |f,l,n,m,e>
    K |f,l,n,m,e> = m x |f,l,n,m,e>
    J |f,l,n,m,e> = (x x d/dx) |f,l,n,m,e> + sigma |f,l,n,m,e>
where
          sigma^2 |f,l,n,m,e> = l(l+1) |f,l,n,m,e>
          sigma_3 |f,l,n,m,e> = n |f,l,n,m,e>
          and n = -l, 1-l, 2-l, ..., l-2, l-1, l.

So, the states are parametrized by a discrete parameter n and a
continuous parameter x; and the state spaces by l, m and e.

The ordinary representation, in this light, which was discussed
in a previous article, is then seen clearly to be just the
representation for 0 mass particles.  The ones above are for
non-zero mass particles of mass m.

Thanks.

Is there a nice way of parameterizing the Poincare group and its physically
relevant unitary irreducible representations in such a way that the
corresponding representations of the Galilei group result
as c -> infinity?

Arnold Neumaier