Hi Emma, thanks for the e-mail and sorry to keep you waiting! I am not an expert on quantum cryptography/optics. However, I might be able to help you on the difference between the massive and the massless irreducible representations of Poicare' group. Ok, let us do it anyway.
Let q_{a} be a 4-vector living on the surface
p_{a}p^{a} = m^{2} \geq 0 , \ \ p^{0}> 0
in momentum space. Let \mathcal{V}_{q} \subset \mathcal{H} be the space of one-particle states having the 4-momentum q_{a};
P_{a}|q \rangle = q_{a} |q \rangle , \ \ \forall |q \rangle \in \mathcal{V}_{q}
Every momentum eigenstate realizes a one-dimensional representation of the translation
U(0,a) |p \rangle \equiv \exp (i a_{a}P^{a}) |p \rangle = \exp (i a_{a}p^{a}) |p \rangle
Therefore, to construct IRs of the whole Poicare' group, it is sufficient to find the related representations of the Lorentz transformations
U(\omega , 0) = \exp (- \frac{i}{2}\omega_{ab}M^{ab}) ,
as follows from the multiplication law
(\Lambda , a) = (1 , a) (\Lambda , 0)
According to Poicare' algebra, the operator U(\Lambda , 0) transforms some state |q \rangle into another |\bar{q}\rangle where
\bar{q}^{a} = \Lambda^{a}{}_{b}q^{b} \ \ \ (1)
Next, we define the set H(q) of group elements (\Lambda , a) such that
U(\Lambda , a) \mathcal{V}_{q} \in \mathcal{V}_{q} \ \ \ (2)
Clearly H(q) forms a subgroup of the Poincare' group. It is called the little group (or the stability subgroup) for \mathcal{V}_{q}.
To find this subgroup, we simply set \bar{q} = q in (1) as required by the condition (2). This leads to
q^{a} = ( \delta^{a}{}_{b} + \omega^{a}{}_{b})q^{b}
or
\omega^{a}{}_{b}q^{b} = 0
which has the general solution
\omega_{ab} = \epsilon_{abcd}q^{c}n^{d} \ \ \ (3)
where n is an arbitrary 4-vector.
Thus, a typical element of the little group H(q) looks like
<br />
u(n,a) = \exp (ia_{a}P^{a} - \frac{i}{2}\epsilon^{abcd}q_{c}n_{d}M_{ab}) \ \ (4)<br />
We are now in good posision to introduce the Pauli-Lobanski vector into our story.It is defind by
W^{a} = \frac{1}{2}\epsilon^{abcd}M_{bc}P_{d} \ \ \ (5)
For later use, we list some of its properties, viz
W_{a}P^{a} = 0, \ \ [W_{a}, P_{b}] = 0 \ \ \ (6a)
[M^{ab} , W^{c}] = i \eta^{ac}W^{b} - i\eta^{bc}W^{a} \ \ (6b)
[W^{a} , W^{b}] = i \epsilon^{abcd}W_{c}P_{d} \ \ \ (6c)
On the subspace \mathcal{V}_{q}, elements of H(q) can now be written in terms of the P-L vector as
u_{q}(n) = e^{i\alpha} \exp (in_{a}W^{a}) \ \ (7)
where \alpha = a_{a}q^{a}. This equation, together with eq(6c), suggests that the components of the P-L vector form a Lie algebra restricted to \mathcal{V}_{q}. And, the little group H(q) acts on \mathcal{V}_{q} as U(1) \times G, where G is a Lie group generated (on \mathcal{V}_{q}) by the P-L vector. Below, we will see that G = SO(3) \approx SU(2) in the massive case, and G = SO(2) \approx U(1) in the massless case.
From (7), we see that (up to phase factors) the little group H(q) is generated by the P-L vector. Since all states in \mathcal{V}_{q} describe particles with the same momentum. Therefore, on physical ground, any two linearly independent states |q,1\rangle , \ |q,2 \rangle \in \mathcal{V}_{q} sould correspond to different spin projections, and transform into each other under the action of H(q) {or G}.This indicates that the spin properties of the particles are determined by the P-L vector.
We know that the spectrum of spin projections contains only finite number of values; particles of infinite spin (= infinite-dimensional IRs) do not seem to exist in nature. Therefore, on physical ground, all subspaces \mathcal{V}_{q} must be finite dimensional and H(q) (or G) acts irreducibly on \mathcal{V}{_q}.
Now one can use Wigner's methods of induced representations to show that the unitarity and irreducibility of the representations of Poincare' group are directly related to the same properties of the little group.[see Wienberg, Vol 1, sec.(2.5), P.(62-66)].
Thus, in order to find physically admissible IRs of the Poincare' group, it is sufficient to construct all unitary finite-dimensional representations of the little group H(q) corresponding to an arbitrary momentum on the surface
p^{2} = m^{2}\geq 0, \ \ p_{0} > 0
1) Massive IRs
we chooce the 4-momentum q_{a}=(m , 0,0,0) of a particle at rest. When acting on \mathcal{V}_{q}, the components of the P-L vector are (using eq(5) and eq(6a))
W_{0} = 0, \ \ W_{i} = m J_{i}
where
J_{i} = \frac{1}{2}\epsilon_{ijk}M^{jk}
Hence
W^{2} = -m^{2}J^{2}
From eq(6b) or (6c), we find the algebra
[J_{i},J_{j}] = i\epsilon_{ijk}J_{k}
This is nothing but the angular momentum algebra su(2) \approx so(3). Thus
H(q) = U(1) \times SU(2)
As you know, the finite-dimensional IRs of SU(2) are characterized by the eigenvalues of its Casimir operator;
J^{2} = j(j+1)I, \ \ j = 0,1/2,1,...
and for a given j, the dimension of the representation is (2j+1):
W_{3}|\vec{q}=0, s \rangle = |m|s |\vec{q} = 0, s \rangle
s = -j, -j+1, ... ,j
It is clear that the projection of the spin is not Poincare' invariant, while the spin value j does not depend on the choice of the frame of reference. Hence, in the massive case, IRs of the Poincare group are classified by mass and spin:
W^{2}= - m^{2}j(j+1) I
2) Massless IRs : P^{a}P_{a} = 0
On the upper light-cone surface p^{2} = 0, \ \ p_{0}> 0
we take the standard momentum
q_{a} = \omega (1,0,0,1), \ \ \omega \neq 0 \ \ \ (8)
For simplicity, we set \omega = 1.
As before, we use eq(5) & (6a) to find the components of the P-L vector on \mathcal{V}_{q}:
W_{0} = W_{3} = M_{12} \ \ \ (9a)
W_{1} = M_{32} + M_{02} \equiv E_{1} \ \ \ (9b)
W_{2} = M_{31} + M_{01} \equiv E_{2} \ \ \ (9c)
Thus, on \mathcal{V}_{q}, we have
-(W_{a}W^{a}) = (E_{1})^{2} + (E_{2})^{2}
So, mathematically, P_{a}P^{a}= 0 does not imply W^{a}W_{a}= 0.
From eq(6b) or (6c), it follows that the (little group) algebra generated by E_{1}, E_{2} and M_{12} is isomorphic to the Lie algebra of the group E(2) of translations and rotations on a 2-dimensional plane;
[E_{1} , E_{2}] = 0
[M_{12}, E_{1}] = -iE_{2}, \ \ [M_{12}, E_{2}] = iE_{1}
The Casimir operator of E(2) is (E_{1})^{2} + (E_{2})^{2}, and in any unitary IR, it takes the form
(E_{1})^{2} + (E_{2})^{2} = \mu^{2} I, \ \ \mu^{2} \geq 0
If \mu^{2} > 0, IRs of E(2) are labelled by the points \vec{e}= (e_{1},e_{2}) of the one-sphere
e_{1}^{2} + e_{2}^{2} = \mu^{2}
where the states
<br />
|\vec{e}\rangle : E_{1}|\vec{e}\rangle = e_{1}|\vec{e}\rangle , \ \ E_{2}|\vec{e}\rangle = e_{2}|\vec{e}\rangle<br />
represent a base in the space of representation. Thus, if \mu^{2}> 0, the generators E_{1} and E_{2} have continuous spectrum, and the representations are infinite-dimentional. They should be interpreted as particles of infinite spin! Since no such particles are found in nature, these representations are of no PHYSICAL interest. As we mentioned before, the subspace \mathcal{V}_{q} should have a finite dimension. Therefore, we are left with no other possibility but \mu^{2} = 0 and E_{1},E_{2} are trivial on \mathcal{V}_{q}. Thus
W_{1} = W_{2} = 0 \ \ \ (10)
and on physical ground, we conclude that W_{a}W^{a}= 0.
Therefore, in the massless case, eq(7) becomes
u_{q} = e^{i\alpha} \exp (i \theta M_{12})
This means that (up to phase factors) the little group H(q) acts on \mathcal{V}_{q} as a group of rotations in the Euclidean plane, i.e., as the SO(2) \approx U(1) group generated by the only non-trivial operator M_{12}.
Since the action oh H(q) on \mathcal{V}_{q} should be irreducible, this space consists of only one non-trivial state, i.e., one-dimensional;
M_{12}|q, h \rangle = h |q , h \rangle \ \ \ (11)
Indeed, since E(2) is non-compact Lie group, its only finite-dimensional unitary representation is one-dimensional.
Notice that \exp (i \theta M_{12}) describes space rotation along the direction of motion. Putting \theta = 2 \pi, we get
e^{2i \pi M_{12}}|q , h \rangle = e^{2i \pi h}|q , h \rangle
The phase factor must equal to (\pm 1) depending on whether the representation is single-valued or double-valued. Thus
h = 0 , \pm 1/2 , \pm 1 , ...
The quantity |h| is called the spin of massless particle, and h is the helicity quantum number.
From eq(9), (10) and (11), we see that the relation
W^{a} = h P^{a} \ \ (12)
holds in the reference system q = (1,0,0,1). Since W and P are Lorentz vectors, eq(12) holds in any coordinates system. Thus, the helicity is a Lorentz invariant and, a fortiori, Poincare; invariant characteristic of massless particles; the massless finite-dimensional IRs of the Poincare' group are classified by helicity.
Please note that the change of sign of h defines, in general, a different particle state which may not exist physically. The 2 helicity states of the photon mean that the photon is described by a reducible representation of the Poincare' group P(1,3); under space inversion, the states with h = \pm 1 transform into one another. These 2 states form an IR of P(1,3) + space inversion.
Now we have all that is needed to prove that the two polarization states of photon can not be described by Lorentz vector, and that the only admissible vector field of mass zero is a gradient of a scalar field of helicity zero.
regards
sam
